Mathematical Tripos: IA Algebra & Geometry (Part I) Contents

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ahmtclTio:I ler emty(atI a I) (Part Geometry & Algebra IA Tripos: Mathematical Numbers Complex 1 Introduction 0 Contents .1MoisTasomtos...... M¨obius Transformations . . . . 1.11 ...... Plane . . . . Complex . . the . . in . . . . Circles . . and . . Lines . . . . 1.10 ...... Powers . . . Complex . . and . . . Logarithms ...... 1.9 . Theorem . . . Moivre’s . . De . . . . . 1.8 . . Unity . of . . Roots ...... 1.7 ...... Function . . . Exponential ...... The ...... 1.6 ...... Representation . . . . . (Modulus/Argument) . . . . Polar ...... 1.5 . . . Diagram . . Argand . . . The ...... 1.4 ...... Numbers ...... Complex ...... of . . . Functions ...... Numbers ...... Complex . 1.3 . . . . . of . . . Manipulation ...... Algebraic ...... 1.2 ...... Introduction ...... 1.1 . . . . . This? . . Study . . . . . Why ...... 1.0 ...... Revision...... 0.6 ...... Acknowledgements...... 0.5 . . . . Sheets . . Example ...... 0.4 . . . Notes . . Printed . . . . 0.3 . . . Lectures . . 0.2 . Schedule 0.1 ahmtclTio:I ler emty(atI) (Part Geometry & Algebra IA Tripos: Mathematical .. ope oes...... powers . . . Complex ...... 1.9.1 ...... 1 . . . . . for . . . . . expression . . . . Modulus/argument ...... form . . . . 1.6.5 modulus/argument . . . . to . . . . Relation ...... functions . . 1.6.4 trigonometric . . . . complex . . . The ...... function . . 1.6.3 exponential . . . complex . . . . . The ...... function . . 1.6.2 . exponential . . . . real . . . The ...... 1.6.1 ...... multiplication . . . of . interpretation . . . Geometric ...... 1.5.1 ...... Modulus ...... 1.3.2 . . . conjugate . Complex ...... 1.3.1 ...... numbers equation . Complex quadratic . a of . 1.1.3 solution . general The . . . 1.1.2 numbers Real 1.1.1 .11Cmoiin...... Composition . . 1.11.1 ...... Circles . . 1.10.2 . . Lines 1.10.1

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This is a supervisor’s copy of the notes. It is not to be distributed to students. ahmtclTio:I ler emty(atI b I) (Part Geometry & Algebra IA Tripos: Mathematical Algebra Vector 2 .0Sffi oain...... Notation . . Suffix . . . 2.10 ...... Co-ordinates . . Polar ...... 2.9 . . . . Identities . Component . . . Vector ...... 2.8 ...... Components ...... 2.7 ...... Components . . and . Bases ...... 2.6 ...... Products . . Triple . . . . . 2.5 ...... Product . . . Vector ...... 2.4 ...... Product . . Scalar . . . . 2.3 ...... Vectors . of . Properties . . 2.2 . . . . Vectors 2.1 . h td hs ...... This? Study Why 2.0 .. yidia oa oodnts...... longitude) . . . . . and . . . . latitude . . . height, . . . . (cf. . . . . co-ordinates . . . . . polar ...... Spherical ...... 2.9.3 . co-ordinates . . . . . polar . . Cylindrical ...... co-ordinates . . . . . 2.9.2 polar . . plane . . . 2D ...... 2.9.1 ...... cosines . 3D . Direction . . in . . . basis . . standard . . . . 2.7.2 or . . . . Cartesian . . . The ...... 2.7.1 ...... spaces . . . . . dimensional . . . . . Higher ...... 2.6.3 space . . . . dimensional . . . . Three ...... 2.6.2 . . space . . . . dimensional . . . Two ...... 2.6.1 ...... product . . . . . triple . . . . scalar . . . the . . . . . of . . . . . Properties ...... 2.5.1 ...... triangle/parallelogram . . . a ...... of . . . . area . . . Vector ...... product . . . . . vector . . . 2.4.2 . the . . . of ...... Properties ...... 2.4.1 ...... rule . . . cosine . . . the . . . . . Example: . product . . . . scalar . . the . . . 2.3.4 . of . . . . property . . Another ...... 2.3.3 ...... Projections . . . . . product . . . scalar . . 2.3.2 . the . . of . . . . Properties . parallelogram . . . a . . . . form . . quadrilateral 2.3.1 . . . . . any . . of . . . . . sides . . . . . the . . . of . . . . midpoints . . . . . the . . Example: ...... scalar . . . 2.2.3 . a . . . by . . . Multiplication ...... 2.2.2 ...... Addition ...... 2.2.1 ...... representation . Geometric ...... 2.1.1 ...... Möbius . map . general . The . . . 1.11.4 . . . Maps Basic . 1.11.3 . . . Inverse 1.11.2 .04Mr nbssvcos...... symbol . . . Levi-Civita . . or . . . tensor . . . alternating . . The . . . . . 2.10.5 . vectors . . . basis . on . . More . . . . 2.10.4 ...... delta . Kronecker . . 2.10.3 . . . . convention . Summation . 2.10.2 equivalents suffix and Dyadic 2.10.1

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This is a supervisor’s copy of the notes. It is not to be distributed to students. ahmtclTio:I ler emty(atI c I) (Part Geometry & Algebra IA Tripos: Mathematical Spaces Vector 3 . clrPout aka ne rdcs ...... Products) . . . . Inner . . (a.k.a. . . Products . . . . Scalar ...... 3.6 ...... Subspaces . . of . . . . Sum . . and . . Intersection ...... 3.5 ...... Components ...... 3.4 ...... Bases . and . . . Dimension . . Sets, . Spanning ...... 3.3 ...... Subspaces ...... 3.2 ...... Space? . Vector . . . . . a . . . is . . What ...... 3.1 ...... Sphere a . . . . in . . . Inversion . . . . 2.15 ...... Inversions . . . and . Isometries . . Maps: . . . 2.14 . . Sections . . Conic and . . Cones . . 2.13 ...... Planes . and Lines . 2.12 . . Equations Vector 2.11 . h td hs ...... This? Study Why 3.0 .. xmls...... Examples ...... for . . . . 3.6.5 . product . . scalar ...... The ...... 3.6.4 ...... inequality . . . . Triangle ...... 3.6.3 . . . . . inequality . . . . Schwarz’s . . . . . product . . scalar . . . 3.6.2 . . a . . . of . . Definition ...... 3.6.1 ...... Examples ...... 3.5.2 ...... dim( ...... 3.5.1 ...... sets . . . . Spanning ...... 3.3.2 . . . independence . . . . Linear ...... 3.3.1 ...... Examples ...... 3.2.1 ...... Examples ...... 3.1.3 ...... Properties...... 3.1.2 . . . . . Definition ...... 3.1.1 ...... Isometries ...... 2.14.1 ...... Planes ...... 2.12.2 ...... Lines . . . . 2.12.1 ...... product . . triple . Vector . . . 2.10.9 . . . . product . triple . . Scalar . . 2.10.8 . . . identity . An notation suffix 2.10.7 in product vector The 2.10.6 U + W ...... ) R n ......

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This is a supervisor’s copy of the notes. It is not to be distributed to students. ahmtclTio:I ler emty(atI d I) (Part 76 Geometry & Algebra IA Tripos: Mathematical Equations Linear and Inverses Matrix Determinants, 5 Matrices and Maps Linear 4 . ovn iereutos...... equations . . linear . . Solving . . . Elimination . Gaussian . 5.5 . Equations: . Linear . Solving . . . . 5.4 ...... 3 . . a . . . of . Inverse . . . The ...... 5.3 ...... 3 . . Unknowns for . Two . Determinants . in . . . Equations . Linear . 5.2 . Two . . . of . Solution ...... 5.1 . . This? Study . . Why . . . . 5.0 ...... basis . . of . Change . . . . 4.8 . Matrices . Orthogonal . . 4.7 ...... Matrices . . . of . . . Algebra ...... 4.6 ...... Maps . . . of . Description . . . . Matrix . . . the . and . . . . Bases ...... 4.5 ...... Maps . . of . . . Composition ...... 4.4 ...... Nullity . and . . Kernel . Rank, . . . . 4.3 ...... Maps Linear . . 4.2 . Deﬁnitions General 4.1 . h td hs ...... This? Study Why 4.0 .. rnfrainmtie ...... 3 . of . . . . Properties ...... 5.2.1 ...... maps . . . . . linear . . . . . representing ...... matrices . . . . . for . . . . . law . . . . Transformation ...... components . . . vector . . . 4.8.4 . . for . . . . law . . . . . Transformation ...... matrices . . . . . 4.8.3 . transformation . . . . of ...... Properties ...... 4.8.2 ...... matrices . . . . . Transformation ...... 4.8.1 ...... 3 . . . . . (for ...... Determinants ...... matrix . . . a . . . . . 4.6.9 of ...... inverse . . . . . The ...... matrix . . identity . . . . . 4.6.8 . or . . . unit ...... The ...... 4.6.7 ...... Trace ...... 4.6.6 ...... Symmetry ...... 4.6.5 ...... Transpose ...... 4.6.4 . . . . . multiplication . . . . . Matrix ...... 4.6.3 ...... Addition ...... scalar . . . . 4.6.2 a . . . by . . . Multiplication ...... 4.6.1 ...... maps) . . of . . dim(domain) . deﬁnitions . . . new . . important . . . 4.5.3 some . . . (including . . Examples ...... 4.5.2 . . . notation . Matrix ...... 4.5.1 ...... Examples . . . . . 4.4.1 ...... Examples . . . 4.3.1 . . . . Examples 4.2.1 .. osrcino h nes ...... inverse . the of . Construction . . 5.3.2 cofactors and Minors 5.3.1 × × i(ag)...... dim(range) 6= arx...... Matrix 3 arcs...... Matrices 3 × eemnns...... determinants 3 × n 2 and 3 × arcs ...... matrices) 2

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This is a supervisor’s copy of the notes. It is not to be distributed to students. ahmtclTio:I ler emty(atI e I) (Part Geometry & Algebra IA Tripos: Mathematical Spaces Vector Complex 6 . ierMp ...... Maps . . Linear . . . . 6.3 ...... Spaces Vector . Complex . for Products . Scalar . . 6.2 ...... Numbers Complex The Over Spaces Vector 6.1 . h td hs ...... This? Study Why 6.0 .. rpris...... Components . . . of . . Terms . . . in . . Products . . . . Scalar ...... 6.2.3 ...... Properties ...... 6.2.2 . . . . Deﬁnition . . . . 6.2.1 ...... 6.1.3 . . . Properties . . 6.1.2 . Deﬁnition . . . 6.1.1 ...... equation inhomogeneous . the . of solution . the . for Implications . . of . 5.5.4 view . mapping Linear . . . 5.5.3 of view . Geometrical . problems 5.5.2 homogeneous and Inhomogeneous 5.5.1 C n ...... A x = A x 0 = ...... 0 ......

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This is a supervisor’s copy of the notes. It is not to be distributed to students. † rnpstos h ino emtto.[5] i [11] I) (Part Geometry & Algebra IA Tripos: Mathematical [5] Beardon F Alan Armstrong M.A. orthogonal the matrices. groups, spin linear Pauli and special quaternions and groups, Lorentz general the the groups, example, unitary groups, for orthogonal groups; special matrix and and of cycles Permutations, (only) theorem. [4] Examples permutation. isomorphism a Intro- the of Examples. and sign homomorphism. groups The a quotient of transpositions. subgroups, kernel normal the to groups, abstract duction of infinity. homomorphisms at and point Isomorphisms the subgroups. of preservation conjugate treatment Möbius cross-ratios, The group; informal and [3] tetrahedron. groups, circles, and cosets Euclidean cube of the stabilizers, polygons, including regular geometry, Orbits, of from groups actions. symmetry Examples group theorem. Lagrange’s and theorem. subgroups Orbit-stabilizer groups; their for and Möbius map. Axioms conics corresponding forms, the quadratic of distinct that points of and fixed discussion 2 real, and Brief for are eigenvalues eigenvectors. forms that Canonical of 3 proof classification. basis real matrices, orthogonal A symmetric diagonalized. give Real eigenvalue. be eigenvalues real cannot that a matrices has of matrix examples diagonalization, of Discussion invariant as significance geometric Eigenvectors, eigenvalues. of existence lines. the for implication its and only), Elimination. Gaussian of unknowns). Discussion 3 simultaneous in of equations interpretation geometric (3 and equations Solution linear inverses. and matrices non-singular Determinants, matrices, by maps linear matrices. of of representation Bases, algebra maps. the linear of Composition rotations). shears, dilations, from maps Linear dimension. term and Michaelmas sets spanning lectures, to 48 tion inversions. and to isometries Introduction Maps: sections. conic and cones spheres, planes, Lines, equations. coor- and Vector convention polar summation cylindrical and including spherical notation: plane, Suffix coordinates; dinates. Cartesian interpretation. Geometrical products. in Vectors complex of treatment transformations. Informal theorem. Moivre’s de and logarithm, argument modulus, numbers, complex of the Review paragraph. in the schedules material in the only material of GEOMETRY the and paragraphs AND to lectured of ALGEBRA devoted end be be the will will at that schedules brackets lectures the square of number in in the numbers material roughly The the schedules. indicate examined. all of be say, will booklet to schedules the is from that copy examining; a is This Schedule 0.1 Introduction 0 1 See http://www.maths.cam.ac.uk/undergrad/schedules/ R n-th 3 C lmnayagbao clr n etr.Saa n etrpout,icuigtriple including products, vector and Scalar vectors. and scalars of algebra Elementary . n ot n ope oes qain fcrlsadsrih ie.Eape fMöbius of Examples lines. straight and circles of Equations powers. complex and roots R iermp n arcs ievle,tefnaetltermo ler (statement algebra of theorem fundamental the Eigenvalues, matrices. and maps linear , ler n Geometry. and Algebra rusadSymmetry. and Groups n R clrpout acyShazieult n itne usae,bifintroduc- brief Subspaces, distance. and inequality Cauchy–Schwarz product, scalar , m to R n ihepai on emphasis with × arcs icsino eainbtenegnauso matrix a of eigenvalues between relation of discussion matrices; 2 U 05( 2005 CUP prpit books Appropriate pigrVra 98( 1988 Springer–Verlag 1 ,n m, ceue r iia o etrn n aia for maximal and lecturing for minimal are Schedules £ . 6 19 paperback, 21.99 .Eape fgoerclatos(reflections, actions geometrical of Examples 3.

c ..olydmpcma.k ihems2006 Michaelmas [email protected], δ ij £ , 30 hardback). 33.00 ijk [5] . £ 8hardback). 48 × orthogonal 3 [2] [3] [5] [2] [3]

This is a supervisor’s copy of the notes. It is not to be distributed to students. . rne Notes Printed 0.3 ahmtclTio:I ler emty(atI ii I) (Part Geometry & Algebra IA Tripos: Mathematical you. of Lectures 0.2 Smart D. Sernesi E. Green J.A. Burn R.P. Kendall P.C. and Bourne D.E. Bloom D.M. 4 3 2 • • • • • • • • • • • • • aigsi ht eerhsget htwti h rt2 iue il tsm on,hv otteatnino all of attention the lost have point, some at will, I minutes 20 first the within that suggests research that, said Having fyutrwpprarpae laepc hmu.Iwl iku h rtoet tyi h i o seconds. 5 for air the of in case stay the to in one miserably first fail the will up I pick But will I up. them pick please aeroplanes paper throw you If paperback). ial,Iwsi orpsto 2yasaoadnal aeu h rps fyufe htthe that and feel come it, you on If week Tripos. a hours the 20-24 up than gave more nearly spending are and you ago or read. years head, cannot your chat. 32 you over position pens going colour your is which course in me was tell I and Finally, come please blind be colour principle is in anyone could If ‘proofs’ outline/sketch the time. all sufficient that given of lecture. said up result next having tightened a the rigour; as of of on beginning levels plough mathematical the between to somewhere at have is problems will course any I The up circumstances clear such will Under I lecture. however a constraints; of time middle the and in you myself both lecture. confuse next may the I Sometimes of end the at me catch or words the use avoid to aim I a pillow. lectures out. your your 6 carried shout under have at them you (especially put because welcome or lost just I . . you hopelessly . into week get diffuse a not will you for will you but around notes clear otherwise notes your . be . distraction. of . to understanding a lecture best An is my next week). chatting do the Such will laid. before I and/or notes learn. drunk to get you not) questions. want answer did I to (or order did in who lectures other or each after results, to front football chat the means at proof/explanation. all long minutes a By few of jokes middle a the and/or for in around rest dead stop stay a to will going for not I am lecture but 11:55, the by finish of to aim middle will I the in break lecture. the minute since throughout aeroplanes 2 time concentrate paper a on and/or have be politics Please to and/or lecture. endeavour last late. will the in I walking of people summary have a to with distracting promptly is 11:05 it at start will Lectures fi snti h oe ro h xml hesi hudntb nteexam. the in be not rather should me it to sheets listen example can the you on lecturing). that or my so of notes . ahead . the . keeping typing in course my not the on is depends for that it out (but If down handed things be scribble will to having notes than printed that hope I straightforward ierAgbaadGeometry. and Algebra Linear ierAgba emti Approach. Geometric A Algebra: Linear rus aht Geometry. to Path a Groups, esadGop:afis orei Algebra. in course first a Groups: and Sets ierAgbaadGeometry. and Algebra Linear constructive or iial ols time last to similarly ekig fIa nuil,ilgbe nla rjs li rn hnplease then wrong plain just or unclear illegible, inaudible, am I If heckling. trivial etrAayi n atsa Tensors. Cartesian and Analysis Vector yes , quietly easy applied . abig nvriyPes18 ( 1987 Press University Cambridge abig nvriyPes18 oto print). of (out 1988 Press University Cambridge 3 , abig nvriyPes17 oto print). of (out 1979 Press University Cambridge obvious myself fIa nla,btpes ontdsus a,ls night’s last say, discuss, not do please but unclear, am I if 2 tdnsse ofidta h ra ae tese to easier it makes break the that find to seem students ; and fi snt mi e( me email not, is it if ; R rs 93( 1993 Press CRC pure Ia o nalbe,admyntb bet extract to able be not may and infallible), not am (I and hpa n alCC18 ( 1988 Hall/CRC and Chapman ahmtc.Hned o lasepc pure expect always not do Hence mathematics. yes

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This is a supervisor’s copy of the notes. It is not to be distributed to students. h olwn oe eeaatd(..soe)fo hs fPtrHye,m semdHa fDepart- of Head esteemed my Haynes, Peter of those from stolen) (i.e. ment. adapted were notes following The Acknowledgements. 0.5 ahmtclTio:I ler emty(atI iii I) (Part Geometry & Algebra IA Tripos: Mathematical Sheets Example 0.4 6 5 • • • • • • • • • • • • • fyural aebe l n antfidacp ftents hncm n e e u rn orsick-note. your bring but me, see and come then notes, the of copy a find cannot pedants. and the ill for been lectures have three really first you the If of exception the With htte ilntb h aenx year). next same the be not mathe- will press-ups, they do that athletes scales, do Your pianists design; by sheets algebra/manipulation. the lectures. do on of maticians not repetition 2 do some week you is of that there- There suggest middle or I the respectively, Personally before 6/12/18/23 supervisions. supervision lectures arranging first when after mind your 1/2/3/4 have in sheets this example bear same do Please the to abouts. about able at be WWW should You the on available be sheets. (see will out ‘study’ They them supplementary sheets. hand example I as main time four be will There ( sheets example and notes in the overwhelming to was corrections courses me lecture email previous notes Please my the of examples. in one worked material to in unlectured with tempted vote unhappy including been have be a of However, would I favour students past lectured. that the not worried In was was notes. that I the in because examples time. these worked include to unlectured not time of number from a down are something There write taught to was on have I course you notes. teaching if the awake my from can stay omitted at you deliberately this that are do textbooks figures/diagrams excellent to exceptions of two number or one a With are notes. there my then of lectures place attend in to want use not do you If independently. used be to meant deliberately are are notes they The unless friends absent your for copies. copies take take lecture not the do at Please students were there as you notes unsound. ask). which environmentally the not from of more Saturday. . do . copies previous . (so many the notes mess) as approximately time’s worse on be last even only of an will copies be There have would not office will my I that (otherwise conclude back-copies may keep only not and do lectures I in available be only will notes Any supervisors 6 ih iet nwta h xml heswl etesm sls er(but year last as same the be will sheets example the that know to like might 5 not eaegigt elafrs si s n aen eiet eeven be to desire no have I and is, it as forest a fell to going are We http://damtp.cam.ac.uk/user/examples/ vial nteWW hyaea duc olcue n r not are and lectures to adjunct an are they WWW; the on available o oLecture To How

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This is a supervisor’s copy of the notes. It is not to be distributed to students. ahmtclTio:I ler emty(atI iv I) (Part Geometry & Algebra IA Tripos: Mathematical functions. cosine and sine The logarithm. The function. exponential The alphabet. Greek The following. the recall you that check should You Revision. 0.6 hr r lotpgahcvrain nesln(i.e. epsilon on variations typographic also are There thstefloigproperties following the has It properties following the has It h oaih of logarithm The M ∆ K Θ H A B Λ E Γ Z I h xoeta ucin exp( function, exponential The α β µ λ κ γ η ζ θ δ ι h ieadcsn ucin r endb h series the by defined are functions cosine and sine The > x uΩ mu abaΨ X Φ Υ T lambda kappa iota theta eta eaΣ P Π Ξ O zeta epsilon delta gamma beta lh N alpha log(exp( sin( exp( x()=1 = exp(0) x()= exp(1) ,ie log i.e. 0, e cos( x log( x + x = ) o()=0 = log(1) log( y = ) exp( x xy = ) x n e exp( X = )= )) ∞ log = ) 1 = ) =0 x x sdfie steuiu solution unique the as defined is , n X = ) ∞ =0 ( e e e y (2 − x x = ) , ≈ e ) ( n , X n y ∞ n − =0 , x 2 1)! + x . 2

) . . x , , c 71828183 2 n n ε x n n x x ,pi(i.e. phi ), ! ..olydmpcma.k ihems2006 Michaelmas [email protected], n +1 ! 2 x log + n . ,i endb h series the by deﬁned is ), , . . y , ψ ω χ σ π φ υ ν τ ρ o ξ ϕ ,adro(i.e. rho and ), omega psi upsilon chi phi tau sigma pi rho omicron xi nu y fteequation the of % ). (0.3b) (0.2d) (0.2b) (0.1d) (0.1b) (0.2a) (0.3a) (0.1a) (0.2e) (0.2c) (0.1e) (0.1c)

This is a supervisor’s copy of the notes. It is not to be distributed to students. ahmtclTio:I ler emty(atI v I) (Part Geometry & Algebra IA Tripos: Mathematical circle. a of equation The line. a of equation The rule. cosine The identities. trigonometric Certain ete( centre asstruh( through passes where by given is line this of equation the form parametric In t ,q p, steprmti aibeand variable parametric the is sgvnby given is ) x 0 y , 0 n2 atsa oodnts ( co-ordinates, Cartesian 2D In sgvnby given is ) cos( cos( n2 atsa oodnts ( co-ordinates, Cartesian 2D In sin( sin( x x x x tan( cos( sin( ) cos( + ) ) sin( + ) − − x x x cos( sin( o hudrcl h following the recall should You x ± ± ± = y y y y y y y ( sin 2 = ) sin( = ) cos 2 = ) cos 2 = ) = ) = ) cos( = ) x x y a − 0 − + sa rirr elnumber. real arbitrary an is p ) y y , t a 2 0 − 1 ( + = tan( oierl as nw stecsn oml rlaw or formula that cosine states the cosines) as of known (also rule cosine at subtended angles A the that suppose Further tively. vertices opposite Let sin 2 ∓ , m x y x tan( B cos( ) x cos( ) − ( ABC x ) = x x x ,y x, and q

± − x x c + + + 2 ) 2 2 y tan( ) 2 ,y x, 0 tan( y + x y 2 ..olydmpcma.k ihems2006 Michaelmas [email protected], y y y ,teeuto faln ihslope with line a of equation the ), = ) + ) 0 eatinl.Lttelntso h sides the of lengths the Let triangle. a be C y ) ± ∓ a ,teeuto facrl fradius of circle a of equation the ), c b r , t m a . cos y 2 2 2 sin cos 2 are cos( y sin( ) sin . ) = = = , α x x A x x x , sin( ) sin( ) , x − − − 2 2 b a a β 2 B 2 − 2 2 2 y y y and + + + and y y y ) ) c b c 2 . , , , 2 2 , γ − , − − C epciey hnthe Then respectively. 2 2 2 be bc ab ac cos cos cos a , b , α . γ , β and c m respec- (0.4d) (0.4b) (0.5b) (0.6b) (0.4g) (0.4a) (0.5a) (0.6a) which (0.4e) (0.4c) (0.5c) r (0.4f) (0.7) and

This is a supervisor’s copy of the notes. It is not to be distributed to students. Additions/Subtractions? Examples. Suggestions. ahmtclTio:I ler emty(atI vi I) (Part Geometry & Algebra IA Tripos: Mathematical 3. 2. 1. 8. 7. 6. 5. 4. 3. 2. 1. s h olwn lentv ro fShazsieult sa example: an as inequality examples? Schwarz’s Imperial of the proof of alternative some following using the sheet, Use study numbers log( complex e.g. a revision, Include covering 0 sheet a Introduce eieaemsaeady(ihmr bar). mars (with day a mistake Deliberate Matrix) (Shear Do elimination. Gaussian doing when elimination. matrix n Gaussian inverse on the notes of simplify construction and Add theorem, rank-nullity of matrix, proof reflection Add rotated (e.g. dilatation). diagonalisation of and example notation). reflection an suffix and or (using pictures add multiplication basis: e.g. of matrices, Change convention. diagonal summation on the section introducing a when Add signs summation grey more Add × n eemnns ness n asinelimination. Gaussian and inverses, determinants, k x n k n emti interpretation. geometric and , 2 k y k 2 | − x · y | 2 = > = = 0 x 2 1 2 1 i ( x . x x i i x i xy y y i

j j c y y log( = ) j − j y ..olydmpcma.k ihems2006 Michaelmas [email protected], j − x + j x y i i 1 2 y )( x x i x x j log( + ) x j i y y j j j y i − y i x − y j ). y x i ) i y i x j y j

This is a supervisor’s copy of the notes. It is not to be distributed to students. oti o qa oayra ubr nodrta ecan we that order In number. real any to equal not is root where equation equation quadratic quadratic the a is Consider of there solution line general a The on number). real points two another always any 1.1.2 is between there (e.g. numbers line real a two any on between lying and as point, numbers another real visualise sometimes We ahmtclTio:I ler emty(atI 1 I) (Part Geometry & Algebra IA Tripos: Mathematical For and by denoted are numbers Complex and introduction numbers the Complex with happy being 1.1.3 us to Subject [numbers]. real now are of roots existence square the where If Remark. If by denoted are numbers real The numbers Real 1.1.1 that FP3 speed and the Introduction FP1 note OCR instead say, asleep; done, fall 1.1 not not have do who please revision, you is of introduced. it few which being a for are for those concepts For revision, new mathematics. new. is throughout be occur section and will this useful this are you they because of numbers: many real study For we as reason same the For This? Study Why 1.0 Numbers Complex 1 β β 2 2 z i < > = sa endi 12.W a that say We (1.2). in defined as is ∈ 4 4 a en blnst’ hshstoroots two has This to’. ‘belongs means αγ αγ + oethat Note 11 a o erewritten be now can (1.1) , ib i hnterosaera teei eetdro if root repeated a is (there real are roots the then ecnnow can we , esmtmswrite sometimes we , ainl,dntdby denoted by denoted rationals, integers, rainl,ters fteras e.g. reals, the of rest the irrationals, i z 1 ssmtmsdntdby denoted sometimes is z = 1 − = always 2 − αz β α β 2 a + b + + m( Im = e( Re = R C ov udai equation. quadratic a solve i p p βz edfieacmlxnme,say number, complex a define We . n oss of: consist and z β 4 2 = + αγ α 2 2 z z z − a α γ h elpr of part real the : ) h mgnr atof part imaginary the : ) − ∈ Q Z + : 0 = 4 , αγ , β C ib, 2 oethat Note . i p/q . . . j = and and yengineers. by where − √ where − ,β γ β, α, 3 , 1 z z − . 2 2 always 2

z = = ,b a, ,q p, c , 1 − − ∈ − ..olydmpcma.k ihems2006 Michaelmas [email protected], and ∈ 1 β 2 R r nees( integers are β z, β , α ov udai qain eintroduce we equation, quadratic a solve − R 2 0 z − , 2 , 4 = α , p n(.)aeo hsform. this of are (1.3) in 1 i β 2 p , z. 0 6= αγ α 2 2 z 4 − . . . , ob ubrwt h form the with number a be to , √ αγ .If ). 2 , 4 2 α αγ ,π π π, e, , − q β 0) 6= β . 2 2 < , 4 αγ 2 . hntesquare the then (1.1) (1.3) (1.2) (1.4)

This is a supervisor’s copy of the notes. It is not to be distributed to students. nodrt aiuaecmlxnmessml olwterlsfrras u digterule the adding but reals, for rules the follow simply numbers for complex Hence manipulate to Numbers order Complex In of Manipulation Algebraic 1.2 ahmtclTio:I ler emty(atI 2 I) (Part Geometry & Algebra IA Tripos: Mathematical a The conjugate Complex number 1.3.1 complex new a function defines complex-valued and A ‘input’ numbers. complex as to number functions complex of a idea the extend may Numbers We Complex of Functions 1.3 closure Remark. Exercise: Hence zero. is zero, to equal or than less 1.2. is Corollary that number a to equal is that Then Proof. unique. 1.1. Theorem Remarks. − 4. 3. 2. 1. ib ope conjugate complex ope ubr eefis sdb atgi 10-57 n adn 10-56.Teterms The (1501-1576). Cardano and (1500-1557) Tartaglia by used real first were numbers Complex + 0 number complex A certain solving makes it e.g. C equation, quadratic a solve easier. to much able equations being differential always to addition ( in real isations from system number the Extending i.e. , a diinsbrcin: addition/subtraction : Assume − otisalra ubr ic if since numbers real all contains C c and z l h bv prtoso lmnsof elements on operations above the All scoe ne diinadmultiplication. and addition under closed is For 1 = = utpiain: multiplication i imaginary ( z a ∃ d 1 − h ersnaino ope number complex a of representation The If + ,b ,d c, b, a, − 1 z ib sdfie n(.) hc that check (1.7), in defined as b 1 nes : inverse ,ads ( so and ), and = of z 2 eefis nrdcdb ecre (1596-1650). Descartes by introduced first were ∈ z z where 2 R = = i.b uhthat such a a c ssi obe to said is + + − z 1 if id ib c z , ) hc suulywitnas written usually is which , where , 2 z 2 1 z = ∈ z + = z 1 − a C 1 − z z z a ∈ 2 1 2 then , ( = d + = ( = ( = ( = ,b ,d c, b, a, ueimaginary pure R − a R ib z 1 + then ocmlx( complex to ) a ac a b ) = e( Re + + z ib 2 then − 1 u h nynme rae hno qa ozero to equal or than greater number only the But . C ib ib a = z a ∈ bd 1 − z ) ( ) euti e lmnsof elements new in result + 1 + c 1 1 R + ) ± e( Re = ) c + ib + 1 = ehv that have we , i. z + (

z c 0 c id. i a a = id + ( ∈ ntrso t eladiaiayprsis parts imaginary and real its of terms in . ..olydmpcma.k ihems2006 Michaelmas [email protected], − − bc a = ) C id C ib ib i. − + z losanme fipratgeneral- important of number a allows ) . ( = ) 0. 2 f = ib. ac ad ) ( z z and (1.6) ; ) s‘output’. as ) a + u oeie as sometimes but a 2 ibc ± + a m( Im c b + + ) 2 a z − = ida 1 i m( Im = ) c a ( b 2 ( + C and ± ib + hsi ecie as described is This . ib d b 2 (1.5) ; ) b ( ) f z . = soeta takes that one is z 2 id ) ∗ d . ) sdﬁe as deﬁned is , . i 2 = (1.7) (1.8) − 1.

This is a supervisor’s copy of the notes. It is not to be distributed to students. 1/03 . h radDiagram Argand The 1.4 Example. ahmtclTio:I ler emty(atI 3 I) (Part Geometry & Algebra IA Tripos: Mathematical The Modulus 1.3.2 Hence Definition. Exercises. Exercises. 4. 3. 2. 1. 2. 1. modulus f ( z = ) Let hwthat Show hwthat Show ie ope-audfunction complex-valued a Given of z p f z 2 ( = z + = ) a z q + pz + ib f 2 hc switnas written is which , r ( + . z f = ) qz ( z + ) f ≡ r ( z with f ) , ( z = ) n ec rm(1.9a) from hence and ,q r q, p, | z | pz = z 2 f 1 | ( ∈ z + h ope ojgt function conjugate complex the , z ± z z a | − 1 C | sdfie as defined is , − 2 z qz z z 1 1 z | 2 + 2 Remarks. number ( and origin the tween an of in parts ordinates imaginary and rep- real can the we Then (2D) each axes. dimensional resent Cartesian two to in referred points space of set the Consider ( = ) hnb sn 19)ad(1.9c) and (1.9b) using by then 2 = + = = = = b 2. 1. 2 r z . z | z z z 1 z z 1 1 = rad(1806). Argand Jean-Robert Caspar by re-invented by and invented (1797), Wessel was diagram Argand The the and plane The | z. / )

z 2 ± c − 2 2 p . z . 1 . z ..olydmpcma.k ihems2006 Michaelmas [email protected], z . uhapo scle an called is plot a Such . 2 2 xy erfrt the to refer We . . z + f = y ln srfre oa the as to referred is plane q ( ai sthe as -axis xy z z x = ) + + lt elblte2 etrbe- vector 2D the label We plot. iy f r. ,y x, ( ∈ z ) ,say ), . C mgnr axis imaginary ytepit( point the by x f ai sthe as -axis OP sdﬁe by deﬁned is z → r iwda co- as viewed are raddiagram Argand ytecomplex the by , . elaxis real ,y x, complex (1.12b) (1.12a) (1.9d) (1.9b) (1.10) (1.9a) (1.11) (1.9c) ,i.e. ), . ,

This is a supervisor’s copy of the notes. It is not to be distributed to students. 1/02 ahmtclTio:I ler emty(atI 4 I) (Part Geometry & Algebra IA Tripos: Mathematical Remark. 1.3. Theorem eut(.3)i nw sthe as known is (1.13a) Result If z 1 z , 2 ∈ C then | | z z 1 1 + − rageinequality triangle z z 2 2 | | 6 > Interchanging that hence and Let (1.13a). z from follows (1.13b) that 11b follows. (1.13b) Addition. conjugate. Complex Modulus. Proof. |

2 0 z | 1 z = | 1 | z + | − | z hc ssmtmscle the called sometimes is which represents and P vector the of nitude em fvco addition vector of terms ycmltn h parallelogram the completing by point the with associated is 1

2 c | ytecsn rule cosine the By + 0 othat so , z 2 z is adi nfc n fmany). of one fact in is (and ..olydmpcma.k ihems2006 Michaelmas [email protected], | z z | 2 z h ouu of modulus The z Let 3 2 , | 2 0 2 P |

= 2 − = . eetdi the in reﬂected z z z z | | x 1 0 z z ( = 6 = 1 1 1 0 z 2 1 1 0 0 z | + and = where , | + 1 − = 6 | OP z | | = z x iy z z z 2 If | → | | z z 1 | 1 1 2 0 z 2 z ( = = 1 1 z 0 | | | 3 2 0 + 2 2 1 0 OP | 1 0 = eascae with associated be > − → + eas aethat have also we + + − − iy P OP x x z z OP | → | 1 | | z z 2 → 0 2 1 0 z z z z represents 1 2 orsod otemag- the to corresponds 0 0 | + 2 2 0 stepit( point the is + 2 2 1 | − | | eascae with associated be hn(.3)implies (1.13a) Then . | > x | | ) + 2 2 + since x y -axis. 2 2 + − | 2 2 | z . OP + ) z z P 2 0 → 2 1 2 2 z 0 0 3 | − | / | | | | 1 0 z z 2 ragelaw triangle 2 . , hti obtained is that i 1 1 = , . ( | | | | P y z z z z 1 1 1 0 z then , 2 2 | 1 P O + x, | | . + cos P y − 2 (1.13b) 2 (1.13a) 2 z Then . y ) P 2 ψ ;i.e. ); 3 , OP . and In . P → 1 0 ,

This is a supervisor’s copy of the notes. It is not to be distributed to students. 1/06 ( that ( pair The ahmtclTio:I ler emty(atI 5 I) (Part Geometry & Algebra IA Tripos: Mathematical of multiplication words: In Hence multiplication of interpretation Geometric 1.5.1 0 to Remark. Representation (Modulus/Argument) Polar 1.5 Exercise. n ∈ 6 − Z θ < π ..teitgr)de o change not does integers) the i.e. , < θ ,θ r, nodrt e nqevleo h rueti ssmtmsmr ovnett restrict to convenient more sometimes is it argument the of value unique a get to order In 2 ideuvln eut for results equivalent Find π 6 speciﬁes ) π r ( arg oeie aldthe called sometimes , | z z 1 1 z z 2 2 r ( arg = ) z | nqey However, uniquely. z = 1 by | z 1 z | | 2 z z scales 2 1 | r ( arg + ) z , 1 rnia value principal /z Then Consider z z 2 . o each For . 1 by z z 2 does (+2 ) z | 1 z z 2 z 2 oethat Note eetpsto nAgn iga.Let diagram. Argand and rep- in to co-ordinates position polar resent plane using by numbers obtained complex is of representation helpful Another 1 | , not n rotates and z z = = fteargument. the of • • • 2 nπ y hr sauiu au fteargument the of value unique a is there rte nmdlsagmn form: argument modulus in written aldthe called for expression The θ Hence z z pcf ( specify

= c 2 1 r r scle the called is with 1 1 r z ..olydmpcma.k ihems2006 Michaelmas [email protected], r r sin 2 2 = = = cs( (cos (cos + r n x θ i stemdlsof modulus the is z | then , r r ,θ r, + nabtayinteger) arbitrary an z (sin 1 2 1 ouu/ruetform modulus/argument θ | (cos (cos θ iy yarg( by 1 = nqey ic dig2 adding since uniquely, ) 1 . θ cos argument + 1 x . θ θ = = θ 2 cos 2 1 θ 2 2 + + + + ) z − r r 2 θ z y i i ). 2 (cos cos 2 sin sin sin i ntrsof terms in sin + i ( sin 1 of / θ θ θ θ 2 θ z 2 1 1 + z ) ) = . + θ (mod( θ ag( (arg sin 1 ir . , 2 i . r . + . sin sin cos θ θ 2 2 z z θ θ )) . o short). for ) o short). for ) θ ) r x 1 . )) . and = nπ (1.17b) (1.17a) θ r (1.15) (1.14) (1.16) cos such to θ is θ θ θ

This is a supervisor’s copy of the notes. It is not to be distributed to students. hssre ovre o all for converges series This ahmtclTio:I ler emty(atI 6 I) (Part Geometry & Algebra IA Tripos: Mathematical Solution. exercise. Worked Definition. exp( function, exponential real The function exponential real The 1.6.1 Function Exponential The 1.6 hneo aigtepositive the taking on Thence For etnt rmapyn (1.19) applying from note Next that have we (1.19) from Also with (1.18) definition series power the From n For exp( 6 ewrite We n − n rce yidcina above. as induction by proceed 2 hr sntigt rv.For prove. to nothing is there 1 = x()exp( exp(1) = ) hwfor Show exp( − x ,p q p, n, )ep1 exp(0) = exp(1) 1) ∈ exp( exp( R n − x sethe (see ∈ x n exp = ) q )= 1) ,i endb h oe series power the by defined is ), Z = ) hroot th q hr ihu oso eeaiy(wlog) generality of loss without where , ie that times e okdexercise. Worked Solution. exp e n nlssI Analysis exp( x + 1 = x()= exp(1) x()=1= 1 = exp(0) exp exp p q n n exp and , − n n x n hneexp( thence and 1) x q 0: = p > q exp x ( exp = + , course). ,aduig(1.19), using and 2, . e = x 2! y (exp n hneb nuto exp( induction by thence and 2

c e · · · e p q = = = exp( = = hwfor Show ..olydmpcma.k ihems2006 Michaelmas [email protected], 0 p . = ) . p q x = (exp ) X n X X r r r X n X ∞ ∞ ∞ ∞ =0 =0 =0 =0 ∞ = e =0 p m r ( e X . x 1 n x x x n − ! y r =0 p q n ,y x, ! n + ! x ( exp = ) + )= 1) . m r X . m ! ( X y r =0 y ∞ r ∈ ) =0 x ) − . r r ( R − 1 e r y m m m m − = that ytebnma theorem binomial the by ! )! x > q r m e + y m ! − m )! 1 y ! ,that: 0, m ) . ! . for x n r = ) − n m = y e m r n − . (1.19) (1.18) m

This is a supervisor’s copy of the notes. It is not to be distributed to students. hssre ovre o l finite all for converges series This ahmtclTio:I ler emty(atI 7 I) (Part Geometry & Algebra IA Tripos: Mathematical for follows then (1.21) result The Next, series). infinite re-ordering and dangerously living mind not do we as long (as for required as is which that have we real, are arguments Proof. 1.4. Theorem functions trigonometric complex The 1.6.3 Remark. Definition. (1.19). for as essentially proof the with Remarks. Definition. function exponential complex The 1.6.2 if that follows it above the From Definition. 7 gi oeta hs entosaecnitn ihtedfiiin hnteagmnsaereal. are arguments the when definitions the with consistent are definitions these that note Again w ∈ is consider First C hsdfiiini ossetwt h ale eut n entosfor definitions and results earlier the with consistent is definition This When define o irrational For For For For z h ope rgnmti ucin by functions trigonometric complex the z z ∈ ∈ ∈ w R cos C C w x ( exp the , ∈ hsdfiiini ossetwt 11) For (1.18). with consistent is definition this and el hnfo sn h oe eisdfiiin o oieadsn hntheir when sine and cosine for definitions series power the using from Then real. w C = iw x z ope exponential complex define , n X = ) 6∈ ∞ =0 R = cos = = y ( w | − edefine we z ∈ x ( exp (exp | 1) complex. n n X X R aansethe see (again ∞ ∞ n =0 =0 1 hni scnitn owieexp( write to consistent is it then , (2 iw z − w w ( ( 1 − iw n 2 (exp ) ) n w + n exp( 2! )! 1) ≡ ! ) 2 i n e n e + sin e x z (2 n sin and + 1 = iw z w z exp( = exp( = = ) w sdfie by defined is 2 4! n 2 , w cos = x ( exp = ) n 4 )! nlssI Analysis . . . n X + ∞ =0 iw x z i ) ) w n X 7 z − n + ∞ w , . =0 n

! + c z w i = . 1 2 ( ..olydmpcma.k ihems2006 Michaelmas [email protected], i 2 − + n sin X w − ∞ 1) course). =0 z − 2 n i . w ( ) w 3! − (2 w z 3! , 3 w 1 1) 3 n z , . . . 2 + n 1)! + n 2 +1 (2 ∈ w 5! y w n 5 = ) C 2 1)! + . . . n , +1 e z y . ∈ . R . (1.20) (1.21) (1.22)

This is a supervisor’s copy of the notes. It is not to be distributed to students. 2/02 Let form modulus/argument to Relation 1.6.4 ahmtclTio:I ler emty(atI 8 I) (Part Geometry & Algebra IA Tripos: Mathematical 0 from hence and 0 Proof. 1.5. Theorem (again) with that (1.14) representation polar the from follows It Remarks. ahrelegant: rather osdrsltosof solutions Consider 1 for expression Modulus/argument 1.6.5 (1.17b). and (1.17a) conﬁrming oto nt saslto of solution a is unity of root A Unity of Roots 1.7 cos that thence and deﬁnition by Since 6 6 2. 1. w < θ < θ rm(.1 n 12)i olw that follows it (1.23) and (1.21) From otherwise, or (1.21), of conjugate complex the taking From = n ouinis solution One 2 2 θ π π where hr are there , othat so r = § hr are There 1.6.5, θ | z ∈ | ,θ r, θ θ and R sntmlivle.Then multi-valued. not is n n sin and 1 = hnfo (1.21) from Then . ∈ r z itntrosgvnby given roots distinct n θ R cos .Se oegnrlsltoso h form the of solutions general more Seek 1. = and 1 = n tflosthat follows it , arg = ouin of solutions w = z z n 1 z z θ 1 2 nti ersnaintemlilcto ftocmlxnmesis numbers complex two of multiplication the representation this In . ,with 1, = θ 2 θ n = .W euethat deduce We 0. = = x ( exp e z iw 2 2 = = e kπ n r θ i θ + 1 − z r e e r e kπ 2 = n iθ e cos = θ i (cos iw r − θ i 1 = with z 1, = 1 cos = z iw with ) π, kπ ∈ = ≡ n θ r C ie hr are there (i.e. θ , = re 2 + e e − k + and θ k θ i θ i r i iw for n sin and ∈ n sin + i 0 = 2 e 1 = sin Z cos = inθ i n = ecnld htwti h eurmn that requirement the within that conclude We . sin θ , k = ) oiieinteger. positive a 1 .

θ r 1 = c n , . . . , ∈ 1 . θ 1 = r ..olydmpcma.k ihems2006 Michaelmas [email protected], w Z 2 re , w . e − n iθ , i ( = θ i ‘ , − 1 n sin + 2 hroso unity’) of roots th 1 1 θ i . 2 . w ) e , iw z − = e − e r iw θ i . ihterestriction the with (1.24) (1.23) (1.25) (1.28) (1.27) (1.26) 2/03

This is a supervisor’s copy of the notes. It is not to be distributed to students. . eMir’ Theorem Moivre’s De 1.8 because ahmtclTio:I ler emty(atI 9 I) (Part Geometry & Algebra IA Tripos: Mathematical (unlectured). proof Alternative Remark. Proof. 1.6. Theorem Remark. ht(cos that suetu for true Assume hn sn h rvdrsl for result proved the for using true Then, is result the Hence ec eMir’ hoe strue is theorem Moivre’s De Hence rm(1.25) From ω lhuhD ovestermrequires theorem Moivre’s De Although n fw write we If 1. = (cos nθ eMir’ hoe ttsta for that states theorem Moivre’s De + θ i + sin ω n i sin = = nθ e > p θ snl aud soevleo (cos of value one is valued) (single ) 2 ) i/n π p + 1 +1 ,ie sueta (cos that assume i.e. 0, 13)i refor true is (1.30) cos hnterosof roots the then , n cos ω o ( cos = (cos = (cos = cos = (cos = + nθ nθ p · · · + θ ,ads od o all for holds so and 1, + > p + + i + ∀ θ. sin i i p θ θ ω sin 0, n sin cos + + 1) + n nθ ∈ − nθ i i θ 1 pθ sin sin Z ) n = n θ θ . (cos = = (cos = = .Nwageb induction. by argue Now 0. = − + ∈ θ θ n Solution. Example. k X z (cos ) (cos ) θ =0 − cos = cos = = = (cos = sin n i R 1 e ∈ θ i ( sin r 1 are 1 = e i ω and θ. + ( θ i R θ k nθ cos (cos ahro orsod oavre fapen- a of vertex a to tagon. corresponds root Each r 1 write are we of If values roots. smaller) (or Larger by given roots distinct ﬁve thus are There

pθ θ sin + i c p = and θ ) n sin + n 1) + θ n pθ θ p i + ..olydmpcma.k ihems2006 Michaelmas [email protected], + θ pθ 1 θ Put sin Solve i 1 ∈ ,ω ω, , θ + i − − + i sin θ θ n n − + + ) ,ω ω, , sin + sin 1 Z 1 p . θ θ + i 2 = ω i ∈ i > 13)hlsfor holds (1.30) , ω i i ) z sin θ cos = i sin sin n n + 1 sin sin i 2 ) pθ Z (sin θ e z = .Nwconsider Now 0. p ω , sin πk/ 5 ) 0 = . 5 2 θ θ i n θ p θ ) ω , . . . , nθ e pθ ) 1. = ) 3 ω θ i − θ θ. p ω , pθ = with 5 ) , p + hnw eur that require we then , n cos 4 e + ω and , (multivalued). 2 ω πki 2 n pθ i − + = sin 1 cos + ω ute,for Further, . e pθ for 3 2 k ,n θ, i/ π + Then . 0 = < n θ. ω 5 k hnteroots the then , ∈ 4 k sin , ∈ ,say 0, 0 = 1 C il onew no yield , Z pθ 2 ntesense the in , . . ) 3 n , n 4 > . = (1.29) (1.30) 2 − p . 2/06

This is a supervisor’s copy of the notes. It is not to be distributed to students. ahmtclTio:I ler emty(atI 10 I) (Part Geometry & Algebra IA Tripos: Mathematical Definition. of definition the Recall powers Complex 1.9.1 log( of value principal The Example. Definition. Remark. Thus let logarithm complex hence the and of nature the understand To Powers Complex and Logarithms 1.9 Example. any for Remark. k ∈ ic arg Since ic log Since If h au of value The Z h rnia au flog of value principal The For . z = z − 0, 6= x z z with smlivle ois so multi-valued is samlivle ucin oi log is so function, multi-valued a is x ,w z, i a i i for , − i sgvnby given is x log x ∈ ∈ slog is ) = = = = e C ,a x, R z u v define , and e e e e log = log = arg = = i i i − ∈ lg1+2 (log (log log “ | w R x 2 x − > x k a , | i | | z + log = < π i z + | > x z = + | 2 1 ssc that such is w z i iπ ki ” ,then 0, iπ − | | = e z arg w π by arg a x . and 0 i.e. , z log + x = z | . r i w iπ/ o any for ) +(2 | log = z x = + θ Im(log = 2) z x ( exp = 2 + i w e a k Definition. tion namely equation if plex that already know We r ( arg w sol endut nabtaymlil of multiple arbitrary an upto defined only is 1) + rainl namely irrational, | log kπ z k w | − + w ∈ z iπ a = .

x z c z i o any for Z log of . ) y u ) arg ..olydmpcma.k ihems2006 Michaelmas [email protected], 6 + log = o any for x wihi real). is (which iv π. ) . z edﬁelog deﬁne We . with k e x y ∈ o ln (or k = Z ∈ ,v u, . x Z e w x . ∈ a nqera solution, real unique a has x = fyuprefer). you if R ∈ z Then . . z for R and z ∈ e u > x C + iv s‘h’solu- ‘the’ as ,tecom- the 0, = z (1.32b) (1.32a) = e (1.31) (1.33) 2 kiπw re iθ , ,

This is a supervisor’s copy of the notes. It is not to be distributed to students. 3/03 3/02 .02Circles 1.10.2 osdra Consider Möbius Transformations 1.11 ahmtclTio:I ler emty(atI 11 I) (Part Geometry & Algebra IA Tripos: Mathematical where Exercise. Solution. exercise. Worked Plane Lines Complex the in 1.10.1 Circles and Lines 1.10 h iepse hog h origin. the through passes line the z ,b ,d c, b, a, 0 w ¯ − fteln asstruhteoii hnput then origin the through passes line the If hwta if that Show map z ¯ 0 w ∈ of ,te h qaino h ieis line the of equation the then 0, = C hwthat Show C r osat.W eur that require We constants. are → z w C ¯ − (‘ C zw z ¯ 0 into w ¯ ,then 0, = − z C ¯ 0 )dfie by defined ’) w z fadol f(ff h ie(.4)pse hog h origin. the through passes (1.34a) line the (iff) if only and if 0 = 7→ z z = 0 = γw f Remarks. numbers complex of set the i.e. nteAgn iga,acrl fradius of circle a diagram, centre Argand the In Thus sa lentv ersnaino h line. the of representation alternative an is λ Remark. plane) (complex through diagram line straight Argand on points the in represents equation the fixed For o some for ( z = = ) • • z λ ¯ w ¯ Since If that hence and , qainfrtecrl is circle the for equation centre with circle a for equation ( the is which ,q p, v − az cz z

( c z zw ¯ S r = + n radius and ) z γ + Since 0 ..olydmpcma.k ihems2006 Michaelmas [email protected], n(.4) n h eutflos If follows. result the and (1.34b), in 0 = ∈ = ∈ | w , | d x b z .Ti sstse by satisﬁed is This 0. = z | R z , + { R − − | , z 2 ∈ . z iy v v λ v ∈ − w ¯ | C | ∈ 2 z 2 and ( = C − vz ¯ ( = − C with w z (¯ = zw ¯ sgvnby given is ) : − z = z r x v 0 | v − z nCreincoordinates. Cartesian in z z = − = = z ¯ w 0 − − z + z p + p 0 z ¯ 0 ,advarying and 0, 6= v ) ¯ v ) + 2 w | ( ) − /w λw ¯ v w | ¯ z ( + = z | iq − z 2 z ¯ 0 uhthat such 0 − r ∈ = z ¯ y n aallto parallel and then } . 0 v − w r R , z ,a alternative an ), 2 tflosthat follows it , q . ,adhence and 0, = ) 2 = | z r r − and 0 6= 2 (1.35b) (1.34b) , (1.35a) (1.34a) λ v (1.36) | ∈ w = . R r . ,

This is a supervisor’s copy of the notes. It is not to be distributed to students. hs h aetknFrhrMteaisAlvlsol hncnld something. conclude then should A-level Mathematics Further taken have who Those Remarks. ahmtclTio:I ler emty(atI 12 I) (Part Geometry & Algebra IA Tripos: Mathematical Exercise. Remarks. versa. vice and (1.36), to i.e. the For Inverse 1.11.2 therefore is Möbius maps all ( that note we where map combined the Then Möbius transformation second a Consider Composition 1.11.1 (ii) (i) 2. 1. • • • α ieetpit a odffrn ons ..if i.e. points, different to map points different c digte‘on tifiiy makes infinity’ at ‘point the Adding ( that require only infinity). need to we hence (ii), condition f of subset a is (i) Condition h nes 13)cnb eue rm(.6 yfra manipulation. formal by (1.36) from deduced be can (1.38) inverse The maps (1.36) = ( and z ,b ,d c, b, a, − aseeypito h ope ln,except plane, complex the of point every maps ) d , o Möbius maps For d β az cz r o ohzr,s httempi nt ecp at (except finite is map the that so zero, both not are = 1 1 z ∈ + 0 + b 7→ , R C d b γ \{− αa si 13) osdrteMöbius map the consider (1.36), in as z 6= = 00 az cz = + c d/c z 2 and 2 βc g 7→ + + ( } ( ) z closed d f b 0 to z δ = ) γb z , 00 , = 00 g C sas Möbius transformation: a also is + = and − reuvlnl ( equivalently or αz \{ γz δd ne composition. under a g 0 a/c hnfo (1.37), from Then . 0 ( ) + + z h − 0 = ) } β δ eosrt httemp are maps the that demonstrate ( hl 13)maps (1.38) while , αb f = = = z complete. where 00 + = g βd ( α α γ γ ( αa ( γa − ( ( f ( ) cz az az az cz az dz cz z ( + + ad γa 1 z 0 ,β ,δ γ, β, α, + + + + + + 0 )) − 6= βc d δc d b b + − + b b a z ) ) + ) + ) b z bc + 2 δc + z z

z 00 c , then )( ( + ( + ( = ) β δ = = β δ ..olydmpcma.k ihems2006 Michaelmas [email protected], z C ∈ ( ( 1 − z αb γb cz cz \{ ecnld ht(.8 sthe is (1.38) that conclude We . C − d/c ad z + + , 1 + + 0 z a/c 2 6= − noaohr( another into , d δd βd z d ) and ) ) bc } z 0 6= = ) ) 2 0 to )( ..w eur that require we i.e. , , − associative αδ , d/c C αδ − \{− ..( i.e. ); − βγ βγ d/c ) ad .Hnetestof set the Hence 0. 6= z ad ..( i.e. , 0 6= } − = . − bc − . bc ) d/c fg 0. 6= ) 0 6= ) h smapped is = . inverse f (1.38) (1.37) ( gh ).

This is a supervisor’s copy of the notes. It is not to be distributed to students. ahmtclTio:I ler emty(atI 13 I) (Part Geometry & Algebra IA Tripos: Mathematical Maps Basic 1.11.3 neso n reﬂection. and Inversion rotation. and Dilatation Translation. h circle The where but radius same the by of oﬀset circles centre a to with map parallel circles to while map lines lines, e.g translation; a represents map This iepse hog h rgn n h apdcurve, mapped the and origin, the z through passes line the through when is passes exception (which centre circle with a origin) is this (1.35a) From as successively ten line The origin the on tred represents map this Hence d line The origin scales map This that ymliligby multiplying By where becomes ¯ 0 w ,s that so 0, = ¯ z − 0 =

z O z v z a 0 Put 0 0 0 w 0 . 1 r z z = = − e ,i loasrih ietruhteorigin. the through line straight a also is 0, = z | = = z −

av 0 az z a v 0 c iθ z = − ¯ 0

z z 1, = 0 z 0 − ..olydmpcma.k ihems2006 Michaelmas [email protected], 0 0 ¯ , z = z and − 0 v az 0 0 w and + + ¯ | z z = r z = z = i.e. z z w 0 − λw λw w ¯ = by O 0 ¯ c 0 | w 0 r etput Next ¯ az z w z 1 w o put Now ¯ r 0 and , w z ¯ requivalently or − − and 0 = z 0 | hsif Thus . z − reuvlnl se(1.34b)) (see equivalently or , where , w | ¯ a | 0 0 = − | ¯ 0 − 2 0 becomes z w 0 a w || zw ¯ w ¯ z t. hseuto a erewrit- be can equation this etc., , w ¯ z b 0 = + ¯ z | z ¯ − w z | ¯ | 0 ¯ . z 0 0 ( = 0 a ¯ n rotates and | = z 0 w w = λaw ¯ aw = | 0 e eeto ntera axis real the in reﬂection = w neso nteui circle unit the in inversion r w i hc sjs nte circle. another just is which ,

| z (arg ,i hc aeteoriginal the case which in 0, = z z z 2 hc sjs nte line. another just is which , z − 0 0 λ d | b + n radius and 0 w a = − w z ¯ = ¯ w oobtain to 1 = ∈ ¯ z 0, = a 1 ¯ . b = − 0, =

0 +arg − z − n arg and R z w 0 0 z re 0 ¯ z z ¯ z ¯ 0 + w − and 0 0 ¯ 0 w w iθ z c . w z , w | ) − w λw z z b ) ¯ and 0 = yarg by . 0 then 0 z z w w 1, = − ¯ ¯ 0 0 0 z z ¯

, w 0 0 ∈ 0 = z v 0 = 0

w | C ¯ 2 a − = w c − ¯ becomes , z bu the about ¯ 0 d and 1 = r arg w 0 (1.39b) (1.39a) so 1 = ,

. The . z. cen- 3/06

This is a supervisor’s copy of the notes. It is not to be distributed to students. Exercises. h esnfritouigtebscmp bv sta h eea ¨bu a a egnrtdby generated the be consider this can Möbius see map To general reflection. the and inversion that and is rotation, above and sequence: maps dilatation basic translation, of the composition introducing for reason The Möbius map general The 1.11.4 ahmtclTio:I ler emty(atI 14 I) (Part Geometry & Algebra IA Tripos: Mathematical straight and circles to so). lines does straight transformation and constituent circles each transforms Möbius (since map general lines a that implies above The (ii) (i) translation rotation and dilatation iaainadrotation and dilatation reflection and inversion translation osrc iia euneif sequence similar a Construct that Show eue to reduces if is exception rgnmpt ice,wiecrlsadsrih ie htd astruhoii a ostraight to map origin through pass do that lines straight and circles while lines. circles, to map origin line. straight a Summary. of equation the is which ¯ centre with circle a for equation the is this (1.35b) From equivalently or Hence map the under Further, requivalently or ne neso n eeto,crlsadsrih ie hc ontps hog the through pass not do which lines straight and circles reflection, and inversion Under | z v 0 | z ¯ 2 0 = − r ( | 2 v nwihcs h rgnlcrl asdtruhteoii,adtemap the and origin, the through passed circle original the case which in , | z z z z z z 2 0 4 3 2 1 v − 7→ = 7→ 7→ 7→ 7→ r z 1 z 2

z z z z 1 ) z c h circle the 5 4 3 2 1 z = 0 and 0 = = 1 = = = 0 z − − cz 0 z z z ¯ v 4 1 0 /z (

bc | + + v = | 2 v | − 2 a/c d | c , r 2 | v ¯ d − vz z ad z − 5 0. 6= r 0 − = r 2 ¯ + (1 i.e. 2 ) v z z v ¯ az cz 3 − | 0 − z = ¯ + 0 + vz + vz | 1 = r 1 (

d 0 b c | 0 becomes − ) v ( ( ( − . | c bc c ..olydmpcma.k ihems2006 Michaelmas [email protected], , 2 vz 1 | v/ 0) 6= 0) 6= v v ¯ − − 6= | z 0 ¯ 2 | ( 0 r | v ¯ = ad v 0 = 1 + 2 z ¯ | ) 0 2 r 2 ) − | z 0 r | = = 2 . n radius and ) r ( 2 | . v z ¯ | 0 2 z r 0 − 2 , r 2 ) r/ 2 . ( | v | 2 − r 2 .The ).

This is a supervisor’s copy of the notes. It is not to be distributed to students. ahmtclTio:I ler emty(atI 15 I) (Part Geometry & Algebra IA Tripos: Mathematical of tion/sense vector a represent We representation Geometric 2.1.1 Remarks. vector Definition. Vectors mathematically. field. them magnetic 2.1 the describe a to particle, and, field, able multiplication electric a subtraction, be an of addition, to force, by are position (e.g. a you quantities wave, the if these a e.g. differentiation) manipulate of term, direction, to propagation next how a of know and direction to need the magnitude You particle, a a temperature both of e.g. have velocity scalar, quantities A other as that). However numbers before such possibly to differentiation) (or ( multiplication, school refer subtraction, position We in addition, of day by number. Such (e.g. first concentration. single scalars your density, such a since temperature, manipulate by to time, how specified e.g. learnt completely have magnitude, be a can have quantities just quantities scientific Many This? Study Why 2.0 Algebra Vector 2 8 • • • • ohsiae ahmtcasueneither use mathematicians Sophisticated aeo u xml ewrite we example our of case force e.g. vector, A i.e. direction, same e.g. bold, vectors in Two vectors vector represent a of will magnitude notes The the under the squiggle course a put this will I of head/blackboard purpose the For . ,y z y, x, uniyta sseie ya[oiie antd n ieto nsaei alda called is space in direction a and magnitude [positive] a by specified is that quantity A v sfrom is srfre oa a as to referred is ) u and v A v saln emn,say segment, line a as u F to hti ucino oiin( position of function a is that , r qa fte aetesm antd,i.e. magnitude, same the have they if equal are sprle to parallel is B ). v swritten is F ≡ clrfield scalar F bold v ( ,y z y, x, n nbt etr r ntesm direction/sense. same the in are vectors both in and nor | v Examples. | (ii) AB . ). (i) squiggles ntecs foreapew write we example our of case the in ; → ihlength with , on ntecmlxpae n ec otepo- the to hence point. unique and that of plane, a vector complex to sition the corresponds in point number complex Every Remarks. vector point Every and ∼ v • • . . r 8 h oiinvco sa xml favector a of example an field. is vector position The by represented usually is by magnitude) represented is than vector rather position the Often ,

= c ,y z y, x, r rmsm hsnorigin chosen some from , OP ..olydmpcma.k ihems2006 Michaelmas [email protected], = srfre oa a as to referred is ) P | v | | n3 o D pc a a has space 2D) (or 3D in r n ieto htof that direction and | . r u vnte h egh(i.e. length the then even but , | u | = | v | T n hyaei the in are they and , hti function a is that , etrﬁeld vector T v O nteover- the On . ≡ with , v T scalars ( tedirec- (the ,y z y, x, r . position nthe in ; r You . = ). OP → x 4/03

This is a supervisor’s copy of the notes. It is not to be distributed to students. 4/02 ubro rprisflo rmteaoedfiiin nwa follows what In definition. above the from follow properties of number A u h poiedrcinsneas direction/sense opposite the but If scalar a by Multiplication 2.2.2 Remarks. ahmtclTio:I ler emty(atI 16 not will I) I (Part but Geometry right, & it Algebra get IA and Tripos: try Mathematical will sometimes and I 0 ‘sophisticated’. mean more sometimes be will to 0 need context on will Depending you always. overhead/blackboard the on However, law: Distributive Addition 2.2.1 Vectors of Properties 2.2 (iii) (iv) (ii) (v) (i) λ 9 aeatmtdt laswrite always to attempted have I ∈ If i.e. (1.13a), to analogous inequality triangle a is There is Addition is Addition Since hsi rvdi naaoosmne o(.3)uigtecsn rule. cosine the using (1.13a) to manner analogous an in proved is This R | a then | ,write 0, = OB → λ = a AC a magnitude has associative commutative → ti lotu that true also is it , a = 0 where , i.e. , i.e. , | λ 0 b || nbl ntents fIhv o twogsmweete laeltm know. me let please then somewhere wrong it got have I if notes; the in bold in 0 + a a | stenl etro eovector. zero or vector null the is sprle to parallel is , if 0 a = < λ ( + b λ ( λ ( , seblwfor below (see 0 b a | + a OA + + → a (vi) n rm(2.3) from and + µ + b c ) b a ( = ) + = ) b | where equivalently or en utato fvcosby vectors of subtraction Define vector the Define ieto/es s hti is the it called that (so direction/sense as magnitude same the the to according add Vectors a 6 AC = n thstesm ieto/es as direction/sense same the has it and , → = | b a = | λ λ + C. OC + a

OACB c → a a a + | + + λ 0 b ..olydmpcma.k ihems2006 Michaelmas [email protected], . . b negative | 0). = λ µ . + ) b a 0 , . + saparallelogram. a is OA c → b b . 9 − = ( − of o l vectors all For + − a a a b a ,µ λ, + OB a + ) ≡ → . ob aallto parallel be to n ssuch is and b a b u ohv h opposite the have to but , ∈ a ( + = = R = aallga rule parallelogram . anti-parallel − 0 C, OC c a . → ) , b . a a .Ti is This ). ohave to , if > λ (2.7b) (2.1b) (2.8b) (2.1a) (2.8a) (2.7a) (2.6) (2.3) (2.5) (2.4) (2.2) : 0,

This is a supervisor’s copy of the notes. It is not to be distributed to students. Further is quadrilateral closed the since Then midpoints. respective n thus and ahmtclTio:I ler emty(atI 17 I) (Part Geometry & Algebra IA Tripos: Mathematical (2.12), using from Similarly, AB by can interpretation quadrilateral and geometric the their of and vertices manipulation the vector parallelogram theorems. Denote of geometric a rules prove the form to that quadrilateral used fact any the be of of example sides (e.g. an the broken This of often midpoints is the that convention Example: a is this 2.2.3 that note but vector, unit a indicate to used see often is ˆ A a vectors. Unit Definition. -1: and 1 0, by Multiplication law: Associative ˆ → stre a termed is , § D 2.7.1). BC ecnas eoe h nlrsl ihu peln ogoer yuig(.) 26,(2.7a), (2.6), (2.4), using by geometry to appealing since without (2.10b) result and final (2.10a) (2.8a), the (2.7b), recover also can We → Let . and SR → Q P a = → h vector The CD , ntvector unit Q P → Suppose → b RS = a → , n let and , A P Since . + c → b and = = = + + a AQ c c → d Q P − − 2 1 since 6= P + = ( 2 1 ersn h sides the represent = c , Q P , 0 ( d → λ Q + ( and a hndefine then , − 1 2 a 0 = , + a d 1) + R 1 0 ) + b SR µ a a a . and ) b 1 2 b aeeulmgiueadaeparallel, are and magnitude equal have sdsrbda a as described is = = = | a . ˆ S ( | − eoethe denote = − a 0 1) A ,

a a | λ , a 1 ic 0 since (2.12) ( B | µ DA since a ˆ

( = (( = ( = ( = = = → , a | = a C ( = ) | , = 0 − | a a | − − − − | a − a − | | 1) 1) 1) + 1 | iercombination linear a . 1 . 1) 0, = a 1 | a a | a | |

a λµ + ( + c a | 1 + | a 1 = 0 ..olydmpcma.k ihems2006 Michaelmas [email protected], ) a a = − a . − | ) . a a a − | ) and a − 1 of < QSR P a 0. and b saparallelogram. a is . (2.10b) (2.10a) (2.10c) (2.11) (2.9) 04/06

This is a supervisor’s copy of the notes. It is not to be distributed to students. ahmtclTio:I ler emty(atI 18 I) (Part Geometry & Algebra IA Tripos: Mathematical Projections 2.3.2 product scalar the of Properties 2.3.1 Product Scalar 2.3 (iii) (iv) (ii) (i) If 0 If is product scalar The modulus: its of square the is itself with vector a of product scalar The a 6 and 0 6= < θ 1 2 π b then , and 0 6= a · commutative b a > · b ,wieif while 0, ,then 0, = : 2 1 a θ < π a and (v) · a a · b b = 6 hr 0 where iial,o yuig(.5,( (2.15), summary using by or Similarly, instead If the head’. to Remark. ‘head or tail’ to ‘tail placed been sense have (anti-clockwise) positive the in sured b Definition. en htcos that being Suppose rmgeometry, From The scalar to projections. the discuss of to jection property need we more product, one deducing Before = utb rhgnl(i.e. orthogonal be must | π sdfie ob h el(clr number (scalar) real the be to defined is a b then , | b o product dot 2 wihhr ewl eoeby denote will we here (which = ·

a c of a 6 . a 2 λ ..olydmpcma.k ihems2006 Michaelmas [email protected], a 0 < λ a . a The θ ∈ = · a · onto a 6 b R The ( · | λ · a then 0 θ ( If . < π . ( b clrproduct scalar λ 0 > | λ | a b a = ) steagebetween angle the is 0. | b b clrproduct scalar b b 0 · ( = ) > λ ) hssince Thus 0). | = ) | b sta atof part that is ======| then 0 λ a | | a λ −| −| | | a a a | ) || cos a θ || λ | | | λ λ cos λ λ a b · λ · = | || a || | || b | b b a a θ a λ a cos · θ | = || 1 2 . sas eerdt as to referred also is · || b b · λ asm o h time the for (assume cos( π b | b b ftovectors two of b | a b b λ . , θ ). | cos a | ) | cos a cos 0 a π . · a sprle to parallel is · b θ 0 hti parallel is that − b ). θ θ a = θ . and ) λ nethey once a b (2.17a) · (2.13) (2.14) (2.15) (2.16) a b mea- pro- and In . b ,

This is a supervisor’s copy of the notes. It is not to be distributed to students. . etrProduct Vector 2.4 product scalar the of property Another 2.3.3 where ahmtclTio:I ler emty(atI 19 I) (Part Geometry & Algebra IA Tripos: Mathematical rule cosine the Example: 2.3.4 with sides both ‘dotting’ before LHS ( the then on (2.8a) use Now (2.13) from Hence Exercise. λ b ˆ + steui etri h ieto of direction the in vector unit the is hwta 21a ean rei cos if true remains (2.17a) that Show µ ) a a | = a · | γ b 2 a a ec ( hence , + a | a · | 2 c λ a + a µ = = = 0 ) | = a | { { a 2 | rjcino ( of projection of projection a · = | ( | | b a a a γ Definition. savco uhthat such vector a is | (iii) || | + · 2 (ii) a (i) b b BC | c 2 | b n so and , ) | ue,ie aeargthn,pitteidxfigri the in finger index then the point hand, of right direction a take i.e. rule’, a a a 0 with . b b < θ 2 a × | × × . = b b b = = = = | ≡ = b 0. a if true [clearly] assume is result The ews oso that show to wish We b a a 6= | a h es/ieto endb h ‘right-hand the by defined sense/direction the has onto spredclrotooa oboth to perpendicular/orthogonal is b 6 × for + λ · The BA BA BA BA | 0 b 2 → b c θ BC + ); BA b → onto ) → b 6

si h ieto ftethumb. the of direction the in is a c 2 2 2 µ a and , · etrproduct vector } a ( = h eodfigri h ieto of direction the in finger second the , 2 + − 2 + a π BA | = ..olydmpcma.k ihems2006 Michaelmas [email protected], + 2 + → .Te rm(.7)(fe exchanging (after (2.17b) from Then 0. 6= ooti h eut(..if (i.e. result the obtain to 2 enda before; as defined γ a AC { = AAC BA AC BA a → BA ). b | rjcinof projection + a · → a } b ˆ · or | × BA ) ( → BA · b b → ˆ · b c AC → + , | · r( or BA = + AC → c cos cos → + = ) a | AC a b → AC b geometry) (by + × α θ || + + b a + c b + AC | 2 | AC → c 2 · → onto sin AC fa ree pair ordered an of for ) AC b a + · , θ 2 2 ,s henceforth so 0, = BA a a → . a } · etc.) , c + λ . a AC a → + and µ · a (2.17b) AC b → (2.18) (2.19) = and , b a γ , (if a b , 5/03

This is a supervisor’s copy of the notes. It is not to be distributed to students. 5/02 .. rpriso h etrproduct vector the of Properties 2.4.1 ahmtclTio:I ler emty(atI 20 I) (Part Geometry & Algebra IA Tripos: Mathematical Remarks. (iii) (vi) (iv) (ii) (ii) (v) (i) (i) Consider product vector the of deﬁnition the from follows It exists there If zero: is itself with vector a of product vector The is product vector The is overhead/blackboard) my on except favour of out falling is (that the notation alternative as An to referred also is product vector The ecnueti emti nepeaint hwthat show to interpretation geometric this use can We geometry By to direction opposite the in looking when (‘anti-clockwise’ direction to orthogonal plane i.e. that geometry by noting ﬁrst by to equated be tion a and 0 6= b 00 { rjcinof projection sotooa oboth to orthogonal is b b' h ln epniua o hog nage anti−clockwiseabout throughanangle the planeperpendicularto 00 λ is theprojectionof = b | b ∈ a ˆ 0 and 0 6= a ˆ | × R = × b uhthat such a | . b a ˆ b b hsvco a ecntutdb w prtos is project First operations. two by constructed be can vector This . | ognrt h vector the generate to not θ sin noplane onto a b' b × θ omttv fo h ih-adrule): right-hand the (from commutative = b a a | = = a ˆ b and × λ 0 onto b ⊥ a then , b b a ). × | b 0 to and , n hshstecretmgiueadsnedrcinto sense/direction and magnitude correct the has thus and , a + ( b × a c a } θ + 0 ( × b λ or 0 = a c b 00 ( = = + b b = ) rs product cross (looking intheoppositedirectionto b'' × = ) a h aemgiueas magnitude same the has 0 = hnrotate then , a { { is theresultofrotatingvector a − b = λ rjcino ( of projection of projection θ × b ( + 0 a =

× b c a . c × π + ) a ..olydmpcma.k ihems2006 Michaelmas [email protected], 0 i.e. , b . a , ) . × b' . b'' b 2 π c a 0 2 π , about c and b noplane onto + b a ˆ c ). noplane onto ) a ˆ r aall(requivalently (or parallel are by b 0 π 2 ute,b construc- by Further, . ⊥ na ‘anti-clockwise’ an in to a b' ) a ⊥ } to a a } b , (2.20b) (2.20d) (2.20a) (2.20c) noa onto a ∧ b .

This is a supervisor’s copy of the notes. It is not to be distributed to students. 5/06 ahmtclTio:I ler emty(atI 21 I) (Part Geometry & Algebra IA Tripos: Mathematical product: triple Vector products. triple two form can we product: product, triple (‘cross’) Scalar vector the and product (‘dot’) scalar the Given Products Triple 2.5 the and Let triangle/parallelogram a of area Vector 2.4.2 .. rpriso h clrtil product triple scalar the of Properties 2.5.1 O rmuig(.5 n (2.20a). and (2.15) using from rmuig(2.20a). using from , A etrarea vector , C and B eoetevrie faprleorm with parallelogram, a of vertices the denote is a × ( a b × . b ) × ( c rao aallga = parallelogram of Area a = × − b c ) h aemgiuea h rao h rage n snormal is and triangle, the of area the to as magnitude same the epciey Then respectively. the be Let 2 1 rao rage= triangle of Area · × a c OAB × ( O = a oueo parallelepipeds. of Volume b , × c A altitude srfre oa the as to referred is ..nra otepaecontaining plane the to normal i.e. , · b ( and = ) a n yte rotating then by and if parallelopi- edges with or parallelopipedon) parallelopiped or pede or parallelipiped (or about clockwise’ × a , − b B b = ) ( through

b eoetevrie fatinl,adlet and triangle, a of vertices the denote c and oue=Bs Area Base = Volume | × ..olydmpcma.k ihems2006 Michaelmas [email protected], a − 2 1 a × A.NB . OA c ( ) b OA b aetesneo h ih-adrule. right-hand the of sense the have → × × B | b , c Denote . a and 00 a = ) = = ( = etrarea vector + oso that show to · c c = c h oueo parallelepiped a of volume The OB × b , → 00 | | a 0 a ( 2 1 , ( ( = a × b A.OB . OA × c sbfr.Then before. as × OA 0 × b → b b n ( and b | | ) a + ) c · ftetinl.I has It triangle. the of ) and · c | c , c cos ) > b | 00 × sin a + OB 0 . φ → and Height , c θ a ) , = 0 by b by b 1 2 and . | a a π 2 (2.23b) (2.23a) × and (2.21) (2.22) ‘anti- c b NB is | . b

This is a supervisor’s copy of the notes. It is not to be distributed to students. 6/03 ahmtclTio:I ler emty(atI 22 I) (Part Geometry & Algebra IA Tripos: Mathematical Uniqueness. Definition. scalars real unique and suitable for origin an vector space, position 2D consider First space dimensional Two 2.6.1 Components and Bases 2.6 vectors. Coplanar Identities. 10 hti motn steodrof order the is important is What n so and Hence that (2.24a) and (2.15) from thus follows parallelepiped; also the It of volume’ ‘negative the equal all must products triple scalar their so ( and triples ordered the Further that follows it parallelepiped same the of volume ( triples ordered ic h oueo h aallppdi eo ovreyi non-zero if Conversely zero. is parallelepiped the of [ volume the since inconsequential. is ‘dot’ and ‘cross’ the notation of the order the hence and a , b , c sueta h ree rpe( triple ordered the that Assume ,then 0, = ] λ ups that Suppose esyta h set the that say We − r λ faypoint any of 0 If = a µ , b b a , − c , and , µ b a λ 0 ( ,ad( and ), and ,since 0, = and a c P × r olnrthen coplanar are ntepaecnb xrse as expressed be can plane the in c c a µ { a ) , a r coplanar. are b , r o-nqe n htteeexists there that and not-unique, are · λ ( , O c c a b b , and , and n w o-eoadnnprle vectors non-parallel and non-zero two and , a } × b ( = a , ,( ), spans b b c and r .Sneteodrdsaa rpepout ilaleulthe equal all will products triple scalar ordered the Since ). . ( b µ ) r a ( = b . · a [ = λ × a c × , , λ , OP b a − b a ( = h e fvcosligi h plane. the in lying vectors of set the → b a b , , ) r o aall(r‘rs’fis with first ‘cross’ (or parallel not are c , λ [ c ) + a · c n ( and ) = 0 a h es ftergthn ue hns othe do so Then rule. right-hand the of sense the has ) b · ) , c ( = ] µ a c b emti construction. Geometric λ × ( = b , = ( = a c c = 0 = ] + a ) a c µ c · × λ × · µ n hence and hnteeexist there Then at tension) to allel , 0 a

( 0 c b b − a b b b ( = , , , ) + ..olydmpcma.k ihems2006 Michaelmas [email protected], ) a µ × · · l aetesneo h ethn rule, left-hand the of sense the have all ) ) µ c c a b c 0 ) . × b OA , = → , ON . a → 10 − ) = N · ( o hsrao esmtmsuse sometimes we reason this For = a b a r alnnprle ie intersect). lines non-parallel (all × . µ = ,λ λ, ointersect to b OP ,µ λ, b → ) 0 rwaln through line a Draw ,µ µ, , a · and = c , ∈ . b λ R a 0 and ∈ uhthat such + a NP a OB → R µ → and n then and b c uhthat such = = . r uhthat such are λ b b a hnthe Then . o t ex- its (or , b (2.24b) (2.24d) (2.24a) (2.24c) P (2.25) ). par-

This is a supervisor’s copy of the notes. It is not to be distributed to students. ahmtclTio:I ler emty(atI 23 I) (Part Geometry & Algebra IA Tripos: Mathematical that conclude we space, 3D spans also of uniqueness The Uniqueness. [ if that conclude We origin an space, [ 3D consider Next space dimensional Three 2.6.2 Remark. and set Deﬁnition. that say we then Deﬁnition. Remark. scalars real unique and suitable for a , b , c since consider and ] ) hntepsto vector position the Then 0). 6= a and { { b a a , , ecnso that show can We ffrtovectors two for If esyta h set the that say We · b b ( b b } , c r ieryindependent. linearly are × onthv ob rhgnlt eabasis. a be to orthogonal be to have not do a } λ and c onthv ob uulyotooa ob basis. a be to orthogonal mutually be to have not do , = ) a µ , b and b c , c are r · ] ( ν · hnteset the then 0 6= b ( mle htteset the that implies b ieryindependent linearly × a λ × λ c { , and .Hne n iial rb permutation, by or similarly and Hence, 0. = ) a c = µ , ( = ) λ α b O and [ [ , a a } b r n he o-eoadnncpaa vectors non-coplanar and non-zero three and , µ = = , , { + r sa is b r b and a and ν faypoint any of , , β , = c c b λ λ b r nqeb osrcin ups that Suppose construction. by unique are basis OP ] λ ] , a a → ,β α, µ , a = c ν { · · } . a + ( ( = 0 , b b sa is o h e fvcosligtei ln fi saspanning a is it if plane in the lying vectors of set the for b ∈ µ λ { × × . = , b a a R c ⇒ emti construction. Geometric c c , basis + } + , [ [ b + ) ) a r P spans ν , , , µ , c c b c b nsaecnb xrse as expressed be can space in α } µ , ) , o some for ist Π intersect will because olw that follows exists to taining ON o Dspace. 3D for a +

c b · is → c = ] ] ( · OC ν b Dspace. 3D ν , ieryindependent linearly ..olydmpcma.k ihems2006 Michaelmas [email protected], ν → β ( c b × isi h ln Π plane the in lies , ∈ 0 = = ,µ λ, × c a r R ) = a c c ,µ ν µ, λ, and , , hsln antb aallt Π to parallel be cannot line This . + ) = uhthat such [ [ b ∈ a r , , and OP b a ν b R → rwaln through line a Draw . , c ab , ∈ b c uhthat such · a at say , ] ] c ( R b . = = e Π Let . r o olnr ec it Hence coplanar. not are × NP → c n hssneteset the since thus and , λ ON ) → ab a = N ab + from , r ON n hr ilex- will there and , + ν → etepaecon- plane the be µ sgvnb (2.27) by given is c a b NP ute,since Further, . , = → + b λ § ν .. there 2.6.1 and a c P , + parallel µ c (2.28) (2.26) (2.27) b (i.e. It . ab 6/02

This is a supervisor’s copy of the notes. It is not to be distributed to students. ahmtclTio:I ler emty(atI 24 I) (Part Geometry & Algebra IA Tripos: Mathematical Definition. Definition. Components 2.7 Definition. in spaces; dimensional higher had to ‘boot-strap’ can We spaces dimensional Higher 2.6.3 where y‘otn’(.0 with (2.30) ‘dotting’ By magnitude, unit have a and define orthogonal to mutually right-handed. said are are are vectors they they basis case the which if in simplified are matters However, that 3D noted in have basis We standard or Cartesian The 2.7.1 ( call we then Remarks. vectors 3D all for Hence (i) n suigta eko h ai etr adrmme htteeaea nonal infinite uncountably an are there that write remember often (and we vectors axes), basis Cartesian the of number know we that Assuming ntrso hsnotation this of terms In etr.Hwvr hsi o eesrl h etwyo okn thge iesoa spaces. dimensional higher at looking of way best the necessarily not is this However, vectors. v x v , y v , ffravector a for If If the define we fact In ,µ ν µ, λ, z { ∈ a , R b the ) edfie( define we , , { c a } , sa is i b v components , , j c v } and basis n atsa basis Cartesian a and onthv ob uulyotooa o ih-add ob basis. a be to right-handed) (or orthogonal mutually be to have not do v k x i v dimension v , o Dsae n ffravector a for if and space, 3D for x (1 = epciey eddc rm(.9)ad(.9)that (2.29b) and (2.29a) from deduce we respectively, v y = v , ( = of v , z rhnra basis orthonormal 0 r ob the be to ) v · , v i ihrsetto respect with 0) r · v , = i = ) , fasaea h ubr f[ieet etr ntebasis. the in vectors [different] of numbers the as space a of i v v λ x ( + y a j i ( = = + (0 = + v v v v omabssfr3 pc satisfying space 3D Then for ˆ. a basis add a to form conventional not is it where xs Let axes. Let µ atsa components Cartesian · { x y b j i · n j v , , , ) j 1 j dmninlsaew ol n httebasis the that find would we space -dimensional + + j OX v , , y , ( + k 0) v , ti locnetoa oodrte othat so them order to conventional also is It . v ν {

} z c c a z , , , k v i ) z , , k ..olydmpcma.k ihems2006 Michaelmas [email protected], OY j × i b , · . = k , k j c ) v , eteui etralong vector unit the be eteui etralong vector unit the along be vector unit the be (0 = } = k OZ . · i i r . k k · · , j i , . , eargthne e fCartesian of set right-handed a be = = 0 , [ 1) j i j j , × · · j . j of k , k k = = 1 = ] v = k k ihrsetto respect with i · · , k i . 0 = 1 = k × , , i OX OZ OY = j , , , , { { (2.29d) (2.29b) (2.29a) (2.29c) i i , (2.33) (2.30) (2.34) (2.31) (2.32) , j j , , k k } } .

This is a supervisor’s copy of the notes. It is not to be distributed to students. 6/06 ups that Suppose Identities Component Vector 2.8 ahmtclTio:I ler emty(atI 25 I) (Part Geometry & Algebra IA Tripos: Mathematical product. triple Vector product. triple Scalar product. Vector product. Scalar identity). vector true one (and components for Addition. identities vector of number a deduce can we Then cosines Direction 2.7.2 (iii) (ii) vr etri D3 pc a euiul ersne ytotrera ubr,s eoften we so numbers, real two/three by represented uniquely be may space write 2D/3D in vector Every point the If rmrpae plcto f(.a,(.b n (2.9) and (2.8b) (2.8a), of application repeated From ( R a 2 × / R b rmrpae plcto f(.) 21) 21) 21) 22a n (2.29b) and (2.29a) (2.18), (2.16), (2.15), (2.9), of application repeated From a rmrpae plcto f(.) 22a,(.0) 22c,(.0) (2.29c) (2.20d), (2.20c), (2.20b), (2.20a), (2.9), of application repeated From 3 ) P = o D3 space. 2D/3D for · c a h atsa oodnts( co-ordinates Cartesian the has a x rm(.9 n (2.40) and (2.39) From ews oso that show to wish We a i (( = = + × a λ b a y a x j a + + b y ( = = = ( = a y b OP µ a · c → z b z b z − a a k + a a = ( = x x , x y a i b ( = = = = a i z r b a × x y λa + z b = b i × b y b − × z a x a a a ) x = x ( a i c y i x x x a + b i x x ( + j i i + b b b z + i + + · × x x x + µb b + a b i y i a + a x a y x c a + x ) · z x a i z i ( = ) i z ) j i a b y + k b × i ( + b b + + y x j y y ) ( + x and where epciey the respectively, epc otebasis the to respect If b + a j − b i ,y z y, x, × z a c y y x a + × a y x t k a j λa i z a + ( · − z b · + b x b j saui etrwt opnns( components with vector unit a is b , γ y c k x z b a x y + b ) a y k i a ) i z ,te h oiinvector position the then ),

z + − r h nlsbetween angles the are b c j α x + x · ) i.e. · a b + j i j z b ( − x µb a ..olydmpcma.k ihems2006 Michaelmas [email protected], b b steagebetween angle the is and z ( + × + . x b y x t a ( c z y j i b x a b y a (cos = ) r z i a + + z i j x − · ) k × x · ( = ( + j t b b b b b b c x z + a ( + y z k ) z y ieto cosines direction y c k j i k = = ,y z y, x, λa − . . . + · + b ) . α, a + x k t c { x z a . . . x b c · i cos . . . + y z b + i z , i y k b j + ) − = . . . , x µb ) − . β, k ) c a k | } y z a t z cos then , ) j ) | | y k b + · b i y ( . | x t γ c c c cos ) t x ) x z and k i k . . and of . + , α . t c j and , y r endby deﬁned are t i j x ec if Hence . + t , y c t , z t k (2.36b) z (2.36a) and ) with ) (2.38) (2.35) (2.41) (2.40) (2.39) (2.42) (2.37) k β ,

This is a supervisor’s copy of the notes. It is not to be distributed to students. ahmtclTio:I ler emty(atI 26 I) (Part Geometry & Algebra IA Tripos: Mathematical Exercise. increasing of directions the dependent. in position vectors are vectors unit unit the the use that is can co-ordinates we Similarly co-ordinates. and Cartesian 2D in use We Remark. the with begin this, prove To parentheses. the outside vector Remark. .. Dpaeplrco-ordinates polar plane 2D space. 2.9.1 in vectors. basis positions orthonormal describing of for sets systems alternative co-ordinates to Cartesian lead naturally to They alternatives are co-ordinates Polar Co-ordinates Polar 2.9 axes. of choice arbitrary the of because components all for true o rce iial o the for similarly proceed Now θ epcieya a ai n2 ln oa oodnts u e ieec ntecs fpolar of case the in difference key a but co-ordinates, polar plane 2D in basis ‘a’ as respectively a i × h uvso constant of curves The ofimthat Confirm and dniy(.2 a ocmoeti h direction the in component no has (2.42) Identity ( b × j rhgnlui etr ntedrcin fincreasing of directions the in vectors unit orthogonal , c ) x ≡ ( a × ( b x y × cmoeto h ethn ieo 24) hnfo (2.40) from Then (2.42). of side left-hand the of -component and e r r c n uvso constant of curves and ) · e z · r i opnns rnt hti t refroecmoeti utbe must it component one for true its if that note or components, 1 = ( = ( = = = = , a a e a a ( y y a θ y · ( ( c · b b · c y x e c ) × + b θ ) c Equivalently stant > x define stant Define that such be should arctan of choice 0 the where hr 0 where ( co-ordinates ( Cartesian co-ordinates 2D polar plane 2D Define 24a tflosthat follows it (2.43a) x b y 1 = c a π < θ < − − − ) z z c ( and 0 r b ( z − , a y a θ ) θ r =

e b c c · e a · ntedrcino increasing of direction the in x θ nesc trgtage,ie r orthogonal. are i.e. angles, right at intersect x b r e ntedrcino increasing of direction the in z b 6 a ..olydmpcma.k ihems2006 Michaelmas [email protected], ) + r ) x sui vector unit as x ( ) sui vector unit as ..n opnn ntedrcino the of direction the in component no i.e. , c − y if b c 2 · < r x a = e + ,and 0, = × a x e e > y j i θ · z b r θ r i y c 0 = x ( cos ∞ 2 ≡ c ) cos = sin = b y x z cos = = 0, 2 1 c − . n 0 and y , θ ( x x θ , a < θ < π a − − θ · and x c s θ b b = s θ sin x ) x e θ e rhgnlt ie fcon- of lines to orthogonal b arctan = 6 rhgnlt ie fcon- of lines to orthogonal c = π c r r i x y θ z − ,y x, cos + sin + − r < θ ) if − i epciey sabasis a as respectively, 2 sin ( cos + sin < x ( a π othat so ) a · y 2 , θ if θ θ θ b b π and 0 y j ) e e rminverting From . θ < y c , θ θ x ,θ r, y + j θ . , x . Then . a ntrsof terms in ) r , . z 0, Similarly, . b y z 0. = ) θ c (2.45b) (2.44b) (2.43b) (2.45a) (2.44a) (2.43a) x if 0 = (2.46) r 7/03

This is a supervisor’s copy of the notes. It is not to be distributed to students. en Dclnrclplrc-riae ( co-ordinates polar cylindrical 3D Define co-ordinates polar Cylindrical 2.9.2 basis the to respect ahmtclTio:I ler emty(atI 27 I) (Part Geometry & Algebra IA Tripos: Mathematical using not for using not for reason 0 that such be should arctan and of choice the where 0 where Example. vector a of components the Given Remarks. (iii) (ii) 11 (i) oethat Note y oet htvr ihpsto,eg rm(.5)tecmoet of components the (2.45a) from e.g. position, { with vary that ponents natraientto o( to notation alternative An point a basis of co-ordinates polar The that note to crucial is It ,and 0, = e r 6 , e o h aeo h oiinvector position the of case the For < ρ θ { r } e and ρ r r (cos are , and ∞ e φ θ θ 0 , r } = npaeplrco-ordinates. polar plane in φ and r ( are π { 6 r ffcieyaissfrthe for aliases effectively are e θ, if r θ < φ , v sin r, < x r sw hl e new e oshrclplrc-riae.IH hr sas odreason good a also is there IMHO co-ordinates. polar spherical to get we once see shall we as e θ ( = )(ntecs fcmoet the components of case the (in 0) } 2 θ ( = = = : i π ). and 0 k x and cos { v v x e v x x j ,θ r, x r ρ (cos i x θ , y −∞ + cos v = e + = s( is ) θ 0. = v ihrsett basis a to respect with } θ y y θ ρ x P O j e < z < aywith vary sin + cos 2 r ,φ z φ, ρ, z ,φ ρ, r ( are + φ − v θ y y , φ y ) sin r ,wiha hl eoecerhsisadvantages. its has clear become shall as which ), sin 2 r e ∞ ,θ r, = r ) and 2 1 11 θ θ ( + rmivrig(.9)i olw that follows it (2.49a) inverting From . ρ x ) e ,wietecmoet of components the while ), π < φ < φ , i ntrso DCreinc-riae ( co-ordinates Cartesian 3D of terms in θ r θ = e θ − hsmasta vncntn etr aecom- have vectors constant even that means This . + + ) r fpaeplrc-riae.Hwvr n3 hr sagood a is there 3D in However, co-ordinates. polar plane of ρ x ( + y sin sin arctan = j v y − tflosfo sn 24a n 24)that (2.47) and (2.43a) using from follows it (sin z , φ

θ v c if x N + e θ ..olydmpcma.k ihems2006 Michaelmas [email protected], > y sin z { θ y z P i eednei hdn’i h basis). the in ‘hiding’ is dependence , e cos j θ r = } 0, + cos + x y e ecnddc h opnnswith components the deduce can we , z θ e φ e < φ < π v ) ρ e φ ρ , y e cos θ θ = e θ θ r y ) ) e e i 2 r r θ π ihrsett h basis the to respect with = . . if OP → < y ihrsett the to respect with 0, φ ,y z y, x, if 0 = othat so ) (2.49b) (2.49a) (2.47) (2.48) > x 0

This is a supervisor’s copy of the notes. It is not to be distributed to students. 7/06 co-ordinates. ahmtclTio:I ler emty(atI 28 I) (Part Geometry & Algebra IA Tripos: Mathematical arctan. of choice the in needed is care some again where 0 where ( longitude) co-ordinates and polar spherical latitude 3D height, Define (cf. co-ordinates polar Spherical 2.9.3 Example. form. Component Remark. Exercise. Equivalently increasing Define Remarks. (ii) 12 (i) oethat Note yidia oa oodntsaehlfli o,sy att eciefli o onalong a down flow UK). fluid the and describe Norway and to between pipe want point gas say, any new At you, the like if (e.g. helpful pipe are cylindrical [straight] co-ordinates polar Cylindrical 22c n 22d,w a hwta nlgu etrcmoetiette o(.8,(2.39), (2.38), to identities component vector analogous that hold. show also (2.41) can and we (2.40) (2.29d), and (2.29c) (2.29d)) and (2.29c) (2.29b), (2.29a), (cf. that e 6 ρ Since , z hwthat Show ihrsetto respect With ρ < r e osat ..tenras r uulyorthogonal. mutually are normals, the i.e. constant, = , φ φ r and ∞ and { and r e 0 , = ρ Since , P θ e z e 6 z φ { x eeare here ntesrae h etr rhgnlt h surfaces the to orthogonal vectors the surface, the on epciey Then respectively. e , 2 θ sui vector unit as e ρ + , z 6 { e } e y φ { ρ π x aif nlgu eain otoestse by satisfied those to relations analogous satisfy 2 , e , e e + not n 0 and ρ = z , φ e } e z , r ρ φ e 2 the sin r ih-addtido uulyotooa ntvcos ..show i.e. vectors, unit orthogonal mutually of triad right-handed a are , z × e e e } 2 1 6 ρ ρ z e θ r omabssw a rt n vector any write can we basis a form θ , } s · · φ r < φ cos and e k e e e e h oiinvector position the j i = rhgnlt ie fconstant of lines to orthogonal φ φ ρ ρ z ,θ φ θ, r, = y , φ 1 = 0 = e arctan = = cos = sin = 2 θ v z = = cos = π fpaeplrc-riae.However, co-ordinates. polar plane of ON , = rmivrig(.5)i olw that follows it (2.55a) inverting From . → , ) , e 12 v [ − k z e e ρ = e + φ ρ e . . e ntrso DCreinc-riae ( co-ordinates Cartesian 3D of terms in sin φ φ , φ ρ ρ × r φ e NP e   e · · + φ sin → ρ i φ ρ e e e , sin + cos + z φ z v − i e x φ θ z cos + 0 = = 1 = 2 e sin 1 = ] sin = ( = + φ

e z r c φ + ρ , φ φ y , z , φ a eepesdas expressed be can , j 2 ..olydmpcma.k ihems2006 Michaelmas [email protected], φ ρ e e v . , ρ, φ φ z e e j e 2 1 e e φ ρ , , , z 0   z z z , + · · . × = e e φ , ) z z z . e r e 1 = 0 = ρ z cos = ρ arctan = , , , , θ v e { φ φ ρ i ncmoetfr as form component in , , φ and j constant, = , sthe is k } ..(.9) (2.29b), (2.29a), i.e. , z x y ntedrcinof direction the in φ , fclnrclpolar cylindrical of ,y z y, x, φ constant = othat so ) (2.51b) (2.50b) (2.55b) (2.54b) (2.52d) (2.52b) (2.51a) (2.50a) (2.54a) (2.52a) (2.55a) (2.51c) (2.50c) (2.52c) (2.53) 7/02

This is a supervisor’s copy of the notes. It is not to be distributed to students. ahmtclTio:I ler emty(atI 29 I) (Part Geometry & Algebra IA Tripos: Mathematical Remark. Exercise. Equivalently increasing Deﬁne Remarks. (ii) 13 (i) h reigof ordering The peia oa oodntsaehlfli o,sy att ecieamshrc(roenc or oceanic, surfaces warming. orthogonal. (or global the mutually atmospheric understand to to describe normals order to in The want e.g. say, earth, you, the if on helpful motion are both) co-ordinates polar Spherical 22c n 22d,w a hwta nlgu etrcmoetiette o(.8,(.9,(2.40) (2.39), (2.38), to identities hold. component also vector analogous (2.41) that and show can we (2.29d), and (2.29c) (2.29d)) and (2.29c) (2.29b), (2.29a), (cf. that e r Since , hwthat Show r e , θ θ and and { e r e , r e φ e , φ θ { e epciey Then respectively. , θ e sui vector unit as e and , r φ , } e i 13 θ k , e e aif nlgu eain otoestse by satisﬁed those to relations analogous satisfy φ φ e k e e x j i sipratsince important is } e φ θ r j r r ih-addtido uulyotooa ntvcos ..show i.e. vectors, unit orthogonal mutually of triad right-handed a are sin = cos = cos = × e e = sin = cos = r r r e s · · θ constant, = O e e − = rhgnlt ie fconstant of lines to orthogonal r θ z sin φ 1 = 0 = φ θ φ e θ θ φ (sin e (sin cos θ cos r φ , , , − i { [ θ φ cos + θ φ e e e sin e e e e i ρ θ r r i r θ , , r r sin + cos + × r θ e e θ cos + · · cos + φ θ e e e e , osatand constant = φ , θ φ e φ θ e j θ . φ 1 = θ 0 = = } . θ 1 = ] θ sin θ ol omalf-addtriad. left-handed a form would sin

e e c e , r , θ θ N φ φ cos + ) , ..olydmpcma.k ihems2006 Michaelmas [email protected], ) P . j − j e e z cos + − φ φ e sin φ · · sin e e e e × θ r e φ φ φ θ e φ θ θ e φ e e 1 = 0 = φ r k k φ φ osata n point any at constant = = , , , r , , , , e θ { θ i y , and , j , k } φ ..(.9) (2.29b), (2.29a), i.e. , ntedrcinof direction the in (2.58d) (2.58b) (2.57b) (2.56b) (2.58a) (2.57a) (2.56a) (2.58c) (2.57c) (2.56c) P are

This is a supervisor’s copy of the notes. It is not to be distributed to students. hr si sue that assumed is is where ahmtclTio:I ler emty(atI 30 I) (Part Geometry & Algebra IA Tripos: Mathematical the omit we when where, i equation vector a is This as equality this express we notation suffix In form component in equivalently or vectors two If equivalents suffix and Dyadic 2.10.1 Remark. vector. position the Example: notation. Suffix vectors manipulate to easier notation generally the introduced is we it (2.30) notation, In suffix with notation. familiar suffix Once using tensors). (and used vectors have we far So Notation Suffix 2.10 Example. form. Component natraiei owrite to is alternative An agstruh1 through ranges 15 14 nhge iesostesffie ol easmdt ag hog h ubro dimensions. of number the through range to assumed be would suffices the view. dimensions that higher to In dissenters are there Although nesadbegvnteaoeepeso o h oiinvco nsffi oain ecfrhwe Henceforth notation. suffix in vector position the use for will expression above the given understandable h s of use The ihrsetto respect With x a and and ee to Refer , Since 2 r , b 14 ydcnotation dyadic x oa ogv he opnn qain.Similarly equations. component three give to as so 3 interchangeably. ahrthan rather , r qa,w write we equal, are { i v e , { r j as , e v , e r α = rt h oiinvector position the Write , θ { e , v and e θ v i } , c φ 1 e ihthe with , i } v = φ + omabssw a rt n vector any write can we basis a form U = r } λ r o vectors. for o h oiinvco ndai oainpsil em more seems possibly notation dyadic in vector position the for , v epciey ag hog (1 through range respectively, , h oiinvector position the a v 2 ( = a j x + i i + = r v + ,y z y, x, µ v = b b v 3 i i = a a a k y 1 = a v 1 3 2 j r ⇔ ⇔ ⇔ ⇔ ( = ) + for OP e = ( = → , uxnotation Suffix r = = = = for v 2 + z , c c c c k x understood. 3 i U α j i { v i b b b b 1 v ( = = v ( = 1 = θ = 1 3 2 = x , 1 = = 1 = r i e , , . , } v , θ λa λa as 2

λa v r λa , c r + 2 x , , x 2 r, v , i a eepesdas expressed be can 2 for j e v , α , U ..olydmpcma.k ihems2006 Michaelmas [email protected], v 3 , + r 0 + 3 3 φ = ) 3 + y , + ) . ti nesodta h n resuﬃx free one the that understood is it , e µb v , 0) µb φ µb µb i z i sa lentv en fexpressing of means alternative an is { . . j 1 = ) α x U i . i ste emda termed then is } , , , 2 . 2 , , 3). 3 v . 15 ncmoetfr as form component in resuﬃx free . (2.61b) (2.60b) (2.63d) (2.63b) (2.61a) (2.60a) (2.63a) (2.63e) (2.63c) (2.59) (2.62)

This is a supervisor’s copy of the notes. It is not to be distributed to students. ahmtclTio:I ler emty(atI 31 I) (Part Geometry & Algebra IA Tripos: Mathematical Remark. suffices. Dummy Remark. ntecs ffe ucsw r suigta hyrnetruh(1 through sum, range explicit they the that summation assuming Einstein’s are Under we so. suffices say free of case the In convention Summation 2.10.2 • • • h resffie hudb dnia nec ieo nequation). an of side each on identical be should suffices free rules above the the but (unless examinations), wrong in gone has something equation, an of term over, one summed sum). in and explicit twice suffix an than dummy is more a there through, appears be ranged suffix to and a taken suffix if is free it twice a appears be suffix to taken a is if it once appears suffix a if term. each in hr the where hr ent htteeuvln qaino h ih adsd a ofe ucssnethe since suffices free no has side hand right examples. Further the on case equation this equivalent (in the suffix dummy that note we where Similarly (ii) (i) hsntto spwru eas ti ihyabeitd(n oad aclto,especially calculation, aids so (and abbreviated highly is it because powerful is notation This same the be must it but suffix, free the for chosen is symbol, or letter, what matter not does It nsffi oainteepeso ( expression the notation suffix In where ( equation the consider example another As ti seta htw sddffrn ybl o ohtedmyadfe suffices! free and dummy the both for symbols different used we that essential is It hr,epcal fe h eragmn,i sesnilta h um ucsaedifferent. are suffices dummy the that essential is it rearrangement, the after especially where, i , k nsffi oaintesaa rdc becomes product scalar the notation suffix In k stedmysffi,and suffix, dummy the is t.aerfre oas to referred are etc. , α a gi ensme out. summed been again has ) ( a · a b a · )( k X · b =1 3 um suffices dummy b c must = i a · ( stefe uxta sasmdt ag hog (1 through range to assumed is that suffix free the is d a · λ = = = = ) k b b )( efloe,adrmme ocekyu nwr (e.g. answers your check to remember and followed, be k = ) c ⇔ a X k X i =1 =1 c 3 3 1 · P i b d 1 = a a a

X i a eoitdfrdmysﬃe.I particular In suﬃces. dummy for omitted be can , becomes ) =1 + i k 3 α X · X b i k X b =1 =1 3 b i 3 =1 k a 3

X j c ) ic hyae‘umdot fteequation. the of out’ ‘summed are they since 2 =1 , 3 c a b a a ..olydmpcma.k ihems2006 Michaelmas [email protected], 2 α i = k a b b + etc. b i i α k ! b d c a i = nsﬃ oainti becomes this notation suﬃx In . c i   3 , j b = d , λ X j 3 , j =1 3 2 d , , i c )wtottene oexplicitly to need the without 3) , j d j   (2.64) , 2 , 3).

This is a supervisor’s copy of the notes. It is not to be distributed to students. 8/03 h rnce delta, Kronecker The delta Kronecker 2.10.3 ahmtclTio:I ler emty(atI 32 I) (Part Geometry & Algebra IA Tripos: Mathematical equation: matrix a as written be can This Examples. o htw aeitoue uxntto,i smr ovnett write to convenient more is it notation, suffix introduced have we that Now vectors basis on More 2.10.4 ntvectors unit Properties. epepaieta h ,2ad3aelbl ahrta components. than rather labels are 3 and 2 1, the that emphasise help 4. 3. 2. 1. sn h ento ftedlafunction: delta the of definition the Using h olwn qain aen sense no make equations following The Similarly ne uxntto n h umto convection summation the and notation suffix Under i , j a and k b k a c k k k δ natraientto is notation alternative An . ij , = = ,j i, d b 1 = j k , eas h resffie r different are suffices free the because because ( 2 a   , a ,i e fnn ubr endby defined numbers nine of set a is 3, · δ δ δ δ δ 11 b + 11 31 21 ii a a a ) i i p c b 1 = δ = δ δ i ij = k δ δ δ 1 = δ pq δ X 12 32 22 i ij ij =1 srpae oeta wc ntelf-adside. left-hand the on twice than more repeated is 3 δ , d c b δ = = = = q if 0 = jk δ δ δ δ = ii switnas written is switnas written is 13 33 23 = 22 a X a a = i a   e =1 3 j 1 1 X j p 1 = (1) δ =1 3 . δ . b 11 = 11 a p , i δ δ , = i δ +   e + ij i 6= (2) 1 δ 1 0 0 0 1 0 0 0 1 a a δ

jk c q 2 22 . j 33 δ and b 21 = ..olydmpcma.k ihems2006 Michaelmas [email protected], q + 1 = a a = δ + δ i i ik e 33 b a +   a (3) , i . 3 · c 3 = b δ hr h s fsprcit may superscripts of use the where , b . j i 31 = = . . d c i j e . 1 , e 2 and e 3 o h Cartesian the for (2.65b) (2.66b) (2.66d) (2.65a) (2.66a) (2.66c) (2.66e) (2.65c)

This is a supervisor’s copy of the notes. It is not to be distributed to students. hr ent htteei n resffi n w um suffices. dummy two and suffix free one is there that note we where notation that claim suffix We in product vector The 2.10.6 ahmtclTio:I ler emty(atI 33 I) (Part Geometry & Algebra IA Tripos: Mathematical Example. Check. Example. the Further Hence even/odd. or is exchanges 3, 2 (or 1 swaps case pairwise this of in number ordering, the original if the of permutation components recover non-zero even/odd to an necessary is transpositions) sequence ordered An symbol Levi-Civita Definition. or tensor alternating The 2.10.5 equivalently Or hence and notation superscript the of terms in Then ecnld that conclude we srqie rm(.0.D ene od more? do to need we Do (2.40). from required as For edefine We ymti tensor symmetric a e e ε ε ( ( ijk ε j i ijk ijk a ) ijk ) · ( · · s e e ij e e ( j r ie by given are ( ,j k j, i, ( ( j ) otherwise 0 = = if 1 = ε i i 0. = ( ) ) ) i ijk a ( a × ( = = = = = − = s × 1 = ij b if 1 ε b e , ) jki a δ 1 ) ( , ,j i, e ij e e j i i 2 = ) ε i ( ( , = , ) ε i j , 132 = × ) )t etesto 7qatte uhthat such quantities 27 of set the be to 3) j ) ijk 123 1 = ε X j 123 j e ε k j i k j i i =1 = 3 s = kij ( k = , ij = a ) 2 k X ε δ 2 =1 , = = ε 3 213 ij sa d emtto f123; 2 1 of permutation odd an is sa vnpruaino 3; 2 1 of permutation even an is i ,sc that such 3, 231 δ b 3 ij . − ε . jik = = = ijk = + . ε = ikj (the s ε ε a ε 132 ji 321 ε ε ε j 312 = ijk ilm ilm b = k a

i = − c hcmoetof component th 3 1 = . − = δ e ε b jl − ..olydmpcma.k ihems2006 Michaelmas [email protected], s kji 2 ( ε ijk j ij 1 δ ijk ) = km = s = l a ij a e 2 s − j , ( ji b b k ε 3 k ) jik evalulate − , m . a 3 b e 2 ( , j ) ) ijk s ij Since . (2.67d) (2.67b) (2.69b) (2.68b) (2.68a) (2.67a) (2.69a) (2.67e) (2.67c) (2.68c) (2.69c) (2.71) (2.72) (2.70) 8/06 8/02

This is a supervisor’s copy of the notes. It is not to be distributed to students. ahmtclTio:I ler emty(atI 34 I) (Part Geometry & Algebra IA Tripos: Mathematical recover we product triple vector the for notation suffix Using product triple Vector 2.10.9 by given is product triple scalar the notation suffix In product triple Scalar 2.10.8 Example. whenever Similarly Proof. Remark. 2.1. Theorem identity An 2.10.7 iial whenever Similarly while nareetwt (2.42). with agreement in ec 27)rpeet 3 represents (2.73) Hence If j hr r orfe ucsidcso ahsd,with side, each on suffices/indices free four are There Take = k ,sy then say; 1, = j = j j p = 6= n(.3 sa xml farpae ux then suffix; repeated a of example an as (2.73) in k k . (or a p 4 × = equations. ( q b .Nx suppose Next ). H = RHS H = LHS H = RHS H = LHS × a ε ε c ipk · ijk ) ( b ε = = = ε i ipq ipq × ======c ε ε       δ = = ) i 312 1 δ ε 12 3 = = − p − 1 i 11 δ otherwise 0 if 1 otherwise 0 if 1 p ε − − ε a δ if 1 jp if 1 ε = ε ( ijk ijk 2 δ j ε ipq ( ε a 3 q 1 j δ ipq b δ δ kji pq q pp i − δ jl · kq and 1 = a a − kq c c a ε δ j j j 0 = δ ε δ ijk p p ) i p p im − ε ( kq 1 δ klm b ( − − klm b 2 = 1 = q 2 = 1 = b 1 δ q

− a − δ − δ a c , jq × × i 2 δ kq a j p ( 1 b q , q , b δ q , q , δ ..olydmpcma.k ihems2006 Michaelmas [email protected], j δ b c a p c j l jm k pq kp j ) b ) i c c 0 = 2 = · i 1 = 2 = l 1 = 2 = k c m ,sy then say; 2, = k sadmysﬃ ntelf-adside. left-hand the on suﬃx dummy a as b δ δ c i . kp il . m ) c ) . δ , a . kq i j , b . l c m (2.75) (2.74) (2.73)

This is a supervisor’s copy of the notes. It is not to be distributed to students. 9/03 ahmtclTio:I ler emty(atI 35 I) (Part Geometry & Algebra IA Tripos: Mathematical Remark. eliminate may We Lines 2.12.1 equations. vector by described be can objects geometrical Certain Planes and Lines 2.12 Solution. Exercise. Worked components, in equation the out write to be might approach one ( equation e.g. vector a with presented When Equations Vector 2.11 hsi neuvln qainfrteln ic h ouin o(.7)for (2.77b) to solutions the since line the for equation ( equivalent an is This ih xettoabtayprmtr nteslto a esalsei h aei 28)blw.An below). (2.81) progress. in make case to the manipulation is vector see use shall to we is (as forward way solution better, the often in and parameters alternative, arbitrary two expect might given For x − a x rirr scalar. arbitrary using Remark. hndtti with this dot then hnsbttt hsrsl notepeiu qaint obtain: to equation previous the into result this substitute then o erag oddc that deduce to rearrange now aallto parallel ) − a qain(.7)hsmn ouin;temlilct ftesltosi ersne yasingle a by represented is solutions the of multiplicity the solutions; many has (2.77b) Equation is xadtevco rpepoutuig(2.42): using product triple vector the expand First a ) a , · n , n c ol become would 0 = ∈ and o h aewhen case the For R o given For λ t 3 . a rmteeuto yntn that noting by equation the from hsi igeeuto o he unknowns three for equation single a is this × b c : sabasis. a as a x , 1 b n , 1 c + a ∈ and x R 2 x x 3 n 1+ (1 − n solutions ﬁnd 2 c a + r o aall ecudhv lentvl ogtasolution a sought alternatively have could we parallel, not are ( ( x x b x a − 3 − · · n x b ( a 3 x + ) x = ) ) = = × × o some for requivalently or by given is line the on point a for equation vector point a through vector line the Consider b x a c a t c 1 ( 1+ (1 ) · x x + a = + n x × − ∈ · 1 a 0 a = b

b + ( t c a R ( a . b = ) n let and , b b = 3 a ..olydmpcma.k ihems2006 Michaelmas [email protected], = λ · · · 2 b · to c c ∈ c λ n c c ) ; ) ) . t 2 R ; ; n hence and , . + . OP x → a P ( = x 3 n eapito h ie hnthe Then line. the on point a be 3 x = = . 1 x , OA a 2 t → x , + 6= 3 λ + ) 0 t ∈ AP r either are , → R 3 A n ec we hence and , aallt a to parallel x (2.77b) (2.77a) = (2.76) a or

This is a supervisor’s copy of the notes. It is not to be distributed to students. 9/02 ahmtclTio:I ler emty(atI 36 I) (Part Geometry & Algebra IA Tripos: Mathematical Remarks. thus is plane the of equation The Planes 2.12.2 Exercise. Worked 2. 1. where point any then plane), the in lie vectors If Solution. 28)i ouint qain(.0) h rirrns nthe in arbitrariness The (2.80b). equation to λ solution a is (2.81) and l and 27)cnol efudu oa rirr utpeof multiple arbitrary an to that up assuming found be only can (2.78) if that observe Finally ..asrih iei direction in line straight a i.e. hsteeaen ouin unless solutions no are there Thus Hence µ ,µ λ, m en htteeuto a ucutby aysolutions. many [uncountably] has equation the that means is dt 27)with (2.78) ‘dot’ First ∈ r w ieryidpnetvcossc that such vectors independent linearly two are o given For R . t · u u , ,is 0, = t ∈ x R 3 saslto o(.8 ois so (2.78) to solution a is t n solutions ﬁnd t × oobtain to u t hog ( through x = · t x t t n x x × · · = = u ntepaemyb rte as written be may plane the in = u x u ( a x .Nx cos 27)with (2.78) ‘cross’ Next 0. = a = = t = ln,i olw rm(.0)that (2.80a) from follows it plane, to plane the of distance to parallel Let otepae Let plane. point vector the unit a to a through to orthogonal goes is that that plane a Consider + | × x · t × t t ( t n | x × λ 2 ∈ d t | · u × t l ( = ) n × = ( Q | u R + x 2 + u − ) t

. d 3 c × / µ etepiti h ln uhthat such plane the in point the be . + ( a | to t m t t ..olydmpcma.k ihems2006 Michaelmas [email protected], t ) | | · µ 0 = ) t · 2 · n , t | . x t n 2 ) ups that Suppose . ) x , x l t t 0 = ec h eea ouint (2.78), to solution general the Hence . · AO + − . → . n µ ( P , two t and 0 = t + ( · eaypiti h ln.Then plane. the in point any be x o any for x n hence and OP − → ) needn rirr scalars arbitrary independent t AP O → a . ute,since Further, . ) · · m · t n n n OQ µ → oobtain to · ∈ n = 0 = 0 = 0 = R a s htboth that (so 0 = n d ..sltosof solutions i.e. , · ; n n n then , . , , = stenormal the is d Q n 2 si the in is d (2.80b) (2.80a) = A OQ (2.79) (2.81) (2.78) sthe is → . d and is

This is a supervisor’s copy of the notes. It is not to be distributed to students. 9/06 .3CnsadCncSections Conic and Cones 2.13 intersect? ymaso rnlto ecnnwgnrls 28b oteeuto o oewt etxa a at vertex a with cone a for equation the to (2.84b) generalise now can we point translation general a of means By exercise. Worked ahmtclTio:I ler emty(atI 37 I) (Part Geometry & Algebra IA Tripos: Mathematical Let by spanned is a ( also is then (2.83) holds, that show can we Further, hnfo h qaino ie 27a,w euethat deduce we (2.77a), line, a of equation required). the from then a : ne htcniin otetolines two the do conditions what Under b t − and a utlei h ln hog h rgnta snra o( to normal is that origin the through plane the in lie must ) u n ec hr exists there hence and , ( x − a ) x · = n b Solution. suﬃcient − a 2 + ( = a nesc sthat is intersect a that deduce we tersects ih iclrcn sasraeo hc every which on surface a (0 is cone point circular right A intersect L hr ysurn h qainw aeicue the included cone. have we ‘reverse’ equation the squaring by where Because is Π plane the on points specifying equation h etreuto o oewt t etxa the at vertex its thus with is cone origin a for equation vector The and a through passes hence lel, h bv description above the Let cone. O λ = ,µ λ, x t 2 h point The . aallto parallel λ < α < − = nescsΠnwee(nwihcase which (in nowhere Π intersects n t ftelnsaet nesc hycno eparal- be cannot they intersect to are lines the If L ∈ odto o h ie oitret o f(2.83) if For intersect. to lines the for condition a + b eaui etrprle oteai.Te from Then axis. the to parallel vector unit a be P 2 0 L

) c R ihnormal with , − 2 1 µ ( : Π L ssc that such is cos L ..olydmpcma.k ihems2006 Michaelmas [email protected], u uhthat such L ( : µ 1 2 2 t 2 .I h atrcs,then case, latter the If ). 2 . π u x and ,or ), 2 ,wt ie xsta assthrough passes that axis given a with ), sprle to parallel is ; . α − u x O e etepaecontaining plane the be Π Let . ( a a ( b u x let ; sapito both on point a is ) srfre oa h etxo the of vertex the as to referred is L necessary x x − × utb ieryindependent. linearly be must · 2 − · n a n isi i hc case which (in Π in lies t OP ) L a t ) 2 and 0 = = ) 2 0 · × = · ( eteln asn through passing line the be t | ae xdage say angle, ﬁxed a makes ( u x x t × L ec rm(.0)the (2.80a) from Hence . | odto o h ie to lines the for condition 2 × cos 2 0 u cos u n hneΠ either Π, thence and 0 = ) L 0 = ) . α 2 2 , α t ( : × . b x L . u 1 − isi and Π in lies .Ti plane This ). L and b 1 ) osnot does × (2.84b) (2.84a) (2.84c) L L (2.82) (2.83) u 2 1 0 = (as in- L L α 1 1

This is a supervisor’s copy of the notes. It is not to be distributed to students. ahmtclTio:I ler emty(atI 38 I) (Part Geometry & Algebra IA Tripos: Mathematical plane. a and cone a of Intersection form. Component hc a esmlfidto simplified be can which becomes (2.86) that so plane that hn(.4)becomes (2.84c) then ute,tasaeteae ytetransformation the by the axes in the is translate Further, cone the of axis the that so in axes polynomial Cartesian quadratic choose case a we that by wlog, defined that, curve suppose a is This Y ( l − x n ecnexpress can we and 0 = − ups hti em fasadr atsa basis Cartesian standard a of terms in that Suppose cos a c ) sin 2 l α ( + ( x β − cos − y sin − a X β ) 2 b l 2 β ) X ( + x cos m ( = = ( + sin y 2 e scnie h nescino hscn ihteplane the with cone this of intersection the consider us Let α − x ,y z y, x, z β − + n b − ) − n ( = m Y , a Y c c 2 ) ncmoetfr as form component in ) − cos n ,m n m, l, , cos cn 2 β = a 2 2 ( = α 2 (0 = ) = = ( nti aeteitreto ftecn ihthe with that cone suppose Next the of intersection the z case this In nti aeteitreto ftecn ihthe with cone the of sin intersection the z case this In i.e. i.e. i.e. that suppose First 0 that suppose the intersection with the cone of the graphs of consider is, this why see To considered: be to need sin that cases three now are There = − x y ,b c b, a, ln ilyedacoe curve. closed a yield will plane 0 = ln ilyedtooe uvs hl if while curves, open two yield will plane 0 = − sin ( 2 β − x X < β a , cos = − 2 b 2 sin )

β 2 ) c + + a , ( + cos ) ..olydmpcma.k ihems2006 Michaelmas [email protected], β, β x cos 2 = sin α + c ( + Y n cos y and 2 c sin cos 2 − α hr ilb n pncurve. open one be will there < α < β < β ( = 2 − α y sin , β sin 6 b 2 β − − cos y ) ) ,m n m, l, β α 2 nodrt ipiytealgebra the simplify to order In . cos . β c 2 sin sin b + ( + 2 − sin α 2 ) X 6 2 > β α β cos > α > β sin 2 sin 1 π 2 2 + z ln,adfrdeﬁniteness for and plane, 0 = β π 2 β π − ) − . 2 cos c 2 , − , sin cos 2 α β 2 1 c α π ) cos β . 2 2 2 − cos β π 2 α cos 2 α n sin and . α . α 2 2 + cos = . α c 2 2 yz cos . α β cos = ln.In plane. z 2 , α .In 0. = (2.87) (2.86) (2.85) 2 α . 10/02 10/03

This is a supervisor’s copy of the notes. It is not to be distributed to students. ahmtclTio:I ler emty(atI 39 I) (Part Geometry & Algebra IA Tripos: Mathematical Examples. n Definition. Isometries Inversions 2.14.1 and Isometries Maps: 2.14 x 1 or 2 = 7→ Translation. Remarks. Define x (ii) (i) 1 0 and n h dnicto ftegnrlqartcplnma srpeetn n fteetretypes three Tripos. these the of in one later representing as discussed polynomial be quadratic will general the as of collectively identification known The are curves three above The Then rnlto sa isometry. an is Translation o h iebig.I atclr suppose particular, In being). time the for 3 = An x 2 7→ isometry ups that Suppose x 2 0 then , of | cos R c n 2 2 α b sampigfrom mapping a is sin | − x ∈ 2 1 0 sin R α − n 2 n that and , x | β 2 0 x | | 1 = = − | A sin sin sin x ( x 2 2 x 1 | β > β < β = + 7→ and cos = R where two the between distance the a of of half equation the is This an of equation the is This hsi h qaino a of equation the is This that follows it (2.88) (from tively and | b x x n cos cos ) X 1 0 0 − to − = semi-major = α α α (

x c x R x cos c . . . x x 2 0 2 2 oi sections conic n + 1 ..olydmpcma.k ihems2006 Michaelmas [email protected], nti ae(.6 becomes (2.86) case this In case this In becomes (2.87) case this In | − sin , + 2 . uhta itne r rsre (where preserved are distances that such x b α a b 2 2 . − ) α ∈ | and sin = cos R X − xso lengths of axes n | 2 2 X x A 2 X A β r apdto mapped are 1 = α 2 2 Y 2 2 − . 2 + − + = x = 2 B Y 2 c parabola hyperbola B Y y | B 2 2 cot . ellipse − 2 2 2 1 = 1 = . b , Y β − . . . c A B < A with where , o 2 cot x 1 0 and vertices , x . β semi-minor 2 0 ). B ∈ B R (2.89b) (2.89d) (2.89a) respec- (2.89c) . (2.88) (2.90) sone is n i.e. ,

This is a supervisor’s copy of the notes. It is not to be distributed to students. ahmtclTio:I ler emty(atI 40 I) (Part Geometry & Algebra IA Tripos: Mathematical Remark. Exercise. .5Ivrini Sphere a in Inversion 2.15 pee n a rmec other. each from far and sphere, Reflection. neso nashr snta smty ..topit ls oteoii a ofrfo the from far to map origin the to close points two e.g. isometry, an not is sphere a in Inversion hwthat Show Hence from perpendicular point the of a For vector. unit = Π oeas that also pose But Then n so and Remark. aigdrvdtempig enwne oso htdsacsaepeevdb h map- the by preserved are distances that show to need now Suppose we ping. mapping, the derived Having Let srqie o isometry. for required as OP x | { → NP 12 x osdrrflcini ln ,where Π, plane a in reflection Consider NP 0 → ∈ = = | OP R = P x → x ( = 3 1 0 0 NP | x x − : slna in linear is → a nes point inverse has + x NP x = · → x · = N P n · OP n 2 → → n | 0 ) = ( and | and n x 0 = N P 7→ −| and , + 12 0 → | NP NP | NP x x } 2 = → 0 12 0 x P and − = ..tempigi ierfnto ftecmoet of components the of function linear a is mapping the i.e. , P | | = = = = x let , 0 = n n P, NP OP = → j otepae Sup- plane. the to → OP x 7→ → n | | x | 1 0 P if if 0 x x x N 12 − x . . 12 1 12 saconstant a is NP NP OP − j 0 x → → − → x | etefoot the be − 2 2 12 = 2 0 2( , 0 hnsince then , P. NP = 2( a h poiesneas sense opposite the has a h aesneas sense same the has − x → x j x x 1 4( Let Definition. ihrsett ison lies Σ to respect with ersnsashr ihcentre with sphere a represents = Σ Let − 12 0 · x = n 2( · 12 ) n OP x n x → ) · j − n −

n c · | x = )( 2 2( n n x 0 ..olydmpcma.k ihems2006 Michaelmas [email protected], { x ) 2 2 x = x x n 12 2( + 1, = and · | ∈ o ahpoint each For x n · ( n 0 ) | j R n x 4( + ) | OP x x 1 = 2 3 . → | · : n OP = n 0 , | i.e. , = ) x 2) x n | k n | x 12 0 x | 2 . i.e. , 2 = | = 0 OP then , · | x x x n k OP | · k ) P n 2 2 ∈ = x n satisﬁes and n h nes point inverse the , > · . 2 R O | n x k ( 0 2 | > k n radius and < 2 x 0 . . 0) } hnΣ Then . x k (2.93) (2.92) (2.91) . . P 0 10/06

This is a supervisor’s copy of the notes. It is not to be distributed to students. 11/02 ahmtclTio:I ler emty(atI 41 I) (Part Geometry & Algebra IA Tripos: Mathematical sphere. a of inversion Example: 29)cntu erewritten be thus can (2.94) hc steeuto faohrshr.Hneivrino peei peegvsanother gives sphere a in sphere a of inversion Hence Exercise. sphere. another of (if equation sphere the is which equivalently or (2.93) From reuvlnl if equivalently or hthpesif happens What a 2 6= r 2 ). a 2 6= r 2 ecncmlt h qaet obtain to square the complete can we | x ups htteshr sgvnby given is sphere the that Suppose a 2 −

k = x a | 4 | x 0 k x | r − − 0 4 0 = 2 | | 2 ? 2 2 a , r a − = 2 k · − 2 2 x | a a x k 0 r i.e. k 4 | · 2 2 2 x

2 ( + 0 | n so and = x k x 0 a 2 2 | ( 2 2 − a − + 2 r 2 − 2 r a a

c k 2 2 · r x ) 4 − | 2 x ..olydmpcma.k ihems2006 Michaelmas [email protected], x = ) + 2 0 r | 2 2 , | a x k 2 0 2 | 0 = 0 = 2 − x r 0 2 . 0 = , , . (2.94)

This is a supervisor’s copy of the notes. It is not to be distributed to students. aifigtefloigegtaim rrules: or axioms eight following the satisfying e.g. braces, set. between empty elements The its placing numbers. by real listed of is set set a Conventionally members. a recap, Sets. field). a of ahmtclTio:I ler emty(atI 42 I) (Part Geometry & Algebra IA Tripos: Mathematical section. the of end operations A Definition 3.1.1 set. every any over indeed or numbers, complex spaces vector Space? consider Vector will We a is What matrices, 3.1 polynomials, etc.. of finance, sets be engineering, thought also science, have can you mathematics, they However, point throughout space. this occur 2D) to (or Up spaces 3D section. Vector in the previous etc.. ‘arrows’ the in of functions, example ‘abstract’ set to the (e.g. a is or in as focus section (groups addition vectors this properties under of to we of inverse those which aim that an is with The of on objects something made, given). existence mathematical just taking properties the is all by key and studying progress identity then done certain associativity, and mathematical is closure, identifying multiplication), say, that this numbers, numbers), ways Often real real of the subjects/results. (e.g. case disparate of with be one familiar to indeed are seem mathematics, what of linking strengths by the of One This? Study Why 3.0 Spaces Vector 3 A(iii) A(ii) 17 16 A(i) etrsaeoe h elnumbers real the over space vector • • h rtfu enthat mean four first The eat a elte aerahdnraaa h tr fti eto;nra evc ilb eue oad the towards resumed be will service normal section; this of start the at nirvana reached have they feel may Pedants ehv eerdt esaray n oeta o aecvrdte in them covered have you that hope I and already, sets to referred have We ne clrmultiplication; scalar under multiplication scalar addition; vector addition vector ..vco diinhsa dniyelement; identity an has addition vector i.e. element an exists there is addition is addition set 16 sacleto fojcscniee sawoe h bet fastaecle lmnsor elements called are set a of objects The whole. a as considered objects of collection a is h mt set, empty The commutative associative eoe for denoted V eoe for denoted san is ..frall for i.e. , over 0 ..frall for i.e. , } { bla group abelian ∈ x V or , h elnmes hr r eeaiain ovco pcsoe the over spaces vector to generalisations are There numbers. real the y aldthe called , field ∅ ∈ a steuiu e hc otisn lmns ti ustof subset a is It elements. no contains which set unique the is , saset a is x ∈ V x , x (see y R ( + , by ne addition. under , y 17 z and y ∈ ∈ rus ig n Modules and Rings Groups, x x ulo eovector zero or null V + V + V + x x z feeet,o vcos,tgte ihtwo with together ‘vectors’, or elements, of + y y ( = ) ∈ where , = 0 V = y x by

+ x c + , x ..olydmpcma.k ihems2006 Michaelmas [email protected], a y x (3.1b) ; x + ) + where , y z uhta o all for that such , (3.1a) ; ∈ V a x ota hr scoueunder closure is there that so ∈ nPr Bfrtedeﬁnition the for IB Part in V ota hr sclosure is there that so ubr n Sets and Numbers x { x ∈ : V x ∈ R } binary (3.1c) sthe is To .

This is a supervisor’s copy of the notes. It is not to be distributed to students. 11/03 B(viii) aeysaa utpiaino etr (e.g. vectors of multiplication scalar namely Properties. 3.1.2 ahmtclTio:I ler emty(atI 43 I) (Part Geometry & Algebra IA Tripos: Mathematical B(vii) (iii) (iv) B(vi) A(iv) (ii) B(v) 18 (i) tityw r o setn h soitvt fa prto,sneteeaetodffrn prtosi question, in operations different two are there since operation, an of associativity the asserting not are we Strictly h diieivreo vector x a of inverse additive The (3.1c) vector zero The since all for i.e. vector, zero the yields 0 by multiplication Scalar to us allows vector negative/inverse unique defining a of existence The of inverse unique the denote We hr stemlilctv dniyin identity multiplicative the is 1 where all for i.e. element, identity an has multiplication scalar ‘associative’, is vectors of multiplication scalar clrmlilcto sdsrbtv vrsaa diin ..frall for i.e. addition, scalar over distributive is multiplication scalar all for i.e. addition, vector over distributive is multiplication scalar o all for then 0 + x ∈ x V = x hr xssa additive an exists there 0 and suiu eas if because unique is x + 0 0 = x x µ x o all for x suiu,frspoeta both that suppose for unique, is n utpiaino elnmes(e.g. numbers real of multiplication and ) by 0 x − eaieo nes vector inverse or negative λ ( y 0 x y x λ ( (0 = 0 = 0 = = 1) + (0 = (0 = = 0 . x − ∈ and R λ + 0 ( = = = = = ( = + ; V x x µ µ x 0 0 1 x n hence and , y ) 0 ≡ x x + x 0 . x ( + x x ( = ) = ) y y 0 z x 0 + ( + + 18 y = y x = + 1 + . + r ohzr etr in vectors zero both are ( + + = 0 − 0 + 0 ( + ..frall for i.e. 0 = x x λ λ 0 z 0 x x λµ x x , x x x ( + ) x =

( + ) c 0 − ( + ) ( + , + ) + + + ) 0 (3.1d) ; x x ..olydmpcma.k ihems2006 Michaelmas [email protected], − z ) µ λ . − − , z ) . x x y x − x x )) x . (3.1g) ; ) x ∈ ) ,µ λ, ) ∈ V subtract V ∈ , x y 0 R ∈ ,µ λ, and λ and λµ V ∈ swl sadvcos by vectors, add as well as ). ∈ R uhthat such z V x R r diieivre of inverses additive are and ∈ hnfo 31)and (3.1b) from then and V x , x y ∈ ∈ V V (3.1h) (3.1e) (3.1f) (3.3) (3.2)

This is a supervisor’s copy of the notes. It is not to be distributed to students. (viii) ahmtclTio:I ler emty(atI 44 I) (Part Geometry & Algebra IA Tripos: Mathematical Examples 3.1.3 (vii) (vi) (v) (i) n hnapa otefc httezr etri nqet ocuethat conclude to unique is vector zero the that fact the to appeal then and λ that Suppose all for i.e. vector, zero that the observe yields ﬁrst vector we zero this the with multiplication Scalar since by multiplication Scalar ti tagtowr ocekta () (i,Aii (v,Bv,Bv) (i)adBvi)are B(viii) and B(vii) B(vi), B(v), A(iv), A(iii) A(ii), A(i), that check Hence to satisﬁed. straightforward is It Let all for since freely commutes Negation if So n titypstv nee.If integer. positive strictly any ,i hc aeteeexists there case which in 0, 6= R and λ n x etesto all of set the be = 0 hneither then λ x R = n 0 savco pc over space vector a is where , − n λ λ − ilsteadtv nes ftevco,ie o all for i.e. vector, the of inverse additive the yields 1 or 0 = 0 x tuples λ sazr etrsince vector zero a is + − λ ∈ x y x x 0 , R ( λ y − x − { and (0 = ( = ( = ( = ∈ 1) 1 x = λ x uhthat such R ( = ( ( 0 0 λ − − n x . − λx x with , + λ λ ∈ x , 0 = ( = (( = = ( = ( = = 1 x ∈ R x ) ) 0 ( 1 1 λ + x x R 1 − , . . . , . V λx , x , , x = ( = 1 = = = λ 1) − y 0 − and n osblt sthat is possibility One . 2 − − − 0 1 x (( = ( = = ( = = = x x − x x , . . . , x ( + x , 2 λ = = 1) + 1 1) 1) = 2 ( + λx , . . . , 0) λ 0 λ . 1) − = saoeand above as , . . . , λ 2 x x x − − x − . 1 0 λ − λ , x − + − 1 1 λ − − λ ∈ ( + + λ λ (( ( x 1 . − ( ( 0

+ ( − c x x ( x 1)( λ λ λ .Te ecnld that conclude we Then 1. = ) n y − − x V 0 0 1) − 2 ) x x ) x : ) , , x x ..olydmpcma.k ihems2006 Michaelmas [email protected], 1)) 1) x x , . . . , ( + + ) ) ) n x λ λ ( + ) ( + . , ) n ) x x x x x x , ) ) ) − j ) . − x ∈ − x n ) x y R )) + ) ( = with y n y ) λ 1 , y , j .Hwvrspoethat suppose However 0. = 2 1 = y , . . . , λ , 2 ∈ n , . . . , x n R ,deﬁne ), ∈ , λ V 0 } , where , = 0 osee To . (3.7d) (3.7b) (3.6b) (3.7a) (3.6a) (3.7c) (3.5) (3.4) n is

This is a supervisor’s copy of the notes. It is not to be distributed to students. 11/06 ht( that each For A(iv). Remark. if only and if i.e. Subspaces 3.2 ahmtclTio:I ler emty(atI 45 I) (Part Geometry & Algebra IA Tripos: Mathematical Examples 3.2.1 0 each For A(iii). in inverse an has element every the of If. Proof. on 3.1. Theorem define to used are as multiplication) subspaces. scalar Proper and addition vector (i.e. operations same the under space Definition. x (ii) (ii) (i) (i) U ti tagtowr oso htAi,Ai) () (i,Bvi n (ii odsneteelements the since hold B(viii) and B(vii) B(vi), B(v), A(ii), A(i), that show to straightforward is It = V r loeeet of elements also are o each for each for that hence and function). satisfied, a are is B(viii) element and vector B(vii) B(vi), B(v), A(iv), function the where b < a set the Consider For Thus of subspace 0 A − Hence . nyif. Only 1) λ rprsubspace proper ecncmie()ad(i stesnl condition single the as (ii) and (i) combine can We ( n x x For . U 1 > = x , subset A scoe ne etradto n clrmlilcto,adi ec subspace. a hence is and multiplication, scalar and addition vector under closed is let 2 − x x 2 0 , If x , . . . , x ,g f, subset A ∈ x x y ∈ Hence . R U U ∈ U ∈ Strictly ∈ U n U ∈ sasbpc hni savco pc,adhne()ad(i hold. (ii) and (i) hence and space, vector a is it then subspace a is scoe ne etradto n clrmultiplication. scalar and addition vector under closed is U U U and . ic,for since, = n F tflosfo i)ta 0 that (ii) from follows it , , tflosfo i)ta ( that (ii) from follows it , F U − x { O sasbpc of subspace a is 1 define U x V λ − fra-audfunctions real-valued of fteeeet favco space vector a of elements the of , + )+ 0) V stezr vco’ gi ti tagtowr ocekta () (i,A(iii) A(ii), A(i), that check to straightforward is it Again ‘vector’. zero the is x ene eosrt htAii (i.e. A(iii) that demonstrate need We . ∈ : favco space vector a of y o each for x ∈ and R ∈ ( = x , U µ , U λ ( . y y x { x , 1 0 ∈ 1 U y , ∈ } x , x U hold. ) 2 U ie h e otiigtezr etrol)aesbpcsof subspaces are only) vector zero the containing set the (i.e. , 2 y , . . . , y ( x , . . . , , and f V ∈ ( ( + − λf hti not is that U ,µ λ, f g V O n )( )( )( n − and ( sasbpc of subspace a is − x x x x 1 1 , ∈ 0 = ) = ) = ) = ) x − , )=( = 0) f )with 0) ,µ λ, R ∈ 1) ( x , x U fara variable real a of ) V ∈ u ic also since but ; ∈ λx f − λf or R ( , U

f 1 x c x ( λ , ( V u ic also since but ; + ) x + j { x ..olydmpcma.k ihems2006 Michaelmas [email protected], ) 0 ) ∈ x µy scle a called is , } , V g . R + ( 1 x λx , fadol fudroeain deﬁned operations under if only and if µ and 0 ) y , sa lmn of element an is 2 ∈ j + x F . U 1 = µy subspace x ∈ savco pc weeeach (where space vector a is ∈ V 2 x , λx , . . . , 2 [ ∈ tflosfo 33 that (3.3) from follows it n , . . . , ,b a, V ,where ], of tflosfo (3.4) from follows it n U − V − ,adAi)(i.e. A(iv) and ), 1 1 if + } Then . U ,b a, µy savector a is n − ∈ 1 , R U (3.8d) (3.8b) (3.8a) (3.8c) 0) and sa is . V V . .

This is a supervisor’s copy of the notes. It is not to be distributed to students. 12/03 12/02 .. pnigsets Spanning 3.3.2 that such zero, are which of all not ahmtclTio:I ler emty(atI 46 I) (Part Geometry & Algebra of IA subspace Tripos: a Mathematical is hence and multiplication, scalar and U addition vector under closed is Definition. Remark. Definition. Remark. be to said are vectors the Otherwise, scalars all Definition. nt ubro etr ti sntawy h case). space vector the always the not that is follows) (this that vectors much of in number (and finite here a assuming implicitly are we v in vectors from ideas Extend independence Linear Bases and 3.3.1 Dimension Sets, Spanning 3.3 (iii) (ii) ∈ scle the called is V that For W For Then is origin, the through pass (wlog not zero does that are hyper-plane which of all not and n so and that see shall we Thus hr xs scalars exist there , sasbpc of subspace a is The n n 0 λ ihti definition this With x W > > j 6∈ W f ∈ e of set A h e fallna obntoso the of combinations linear all of set The of subset A ∈ λ osdrteset the consider 2 osdrteset the consider 2 W scoe ne etradto n clrmlilcto,adi ec usae Later subspace. a hence is and multiplication, scalar and addition vector under closed is f j span W f R antb usaeof subspace a be cannot r o eesrl nqe(u e eo o hntesann e sabss.Further, basis). a is set spanning the when for below see (but unique necessarily not are , ( or but j of 1 = consider n x W S + , vectors V 2 S (written sahprpaetruhteoii se(3.24)). (see origin the through hyper-plane a is x λ n , . . . , j X = ic,for since, i U 6∈ =1 n ∈ λ R { x W f = x X i u λ { 3 R W =1 n W f . ∈ 0 + { i 1 ( ), v since v , ( } W α U f j u u i = µ 1 = salnal eedn set. dependent linearly a is i , v = y 2 1 = : ( span = v ierydependent linearly { . . . , x { α λx uhthat such u = ( = x 2 0 x , P 1 . . . , y , = i ∈ λ R ∈ ,i o ere h ubrn fteaxes). the of numbering the reorder not if 0, 6= 2 i n λx + =1 ∈ 1 u n , . . . , n R X i ⇒ R u =1 n . n µy W 1 n S v 1 n α } x X i + n ,adw a that say we and ), + : =1 i λ n : i fvcosin vectors of ( = ( } = ) and P i P x µy λ , λ u ,sc that such ), λ i i 2 v i n i i n + α i =1 1 λ , u =1 for 0 = v λ j λx , 1 − ,µ λ, x 2 i X ∈ i 1 α i α =1 + j = n , .Thus 2. = ) u i i 0 2 V ∈ x j x . . . ic hr xs scalars exist there since ∈ 0 , . . . , α + i i R ( . i ( 0 =

R j 1 = x c j not µy + j , V , i 1 = 1 = ..olydmpcma.k ihems2006 Michaelmas [email protected], + 0) λ 2 i } 1 = } sa is λx , . . . , n 1 = usaeof subspace a o ie scalars given for µ , . o ie scalars given for u , 2 S 2 X , n i n , . . . , W =1 f n , . . . , 2 n pnigset spanning , . spans n , . . . , 2 . n , . . . , α sntcoe ne etraddition, vector under closed not is n i y + ,i.e. ), i ,is ), U 0 = ) µy . n R ieryindependent linearly . ) n , for α osethis see To . λ α j j j V ∈ ∈ V ∈ V a esandby spanned be can R R h etrspace vector The . R ffreeyvector every for if ( ( ( j j j W f 1 = 1 = 1 = hc sa is which , either , , , 2 2 2 n , . . . , n , . . . , n , . . . , (3.10) ffor if (3.9) note ). ), )

This is a supervisor’s copy of the notes. It is not to be distributed to students. 12/06 V ahmtclTio:I ler emty(atI 47 I) (Part Geometry & Algebra IA Tripos: Mathematical Definition. Example. V Since Proof. 3.2. Theorem Example. ups that Suppose reduce slnal independent. linearly is , If . S eedn since dependent ieryidpnetoeta tl spans still that and one independent linearly a S n S If spans − • • • • oalnal needn e lospanning also set independent linearly a to S 1 { osdrteset the Consider Consider h set The set The set The set The then of hsstspans set This hsstdoes set This then u slnal needn hnw r oe fntrpa until repeat not If done. are we then independent linearly is slna eedn hnteeexists there then dependent linear is u 2 A u 1 , λ V u 1 (1 = n basis , λ λ and , faset a If 3 i o ere h etr) Then vectors). the reorder not (if 0 6= u 1 1 , u 4 = = { { { { , , 4 R u u u u 0 } u λ λ o etrspace vector a for 3 u 1 1 1 1 , u n let and , 5 , , , , 0) 2 2 l span all n 4 Testde pntepaecontaining plane the span does set (The . u u u u S not = = a eepesdi em of terms in expressed be can , 4 2 2 2 = R = , , , , λ λ u u u u { 3 u span 3 4 u u { 5 4 3 3 1 ic o any for since u } } } , .Ti e spans set This 0. = .Ti e spans set This 0. = 1 2 + u , 1 R (0 = slnal eedn since dependent linearly is if since independent linearly is if since independent linearly is u , 4 λ u ( 3 R u , 2 1 x 2 u , 2 u 3 n r ieryindependent. linearly are and 1 + u , . . . , 5 ..tevco (0 vector the e.g. , , 1 − } 3 1 u + , , slnal eedn since dependent linearly is 3 x u 0) hr r ubro asb hc hsstcnb eue to reduced be can set this which by ways of number a are There . V λ 3 4 u ) 2 } , u n u salnal needn pnigsto etr in vectors of set spanning independent linearly a is for , x u } 1 X i 2 n =1 ( = 3 u ( + X slnal eedn,adsastevco space vector the spans and dependent, linearly is i + u X i =1 = n 3 =1 1 3 λ u λ u (0 = x x + − i R j λ λ 4 u 1 1 x 2 R R u i j x , 3 sdfie n(.1.Ti e spans set This (3.11). in defined as u n X i i u i − − , ..tesets the e.g. , 4 3 3 u =1 − 2 ( = i , j 2 V i ( = 1 a above). (as ic o any for since u , x u 0 + x , = 1 = 1 ( = , . 3 4 λ 1 λ , λ 1) ) , . . . , u 3 n 0 + i λ − u 1 ) 3 u x 1 n , . . . , λ , . , 2

)cno eepesda iercombination linear a as expressed be cannot 1) u ∈ c − 1 i + + 5 x , 2 . R ..olydmpcma.k ihems2006 Michaelmas [email protected], u u λ , u λ = x 2 3 n 4 4 4 3 x , , 3 − λ , 0 u o l eosc that such zero all not , = (1 = = ) 1 { 4 3 . u h set the , 2 u x ) 0 1 ( = + . 1 , . 0 ( = , , u 1 λ u , x , 4 4 2 1 x 1) S λ , , x , and 1 u p x , , S 4 3 2 = = ) n } x , 2 − , u x , u { 1 3 { 5 u 5 ) 0 u .) 3 = 1 (0 = . ) 1 , . . . , , , ∈ { u u R 2 1 , R , , . . . , 3 1 u u , 3 , p 4 u slinearly is but , 1) } } hc spans which , , . u { n u V − V 1 1 , ecan we , . } u (3.11) 3 spans , u 4 }

This is a supervisor’s copy of the notes. It is not to be distributed to students. n ec that hence and ahmtclTio:I ler emty(atI 48 I) (Part Geometry & Algebra IA Tripos: Mathematical Deﬁnition. basis. a than members more have Remark. Proof. elements. 3.4. Theorem that (3.12) from follows It Proof. 3.3. Lemma set (3.12) from Then if elnal eedn n ovoaetehptei htteset Hence that that the deduce mean that would to this Continue hypothesis basis. the a violate forms so and Assume dependent the linearly of be one least at Moreover, with But hr exists there independent. Since λ 1 { { u λ 0 6= u i λ 1 m i , 1 Let ups that Suppose 1 i o ere the reorder not (if 0 6= , . . . , µ . 1 = = ,w euethat deduce we 0, 6= 2 naaoosagmn a eue oso htn ieryidpnetsto etr can vectors of set independent linearly no that show to used be can argument analogous An x ohrierodrthe reorder (otherwise 0 6= n µ n , . . . , . u h ubro etr nabssof basis a in vectors of number The ∈ i , If n V i { } Let w { 1 = hnsince then , x sabssso basis a is u 1 } 1 = , , . . . , n , . . . , u V { sabssteeexist there basis a is u 2 { µ λ , . . . , w 1 1 1 ea be , . . . , 1

u , n w w a es n fwihi o-eo uhthat such non-zero) is which of one least (at } u ﬁnite-dimensional 1 2 { u n , . . . , sabssfravco space vector a for basis a is µ u − λ } u n i 1 1 } , i ν X spans i , . . . , ν , =2 i n i 1 and i w 2 = for 0 = λ 0 = 1 = λ m 1 i u ν u { } u V n , . . . , 1 n n , . . . , ν , w i w { u i } ! a ierydpnet otaitn h rgnlassumption. original the contradicting dependent, linearly was ute,spoethat suppose Further, . ν , w 1 1 2 λ 1 spans i , . . . , + w 1 + i = 2 = , 1 , w w i x 1 λ X w i i X i µ .Teset The ). =2 n =2 1 1 1 = i + utb o-eo tews h subset the otherwise non-zero, be must n = 1 = 2 1 etrsae hneeybsshstesm ubrof number same the has basis every then space, vector n , . . . , + w = = , . . . , w V X ( µ i V X i =2 m ν n =1 ν 1 n i X X n , . . . , n , . . . , i i 1 u i =1 =1 hr exists there } n n + sthe is λ i ν 0 = µ i w r w ae for bases two are i = X λ λ i i .W a hnddc that deduce then can We ). u + =2 u n n i i i u u { } i i , µ λ ν n ec that hence and , = w . µ i i uhthat such

i 1 1 dimension V c om ai,adhnespans hence and basis, a forms ) i w 1 u 0 u hns is so then , , ..olydmpcma.k ihems2006 Michaelmas [email protected], i 1 2 = i . u . = + 2 µ , . . . , . n , . . . , 0 i X i =2 n uhthat such . { u of w µ n V i } V i { i , − w rte dim written , sa lentv ai.Hence, basis. alternative an is with 1 1 = λ { , λ w i u µ 1 1 2 m 1 m , . . . , , , . . . , u > 2 u { , . . . , i w n u , 1 n wo) ic the Since (wlog). } , V { } w u w om basis. a forms . for n 2 1 V , } , u w If . slinearly is 3 2 , . . . , } n > m would (3.12) u n }

This is a supervisor’s copy of the notes. It is not to be distributed to students. 13/02 eepesda iercmiaino the of combination linear a as expressed be ahmtclTio:I ler emty(atI 49 I) (Part Geometry & Algebra IA Tripos: Mathematical the where 3.6. Theorem Example. spaces). say subspace, proper some a spans for it not if done, are we spans it If independent. in vectors are there Proof. 3.5. Theorem Examples. Exercise. Remarks. o choose Now . Components 3.4 (ii) (i) • • Since subspace The for set The i)o ae4 with on 46 restriction page on (ii) [ on defined functions real encoun- of already space have vector we the (although i.e. dimension, dimension infinite finite of of space spaces vector vector one to tered ourselves restrict shall We space vector The ec ai for basis a Hence and dim But three. dimension has which { any choose Now (1 ( Unlectured. , R − m { v Let n hwta n e fdim of set any that Show (1 i { 1 n hsdim thus and , spans , w e , r nqefreach for unique are , { } 0) 2 − V e slnal independent, linearly is ∈ , 1 1 If Let (0 = , , V e 0) U , 2 R V U Let ) , x 0 V , . . . , , V w 3 , 3 (0 sapoe usaeof subspace proper a is u o in not but ⊂ 1) for , n suppose and 2 eavco pc n let and space vector a be wihms epsil ic eaeol tdigfiiedmninlvector dimensional finite studying only are we since possible be must (which , { y V U } 0 R 6∈ 0 { , ..(dim i.e. , e , } uhthat such n is u y 1) n U R x α 1 } ssi ohv ieso 0. dimension have to said is } ossigo etr ( vectors of consisting 1 { n , 1 , with , ∈ hnteset the Then . (1 y slnal needn n om ai o span for basis a forms and independent linearly is u = = } 2 U , , . . . , U − n sa xeso fabssof basis a of extension an is v α ecnwrite can we Choose . . 1 . 2 V , e U 0) 1 = y 1 V x u = span = (1 = , ` ∈ ieryidpnetvcosof vectors independent linearly (0 } { ` α , { ( = = e )w r oe fnti pn rprsbpc,say subspace, proper a spans it not if done, are we 1) + eabssof basis a be , ( R } R 0 { x , w { e 3 , (1 3 sabssadtu dim thus and basis a is 3 0 1 , u i 1) V x 1 x , , . . . , v i.e. , ,hence 3, = x ) Since 0). = , 1 1 y n ai of basis any , ∈ } − 1 , (1 = 2 . , u S x , U 1 V 6∈ − , 2 X ,x x , . . . x, x, , i 2 0), − = =1 , . . . , , 3 n 0) x of , U ) w 1 1 { , , x , ∈ v e.g. , e 1 (0 U e )+ 0) i V 2 1 e R 3 6∈ u , , . . . , (dim i o eetuntil repeat Now . ) (0 = 0 3 ` , U

, , x c : x w 1) hnteset the Then . 1 y with ) x 3 U ..olydmpcma.k ihems2006 Michaelmas [email protected], 1 , , U + e (0 U 1 x , 1 y (1 = n + w , . . . , 1 a eetne noabssfor basis a into extended be can , } x } oabssof basis a to = x , 0 2 2 x = , eabssof basis a be } ` 1) 2 x 3 mle that implies 0 = .If ). , U slnal needn.I tspans it If independent. linearly is R 0 = 0 ) , . . . 0), ∈ ∈ . , 3 1. = ) r(0 or 0), R R , V U } . ssandby spanned is sabasis. a is { sapoe usaeof subspace proper a is U { u u e 1 R sasbpc seexample (see subspace a is 1 , n , 3 u V u (0 = , . { 2 1 hneach Then . 2 , . . . , (1 , , . . . , ) r(1 or 0), , , − x 0 1 , . . . , u 2 u , e ` 0) = ` , , (1 = w , w − (0 ) sabasis a is 1), 1 , 1 } 1 x , , . . . , v , , 1 slinearly is 0 ) Then 0). 1 U , ∈ V ihno with , . . . , 1) 1 V . V of , (3.13) w ,b a, } then , m can y 1). V ]). V } } , . ,

This is a supervisor’s copy of the notes. It is not to be distributed to students. 13/03 . nescinadSmo Subspaces of Let Sum and Intersection 3.5 uhthat such ahmtclTio:I ler emty(atI 50 I) (Part Geometry & Algebra IA Tripos: Mathematical Remark. Proof. 3.8. Theorem Remark. Proof. 3.7. Theorem Remark. v Definition. Remark. Definition. the But Then exist there also Suppose Proof. v λ n thus and of x 2 ∈ V = + U W . u µ rttpcleapeo real a of example real every prototypical that show to possible dimension ( components with associated ihauiu e fra ubr,nml t components its namely numbers, real of set unique a with λ and . y 2 If If If x + e ∈ Since U U U S o ie ai,w ocueta ahvector each that conclude we basis, given a For v i x + w 1 W , W r ieryidpnet ec ( hence independent, linearly are λ ∩ ∩ + sabsste,fo h ento fabss o all for basis, a of definition the from then, basis a is , µ y v 2 v U the call We W W y o any For . 1 W esbpcso etrspace vector a of subspaces be sneboth (since 2 ∈ ( = ) U + ∩ ∈ U U n sas usaeof subspace a also is least at contains sasubspace. a is U ⊆ W µ U ∩ + etrsae vrtera ubr r h sm as’ ‘same the are numbers real the over spaces vector v λx ∩ U 2 W W sthe is + W = + 1 W ,µ λ, + sasbpc of subspace a is sasbpc of subspace a is then , W λ v v i hnteeexist there then , 0 ( µy U i u ∈ intersection and ∈ the 1 1 and R λx , . . . , + R , , x X components i w W i =1 n λ , W 0 1 = 1 y u y + ) ⊆ v 1 oi o empty. not is so n r usae) Hence subspaces). are ( = i ∈ + e n n , . . . , U U dmninlvco space. vector -dimensional n i µ + dmninlvco space vector -dimensional U µ − y + of ( n usaeof subspace a and V V u u 1 µy y , . . . , X . and i W . 2 2 U =1 n uhthat such , n + v ∈ u of both , i ) v v V and 1 v U w − ∈ , x i v 0 = = e u over 2 , n v and R i ihrsett h basis the to respect with ( = ) 2 y ) X X i 0 i i W = =1 =1 n 0, = ) steasis n h riae while ordinate, the and abscissa the is U Remark. u Definition. 2D-plane. and n n ∈ ∈ U itemr okte eosrtsta htall that that demonstrates then work more little A . ∈ X ∈ R i span = n ossso l vectors all of consists and λ R v v U =1 n λ and W i i . 0 w U n W e e u x ( and i i 1 1 v ec o any for hence , hn( then i navco space vector a in and . . n ossso l vectors all of consists and , i

+ + 1 = c W W λ − { µ x U µ w ..olydmpcma.k ihems2006 Michaelmas [email protected], . (1 v w r usae of subspaces are w n , . . . , u + i 1 0 + λ , ) 2 2 , U e 0) x ( + ) ∈ µ w x v W ∈ V i y + } + 2 W = ( = ∈ W is ∈ and W n ec the hence and , ∈ µ λ . snttesm as same the not is 0 V Hence . y isomorphic U w x W . scle the called is sascae ihcomponents with associated is ) 1 hr exist there 1 ∩ W x , . . . , ,µ λ, + V R uhthat such S W span = n . µ fdimension of and , ob oepeie tis it precise, more be To . w ∈ U n 2 v ) ) + R to ∈ ∈ uhthat such { , U W v sum (0 U R i λ R v v ∩ i u . x n , ∈ 1 n + 1) ute,if Further, . r unique. are W Thus . + = + R , W fsubspaces of } n , U w then , µ sasubspace a is u i sassociated is y U 1 + v ∈ 1 = + ∈ R W ∪ ∈ V w n n , . . . , U U U W where 1 sthe is sthe is ∪ y If . and and and W U is , 13/06

This is a supervisor’s copy of the notes. It is not to be distributed to students. 14/02 λ ahmtclTio:I ler emty(atI 51 I) (Part Geometry & Algebra IA Tripos: Mathematical Examples 3.5.2 Thus Deﬁne for Proof. 3.9. Theorem Let 3.5.1 tflosthat follows It U Similarly But that deduce We in vector a is sum ﬁrst the where dim( hence and i (i) ∩ ∈ U { W { R u rmeape(i npg 6.Te,say, Then, 46). page on (ii) example from ups that Suppose nteohrhand other the On Thus If e n basis a and , 1 1 ). v Let x , . . . , , . . . , i U dim( by 1 = dim( ∈ µ ∩ { i U U e W 0, = U u e r , . . . , 1 U r + ∩ ` , . . . , + , } { span = U g W + W e and W 1 { 1 + i , . . . , e W , . . . , , U 1 = then e W 1 ) W U µ r , . . . , = ) span = span = { } = i { ) w s , . . . , ∈ g (0 eabssfor basis a be e 6 { span = span = 1 t r r x R , . . . , , } e , x dim 0 + f , r ∈ sabss(for basis a is 1 , 1 dim( , i 1 , . . . , s g and 0 = Further . R { { 1 = v , 1 + w U (0 (0 0) 4 , . . . , U = m U t : , , , dim + { { s , . . . , v + U f 1 1 (0 } x (0 ( ( = X i s X i + , , − =1 1 r hl h H savco in vector a is RHS the while , , = =1 , W r 0 0 ebssfor bases be , g g X 2 i 0 0 = U , , 1 W =1 r t r 1 α , X , i 0) 0) , λ } =1 , . . . , λ 1 span = + 0 s ∩ 0 W i dim( + ) i ( , e and , , , i for , e } α x W 0 W (0 (0 1) i s 0) µ i 0, = oee,w a a oeta this. than more say can we However, . 1 , i ( + ) and + i , , } 0) , + n ec ieryidpnet n so and independent, linearly hence and ) 2 + W f g O fepy.Etn hst basis a to this Extend empty). if (OK 0 0 o some for (0 i ν span = X t , , , i i λ = { } =1 Then . (0 1 1 s , i r W x i ∈ u , , 0 ) U slnal needn n stu ai for basis a thus is and independent linearly is , − 2 1 = 0) 0) + U 1 , e µ R 0 , . . . , 1 i 0 = = i ∩ , X , , i t , , f and + =1 1 (0 (0 ) r i 0) { r , . . . , i W , { − + (0 { X α , , 1) i 1 = x , , λ =1 e 0 0 u t dim = ) (0 i , r X i i W 1 , , , ∈ ` 0 =1 e ∈ , . . . , (0 t 0 0 , , dim( = ν n hence and t , . . . , , i

, , 0 R w i 1 c R , − epciey Then respectively. 1) 1) ν , g rm(.5 since (3.15) from , 4 0 1 0 i , i 1) ..olydmpcma.k ihems2006 Michaelmas [email protected], , , . . . , g , , X i , : i e = =1 1 ( ( 1) i t , x − − r U , 1 = eksltosto solutions seek , (0 = , 1 − U 0 ⇒ } 2 2 ν f dim + 2 + , 1 , , 1) i W . dim( + ) w 0 0 g r. , . . . , , . . . , 1 1 , . ⇒ } i m , , Hence . 1 x 0 0 , x } , , , 2 1 − 0) 0) , . W f 0 = = s 1) } , dim , (0 W g , } x } 1 , 2 v ) , . . . , { dim 0 bt r usae of subspaces are (both U 0 = e n dim and − , ∈ 1 1 3 = , . . . , dim( , U W 1) { . g e ∩ t , , 1 } 3 = ν (0 W , . . . , U e i spans , r 0, = ∩ n so and , 0 } . U , W sabss(for basis a is 1 e , ∩ r i − ) U , W 1 = . f 1) + 1 , . . . , U } 2 = W t , . . . , (3.14) (3.15) + For . . W f R s } 4 . .

This is a supervisor’s copy of the notes. It is not to be distributed to students. 14/03 ahmtclTio:I ler emty(atI 52 I) (Part Geometry & Algebra IA Tripos: Mathematical notation. Alternative Remark. space and vector linear three-dimensional product The scalar a of Definition Products) Inner 3.6.1 (a.k.a. Products Scalar 3.6 vrterasb sinn,freeypi fvectors properties. of following pair every the for with assigning, by reals the over (iii) (iv) (ii) (ii) (i) v Then Let (3.8d). and Let (3.8c) (3.8b), used (3.8a), not in deliberately as that etc. such multiplication, function scalar addition, Define Let Hence set. spanning independent linearly a to set spanning dependent dim linearly the reducing after Non-degeneracy write to us allows This Non-negativity Linearity Symmetry (3.14). verifying again vector a ecmie ofr a form to combined be can V U rpris(.6)ad(.6)ipylnaiyi h rtagmn,ie for i.e. argument, first the in linearity imply (3.16b) and (3.16a) Properties U U + etevco pc fra-audfntoson functions real-valued of space vector the be + ∩ v . W W W i.e. , ntescn ruet ..for i.e. argument, second the in W U .Hne nareetwt (3.14), with agreement in Hence, 4. = = = ..asaa rdc favco ihisl hudb oiie i.e. positive, be should itself with vector a of product scalar a i.e. , = = ..teol etro eonr hudb h eovco,i.e. vector, zero the be should norm zero of vector only the i.e. , natraientto o clrpout n om is norms and products scalar for notation alternative An { V f e { { 1 ( : f f 1 1, = (1) odim( so bold f ( : ( : v (1) f f · v dim( (1) (1) f , oain.Then notation). scalar = f , f , (2) e | u U U v ( 1 (2) (2) λ f , 2 and 0 = (2) | · + 2 ∩ u ( hr h elpstv number positive real the where , u f , f , (3)) λ W 1 W · v + h v (3)) (3)) 3 = ) 1 V dim( + ) dim u | hscnb eeaie oan to generalised be can This . µ v + ∈ u | | = k v R µ ∈ ∈ 0 = 2 v ,µ λ, U u ) v , R 3 | ≡ k ≡ i R R · 2 { v dim + 3 · = ) e v 3 3 e v u U · 1 with a h diinlpoet htaytovectors two any that property additional the has i ∈ i , , = = = 3 ;define 0; = (3) u v = v ⇒ + v R with with λ · | > 1 = v λ λ W v v ∈ ( = u W u v · 0 · , 1 { 6 = 4 + 2 = ) V

( · · f , u v c . 1 λ v v · u 2 1 = (1) a , , u . v = 1 , f f 1 ..olydmpcma.k ihems2006 Michaelmas [email protected], 6 = 3 + 3 = 2 · 3 1 2 0 = (2) 0 = (1) + , v + + } clrproduct scalar 0 3 + ) } µ µ µ . sabssfor basis a is 2 1 i.e. , µ f u v u . u 2 0 = (2) 2 · e · 2 · u } } 2 v ) v V , , 2 2 and . . | = v hndim then dim then } n | { , e dmninlvco space vector -dimensional sa is . , f or , 3 V ( : iial weew have we (where similarly odim so n dim and , norm f ne product inner (1) f , W U ,µ λ, c.lnt)o the of length) (cf. (2) 2 = U 2 = ∈ ∩ V f , , W R . 3. = (3)) , e 1 ; 1 = u (3.18b) (3.16d) (3.16b) (3.18a) (3.16a) (3.16c) · ∈ ethe be v (3.17) R ∈ 3 V } R u .

This is a supervisor’s copy of the notes. It is not to be distributed to students. 14/06 ahmtclTio:I ler emty(atI 53 I) (Part Geometry & Algebra IA Tripos: Mathematical exercise. and Remark Proof. Remark. Proof. when only equality with that states inequality Schwarz’s inequality Schwarz’s 3.6.2 hstinl nqaiyi eeaiaino 11a n 25 n ttsthat states and (2.5) and (1.13a) of generalisation a is inequality triangle This inequality Triangle 3.6.3 ehv w ae oconsider: to cases two have We nwihcs 31)i aifida nequality. an as satisfied is (3.19) case which in of choices certain that for conclude since in contradiction quadratic a a have from simplifies then right-hand-side that rm(.c and (3.1c) from For hsflosfo aigsur ot ftefloiginequality: following the of roots square taking from follows This oedrc a fpoigthat proving of way direct more A 0 λ 6 ∈ k k R u u + consider + λ v v k 0 ntesm a ht(.3)cnb xeddt 11b ecnsmlrydeduce similarly can we (1.13b) to extended be can (1.13a) that way same the In k 2 2 0 = = 6 6 6 u ( = = = k k k ( sasaa utpeof multiple scalar a is v u k k u u u u j u u u k k k · k k rm(3.3), from + 2 2 2 u · 2 + 0 2 + 2 + 2 + v + λ 2 + = v k = λ v | k ) u u u λ u u k · u 0 · ) · k k ( · · 2 v u v v (0 · and . k | v v u + + + v u + k k | + · 1 k λ u u u + λ + v k v 0 + v v v λ v − · · k v rm(3.18b) from ) if 0 = k 2 v 0 k 2 · 6= k v k v 2 | v u st set to is 0 = λ v k k 6 6 v 2 0 2 k + ( 0 = ) cwr’ nqaiyfloso aigtepositive the if only equality taking with root, on square follows inequality Schwarz’s a tms n elro.Hne‘ Hence root. real one most at in has quadratic a then that suppose Second, . hssmlfidepeso a engtv.Hnewe Hence negative. be can expression simplified this > λ 2 is,spoethat suppose First, . k k λ u u oa xrsinta slna in linear is that expression an to

2 k k k k v u v + · u v k − k = v

k k c · v v , 0 k ..olydmpcma.k ihems2006 Michaelmas [email protected], λ 1 , v . ( 0 + ) = k

µ (2 rmaoewith above from rm(3.19) from . n(.6) hn since Then, (3.16b). in 0 = u u · rm(.6)ad(3.17) and (3.16b) from rm(3.16a). from · v λ v ) 2 2 v v ht ic ti o negative, not is it since that, 0 = ) 6 6= = 4 k 0 0 u . h ih-adsd is right-hand-side The . othat so , k 2 k λ u v b 1 = = k 2 λ 2 6 − If . . k λ 4 v v ac u k o some for ,i.e. ’, · .The 0. = 0 v = we 0 6= (3.21) (3.19) (3.20) 0 + λ 0 .

This is a supervisor’s copy of the notes. It is not to be distributed to students. 15/02 .. Examples 3.6.5 ahmtclTio:I ler emty(atI 54 I) (Part Geometry & Algebra IA Tripos: Mathematical in Hyper-plane in Hyper-sphere Remarks. µ λ, Exercise. all the of define space We the that saw we over 44 space page vector on (i) example In for product scalar The 3.6.4 • • ∈ Remark. for that is news over good space The matrices). vector need a we on which for (and in (3.22) product than scalar the basis), the non-orthogonal when case the For the the to that respect appreciate with to product scalar important the is as It (3.22) (3.22). view might definition we the the when with only careful (2.13) little for a product be scalar to need We the while The Remark. by given R , length hog h rgn hnΠi usaeo ieso ( dimension of subspace a is Π then origin, the through ofimta 32)stse 31a,(.6) 31c n 31d,ie for i.e. (3.16d), and (3.16c) (3.16b), (3.16a), satisfies (3.22) that Confirm clrproduct scalar sntasbpc of subspace a not is Σ If R neirangle interior or , R n b R n . . eoe by denoted , · h hpr)ln in (hyper-)plane The ulda norm Euclidean n h hpr)peein (hyper-)sphere The ota = Π that so 0 = R e n = Π 1 x (1 = sdfie y(.2 scnitn ihtesaa rdc for product scalar the with consistent is (3.22) by defined as i R x on x θ and = Σ i rhnra ae lasexist. always bases orthonormal , { · and ewe w vectors two between x , R R y k 0 x n ∈ n x y , . . . , = R { i k · x s(f (2.39)) (cf. as favector a of , R adrfre oa elcodnt space). coordinate real as to referred (and y n 2 ( X r opnnswt epc oa rhnra ai.T hsend this To basis. orthonormal an to respect with components are i i R n =1 λ ∈ n k r opnnswt epc oannotoomlbss(..a (e.g. basis non-orthonormal a to respect with components are n y = 0) ( : R x . x + { k , n x i R x x x θ y 0 = e µ : − R i n · · ∈ 2 arccos = k z k = y n x x = ) x b htpse through passes that (0 = R k ihcentre with ) x − n x · 1 = ⇒ > = : y n a R , ∈ 1 P k

1 n 0 = + x , . . . , R = X i i n 0 y x λ qiaett 21)hsamr opiae form complicated more a has (2.13) to equivalent =1 =1 n n n and x , k x · > r , = tpe fra ubr om an forms numbers real of -tuples 2 x sdﬁe as deﬁned is x x x x · y b 0)

kk 0 i 2 2 y c i , · , n y ! a . 0 y , . . . , n + y + i ..olydmpcma.k ihems2006 Michaelmas [email protected], r , ∈ 2 1 sdfie obe to defined is k 0 = ∈ . . . µ , R R x ∈ n + . } n n · e R ..s httehprpaepasses hyper-plane the that so i.e. , b n radius and z n x with − , tnadbasis standard , n ∈ (0 = a )(e xml i)of (ii) example (see 1) y n R ∈ . k n , R n 0 k n a normal has and n , . . . , any } 1 = . r ∈ } 1) clrpoutdefined product scalar . R . sgvnby given is x , y n R , -dimensional z 3 n § ∈ endin defined 3.2.1). ∈ R R n (3.23) (3.22) (3.24) n and is ,

This is a supervisor’s copy of the notes. It is not to be distributed to students. ahmtclTio:I ler emty(atI 55 I) (Part Geometry & Algebra IA Tripos: Mathematical if equivalently or mation Definition. Remark. space vector a for from maps consider shall We Maps Linear 4.2 Remark. Definition. Definition. Definition. Definition. Examples. write We Definition. Definitions General framework general a 4.1 construct to is section this of and aim com- problems. The (since equations, computational linear process problems). many such linear the linear Moreover, viewing in solving satisfy for point booms). at waves some sonic good at electromagnetic being ‘linearisations’ are involve (exceptions e.g. puters problems ‘linear’ nonlinear linear, are solving are to hear approaches world you sounds real all the almost in problems Many This? Study Why Matrices 4.0 and Maps Linear 4 (ii) (i) ,n m, rcl htayeeeto etrspace vector a of element of any that (recall if R ∈ hsi o nuyrsrciesneeeyreal every since restrictive unduly not is This f n Z ( ). ¨ bu aso h ope ln,tasain neso ihrsett sphere. a to respect with inversion translation, plane, complex the of Möbius maps A + f f B A Let Let ) . ( ( ⊆ A x sthe is sthe is = ) sthe is ) ,B A, ,W V, B u hr a eeeet of elements be may there but , T x ( domain 0 range est.A sets. be λ evco pcsover spaces vector be sthe is a mg of image + or , µ b mg of image of = ) codomain T T f map f ( . ( A a λT λ : a + A under = ) ( f a b x → of + ) = ) of , under λT B A f µT V T ( ..testo l mg points image all of set the i.e. , into R f a ( and/or . h map The . oavco space vector a to ( a o all for ) f b + ) . o all for ) B V B T sarl htasgst each to assigns that rule a is ihdim with htaentiae fany of images not are that ( b n x o all for ) dmninlvco space vector -dimensional 7→ T

a c : ∈ x a ..olydmpcma.k ihems2006 Michaelmas [email protected], V 0 , V V b = → W ∈ and = f W ( ousn on focussing , V a x n , ) λ b and sa is orsod oauiu element unique a to corresponds ∈ ∈ R , V iermap linear ,µ λ, , x 0 ∈ ∈ x R x V B V ∈ ∈ . = of sioopi to isomorphic is A A or R . x unique a n iertransfor- linear ∈ and A . W x = (4.1b) 0 (4.1a) (4.2) ∈ R R B m n , .

This is a supervisor’s copy of the notes. It is not to be distributed to students. 15/03 .. Examples 4.2.1 ahmtclTio:I ler emty(atI 56 I) (Part Geometry & Algebra IA Tripos: Mathematical Remark. element. zero The Property. (iii) (ii) (i) etcnie h smti map where isometric the consider Next in translation Consider saohreapecnie rjcinot iewt direction with line a onto projection consider example another As ecnld htti salna map. linear a Remarks. is this that conclude we for Hence is This rmteosrainthat observation the From n ofrall for so and hs(.)i aifid n so and satisfied, is (4.2) Thus ht Set that. hsfo h nqeeso h eoeeeti olw that follows it element zero the of uniqueness the from Thus o pl hoe . npg 45. page on 3.1 Theorem apply Now (ii) (i) T T rjcini o nisometry. an not is projection A of subspace of range The ( ( x b V not , = ) n sasbpc of subspace a is ) b x ∈ 1 iermpb the by map linear a = 0 , R Since x ∈ ,µ λ, 0 3 2 W ∈ and ∈ R V ∈ H R does P 3 P T T . Π R 3 λT x n(.a,t euethat deduce to (4.1a), in | ( ( sgvnby given is n ( : V x ne h map the under 7→ λ R | ( + ) R sasbpc,i olw that follows it subspace, a is ) x not a ;te rm(2.91) from then 1; = T 3 3 1 x + ) ( → 0 + a a smty,ie consider i.e. isometry), (an W T = P = ) imply ( µ R µT ic for since , y x x ( H T x λ 3 1 2 = ) 2 Π strict x x ( T ( 7→ 7→ ( = ) x b P H 1 ( 7→ salna map. linear a is x = ) = ) a ( + b x Π x x R = = 7→ + x + 1 0 2 0 = µ 3 : H ento falna a since map linear a of definition x 0 = ) T x 0 = = R x a 0 Π = = ) 2 T λ λ + ( 0 3 λ + (( = ) λ x x ∈ ( H ( = H x x → a a 1 2 a { y Π V 1 1 x ) Π T − − , P + = = T , ( . + − + ( R ( x ( ∈ x 2( 2( µ a x 1 3 where ( µ 2( a = ) + ) b + ) ( = ) R λP λ b x x x ossigo eetosi h ln : Π plane the in reflections of consisting ) = x ( 1 2 ) 3 λ 2 x ∈ 1 ∈ · · ) x ( T x : µ T 1 x

n n · − 1 c T x x ( H · T − ( n 1 ) ) 0 + x t ( + ) n n 0 · (( 2 ) ( Π ..olydmpcma.k ihems2006 Michaelmas [email protected], = o all for ) V a ) 2( n V t + t ( µ = = ∈ ) o all for ) + ) x ∈ λ x ) x t + λ µP y 2 t T H H 2 x R · ) + ) µ ) n ( for 1 µ . Π Π T 3 where ( ( V · ) x ( + x ( ( ( t n x x x .Hwvr ecnsymr than more say can we However, ). a 0 2 2 ) λ and 2 µ t ) 1 2 . = ) t · 6= ) ) x − t ∈ a ∈ , , 2 ) T t 2( ∈ ) ,µ λ, R 0 R t ( · x a } x . V ∈ 3 · n hc sa1-dimensional a is which , 2 t sdﬁe y(f (2.17b)) (cf. by deﬁned as + 6= ) W ∈ · 1 = n n y 0 R . ) ) . . n . . ) x . n (4.3) 0 =

This is a supervisor’s copy of the notes. It is not to be distributed to students. ahmtclTio:I ler emty(atI 57 I) (Part Geometry & Algebra IA Tripos: Mathematical (iv) (v) saohreapeo a hr h oanof domain the where map a consider of example another As sa xml fampweetedmi of domain the where map T a of example an As that observe We T Remarks. Remarks. T : salna a since map linear a is • • • • • R ( λ otevectors the to nti aew bev httesadr ai etr of vectors basis standard the that observe we case this In 1-dimensional that observe also We that observe We for vectors basis standard The origin. subspace The all for Hence n hc omabssfor basis a form which and 2 x → 1 T + T R ( T µ e : 4 T λ 1 x ( R where ) λ ( 2 3 R ∈ x − (( = ) T T → 3 1 R = = T T = ) R ( ( + 3 e e T T T ( R hstewoeo h usaeof subspace the of whole the Thus . 1 2 span = λ λT salna a since map linear a is T e ieseie by speciﬁed line µ ( ( ( 2 = ) = ) 2 ( R e e e ( λx x x 2 + ) R where 1 2 3 ( 2 2 1 x (1 = ) (1 = ) (0 = ) ( . (( = ) 2 1 ,y x, + 1 span = ) + + ) { T y = = e T T µx ( ) 1 ( ,y z y, x, 1 x , e ((1 ((0 7→ µT , 3 2 e λ λT 1 = ) ( + ) 2 , , ( y , ( T ( , , , λx , x ,t ,v u, t, s, ) (1 = 0)) ) (1 = 1)) x { 2) 0) − ( e ( ) 1 1 2 (1 x R R 3 0 1 1) ) 7→ − λy + 1 λ } 2 3 , + ∈ . + ) ( ). 1 i.e. , 3 y 1 e    ( , R µx 1 x ,v u, 1 and + − , = ) 1 2 µT − 2 y , 3 2 µy e x ( + ) , , , = ) hc r ierydpnetadspan and dependent linearly are which e 1 1 1 0) ( 1 0 hc en that means which T 2 2 x + ) , , T (1 = − T ) , − 1 2 2 + ( λx , (1 T ,y x, ) , ( λy a oe ieso hnterneof range the than dimension lower a has z 3 R 1) . µ ( 1 , , x e + ) , 3 1 ( 0 0) 1 ( = ) 0 x 3 = ) ( = ) , + + ,i apdonto mapped is ), T , 2 1

0), c , µ + µy µx 1) a ihrdmninta h ag of range the than dimension higher a has R ( ..olydmpcma.k ihems2006 Michaelmas [email protected], x x x y 2 e } 2 2 R 2 2 ) + 2 hc r ieryindependent, linearly are which + λy , x , , span = of 3 + (0 = 2( ,x y x, y, y, 2 pne by spanned y R λx 1 y , 2 2 4 + , R x , 2 1 2 T 1 sa2Dhprpaetruhthe through hyper-plane 2-D a is 2 − − µy { x + , ( − T 2 r apdt h vectors the to mapped are λ )and 0) z 3 2 µx ( ( − 3 x ) e e − ,y x, . 2 1 1 z 2 0 y , ) ) 2 3( − T , ∈ ) − ) 2 { λx e e ) . R e ( 3 ( 2 1 e 2 λz 1 (0 = 2 + 2 . − + ) 1 T , e + µx e 2 , 3 ( 0 µz e )= )) 2 + 2 R , 3 ) )aemapped are 1) 2 ) λy , 2 } . )) e . 0 3 T 1 } ∈ ..the i.e. , + consider R µy 2 , 2 ) T

This is a supervisor’s copy of the notes. It is not to be distributed to students. 15/06 ahmtclTio:I ler emty(atI 58 I) (Part Geometry & Algebra IA Tripos: Mathematical Examples 4.3.1 Examples. Definition. Remark. hence and Proof. 4.1. Theorem i.e. Definition. Examples. Definition. of Nullity Let and Kernel Rank, 4.3 (i) W T . osdrtemap the Consider Hence : ups that Suppose V Since → λ nbt xml i)adeape()of (v) example and (iv) example both In neape(v npg 57 page on (iv) example In u T W h ustof subset The The The ( + R T ealna a.Rcl that Recall map. linear a be µ 2 K ( steline the is ) v rank ult of nullity 0 ( = ) T ( ∈ ,y x, ) u K , sasbpc of subspace a is 0 of v T ) ( ∈ T 7→ ∈ T : .Tepofte olw rmivkn hoe . npg 45. page on 3.1 Theorem invoking from follows then proof The ). W R V K T stedmnino h mg,i.e. image, the of dimension the is T 2 , x ( ( htmp otezr lmn in element zero the to maps that sdfie ob h ieso ftekre,i.e. kernel, the of dimension the be to defined is T → 0 ,y x, = and ) ∈ R λ K (2 = ) T K (1 3 ( ( uhthat such ( , T λ T 2 ,µ λ, u n V = ) , ) ( x − . + T ⊆ n 3 + 1) r r ∈ ,wiei xml v npg 57 page on (v) example in while 1, = ) { µ ( V ( ( T v T v T T ∈ R so , ( y, = ) dim = ) ∈ dim = ) dim = ) V R hnfo (4.2) from then 4 steiaeof image the is ) V 3 x = = K n otern ftempi uhthat such is map the of rank the so and , 6 + : ( T § T 0 λ λT 4.2.1, ( otisa least at contains ) y, T 0 K v T , ( = ) ( + ( − ( V u R T

2 + ) c µ ) 2 ) x r 0 1 = ) 0 . . ( ..olydmpcma.k ihems2006 Michaelmas [email protected], − T ∈ µT 2. = ) 3 W W y V ( . (2 = ) v } scl the call is under ) . x 0 T 3 + . n that and y kernel )(1 , n 2 or , , ( − T T 0. = ) 1) ( V ulspace null . sasubspace a is ) of , (4.4) (4.5) (4.6) T ,

This is a supervisor’s copy of the notes. It is not to be distributed to students. 16/03 ealna a,then map, linear a be ahmtclTio:I ler emty(atI 59 I) (Part Geometry & Algebra IA Tripos: Mathematical Examples 4.4.1 that Suppose Maps of Composition 4.4 Proof. Theorem). Rank-Nullity (The 4.2 Theorem (ii) etcnie rjcinot line, a onto projection consider Next Further, o uuerfrnew oethat note we reference future For in line a is which where hc saln in line a is which hc sapaein plane a is which See ierAlgebra Linear n S safie ntvco.Then vector. unit fixed a is x : ( = U ,y x, → V R R ) ∈ 2 3 R and Thus . thus ; 3 K nPr IB. Part in thus ; ( T T r r ( ( f2 if ) P : T r ( V + ) + ) n P u ( → .Frhr h enli ie by given is kernel the Further, 1. = ) P x P 7→ n n .Aanw oethat note we Again 2. = ) K 3 + ( W ( ( R K T P v ( r x T 3 ieso fdmi,i.e. domain, of dimension = 2 = ) ( ieso fdmi,i.e. domain, of dimension = 3 = ) y ( = r iermp uhthat such maps linear are = ) T P = ) P n 7→ ,so 0, = + ) ( S = ) T : ( x { R { u dim = ) x 0 n x 3 (i) ) { = ( ∈ , x → ( = T Let P R ∈ dim = ) where hnb emty(reecs n ler)i follows it algebra) that and exercise (or geometry by Then range the Since Let Definition. ob eldfie h oanof of domain image the well-defined be to that note we where map the R v ( − 3 R x V 3 : 7→ 3 ( = ) K hr si (2.17b) in as where , 3 H s, x and : ( Π w

= 2 T x I c s 1 = ) x erflcini plane a in reflection be . V · steiett map. identity the is : ) = λ S T ..olydmpcma.k ihems2006 Michaelmas [email protected], S W n n · . T n s λ , 0 = era etrsae n let and spaces vector real be ( ) : ∈ The v n . U ) R } ∈ , ⊆ . u H → , } R oan emyapymptwice. map apply may we domain, composite H 7→ Π , } W S Π H w csfis,then first, acts Π : uhthat such R R R = = 3 2 3 T . . H → ( Π 2 S or R ( = u 3 product )) T . , I , uticuethe include must T o h map the For . T map : V (4.10) S T → (4.8) (4.9) (4.7) W is

This is a supervisor’s copy of the notes. It is not to be distributed to students. 16/02 ahmtclTio:I ler emty(atI 60 I) (Part Geometry & Algebra IA Tripos: Mathematical tflosta o general for that follows It j Maps of Description Matrix Let the and Bases 4.5 where of action the consider Now that follows it (4.2), map, linear a of definition the From same). Let the where namely basis, this of terms e (iii) 1 1 = (1 = M e , Let epciey Then respectively. and Remark. i 2 , e , , : j 0 i ) n ehv set have we and 3), 0 R S , steiaeof image the is 1 = x ) t..UigTerm36o ae4 ent htany that note we 49 page on 3.6 Theorem Using etc.). 0), 3 : j → R r h opnnsof components the are , ST 2 2 R , → ,b ai for basis a be 3, 3 o eldfie eas h ag of range the because defined well not R ealna a weefrtetm en etk h oanadrnet ethe be to range and domain the take we being time the for (where map linear a be 3 and S T e M x j T M ( , : ∈ ( : e R e x R j R 2 nabssvco,say vector, basis a on j 0 M ) ) 3 7→ 3 → i ≡ → ( sthe is x x R e R ( = ) x 0 ,y z y, x, j 0 R 3 = ( ssc that such is = ihrsett h ie basis. given the to respect with ,v u, i a ept hn fi stesadr rhnra basis orthonormal standard the as it of think to help may (it elna assc that such maps linear be ( M ,v u, M M X i i =1 ij th 3 ) ) (   ) ( = x ( opnn of component e x 7→ = ) X 7→ 7→ j j 0 =1 3 = ) e (ii) i T j 0 = = e x ) X j ( S T i i =1 j 3 − e ( ( = = ( ecnso b emtyo ler)that algebra) case or this geometry in (by and show twice, can map we the apply may we Again Let ,v u, ,u u u, v, X X X ,y z y, x, j j j i x e =1 =1 =1   3 3 3 X i j j M =1 3 Then . e P ( = ) = x x   T j j j ( M ,

X j = ) epoeto noaline a onto projection be M e X j c snttedmi of domain the not is e + X i =1 3 =1 =1 j 3 j 0 ij 3 )) − ..olydmpcma.k ihems2006 Michaelmas [email protected], v ( ihrsett h basis the to respect with M e e x x 2 = ) i M ,u u u, v, i j j . + ij ) j , M ij x y e j ( . u i   + e + 1 = j x P P , z ) e P v i . ∈ ) . , : 2 R = R , 3 3 3 P , a nqeepninin expansion unique a has → 2 S = R . 3 . P . { e i } ( i 1 = (4.11) (4.15) (4.14) (4.13) (4.12) , 2 , 3,

This is a supervisor’s copy of the notes. It is not to be distributed to students. ahmtclTio:I ler emty(atI 61 I) (Part Geometry & Algebra IA Tripos: Mathematical as elements matrix of terms in defined been has rule multiplication matrix a as where notation matrix in expressed be now can (4.16b), equivalently or (4.16a), Equation form component in Thus s netesadr ai o n te ai)hsbe hsn opeeyseie yte9quantities 9 the by between specified relation completely chosen, explicit been The has basis) other any (or basis M standard the once is, earlier introduced convention summation and notation suffix the of terms in Alternatively, eut ftemapping the of results Since epciey n let and respectively, Let notation. matrix using by form convenient more a in written the be be can equations above The notation Matrix 4.5.1 Remarks. ij • • • • • , i nacpe ovnini ouetesadr oainfravco,i.e. vector, a for notation standard the use to is convention accepted an introduced have we i.e. confusion matrix, avoid and try To x have now We suffix first The (a) write we Sometimes the call We hr eda teto otecma nteRHS. the on commas the to attention draw we where x 1 = ounmatrices column eoigaclm arxo components. of matrix column a denoting a nabtayvco,wa hsmasi htoc eko the know we once that is means this what vector, arbitrary an was , 2 , 3, j x 1 = M oee,i h aeo ounmti fvco opnns ..aclm vector, column a i.e. components, vector of matrix column a of case the in However, . odx bold ij M i , , 2 sthe is ,j i, M or , , ete3 the be 3. x 1 = : 0 ounvectors column eoigavector, a denoting R = row 3 x M , X i 2 x → i 0 =1 3 , × , 0 = the 3 i ubr hl h eodsuffix second the while number, = R x x x 3 x 1 = { 1 0 2 0 3 0 i 0 3 Mx M qaematrix square e x o l lmnsof elements all for i M = , ij = = = elements 2 } , = where   ,and 3, x n (b) and , x x x M M M   requivalently or x = 3 2 1 M M M 11 21 31 i 0   x   = italic i 0 x x x 11 21 31 x x x 1 1 1 = x ftematrix the of X x j 3 2 1 and =1 + + + j 3 M i 0   M , M M M = M M M j x ij M ij ( = i 12 22 32 1 = o hr while short a for X j 12 22 32 x ( = ij =1 3 eoigacmoeto etr and vector, a of component a denoting x j x x x R x x 0 . , 2 2 2 M j 1 M M M M 3

= 2 c x , nohrwrs h mapping the words, other In . . + + + , ij ) 13 23 33   is 3 ..olydmpcma.k ihems2006 Michaelmas [email protected], ij 2 M M M x   x , . x x x M j x 13 23 33 3 0 2 0 1 0 i , 0 3 .   . = ) x x x j , 3 3 3 sthe is M , , . 1 = x pcﬁ oainfracolumn a for notation speciﬁc a , , 2 column , 3 . x M ec enwhave now we Hence . ij number. ecncluaethe calculate can we M : R asserif sans x 3 (4.18b) (4.16b) (4.17b) (4.18a) (4.16a) (4.17a) and → R x 3 0

This is a supervisor’s copy of the notes. It is not to be distributed to students. .. xmls(nldn oeipratnwdfiiin fmaps) of definitions new important some (including Examples 4.5.2 ahmtclTio:I ler emty(atI 62 I) (Part Geometry & Algebra IA Tripos: Mathematical Remarks. (iii) (ii) (ii) (i) (i) hsi h rtclm of column first the is This that follows it of action the consider that have Reflection. map a with of elements The (4.14) From For osdrtemap the Consider matrix construct to wish We hni em ftesaa rdc o vectors for product scalar the of terms in Then o s oml 41)adsmlrepesosfor expressions similar and (4.19) formula use Now matrix map’s the construct to order In the where Hence lentvl ecnoti hssm eutuigsffi oainsnefo (2.91) from since notation suffix using result same this obtain can we Alternatively i 1 = , 2 e , osdrrflcini h ln = Π plane the in reflection Consider i 0 M let 3 nteRSaet eitrrtda ounvectors. column as interpreted be to are RHS the on ti eesr ogv h ai ihrsett hc thsbe constructed. been has it which to respect with basis the give to necessary is it , M r H i P Π eedo h hieo ai.Hnewe pcfigamatrix a specifying when Hence basis. of choice the on depend etevco ihcmoet qa oteeeet fthe of elements the to equal components with vector the be b ( e : H 1 R = ) Π 3 M nec ebro h tnadbss hn ealn that recalling Then, basis. standard the of member each on → H P i e b iial eobtain we Similarly . ( = th H 1 0 H R ( H r e ihrsett h tnadbss htrepresents that basis, standard the to respect with , =   ) o of row 3 i 1 ij x = ( = ) ( = ( ( ( endby defined e ( e e e = →H 7→ 1 H   1 0 1 0 1 0 P − ) ) ) M x Π H − − 1 2 3 1 b b ( x 2 7→ i ij 1 − 2 2 x = 1 Π 0 n b , P n n M , )) ( ( ( = ( = 1 2 e e e 1 1 x   x 2 n ihrsett h tnadbss rtnt that note first basis, standard the to respect with i n n n 2 0 2 0 2 0 0 b , = ) x δ i ) ) ) − 1 2 3 2 = = 2 ij 1 2 3 b i 0 3 0 M , = = ( = ≡ b 3 = ) 2 − P   x − − 1 × ( ( ( b r i 0 0 0 1 e e e 2 − 2 2 3 x δ H ( i { (1 −   = n 3 0 3 0 3 0 n n ) δ x ij b i x · ) ) ) 0 2 ij ij i , 1 2 b 1 , = ) 1 2 3 P − x n x x − n 3 n n ∈ 0 x  

j − j b c 2 2 , 2 3 , − j 2 2 ( for − ,j i, , R )=(0 = 0) x = . e b 2 n − ..olydmpcma.k ihems2006 Michaelmas [email protected], 2( 3 j b n 2 2 1 0 − − for 1 b × n 2 and ) : i x i x   1 n − e 2 2 j x j   1 = x n n n 1 0 · j n n n n i 2 ) · i n 1 2 1 = . j 1 3 2 b , x 1 = . n n n n n   ) , j e 3 2 n 3 3 P 3 i 2 and 0 = 2 0   , , , = b . , − 2 3. 2 ( , . e b   , e 3 2 3 3. 3 0 . ) oddc that deduce to ) − − 1 . − 2 2 , n n | n 2 1 1 n n n | 1 2 3 2 1 =   } . i rm(.1 we (2.91) From . hrwof row th H Π M oti end this To . e associated j · n (4.20b) M (4.20a) (4.22a) (4.19) (4.21) = i.e. , n j , 16/06

This is a supervisor’s copy of the notes. It is not to be distributed to students. ahmtclTio:I ler emty(atI 63 I) (Part Geometry & Algebra IA Tripos: Mathematical (iii) (iv) (v) Dilatation. Rotation. elements The a,the say, the e.g. direction, one Shear. (4.22a), with agreement in Hence, n otempsmti ihrsett h tnadbss say basis, standard the to respect with matrix map’s the so and Then ipeseri rnfraini h ln,eg the e.g. plane, the in transformation a is shear simple A x 1 osdrrtto ya angle an by rotation Consider ai.Udrti transformation this Under -axis. osdrtemapping the Consider e { x 1 P 1 0 ij 7→ = } λx e of 1 0 1 x = P x , 1 b e ieto,b naon rprinlt h itnei htpaefrom, plane that in distance the to proportional amount an by direction, 1 ol lob eie sflos rm(.6)ad(4.21) and (4.16b) From follows. as derived be also could , 2 0 = e P e 2 µx ij 1 0 7→ R x = 2 j 3 x , e λ P = → 2 0 e ij 1 = x R θ , 3 0 = i 0 e 3 = bu the about = 2 ε endby defined e ikj + ε νx 2 0 ijk hstertto matrix, rotation the Thus ne uharotation a such Under h ffc nteui ue0 cube unit 0 the on effect The 0 hntetasomto scle uedilatation. pure a called If factors is axes. transformation different the Cartesian then by different contracted the along or stretched been λ = b 3 k 6 6 e b µ j 1 = x x x e where , k 3 2 0 2 −

x c , 6 6 ( = ε e 3 R e e e x ijk 3 ..olydmpcma.k ihems2006 Michaelmas [email protected], ,o hsmpi osn tt 0 to it send to is map this of 1, µ ( axis. 3 2 1 e ε θ 7→ 7→ 0 , ,µ ν µ, λ, b 7→ 7→ 7→ ikj 3 0 = ) k = e x . b e e e 6 3 0 0 k   ν 3 0 2 0 1 0 ) where = D x e x x cos sin ∈ 1 3 3 0 D j = e x = = = 1 0 0 , . R 6 sgvnby given is , 3 2 θ θ   pae htdslcspit in points displaces that -plane, , and ν e − e λ 0 0 0 ..t uodta has that cuboid a to i.e. , 3 1 − e where cos . cos 1 R sin ,µ > ν µ, λ, µ 0 0 sin ( θ θ θ θ 6 ,i ie by given is ), ν θ 0 + λ   + x e 0 0 ∈ 1 2   e . 6 R sin 2 0 . . cos (4.25) . ,0 1, , θ λ , θ = 6 6 x x (4.22b) µ 1 0 2 (4.23) (4.24) 6 = 6 λ 1, ν , 17/03

This is a supervisor’s copy of the notes. It is not to be distributed to students. ofrw aecniee arxrpeettosfrmp hr h oanadterneaetesame. the are range the map and a domain that the found where have maps for we representations instance, matrix For considered have we far So 4.5.3 ahmtclTio:I ler emty(atI 64 I) (Part Geometry & Algebra IA Tripos: Mathematical i.e. write we before as multiplication of rules same the Using thus and Hence Let and map Let { { N f e o hseapetempsmti wt epc otesadr ai) say basis), standard the to respect (with matrix map’s the example this For j k } } : dim( eabssof basis a be R eabssof basis a be n → domain R m clm vector (column with m where ) × R      | dim( 6= R m matrix 1 x m n x x {z . . . hnteeexist there then , m 0 so , 2 0 1 0 x ,n m, rows)      j 0 } x range ( = 0 ∈ x = Z N 0 N + N ) ( ..amap a i.e. , = = x ( e ( )) x N k j = ) = ) M |      N ( x x N x N N N = = = ) : ( m . . . S m 21 11 R jk λ where 1 = 3 X k k X X X j j m ∈ rows, = m m =1 =1 → =1 =1 n n x N R N N ×   k k X X 7→ N

N x R m . . . =1 =1 n n 22 12 ( 1 0 0 0 1 0 1 k n k X 3 jk jk 2 j {z x n =1   n N x x matrix ed oa3 a to leads 1 = 0 f x k k λ columns) X j j = N . . . N . . . N . . . . = N k e m =1 N . . k

. ,k m, , . . . , jk c N { = ( 0 , N N e   ..olydmpcma.k ihems2006 Michaelmas [email protected], x ( k jk x ij k N ) ,where ), mn ! . . . . } 2 1 f . jk n n j . f        j x × } , k 1 = matrix 3 clm vector (column with n (s.c.) x n , . . . , × ∈ |      matrix 1 x n x x R {z . . . . n 2 1 n rows) M uhthat such )      } and osdrnwalinear a now Consider . S λ x sgvnby given is , 0 ∈ R m . (4.29b) (4.28b) (4.27b) (4.27a) (4.29a) (4.28a) (4.28c) (4.26) 17/02

This is a supervisor’s copy of the notes. It is not to be distributed to students. ahmtclTio:I ler emty(atI 65 I) (Part Geometry & Algebra IA Tripos: Mathematical product matrix the of elements the defining as (4.35) interpret We arbitrary for identical must (4.34b) and (4.34a) because hence and However, thus and Let scalar a by Multiplication 4.6.1 Matrices of Algebra 4.6 o osdrtecmoiemap composite the consider Now of matrix the consistency for Hence, Let map. linear a also is This hnfo 42c,aduigtesmainconvention, summation the using and (4.28c), from then if Similarly, Addition scalar. a by 4.6.2 multiplied matrix a of definition the as use we which of basis given a or Let multiplication Matrix 4.6.3 define we consistency the A S : : ik R R th n n → lmn of element → A R R = m m { T a S and ealna a.Te o given for Then map. linear a be R ij = = n } { { TS if T and S T m ij ij : seult h clrpouto the of product scalar the to equal is R } } B = x m ( 00 = λ n A ethe be ethe be → = A .Te rm(4.28b) from Then ). { = )( x b S T x R ij W j 0 i 00 x { ` } = a = ) = ( elna as Let maps. linear be x λ = ij x r both are m ` 0 S A T = ) } × = S T jk × ij etemti of matrix the be utbe must W m λ x ( A ( S n S λ W A k ik ( : + jk arxof matrix A x arxof matrix ( R ( λ x and ) = )( x x B and n m A = ) k and ) x T = ( = ) → = = ) × ij = { S λ { R n a λa x jk ` λ ij ∈ T arcsascae ihmaps with associated matrices i 00 ihassociated with , T X λ j ( S ij x ij m =1 + A R =   S 00 } x ihrsett ie basis. given a to respect with (s.c.) ihrsett ie basis, given a to respect with ( b jk en ( define A T k X x i 00 X j , =

ij m c =1 =1 n ij )) ) = } wt epc ogvnbssof bases given to respect (with x T x ..olydmpcma.k ihems2006 Michaelmas [email protected], ( . . k X k λa (4.35) ; j 0 W ( =1 n i x th ik 0 jk (s.c.) λ ) a (s.c.) x o of row , A jk ) k TS x x uhthat such ) x k tflosthat follows it , k f . nwords, In . , j (s.c.) f ` j .   × T n ihthe with , matrix R W k th n = → ounof column { W R R m ij n hnfor then , } and If . (4.34b) (4.33b) (4.34a) (4.33a) (4.30) (4.31) (4.32) S R . m ,

This is a supervisor’s copy of the notes. It is not to be distributed to students. 18/03 ahmtclTio:I ler emty(atI 66 I) (Part Geometry & Algebra IA Tripos: Mathematical Remark. Definition. Transpose 4.6.4 Proof. that such matrices are 4.3. Lemma Remarks. elements (iii) (iv) (ii) (i) while instance For columns, if Even when case If special the the with i.e. consistent (4.18b), is and multiplication (4.18a) matrix in of of columns definition of above number The the defined, well be of to rows product scalar the For while instance For ntrso uxntto adtesmainconvention) summation the (and notation suffix of terms In A sa is If S p p hsi h aeaoesince above case the is this ; h utpiaino arcsi soitv,ie if i.e. associative, is matrices of multiplication The A = × = q q arx and matrix, { = a ij AB r }   = (( ( sa is r q p A u t s and b a f e d c AB ( s BC BA AB ota both that so , m ) BC C   )) B × ) xssol if only exists if only exists ij ij n sa is xs,then exist, r q p u t s   arx hnits then matrix, = ( = 0 0 1 0 1 1 0 b a f e d c r × ( − a A AB s ik 1 A AB T   T AB ( arx then matrix, ( ) BC ) BC sa is ij s q ik 0 0 1 6= = = = = ( = c 0 1 1 0 ) and ( ( = ) kj kj BA A − ` p r T A n ste a then is and , × n ste a then is and , 1 =   = ) ) BA transpose T pa sa AB m ji ep ap cp a a S ngeneral in i` = ik = + + arx while matrix, + saclm arx(rclm vector). column (or matrix column a is + ) = = + b xs n aetesm ubro osand rows of number same the have and exist b C A `k

k` a c tc qc seq fs scq ds saq bs . ji . c c ..olydmpcma.k ihems2006 Michaelmas [email protected], kj + `j + . 0 1 0 − = esb ue A = 1 0 epb re 0 1 . T a a − + + + p r i i sdﬁe ob a be to deﬁned is £ £ A 1 × × ter ft tcr dt tar bt b b = £ £ S q + + s , . U U { matrix. matrix; sa is c td qd c a U U n ij T j j + + + + + pca aeconsidered case special 1 = } . , m uteultenme of number the equal must , uf fu du rf bu B ×   = n , . { matrix. b n ij × } m and arxwith matrix C = (4.38b) (4.38a) (4.36) (4.37) { c ij }

This is a supervisor’s copy of the notes. It is not to be distributed to students. 18/02 ahmtclTio:I ler emty(atI 67 I) (Part Geometry & Algebra IA Tripos: Mathematical Remark. Deﬁnition. Deﬁnition. Symmetry 4.6.5 Example. Proof. 4.4. Lemma Remark. Examples. sa1 a is lmnso natsmercmti r eo i.e. zero, are matrix antisymmetric an of elements (ii) (i) o natsmercmatrix, antisymmetric an For important: sometimes are commas that Recall Let × square A square A arx ..asaa.I olw that follows It scalar. a i.e. matrix, 1 x If and A = If y { x a n n e3 be = ij × × }      n n and × x x x ( . . . matrix matrix n 2 1 x ounvcos n let and vectors, column 1      T B A saclm vector, column a is = y ) x T { A A A (( b T x T = ij A = = AB a x } T 11 = y a { { = = T ) r arcssc that such matrices are = T 11 a a A T = − ij ij ( A ) y AB   A A = ij } } T −      = x , , 6 5 4 3 2 1 x a is is a x x x ) 1 11 . . . T 22 n = 2 1 x ( = ( = = ( = ( = antisymmetric symmetric i.e. i.e.      i i.e. , = a x y =   ij i 2 B a a T ( = . . . x y AB B B B ji jk = T j a T a a ) T T x . . . x A = 11 A ji ji b ki = ) A = ) j ki x T ik ji

( = = T = = 6 4 5 2 3 1 x c 1 .Smlryw eueta l the all that deduce we Similarly 0. = A . a ( x , £ ) nn x A if ) ij { ..olydmpcma.k ihems2006 Michaelmas [email protected], y − a a jk 1 n a U 2 T ij £ a . x , . . . , ij ) 0 = a ij U . kj if £ } . AB x y . U 2 U ea3 a be . x . xss then exists, £ . n x . . . ) = , × x T arx Then matrix. 3 n A y sarwvector. row a is . diagonal (4.41b) (4.41a) (4.39) (4.40)

This is a supervisor’s copy of the notes. It is not to be distributed to students. ahmtclTio:I ler emty(atI 68 I) (Part Geometry & Algebra IA Tripos: Mathematical 1. are that elements diagonal Example. the for except 0 are elements the all i.e. Definition. Remark. i.e. Definition. Examples. .. h nto dniymatrix identity or unit The 4.6.7 .. Trace 4.6.6 (ii) (i) Remark. elements. independent three has it i.e. 3 antisymmetric An elements. independent six has it i.e. 3 symmetric A n hence and if (even equal usually not are but Let h 3 The hsec niymti 3 antisymmetric each Thus The The B Let = r T × trace unit { dniymti sgvnby given is matrix identity 3 a b ( BC ij × = } or matrix 3 = ) v fasquare a of ea be 3 , identity × b r T matrix 3 = m ( CB v × 2 S n ee if (even ) and n a h form the has r T r T arxand matrix n × × ( ( A BC CB × c n n a h form the has = m arxcrepnst nqevector unique a to corresponds matrix 3 arxi endt be to defined is matrix I ( = ( ) = ) matrix r T = v I = A 1 m = hn(f 42a n (4.22b)) and (4.22a) (cf. then ,      ( n = A S   A 6= .However ). 0 0 1 0 0 1 . . . C = ) =   1 0 0 0 1 0 0 0 1 = n A = CB BC −   . . . 0 b { ota h arcsaeo ieetsizes). different of are matrices the that so = a { a a c b a f e c e d b ) ) ii c ij ii ii { ...... ij . . a } − } . a ij 0 = = (s.c.)   = c ea be }

1 0 0 . . . = c { c b seult h u ftedaoa elements, diagonal the of sum the to equal is      − ε ij ij   0 c ..olydmpcma.k ihems2006 Michaelmas [email protected], ijk { . c b b n , δ ji ji   , ij × v , k } , m = } . . arx then matrix, b ij c ji , v BC in and R 3 . CB ohexist, both (4.44) (4.45) (4.43) (4.42)

This is a supervisor’s copy of the notes. It is not to be distributed to students. if ahmtclTio:I ler emty(atI 69 I) (Part Geometry & Algebra IA Tripos: Mathematical Proof. 4.6. Lemma Property. Definition. Remark. Proof. B 4.5. Lemma matrix Definition. a of inverse The 4.6.8 Property. h matrix The AC = C = ievra oee,i h aeo qaemti,teeitneo etivrede ml the imply does matrix. or inverse same inverse, left IB the a right Part are of a (see they existence of that versa the implies vice existence matrix, then and the square lemma inverse, guarantee a above right necessarily of a case not of the does existence in inverse However, left versa. a vice of existence the Let i.e. = rm(.7,(4.47), (4.37), From rmuig(.7,(.7 n 44)i olw that follows it (4.48) and (4.47) (4.37), using From I . A h em sbsdo h rms htbt etivreadrgtivreeit ngeneral, In exist. inverse right and inverse left a both that premise the on based is lemma The A rm(.8 tflosthat follows it (4.48) From in delta Kronecker the Define − = 1 B . Let Let { scle the called is ups that Suppose If a ij A A B } ea be ea be ea be salf nes of inverse left a is n ( B square n AB − × inverse × 1 BA A n ) A B n − matrix. and − = 1 arx then matrix, n 1 ( δ A AB × I B ij B − of and = n 1 = ) = r ohinvertible both are A A matrix. A BI suiu seaoe n sdntdby denoted is and above) (see unique is , = AC = ssi obe to said is R = A ( ( if 1 if 0 ( IA AI n B AB = A B B − and uhthat such ) ) IA BA − ( ( 1 ij ij I BB AC ) B 1 − tflosthat follows it i diinto addition (in i i ( A 1 A sa is = = 6= C ( = ) − = = = ( = − = 1 AB j j sargtivreof inverse right a is 1 ) A A B etinverse left invertible A δ a AI BA − ) − ik ik = − B 1 1 n a 1 δ for A = ) = = I ) kj kj C × −

− . c B AIA 1 1 = ,j i, A n = = . ..olydmpcma.k ihems2006 Michaelmas [email protected], − . . IC arcs Then matrices. 1 B ierAlgebra Linear fteeeit a exists there if − 1 = IB a a 1 of = = ij ij = = , , , A A C 2 − B AA n , . . . , . if 1 − .Hence ). BA 1 − B 1 A = = = (4.46) . then I I I , . . o eea ro) The proof). general a for n C × A B sa is − n 1 = matrix seabove). (see ih inverse right C n ewrite we and B uhthat such (4.47) (4.50) (4.49) (4.48) of A

This is a supervisor’s copy of the notes. It is not to be distributed to students. 18/06 19/03 ahmtclTio:I ler emty(atI 70 I) (Part Geometry & Algebra IA Tripos: Mathematical notation. of abuse An 2 Exercises. Remarks. notation. Alternative Definition. by given (4.28c), and (2.67e) of aid the with is, cube mapped vectors basis orthonormal the of volume b signed the that Recall (for Determinants 4.6.9 osdrteeeto iermap, linear a of effect the Consider × and 2 c.(.3) ec o 2 a (4.52c)) for Hence (4.23)). (cf. hr eaetetn h vectors the treating are we where map A matrices. (ii) c (i) • • r ih-add eaiei left-handed). if negative right-handed, are det is hwta h eemnn fterflcinmatrix reflection the of matrix determinant the rotation that the Show of determinant the that Show If map linear A epc oa rhnra ai is basis orthonormal an to respect that). prove can we before machinery extra some e − 1 0 R The { A e snerflcinsnsargthne e fvcost ethne set). left-handed a to vectors of set right-handed a sends reflection (since 1 2 i map A e } → 2 0 = = = = = determinant sasadr ih-addotoomlbsste h set the then basis orthonormal right-handed standard a is R e 2 a ε a a ε lentv oain o h eemnn ftematrix the of determinant the for notations Alternative 3 0 ijk ijk hsi o o h an-ere.I o r noteaueo oainso that show notation of abuse the into are you If faint-hearted. the for not is This 11 11 11 sae rsrigi det if preserving area is

R > R a ( ( 3 a a a a 22 3 1 i e 22 22 → ,adlf-addif left-handed and 0, 1 i e → a i a a Let . a a 1 0 33 2 j 3 R 33 33 det 2 j · R fa3 a of × a a 3 + e det × − − 3 3 k 2 0 seetvl w-iesoa if two-dimensional effectively is k 3 3 A a is × matrix 2 A a a 12 and A 23 32 = ouepreserving volume A e = R a × = a a 3 0 23 | 3 32 23 : A matrix 3 { i a | aallppddfie by defined parallelepiped , R 2 a + ) + ) A = = = = | 31 ij j 3 × = a | and } + A → = 2 × a a

ε ε ε ε etemti soitdwith associated matrix the be ± 12 21 a a a a edﬁetedtriatt egvnb se(.2)or (4.52b) (see by given be to determinant the deﬁne we ijk ijk ijk ijk matrices)

A b R 13 11 21 31 srcl a’sol erpae y‘n’ u eneed we but ‘any’, by replaced be should ‘an’ (strictly 1 ( ( k a a

3 A a a = = a ( a a e 11 21 a s‘components’. as , ntevlm fteui uedﬁe ystandard by deﬁned cube unit the of volume the on , e 23 32 i` i` i 1 0 21 1 sgvnby given is 1 0 ±

a a a ( δ a a a ) a a e b 1 12 22 32 i j 31 13 1. k j i a a 32 1 e ` 1 ( 1 2 12 22 e ) 2 0 a a fadol ftedtriato t arxwith matrix its of determinant the if only and if − − ` 2 0 − jm k )

a a a a a b 3 j a a

2 e jm 13 23 33 2 = a . c δ ( 21 12 3 0 11 e 2

m a ( ..olydmpcma.k ihems2006 Michaelmas [email protected], 3 0 a a R a a e b < ) 11 = 33 33 endi 42a,o qiaety(4.20b), equivalently or (4.20a), in deﬁned 3 23 k a 3 2 endi 42)i +1. is (4.23) in deﬁned a ) km + ) + )

a m 22 a e , 32 a 33 1 0 δ a , − 3 a a kn − n b and 1 = 13 31 a e ( a and ( ( 12 2 0 e 12 a a 3 a 21 12 ) a 21 n e 21 a a c 3 0 32 23 .

A a is a 33 . A hntevlm fthe of volume the then , − − 13 a { − include a a · = e 22 22 ( i 0 a b } a 13 a a 31 × 31 13 srgthne if right-handed is a 22 ) ) c = pstv if (positive ) a a 31 23 . = a (4.52d) (4.52b) (4.52a) (4.52c) (4.52e) 32 (4.51) (4.54) (4.55) (4.53) 0 = a ,

This is a supervisor’s copy of the notes. It is not to be distributed to students. 19/02 ahmtclTio:I ler emty(atI 71 I) (Part Geometry & Algebra IA Tripos: Mathematical Examples. basis. orthonormal of map Property: columns. and rows orthogonal Property: if i.e. Definition. Matrices Orthogonal 4.7 mle htterw of rows the that implies notoomlbss hnfo 41)w ealthat recall we (4.19) from Then basis. orthonormal an of columns the so and the of product scalar the Thus addo ethne eedn ntesg fdet of sign the on depending left-handed or handed if Thus A (ii) (i) sivril and invertible is a,i.e. map, R angle an by rotation basis, (4.23)) standard (see the to respect With of matrix the (4.20b) From map reflection a of application the that seen already have We rm(.7,o itemnplto,i olw that follows it manipulation, little a or (4.57), From Thus Moreover sotooa ic ohterw n h oun r rhgnlvcos n thus and vectors, orthogonal are columns the and rows the both since orthogonal is An A sa rhgnlmti the matrix orthogonal an is H n × sorthogonal. is H n ssmerc hence symmetric, is matrix A − 1 A = A lofr notoomlset. orthonormal an form also A A oma rhnra e.Smlry since Similarly, set. orthonormal an form = T . e e e { H 1 3 2 a i ups httemap the that Suppose ij th = H } 7→ 7→ 7→ and H Π ncmoet 45a becomes (4.56a) components In ( ( is A A T ihrsett tnadbssi pcfidby specified is basis standard a to respect with T ) AA orthogonal , R ik A A A { ) j e ik e e e th ( = ( T i A H 1 3 2 n so and ( } A T   osof rows ) = RR rnfr oa rhnra e wihmyb right- be may (which set orthonormal an to transform ij ) ) h rtclm of column first the h hr ounof column third the of column second the cos sin kj kj I = T 1 0 0 = H = = H θ θ = { A Π 2 if 2 δ a a H T ij A R = ki ik = − 2 A A T

cos − szr unless zero is a a c ). = I R . I sin , kj jk A . 2 ..olydmpcma.k ihems2006 Michaelmas [email protected], = HH θ n : = = θ i R I n . T δ δ j 3 ij ij 0 0 } = →   . , . θ H A . R bu the about A T , 3 . A H H i a matrix a has , = A Π = T j wc eut nteidentity the in results twice I A . nwihcs ti .This 1. is it case which in = x I 3 , xshstematrix the has axis A ihrsetto respect with (4.59b) (4.56b) (4.58b) (4.59a) (4.56a) (4.58a) (4.56c) (4.57)

This is a supervisor’s copy of the notes. It is not to be distributed to students. o oenumbers some for a evee steetiso matrix a of entries the as viewed be can the since Similarly, the hr ntefia arxthe matrix final the in where A numbers some for space ahmtclTio:I ler emty(atI 72 I) (Part Geometry & Algebra IA Tripos: Mathematical the Let from basis matrices of Transformation change a of 4.8.1 think to help may it ideas fix To in basis. work of in basis change orthonormal a standard consider to wish We basis of Change 4.8 maps. Isometric product. scalar the of Preservation ij { { a erpeetdb square a by represented be can e e Remark. since isometry an is map the then basis, form, component in that, suppose For preserved. is and product scalar a basis, orthonormal Hence V i i } : ic the Since . R ai c.(4.19)). (cf. basis i n y 1 = eetosadrttos(hc ehv led enaeascae ihotooa matrices). orthogonal with associated are seen already have we (which rotations and reflections 7→ n ilntasm htete ai sotooml(nessae otherwise). stated (unless orthonormal is basis either that assume not will and | x n , . . . , 0 y h nylnt rsrigmp of maps preserving length only The − 0 = falna a srpeetdb notooa matrix orthogonal an by represented is map linear a If y B A { 0 Ay { ki | ij e } e e = i where , where , i and } } nt h s of use the (note | sabss h niiulbssvcoso h basis the of vectors basis individual the basis, a is x sabss h niiulbssvcoso h basis the of vectors basis individual the basis, a is B A x − { = = 0 e e y · i B           y e e | A : ..lntsaepreserved. are lengths i.e. , R j 0 ki i B ij A B B A A 3 r ob nepee sclm etr ftecmoet fthe of components the of vectors column as interpreted be to are 1 = ...... n n 21 11 21 11 sthe is = = = ( = = sthe is oanwbsswihi o eesrl rhnra.Hwvr ewill we However, orthonormal. necessarily not is which basis new a to 1 1 e ne a ersne ya rhgnlmti ihrsett an to respect with matrix orthogonal an by represented map a Under n e e i n , . . . , j | × = B A B B A A x x x x x = x T T 0 0 ...... n n asserif sans 22 12 22 12 T k n · T k X i − y 2 2 B y I y hcmoeto h vector the of component th =1 y hcmoeto h vector the of component th n X i A rnfrainmatrix transformation =1 0 n y } T e e · · · · · · · · · · · · · · · · · · 0 . . )( etost fbssvcosfran for vectors basis of sets two be | e k . . 2 . . i B Ay A ki ,then ), ) ( = = ( = ij A B A A B B snetebssi orthonormal) is basis the (since snetebssi orthonormal) is basis the (since nn . . . nn . . . 2 1 2 1 R ( n n n n i | ( 3           x x x j 1 = r rnltos(hc r o iermp)and maps) linear not are (which translations are 0 − 1 = − = = − y , y

y c 2 n , . . . , | ) e e e 2 n , . . . , 0 ) 1 1 · ..olydmpcma.k ihems2006 Michaelmas [email protected], . ( · x ( e e e x − 2 2 A 0 ) ) − , y , ) ...... e y e e i 0 j A ) ntebasis the in ntebasis the in ihrsett northonormal an to respect with e e e n n { { e e e i i , , } } n dmninlra vector real -dimensional a ewitnas written be can a ewitnas written be can . { { e e e k i } } gi the Again . h numbers The . x 7→ x 0 (4.62) (4.60) (4.63) (4.61) = e e j B Ax in ki

This is a supervisor’s copy of the notes. It is not to be distributed to students. ic ai ersnaini nqei olw that follows it unique is representation basis a Since iial,i the in Similarly, vector a components Consider vector for law Transformation 4.8.3 that conclusion the to leads notation, (4.62) matrix in Hence that follows it otherwise, or oee,teset the However, ahmtclTio:I ler emty(atI 73 I) (Part Geometry & Algebra IA Tripos: Mathematical the to of respect components with the relate (4.68c) and (4.68b) Equations applying By ambiguity). serious is a there used otherwise deliberately have we where that i.e. the matrix the ﬁnal the in where rmsbtttn 46)it 46)w aethat have we (4.60) into (4.62) substituting From matrices transformation of Properties 4.8.2 { e e i } basis. { { e e v e e i j hni the in then , { v } } e e ai ecnwrite can we basis = = = i sabssads ieryidpnet hs rmntn that noting from Thus, independent. linearly so and basis a is } X X X j j i basis. =1 =1 =1 n n n BA e v v e e e e j j i j e e e X j X i = j =1 =1 j = n n r ob nepee sclm etr ftecmoet fthe of components the of vectors column as interpreted be to are { X i ( I =1 e e n where , A i AB asserif sans i A ij } ij v e ai tflosfo 31)that (3.13) from follows it basis k X j = =1 n wpsmainorder. summation swap ) I e e v atraieyageb eaeigsmer) Thus symmetry). relabeling a by argue (alternatively I X i k v i =1 e e n e steiett arx ovrey usiuig(.0 into (4.60) substituting Conversely, matrix. identity the is v B ott niaeteclm arcso opnns(since components of matrices column the indicate to font j v ki = = = B = B = ki A = k X X A i A =1 n =1 X A j ij n A − A =1 n ij e v − 1 e e = − v , v 1 k = A v 1 i . e k X δ oete ieo 46b tflosthat follows it (4.68b) of side either to . ij ihrsett the to respect with =1 n i δ kj kj . v e

c j , e e . , k rm(4.60) from ..olydmpcma.k ihems2006 Michaelmas [email protected], X i =1 n B ki A ij . { e i } ai otecomponents the to basis (4.68b) (4.68a) (4.68c) (4.67) (4.65) (4.66) (4.64) e j in

This is a supervisor’s copy of the notes. It is not to be distributed to students. hni olw rmfo 46b that (4.68b) from from follows it Then ahmtclTio:I ler emty(atI 74 I) (Part Geometry & Algebra IA Tripos: Mathematical of matrix the that deduce We Let { where map linear maps a linear consider representing Now matrices for law Transformation 4.8.4 Example. Worked e i } e v and , 0 n hsfo 46c,i.e. (4.68c), from thus and that (4.62) from follows it (4.60) with comparison from Hence Answer. vector arbitrary an of components in Thus vector arbitrary an consider Now Moreover iial,since Similarly, i.e. v and 0 R and 2 M idtetasomto matrix transformation the Find . e v stemti of matrix the is etecmoetclm arcsof matrices column component the be v ehv that have We r h opnn ounmtie of matrices column component the are A − Let 1 A = { e AA 1 (1 = M − 1 M A : = M e e , 11 R 0) 2 1 e e e e e v ihrsett hsbasis. this to respect with I n . 2 1 A v e (0 = (1 = ihrsett h lentv basis alternative the to respect with 1 = , = 1 e e v → ( = 1 ( = 0 2 = A = v (0 = A , R − , , − 2 1 )= 1) )= 0) v 1 = = = n MA ( 1 v v Then . v , , B ne which under eddc ht(sabove) (as that deduce we , v 1 1 2 2 1 , )= 1) )=(1 = 1) ntetocodnt systems. coordinate two the in 21 1 e 1) v ( v = A + e A v 1 2 2 1 1 , M − e 1 } 1 1 = ( A (1 ( (1 ( A v htcnet hm eiytetasomto a o the for law transformation the Verify them. connects that e e 1 + + and 2 v − − = 1 = i.e. and ) = 0 1 v − (1 1 v , , A , = A 2 2 )+( + 1) 1) = 2 1 { e ) e e − v Mv e e 2

e e 2 , , 1 − 1 1 1 e 0 2 1 v 1 + ) )+(0 + 0) (0 + 0) 1 MA v 0 12 and −

( (1 = v − , ( = − − − 7→ 1 1 1 1 0 = − v e 2 1 . 1 1 1 2

and 1 2 c 1 1 1 1 v A , , ( v v − , 2 v ))= ) 1) ))= ) 1) = − , , 1) 0 ..olydmpcma.k ihems2006 Michaelmas [email protected], ( 1 and 1 ihrsett h lentv basis alternative the to respect with )= 1) = 1) e e 1 = ! − − v , . MA 1 e e epciey ihrsett h basis the to respect with respectively, , M ! 2 1 + . 2 v ( 2 ) v ( = − e e 2 2 1 1 . ) ( e 1 v 2 v ( ( e e e e e e e e ) . − n i em fclm vectors) column of terms (in and ) 2 1 1 − 1 1 A { . + + v 1 e + − e 22 2 , i e e ) 1) } e e e e 2 2 1 = , 2 2 sgvnby given is } . , ) ) etost fbssvectors basis of sets two be , , , (4.69) { e e i } .

This is a supervisor’s copy of the notes. It is not to be distributed to students. 20/03 ahmtclTio:I ler emty(atI 75 I) (Part Geometry & Algebra IA Tripos: Mathematical Example. from Maps h arxo h ha a ihrsett e ai stu ie by given thus is basis new to respect with map shear the of matrix The hence and (4.59b), and (4.59a) see orthogonal, are matrices rotation that deduced already have We rm(.6 h arxo hsmpwt epc otesadr basis standard the to respect with map this of matrix the (4.26) from where (4.29b) from then bases. two Suppose N Then where bases new consider Now respectively. where epciey Hence respectively. tflosthat follows It : R osdrasml ha ihmagnitude with shear simple a Consider n R e A v v n → and and { sa is to e R i } m R e v v n m sabssof basis a is 0 0 S e (where C × γ r opnn ounmtie of matrices column component are r opnn ounmtie of matrices column component are − ( unlectured 1 n NA = = = arxo opnns(e 46) and (4.61)) (see components of matrix m smpsmti ihrsett h e bases new the to respect with matrix map’s is A     − { 6= + 1 − A e e 1 cos R i S n sin = 1 0 0 } ) − γ ihrsett e ae fboth of bases new to respect with ) n . γ θ A of , γ C θ iia prahmyb sdt euetemti ftemap the of matrix the deduce to used be may approach similar A sin e e { sin e v 1 0 0 1 R f 0 A i cos θ n sin } = 2 − cos . . . θ and 1 sabssof basis a is v NA θ θ = = γ θ e v   A { 0 0 e e , e f e v   n 1 i N v e } − , − cos   7→ n so and = of sin γ 1 0 0 n thus and adbssb nangle an by basis dard Let γ cos and and C v nthe in θ R cos sin 0 − θ R m = θ 1 { m sin 2 NA n let and , θ e e + v v v N cos sin

and , θ 1 0 0 } v C cos c e v 0 γ θ x and and 0 etebssotie yrttn h stan- the rotating by obtained basis the be . = = ..olydmpcma.k ihems2006 Michaelmas [email protected], θ 1 sin θ = θ ieto ihnte( the within direction e e e C e e e A e C f N 3 2 1 θ e v v v 0 0 1 = 0 0 − 0 0 0 C   smti of matrix is , 1   S ihrsett bases to respect with ihrsett bases to respect with = = cos =   − NA . . . γ sa is . . sin = R cos e v sin e − n .   1 0 0 m 3 θ cos e f sin θ and m θ , + θ θ { × 1 0 0 0 1 0 1 { e e e θ bu the about γ θ e . i − m 1 } i e R cos cos γ N } sin + 1 sin m and arxo components. of matrix is cos + . θ θ ihrsett these to respect with 0 θ x { θ   0 0 1 e f e 0 0 x , i   θ x } 2 . e i.e. , 3   2 , 2 { { ln.Then plane. ) xs Then axis. e e e , . i i } } and and (4.70) { { e f f i i } } 19/06

This is a supervisor’s copy of the notes. It is not to be distributed to students. ahmtclTio:I ler emty(atI 76 I) (Part Geometry & Algebra IA Tripos: Mathematical where equivalently or Unknowns unknowns: two Two in in equations Equations linear two Linear Consider larger Two much of normally Solution are world linear real three 5.1 the of of case case complicated the the more on in magnitude). these concentrate However, of systems of by will here. orders the solution We many studied described but the equations. by type be unknowns, (often linear models, the often three of complicated of in solution can more equations equations the in world linear involves results ‘real’ in often world The result models real mathematics. models the Some linear when sorts. even of various study of our equations continues section This This? Study Why Equations Linear and 5.0 Inverses Matrix Determinants, 5 Exercise. that conclude therefore We by (5.2a) of multiplication left from However, that have we (4.54) From (e.g. equations two the of combinations linear suitable forming by solve Now o 3 a For for Determinants 5.2 det if Thus, i.e. × det hc that Check matrix 3 A A ,teeutoshv a have equations the 0, 6= | ≡ A A x ≡ | AA earayhv ht(e 45a n following) and (4.52a) (see that have already we = = = = = − 1 x x 2 1 = a a ε ε

ijk ijk a a a 11 11 A 11 21 31 ( ( , 3 ( ( ( a a − a a a a a 1 i 1 22 22 × 1 11 21 11 i A a a a a a d x x a a x x a a a 2 j 12 22 32 3 2 1 33 33 = 1 2 2 j = 22 12 22 A a a Matrices − I − − 3 k − − − . = k 3 ( = ( = 1 a a a unique d d a a a a 13 23 33 a a = a 2 1 21 11 det 23 23 22 21 21

x x 1 a a det a a a a − 1 1 , 32 32 A 11 12 12 22 1 x a A + + ) ) 21 ) ) A det = ) d − solution A = x x and − − a a 1 − 1 d x 2 1 22 12 a − A 1 a a a i teit)w aethat have we exists) it (if 22 − = x x 21 12 − + 21 a a = = a 2 2 22 1 12 d ( ( 21 A a d a a A 11 d 12 21 . − = = = 2 a a = a d

a a ) a 21 22 c 11 − 2 / { 33 33 12 a

) a d d a det d d 11 / ..olydmpcma.k ihems2006 Michaelmas [email protected], a a ij 1 1 12 − − 2 1 21 11 det } − − a a A 13 23 d d a a A , . a a 2 1 a2 (a 11 12 a a 22 12 , 32 31 d d 2 2

. + ) + ) × . , . arx.(5.2b) matrix). 2 a a 31 13 a ( ( 22 a a 12 21 × a a 23 32 (5.1a) − − a a 22 22 − a a a 13 31 12 ) ) × . (5.1b)) (5.1b) (5.5d) (5.5b) (5.3b) (5.1a) (5.5a) (5.2a) (5.3a) (5.5e) (5.5c) (5.4)

This is a supervisor’s copy of the notes. It is not to be distributed to students. 20/02 .. rprisof Properties 5.2.1 ahmtclTio:I ler emty(atI 77 I) (Part Geometry & Algebra IA Tripos: Mathematical rewritten be can (5.5e), i.e. line, last the that observe We Remarks. (iii) (iv) (ii) (i) • • Let Since ups htw osrc arx say matrix, a construct we that Suppose of two any interchange we of If rows the between dependence linear is Remark. there only and if (5.7) det Similarly Now of row first the of elements the of terms in (5.6) 2 to equivalent of expansion determinants an that means This hsflosfo h atthat fact the from follows This Then columns. other the of Remark. det of sign the change of columns two any rmivkn 57 ecnddc htasmlrrsl ple orows. to applies result similar a that Remarks. deduce can we (5.7) invoking From Remark. vlain fdtriat olre ( larger to determinants of evaluation, det of 2 expansion an is (5.6) (5.6). in pattern sign the Note × • u-arcs hsosraincnb sdt eeaietedfiiin(sn euso) and recursion), (using definition the generalise to used be can observation This sub-matrices. 2 a hspoet eeaie to generalises property This α i 1 ε ijk · = ( β hspoet eeaie to generalises property This hspoet eeaie to generalises property This 57 eeaie to generalises (5.7) a α × i i 1

, a γ α α α A a j fadol if only and if 0 = ) 2 1 2 3 j det a 2 fadol fteei ierdpnec ewe h oun of columns the between dependence linear is there if only and if 0 = + + + k = det 3 × λβ λβ λβ A 3 = β A × u-arcsas xss ec rm(.d,o otherwise, or (5.5d), from Hence exists. also sub-matrices 2 = j 1 2 3 A ε , ecag h ino det of sign the change we A 3 + + + ijk a = a . det 11 determinants k µγ µγ µγ a 3 a

1 11 = 1 2 3   i a a a n A α

2 22 32 γ , × j β β β α α α k ntrso lmnso h rtclm of column ﬁrst the of elements of terms in a a a β hnfo (5.5b) from then , 1 2 3 3 2 1 22 32 3 n α k n a a and , eemnns(u slf unproved). left is (but determinants 23 33 γ γ γ β β β × β 1 2 3 a a 3 2 1 n n 23 33 n

and n γ − × × eemnns(u slf unproved). left is (but determinants γ γ γ × ecag h inof sign the change we

= = ( = det A 3 2 1 e a det n n − γ n yadn oagvnclm of column given a to adding by , 21   arcs(u slf undone) left is (but matrices ) eemnns(u slf unproved). left is (but determinants eemnns(u slf unproved). left is (but determinants a r olnr ..if i.e. coplanar, are A α α

A e 12 α = det = a a · · A det =

+ 12 32 ( ( iial fw necag n w osof rows two any interchange we if similarly ; β β ijk a a λ 21 31 × × β α a a

A c i A γ γ 13 33 + β T ) + ) a a ..olydmpcma.k ihems2006 Michaelmas [email protected], . j µ . . 23 33 γ

γ k + λ ) =

β · a + ( 31 α · β α a ( α

· β 13 A × · ( , ( a a . β ×

β β γ 22 12 × ) a a × γ and 31 21 + ) γ γ a a ) .Hnei einterchange we if Hence ). 23 13 γ . µ a a 32 22 γ A r ierydependent. linearly are

A . · n eemnnsof determinants and

( iercombinations linear β . × γ ) A rfrom or , A A (5.7) (5.6) (5.8) and we

This is a supervisor’s copy of the notes. It is not to be distributed to students. ahmtclTio:I ler emty(atI 78 I) (Part Geometry & Algebra IA Tripos: Mathematical Proof. 5.2. Theorem Remark. becomes side that Proof. 5.1. Theorem otesg frgthn ieas eess iial o wp of swaps for Similarly reverses. also any side for right-hand holds (5.10a) of sign the so ethn iei eo hl h ih-adsd is side right-hand the while zero, of is more) side (or left-hand two that now Suppose hc sas eo aigcvrdalcssw ocueta 51a strue. is (5.10a) (5.5b) that from conclude starting we (5.10b) cases for all Similarly covered Having zero. also is which (vi) (v) p since of column or row single any multiply we If fw utpytematrix the multiply we If Remark. u slf unproved. left is but Remark. since and tr ih(.0) n ups that suppose and (5.10a), with Start epoetersl nyfr3 for only result the prove We • hstermgnrlssto generalises theorem This hspoet sueu neautn eemnns o ntne if instance, For determinants. evaluating in useful is property This h rtrwt usqetrw omake to rows subsequent (5.13b)). to row ﬁrst the q r wpe.Te h ino h ethn ieo 51a eess hl h right-hand the while reverses, (5.10a) of side left-hand the of sign the Then swapped. are hspoet eeaie to generalises property This hspoet eeaie for generalises property This If If A A = and { pqr { a B ij } } r ohsur arcs then matrices, square both are hti emtto of permutation a is that ijk ijk is A a 3 a det qi 1 × by i a a 3 pj 1 p AB j then , λ , λ n a a × pqr pqr ( then , q α rk rk λ × det arcs rm(.b n (5.10b) and (5.5b) From matrices. 3 α and · det det n = = = = = det = = ( ) λ AB n · eemnns(u slf unproved). left is (but determinants p n β det( A A det( ( jik jik r × 1, = β × × det ijk ijk ijk pqr (det = n(.0)aeeul lgtake Wlog equal. are (5.10a) in n a a × n = = λ λ A λ qj 1 A eemnns(u slf unproved). left is (but determinants j ( a a A γ A det γ A b eemnnsto determinants q a a AB by ip ip = ) = ) = ) = ) det pi 1 2, = = b a i { a ijk ijk a a A A p ) jq 123 i λ i λ 1 rk rk 1 )(det B 1 λ λ ogive to , λ λ b a a a a det for 0 = p ( n 3 3 ( r kr . ip pi jq } = = 1 AB α (

det det . .Te 51a sjs 55) etsuppose Next (5.5c). just is (5.10a) Then 3. = b c α a a b b q − − A · q jq qj p 2 B ) ..olydmpcma.k ihems2006 Michaelmas [email protected], · 2 ( 1 b j A , A β a a ( ) b 2 a r ijk ijk β kr rk q . 3 kr , ( , × 2 A b AB i × a a b b . , then , r 1 pi γ 2 = r 3 i γ 3 p ) )) a a k )) 1 qj 3 and . n , . . . , j . a a rk rk r , , or , hndet then ; a p 11 q = and d utpe of multiples add 0 6= q ,sy hnthe Then say. 1, = r A tflosthat follows It . = a 11 ∆ (5.10b) (5.10a) 11 (5.9b) (5.11) (5.9a) (see 20/06

This is a supervisor’s copy of the notes. It is not to be distributed to students. 21/03 ec det Hence ahmtclTio:I ler emty(atI 79 I) (Part Geometry & Algebra IA Tripos: Mathematical that columns of interchanges an from noting after Similarly, Definition. the eliminating by obtained matrix square Definition. the and row square a For cofactors and Minors a 5.3.1 of Inverse The 5.3 Remark. Proof. 5.3. Theorem ehv that have we (5.13b) and (5.6) for from apply definitions above The If o aearayvrfidti o oerflcinadrtto matrices. rotation and reflection some for this verified already have You A A sotooa then orthogonal is = j n en the Define the Define th ± × If ounof column 1. n A matrix A sotooa then orthogonal is det ij det = minor cofactor A A (det           A A det a a Hence . ======( ( a 3 a , i i AA A − +1)1 n 11 { ...... M A × ) 1 1)1 n a a a a 2 − a a ∆ ∆ ij T ij = 11 j 11 × j 12 3 (det = a 1 ij ij 2 } fthe of , = ∆ ∆ 12

∆ ∆ deﬁne ,

n Matrix a . . . a . a . . . . . a . . . a a a ( = fthe of . . a a j 11 j I 12 . .

11 21 31 1 22 32 tflosfo 57 n 51)that (5.11) and (5.7) from follows It . arcs u ecfrhasm that assume henceforth but matrices, 2 a a . . . + . 21 31 + − A 1) a a a a a a a i ij )(det ( ( th 21 12 22 32 ij 22 23 33 A M i i i a a th − +1)( + n 1( det ∆ ij 23 33 th ∆ o n the and row

( 1)( j ij j j ...... lmn fsur matrix square of element M 21 −

− 22 − ob h qaemti bandb lmntn the eliminating by obtained matrix square the be to a a a lmn fsur matrix square of element A j j det = + A 1) 13 23 33 1) − − ij T a + + 1) 1) 21 a = det( = )

( = a 22 a =

31 ± 32 a a

A a a − − ∆ 1 a a ∆ 12 32 ( ( ij a 11 31 a 1) i i . 31

− 32 +1)(

n 1( . c a a a i ( 1)( j AA + j j ...... 32 22 12 a a th +1) +1) ..olydmpcma.k ihems2006 Michaelmas [email protected], a a j 13 33 j j 13 33 +1) +1) T det ounof column

det = )

a a a + − 31 21 11 A a ij a a . . . a . . . a . . . a . . . . . 31 32 . . . a a a . .

33 23 13

I a a a a 1 = A 22 12

21 11 A A ( ( , i.e. , i i ob h eemnn fthe of determinant the be to − +1) nn 1 ...... as . n 1) a a A a a 23 13 23 13 n n sa3 a is          

. × arx Then matrix. 3 (5.14b) (5.13b) (5.14a) (5.13a) (5.12) i th

This is a supervisor’s copy of the notes. It is not to be distributed to students. ahmtclTio:I ler emty(atI 80 I) (Part Geometry & Algebra IA Tripos: Mathematical Hence Proof. then 5.5. Theorem inverse the of Construction 5.3.2 Proof. 5.4. Lemma consider end this To that show to wish we Next rows, interchanging subsequently and (5.8) from starting but Similarly, symmetry) relabelling a by (or Analogously ic w oun r ierydpnet(culyeul.Poedsmlryfrohrcocsof that choices show other also for can similarly we Proceed Further, equal). (5.16a). (actually obtain dependent to linearly are columns two since ecncmiealteaoerslsi lemma. a in results above the all dependent. combine linearly can be We to out turn rows two since AB rm(.7)ad(5.18a), and (5.17b) From (5.16b). and (5.16a) (5.15), (5.14c), (5.14b), (5.14a), See = I iial rm(.7)ad(5.18a), and (5.17a) from Similarly . ie a Given a j i 2 and 2 = ∆ 3 i 1 × 3 = = = 0 = matrix k ,then 1, = a a

det a a a 12 12 , 12 22 32 ∆

A a a A 11 22 32 = a a a a a with ( ( + a a 12 22 32 ji ij B AB a ij ji AB ∆ ∆ 1 ) a a a det ∆ ∆ j ij 22 23 33 ki ik ) ∆ det a a a ij ik ki ∆ 13 23 33 1

A j = = if 0 = if 0 = 21 − = =

A BA ======a + 0 6= a a 22 BA det δ δ 2 j = a δ a 3 j jk jk δ a

1 32 ij deﬁne , ∆ ∆ det ik ij ik a a I det A tflosthat follows It . j j ∆ = 12 32 ( det det . 2 j ∆ det B 3 ∆ j 6= 6=

31 A c jk . A ) = ji I kj A A k k A ..olydmpcma.k ihems2006 Michaelmas [email protected], . a a , (5.16a) . (5.16b) , a 13 33 . , 3 B j

∆ by + 3 j a . 32

B a a 22 12 = A − a a 23 13 1 and

A sinvertible. is j (5.18b) (5.17b) (5.18a) (5.17a) (5.14c) (5.15) and k

This is a supervisor’s copy of the notes. It is not to be distributed to students. 21/02 ahmtclTio:I ler emty(atI 81 I) (Part Geometry & Algebra IA Tripos: Mathematical obtain to as so eliminate to (5.21a) use Now solution). unique no then is there (since Assume Example. Remark. where equations of system the solve to wish we Elimination that Gaussian Suppose Equations: Linear Solving 5.4 suigta det that Assuming etrmto sGusa lmnto,wihw ililsrt o 3 for illustrate will we which solve elimination, to Gaussian wish is method calculate to better be A would method one then numerically, form (5.20a) then solve and to wished we if Hence od for holds hndet Then Hence hc ae hsclsnei htteeeto shear a of effect the that in sense physical makes which A a sa is h oml o h lmnso h nes of inverse the of elements the for formula The 11 osdrtesml shear simple the Consider ,ohrier-re h qain othat so equations the re-order otherwise 0, 6= n × A n S γ − n × 1 ,adatraltl manipulation little a after and 1, = matrix, A d n oee,ti satal eyinefficient. very actually is this However, . ,ti a h omlsolution formal the has this 0, 6= arcs(nldn 2 (including matrices a a (5.21b) 32 22 x − − sa is a a a a 11 31 11 21 − S n γ − a a a a × 1 12 12 a a a 11 21 = 31 21 11 ounvco fukon,and unknowns, of vector column 1 ∆ × x x x x x   21 ∆ ∆ 2 2 1 1 1 52a n (5.21c) and (5.21a) ∆ ∆ ∆ 11 31 + + + + + = 11 12 13 × a a a 1 = 0 = ( − 32 22 12 S A arcs o utbydfie cofactors. defined suitably for matrices) 2 a a , γ x γ 33 23 − ∆ ∆ ∆ x x x A , , = 1 = 2 2 2 21 22 23 x − − ) ij + + +   A ∆ ∆ ∆ = a a a a − = a a a 1 0 0 0 1 0 1 12 22 32 31 21 11 11 ∆ ∆ ∆ d 1 33 23 13 d 31 32 33 det a a , 0 = 1 = 0 = γ x x x A 13 13 .   1 3 3 3 i.e. ,

a A c , , , = 0 11 x x ∆ = = =   ..olydmpcma.k ihems2006 Michaelmas [email protected], 3 3 γ   ∆ ∆ ∆ ji ,adi hti o osbete stop then possible not is that if and 0, 6= − . 1 0 0 1 0 0 1 srvre ycagn h inof sign the changing by reversed is 13 23 33 , d d d = = a a 3 2 1 11 31 0 = 0 = 1 = . , , − d d γ × x 3 2 , , . d × 1 (5.21a) − − 0 yforming by sagiven a is arcs ec ups we suppose Hence matrices. 3   a a a a 11 31 11 21 , d d 1 1 , . , n × A ounvector. column 1 − 1 sn (5.19), using (5.22b) (5.21b) (5.20b) (5.22a) (5.21a) (5.20a) (5.21c) (5.19) γ .

This is a supervisor’s copy of the notes. It is not to be distributed to students. 21/06 ahmtclTio:I ler emty(atI 82 I) (Part Geometry & Algebra IA Tripos: Mathematical of system a be to said is Definition. Definition. det if deduced be can what A about is section This solution unique a has that det If problems homogeneous and Inhomogeneous 5.5.1 equations linear Solving 5.5 all). at solution Remark. a is there (if solution unique no is there gives (5.24a) that providing Now, obtain to as so eliminate to (5.23a) use Now solution). unique no then is there (since Assume become (5.22b) and (5.22a) that so let notation simplify to order In ssi ob ytmof system a be to said is sa3 a is A A sivril and invertible is (and 0 6= × a ti osbet hwta hsmto al nyif only fails method this that show to possible is It 22 matrix. 3 ,ohrier-re h qain othat so equations the re-order otherwise 0, 6= h ytmo equations of system The If x a 3 a 22 0 d hn(.3)gives (5.23a) then , 32 0 6= A = = 0 sa2 a is x hntesse fequations of system the then a a 22 A 32 = homogeneous inhomogeneous − − × − A 1 − xss nsc icmtne tflosta h ytmo equations of system the that follows it circumstances such In exists. ra3 a or 2 a a a a 1 11 21 11 31 d a . a 12 12 x a × 2 a , 33 0 a , n nly(.1)gives (5.21a) finally and , qain.I det If equations. arx hnw nwfo 54 n hoe . npg 80 page on 5.5 Theorem and (5.4) from know we then matrix) 3 (5.23b) a a − equations. 32 0 22 0 23 0 33 0 a a a a x x 22 0 32 0 33 0 33 0 2 2 = = + + a − − − 23 0 A A a a a a a a a a 33 0 23 0 x x A a a 23 33 22 0 22 0 32 0 32 0 22 0 32 0 x x x = = ,weea sa eseils otecs where case the to specialise we usual as where 0, = − a a − 3 3 3 23 0 23 0 × d 0 = a a a a A = = (5.23a) 11 21 11 31 d hnteuiu ouinis solution unique the then 0 6= 0 = 0 6= 3 0

a a a c 13 − 22 13 d d ..olydmpcma.k ihems2006 Michaelmas [email protected], A 3 0 2 0 , , a a ,adi hti o osbete stop then possible not is that if and 0, 6= . , sntivril,ie nyi det if only i.e. invertible, not is 22 0 32 0 , d , x d , 1 d If . 2 0 . 2 0 3 0 = x = 2 d d yforming by 2 3 − − a a a a 11 21 31 11 d d 1 1 , , A − 1 0 A = A (5.23b) (5.25b) (5.24b) (5.24a) (5.23a) (5.25a) 0. = x 0 . = d

This is a supervisor’s copy of the notes. It is not to be distributed to students. 22/03 ecnie aho hs ae nturn. in cases these of each consider We ahmtclTio:I ler emty(atI 83 I) (Part Geometry & Algebra IA Tripos: Mathematical three are There O. origin, the through pass each planes three the possibilities: (5.27b) planes. equations in homogeneous 3 the plane of For intersection a the represents is equations equations individual 3 these of of each Since respectively. as expressed be then may (5.25b) and (5.25a) Equations case For of view Geometrical 5.5.2 (iii) (iii) (ii) (ii) (i) (i) i h ouinsaetu a ieso two. dimension has thus space solution The (5.27b) by to space, solutions solution i.e. the thus plane, and a plane, is the planes specify three may the we If of then plane. intersection a the on Thus lie other. each imply all vectors planes the of intersection the span hence vectors; det If coincide. planes O); three (including the line common a have planes three the O; at only intersect planes three the ial ene ocnie h aewe h ieso fspan of dimension the when case the consider to need we Finally det that suppose Next Since that imply again and span of dimension. hence and condition final The lnsi ie ..teslto for solution the i.e. line, a is planes 1 = , r 2 2 A r , 1 let 3 { r h w ieryidpnetvcos hna bv h rttoeutoso (5.27b) of equations two first the above as Then vectors. independent linearly two the are r r then 0 6= · 1 1 ( , , r x r r 2 2 r 2 = × i , and r etevco ihcmoet qa oteeeet fthe of elements the to equal components with vector the be 0 r 3 ..tetrepae nesc nya h rgn The origin. the at only intersect planes three the i.e. , } 3 x ,alpit nti iesatisfy line this in points all 0, = ) a r en qa o2o ;fis ecnie h aewe ti .Asm lgthat wlog Assume 2. is it when case the consider we first 1; or 2 to equal being r 1 r 3 { utleo h line the on lie must and 3 r · utte l eprle.Ti en that means This parallel. be all then must · 1 A ( x , r r 2 .I hscs h set the case this In 0. = b hnipisthat implies then 0 = 2 × , A r n w ieryidpnetvcossc that such vectors independent linearly two any are r r r 3 x r i T i T { } 3 r x = ) x x { { 1 = x x n h set the and 0 6= ∈ · 0 = = R x ∈ ∈ R 3 and 0 = R R 3 x h rttoeutoso 52b ml that imply (5.27b) of equations two first The . r r i i 3 3 : saln.Teslto pc hshsdmninone. dimension has thus space solution The line. a is · · x : : x x A x x = = = = λ ( 0 = = r a    λ λ 2 t t λ + r r r · λ , λ , { d 3 T 2 T 1 T x snew aeasmdthat assumed have we (since 0 = r µ i {    1 r b ∈ ∈ ,i.e. 0, = , 1 r . , R R where

2 r ( r c i i , 3 , , 2 r t t 1 = 1 = , · ..olydmpcma.k ihems2006 Michaelmas [email protected], 3 r x = = } 3 } .Hneteitreto ftethree the of intersection the Hence 0. = slnal eedn ihtedimension the with dependent linearly is x , , ,µ λ, r r 2 2 1 1 ossso he ieryindependent linearly three of consists utleo h line the on lie must , , × × 3) 3) ∈ r r r , , 2 2 1 R } } · { } . . x r . 1 0, = , ouinspace solution R r 3 2 h ouino ahset each of solution the , , r r 3 2 i } hrwof row th · x s1 h he row three The 1. is a and 0 = · r 1 r 1 · ( x = hshszero has thus r 2 A utleon lie must × b nwhich in , r r · 3 3 r (5.27b) (5.27a) · ) 1 (5.28) (5.26) x 0), 6= 0, = 0 = r 1

This is a supervisor’s copy of the notes. It is not to be distributed to students. (b) Remark. ahmtclTio:I ler emty(atI 84 I) (Part Geometry & Algebra IA Tripos: Mathematical (a) if that recall Next map linear the Consider of view mapping Linear 5.5.3 enwcnie h he ieetpsiiiisfrtevleo h ult nturn. in nullity the of value the for possibilities diﬀerent three the consider now We i.e. of kernel the (4.5), deﬁnition earlier our From basis. standard the to Remarks. our K (iii) (iii) (ii) (ii) (i) (i) ( A equal by choose spanned we now in Suppose those with section this in results the of comparison from Theorem have (see also Theorem we rank-nullity case the each of In accordance in have we (iii) and 59) page (ii) on (i), 4.2 cases of each In emti view geometric steslto pc of space solution the is ) ocuethat conclude some for that claim We possible). always is this v If so and If n n n If span of dimension the (since that w ( ( ( , n A A A n n w 6∈ ( ( ed o edt osdrtecase the consider to need not do We ( 2. = ) 1, = ) 0, = ) A A A r K 6∈ ( hoenon-zero choose 2 = ) then 0 = ) hoenon-zero choose 1 = ) A ( λ K A .I olw rm(.4 n 41)that (4.19) and (4.14) from follows It ). = α ( oetn hsbsst omabssof basis a form to basis this extend to ) A {A µ ∈ oetn hsbsst omabssof basis a form to basis this extend to ) r ( = { ( R A of A e u Hence . 1 ( ν r , i span dim = ) v ) § ( v { {A , since 0 = .. orsodto correspond 5.5.2 A and ) λ , A ubro ieryidpnetclmsof columns independent linearly of number = A w A i span dim = ) ( ( u e } ( : u 2 ) ubro ieryidpnetrw of rows independent linearly of number = R , sabssfor basis a is ) + ) − A { , A { 3 u A A α µ ( ( → , w x A u v ( µ v { i span dim e {A u r ieryidpnet n httern of rank the that and independent, linearly are ) ) A u A ( = , + R 3 u , , v w , ) A A ( ( v 3 ( + ) } v R µ 0 v ∈ } uhthat such , x u { K ( n h ubro ieryidpnetvcosi hsstmust set this in vectors independent linearly of number the and , , v ∈ ihadmnindntdby denoted dimension a with , + ) 3 w A ) w K = span = ) ob h tnadbasis standard the be to ( , e + ν K A ) A } { 1 } ( A 0 ν {A r = ) , A ( R ν ( sbss tflosta h akof rank the that follows It basis. is 1 salnal needn e since set independent linearly a is A {A A ( A v w 3 ); w , r e ) ( ( hnteiaeof image the then , uhta h set the that such ) r ( , n u w 2 = ) { ( 2 = u A A , v ( x , ) {A = ) A A , r + ) ) ( x sabssfrtepoe subspace proper the for basis a is 0 A ∈ 3 , w e sln sw xld h a with map the exclude we as long as 3 = ) 0 n so and , } A 3 7→ ( ( {A ⇔ } ) R u } 0 v n } ( 3 {A ⇔ } w ) ) ( x , stwo). is , A : { ⇔ ) 0 A A R } A 3 = ) = { ⇔ ( ( 3 x

r ieryidpnet oseti oethat note this see To independent. linearly are v R w Since . c A ) α 3 = ) , x µ ..olydmpcma.k ihems2006 Michaelmas [email protected], } A rcl rmTerm35o ae4 that 49 page on 3.5 Theorem from (recall = . where , v 0 λ ( = ( µ } u + w { ( µ v . λ r u A + A ) { ν ( = u + } , e A u w v µ ssandby spanned is 1 . + A ν 1 = ) ν } , v = A = n w e µ since 0 = sabssfor basis a is + sgvnby given is ( 2 A v = ) stemti of matrix the is A A A α , v e ν u + .Tetrecsssuidin studied cases three The ). . ( ( 3 w } o rank row ounrank column = } 0 ν , hnteiaeof image the Then . } = w 0 A = ) 0 and stre i.e. three, is } A { u , {A 0 sto i.e. two, is K , K ). } A v ( ( w ( , u A A w ). ) A .Nx choose Next ). , 6= .Nx choose Next ). } § A 5.5.2, ihrespect with sbss We basis. is 0 ( A v tfollows it , r ) = ( r , A ( A A 0 3. = ) (5.29) ( . 2 = ) w A ) } is ,

This is a supervisor’s copy of the notes. It is not to be distributed to students. 22/06 h atrcs sdsrbdb hoe . below. 5.6 Theorem by described is case latter The ahmtclTio:I ler emty(atI 85 I) (Part Geometry & Algebra IA Tripos: Mathematical Example. Proof. of solution ﬁxed particular a is 5.6. Theorem equation inhomogeneous the det of If solution the for Implications 5.5.4 fdet If exist must there ute,from Further, (iii) (ii) (i) • • n n n or either ic det Since n so and Hence If A A ( ( ( is ent that note we First A A A a then 0 = then 0 6= • • d and 2 = ) and 1 = ) and 0 = ) ,te det then 1, = osdrte(2 the Consider If If ouinis solution h eea ouini then is solution general The where ∈ d r b I b ( ( § ∈ then 1 6= / A A A then 1 = .. and 5.5.2 If I and 1 = ) x ,i hc aeteei tlatoeslto n h qain are equations the and solution one least at is there case which in ), λ (1 = r ( r ( A d ∈ ( ∈ A r r r A A ,i hc hr r osltosadteeutosare equations the and solutions no are there which in ), ( ( ( ∈ R ) R and 3 = ) A A A − − 3 < I . 1 A ,then 1, = ) then 2, = ) then 3, = ) ( o which for a = A § and 3 ,and 0, = ,if ), 1 b x ..,if 5.5.3, × ) n 1 b I 1 = ( )ihmgnoscs of case inhomogeneous 2) hntegnrlslto to solution general the then ( ∈ A − / A 1 a x ∈ .Wehrteei ouinnwdpnso h au of value the on depends now solution a is there Whether 1. = ) span = ) I A I a 0 hndet then 1 6= I ( ( x I + A ( A ( A d = A y y y − ,adteeaen ouin eas h qain r inconsistent. are equations the because solutions no are there and ), sapoe usaeof subspace proper a is ) y 1 = ) steiaeunder image the is n ouin xs teeutosaecnitn) particular A consistent). are equations (the exist solutions and ) a = = = d saslto since solution a is x and λ λ 0 A R − a t ( 1 = n h ouini unique. is solution the and 3 x 1 and + 0 1 1 x (where y a 1 1 + µ A 0 , A stegnrlslto of solution general the is b x + x x and 0 6= n h nqeslto is solution unique the and y 1 = and = ) = y = where , and x I x ( 0 1 d 0 0 x x x x x A 2 1 1 1 = + + A = sntto o h mg of image the for notation is ) + + + A A x A A K λ 0 − x = x i.e. ,

= t x x x λ y 0 1 c = ( 0 1 0 2 2 A rpeetn line). a (representing = + d sayvco in vector any is xssadi nqe Speciﬁcally unique. is and exists R ..olydmpcma.k ihems2006 Michaelmas [email protected], = d − span = ) 1 b . 3 1 d λ where x . . d 1 Then . a a ewitnas written be can = . and + , A µ − b A 1 d y rpeetn plane). a (representing A x = xssadunique. and exists − x 1 = 1 0 = K inconsistent n thus and , 0 A A ( . x A − , 1 ,i.e. ), = x d 1 b = consistent A x .Since ). . 0 ; + y where . b d . (5.30) (5.31) ∈ R x 3 0 ,

This is a supervisor’s copy of the notes. It is not to be distributed to students. B(viii) aifigtefloigegtaim rrules: or axioms eight following the satisfying ahmtclTio:I ler emty(atI 86 I) (Part Geometry & Algebra IA Tripos: Mathematical B(vii) ‘complex’ for ‘real’ exchanging by numbers operations real binary the two over with space together vector a ‘ for deﬁnition and the adapt just We complex Deﬁnition with spaces vector 6.1.1 to generalise now We scalars. real with spaces scalars. vector Numbers considered Complex have The We Over solve Spaces to Vector us solve allow to and us 6.1 numbers, allow real and of numbers, extension real natural of a extension natural are a spaces problems. are vector more numbers complex complex problems, that more way same the In This? Study Why 6.0 Spaces Vector Complex 6 A(iii) B(vi) A(iv) A(ii) B(v) A(i) • • R ne clrmultiplication; scalar under multiplication scalar addition; vector addition vector o ‘ for ’ clrmlilcto sdsrbtv vrsaa diin ..frall for i.e. addition, scalar over distributive is multiplication scalar all for i.e. addition, vector over distributive is multiplication in scalar identity multiplicative the is 1 where all for i.e. element, identity an has multiplication scalar all for i.e. ‘associative’, is vectors of multiplication scalar o all element; for identity an has addition vector i.e. element an exists there is addition is addition C x .Hneavco pc vrtecmlxnmesi set a is numbers complex the over space vector a Hence ’. ∈ ne alia Inter V commutative associative hr xssa additive an exists there eoe for denoted hyaeiprati unu mechanics. quantum in important are they eoe for denoted ..frall for i.e. , 0 ..frall for i.e. , ∈ x V , y aldthe called , ∈ a x ∈ V x , x y C ( + , by , eaieo nes vector inverse or negative λ ( y z λ ( and x y ∈ C λ ∈ x + x ulo eovector zero or null ( + ; + V + x V + µ µ x x 1 x y ) z + + y y x ( = ) ( = ) = ) x ∈ where , x = 0 = = 0 V = y = λ λ λµ x x x x by

+ x c 0 , + + + ) , (6.1d) ; x x ..olydmpcma.k ihems2006 Michaelmas [email protected], a y x µ λ (6.1b) ; x , + ) x y x + where , ,µ λ, . (6.1g) ; ∈ y z uhta o all for that such , V (6.1a) ; ∈ ∈ V C x a 0 x and ota hr scoueunder closure is there that so ∈ ,µ λ, λ ∈ V V ∈ V x ∈ feeet,o ‘vectors’, or elements, of C uhthat such ∈ ota hr sclosure is there that so C and V and x x , ∈ x y V ∈ ∈ V V (6.1h) (6.1e) (6.1c) (6.1f)

This is a supervisor’s copy of the notes. It is not to be distributed to students. ngnrlsn rmra etrsae ocmlxvco pcs ehv ob aeu ihscalar with Let careful be to Definition have we 6.2.1 spaces, vector complex to spaces vector real products. Spaces from Vector generalising Complex In for Products Scalar 6.2 ne rdc,o h ree aro vectors of pair ordered the of product, inner 6.1.3 ahmtclTio:I ler emty(atI 87 I) (Part Geometry & Algebra IA Tripos: Mathematical Remarks. satisfied. exercise. by Proof 6.1. Lemma numbers Definition. of theorems and properties the of Most Properties 6.1.2 en etradto n clrmlilcto by multiplication scalar and addition vector define (iii) (iv) (ii) (ii) (i) (i) V losre sasadr ai for basis standard over a as serves also ‘ and ‘complex’ for subset a ‘real’ when on 3.1 Theorem unique; if is vector a of inverse additive the vector zero the h tnadbssfor basis standard The number. complex arbitrary an by multiplication R ea be n x C sasbe of subset a is z ∈ n i C V ∈ ute ecnepesany express can we Further . n o xdpstv integer positive fixed For dmninlvco pc vrtecmlxnmes ewl eoeasaa rdc,or product, scalar a denote will We numbers. complex the over space vector -dimensional C and C , n i savco pc over space vector a is 1 = λ hc htAi,Ai) (i) (v,Bv,Bv) (i)adBvi)of B(viii) and B(vii) B(vi), B(v), A(iv), A(iii), A(ii), A(i), that Check ∈ 0 n , . . . , C C sunique; is n then u tis it but , For . R R n u u o ‘ for ’ i.e. , + ( = λ e λ 0 1 u v ∈ not U x u (1 = C C 1 = ’. favco space vector a of u , . . . , n ( = ( = § usaeo h etrspace vector the of subspace a n ope vectors complex and C define , C 0 olwtruhmc sbfr,e.g. before, as much through follow 3 z , z , n 0 ( = ∈ Hence . , . . . , λu u u n h C ( 1 z − , and ) u 1 + n 1 v λu , . . . , 1) z , . . . , , C 0) ntrso opnnsas components of terms in v ∈ v x n 1 , . . . , u , . . . , ∈ i C V = ob h e of set the be to n n − by C a dimension has = ) n v V x ) e , ( = n

∈ n sasbpc of subspace a is c X + i and =1 (0 = C n ..olydmpcma.k ihems2006 Michaelmas [email protected], v u v n 1 , n z v , . . . , . v i ) , e ∈ 0 λ ∈ i , . . . , . 0 C C n = n n n tpe ( -tuples n ) where , C , , 0 1) hnvee savco space vector a as viewed when n ; since , , V tl ple fw exchange we if applies still z 1 z , R 2 n z , . . . , sntcoe under closed not is n fcomplex of ) § .. are 6.1.1 (6.3b) (6.2b) (6.3a) (6.2a) (6.4) 2

This is a supervisor’s copy of the notes. It is not to be distributed to students. hn for Then, ups htw aeasaa rdc endo ope etrsaewt ie basis given a with numbers space complex the vector define complex first this a see on To vectors. defined basis product of pairs scalar a have i we Components that of Suppose Terms in Products Scalar 6.2.3 Properties 6.2.2 ahmtclTio:I ler emty(atI 88 I) (Part Geometry & Algebra IA Tripos: Mathematical inequality. triangle the and inequality Schwarz’s argument. first the in Anti-linearity with product Scalar are notations alternative where ubr uthv h olwn properties. following the have must numbers (iii) (iv) (ii) 1 = (i) n , . . . , Non-degeneracy Non-negativity Linearity u symmetry Conjugate hsalw st write to us allows This the i.e. space vector complex a for for (whereas that important is assumption product the scalar the is in equation vectors this in Implicit a where vector iheult nywhen only equality with for i.e. argument, first = h v any v hn(.a mle that implies (6.5a) then , ecamta h clrpouti eemndfralpiso etr yisvle o all for values its by vectors of pairs all for determined is product scalar the that claim We . , v v . ∗ w vectors two i ntescn ruet ..for i.e. argument, second the in sa lentv oainfracmlxcnuae(esalsa ewe and ¯ between swap shall (we conjugate complex a for notation alternative an is is real ..asaa rdc favco ihisl hudb oiie i.e. positive, be should itself with vector a of product scalar a i.e. , 0 ..teol etro eonr hudb h eovco,i.e. vector, zero the be should norm zero of vector only the i.e. , . . ecnaanso that show again can We i.e. , ,µ λ, h u h v ( u h λ sasaa utpeof multiple scalar a is ∈ , u · u v v C G v , 1 i = ( rpris(.a n 65)ipys-ald‘nilnaiy nthe in ‘anti-linearity’ so-called imply (6.5c) and (6.5a) Properties ij + k and λ = v X v = i µ =1 n k 1 u k 2 h |h + v h k 2 v e h ≡ u ) u k h u i i µ e 2 , u h h , + , v | hr h elpstv number positive real the where , i v u v e v v , v 2 ,µ λ, v j i , , ) 0 h , i| i and ti gi rethat true again is It k i i v v = = = clrpouto etrsaeoe h complex the over space vector a of product scalar A . v i v = i i = , i h λ λ ∈ ( 6 6 = = v v 0 = ∗ ∗ ,j i, λ h C h h i w , 0 h h h v v u > u ( v v k k . , 1 = λ 1 = , u u u , , ,

u 0 , c u k k k v u v ⇒ X j i 1 v . n , . . . , R 1 =1 + 1 n ..olydmpcma.k ihems2006 Michaelmas [email protected], i i 0 = + v i i n i ∗ ∗ ∗ k k + w + , , µ hsi o motn) ute,i elet we if Further, important). not is this v + , v j u . µ k µ e 2 µ ) = ∗ . j G ) h . ∗ h , i u ij 0 h u ∗ v , . 2 by v , , 2 u v i 2 i . i . ∗ k v k sthe is ordering norm ∗ freely). fthe of fthe of (6.8b) (6.5b) (6.5d) (6.10) (6.8a) (6.5a) (6.5e) (6.5c) { (6.9) (6.6) (6.7) e i } ,

This is a supervisor’s copy of the notes. It is not to be distributed to students. ahmtclTio:I ler emty(atI 89 I) (Part Geometry & Algebra IA Tripos: Mathematical Exercise. Remark. Let conjugates. Hermitian that have we hni em fti oaintesaa rdc 61)cnb rte as written be can (6.11) product scalar the notation this of terms in Then let and the of terms in product scalar the definition. a determines have (which to expression helps this simplify can We where w hn(.1,o qiaety(.5,rdcst c.(3.22)) (cf. to reduces (6.15), equivalently or (6.11), then (6.12), from Also, Properties. Example. before, as where, where Definition. rdc,nml 65) 65) 65)ad(6.5e). and (6.5d) (6.5c), (6.5a), namely product, eteclm arxo components, of matrix column the be G v † stemti,or matrix, the is fthe If ofimta h clrpoutgvnb 61b aifisterqie rpriso scalar a of properties required the satisfies (6.16b) by given product scalar the that Confirm eteajito h ounmatrix column the of adjoint the be a ij h { ∈ v e o matrices For , C i } The w sdfie obe to defined is , oma rhnra ai,ie r uhthat such are i.e. basis, orthonormal an form i = = = T emta conjugate Hermitian * X X i i eoe transpose. a denotes =1 =1 metric n n X X X j j i =1 =1 =1 n n n A ihentries with , v v v If and i i i ∗ ∗ e G w v i † j ij A ≡ B h , w e = j ealta ( that recall ( i v X . j , ∗ =1 n e ) a a h G T j A ( 11 21 v h w AB i w or † ij v = , G j A ( = v e = w , = ) †† ij i.e. , ojgt transpose conjugate a a j † w v . 12 22 i      h A 1 ∗ + i = e AB w T w w = = i = . . . ) v n 2 1 v , v ∗ †      ) † 2 ∗ e X i T w G ( = then =1 B sterwmatrix row the is n j A , † i =

v . . . c A ∗ v A = T , i ∗ † B ∗ ..olydmpcma.k ihems2006 Michaelmas [email protected], w . δ T ) T T A ij i A ∗ . † , T , n ∗ = = ec ( Hence . A rm(.c n (6.7) and (6.5c) from . or . a a 12 ∗ 11 ∗ adjoint AB a a 22 ∗ 21 ∗ ) T famatrix a of ∗ . = B T G ∗ A ij T ,btﬁs it ﬁrst but ), ∗ n so and , A = (6.16b) (6.13b) (6.14b) (6.16a) (6.13a) (6.14a) (6.12) (6.11) (6.15) { a ij } ,

This is a supervisor’s copy of the notes. It is not to be distributed to students. ahmtclTio:I ler emty(atI 90 I) (Part Geometry & Algebra IA Tripos: Mathematical Example. bases. of Change transformations. linear Real Example. of theory the numbers. extend can we Similarly, Maps Linear 6.3 and Remark. map linear a for through follows where range; and domain both of bases standard where under Then sadaoa arx n a btepaial o h etwy opoedwudb o[partially] obtain to to be would (6.18) proceed of to way) side best right-hand the the not emphatically out (but expand way One matrix. diagonal a is which under basis might we of invert) change to a straightforward are is they there (e.g. whether properties ask desirable have matrices diagonal Since as such Maps Thus allowed. also now are of components complex but before, if as transform components real Further C { n so and f n i v v } osdralna map linear a Consider . 0 ∈ osdrtemap the Consider map linear the Consider of el ..ti ilntb h aei etasomfo tnadbsst ae ossigof consisting bases to bases standard from transform we if case the be vectors. not complex will this e.g. real, ihrsett h tnadbasis standard the to respect with C T N b C n ij m a h aeeetas effect same the has hntecmoet of components the then , If . A ∈ ( N N ne hneo bases of change a Under A − N C 1 − T ili eea eacmlx( complex a be general in will b RA , sbfr hsdfie matrix, a defines this before As . sara iertasomto othat so transformation linear real a is 1 RA r eerdt as to referred are = ) 12 det 1 R ehv bevdta h tnadbasis, standard the that observed have We = = x y A : R T 7→ − − det 2 a N : a → det 22 1 sin R 21 ( T A : T x b y N n R C A θ 0 0 ) ( § nra etr.If vectors. real on ellna transformations linear real 2 → ij e a v nlna asadmtie ovco pcsoe complex over spaces vector to matrices and maps linear on 4 n − : j 22 a ossigo oainby rotation a of consisting = C a ( = → a ihrsett h tnadbasis standard the to respect with { 11 → R 12 2 ( 12 n e a m i codn to according C 12 e → + } e cos sin n let and , T i 0 b j 0 m ) N e cos to T a R e j C = Let . : m 22 2 a a θ θ v = m = sara arx Extend matrix. real a is C 11 11 { 0 N θ × n hence and e e i.e. , n = C A cos sin − − , i e N − } cos → n − { j Tv sin 1 a matrix. ) e 1 T and θ = = NA θ 22 i RA C θ

} . + e − c eteascae arxwt epc othe to respect with matrix associated the be θ X m { i sin N i =1 m eabssof basis a be N . a ..olydmpcma.k ihems2006 Michaelmas [email protected], a → { , sra,i sntncsaiytu that true necessarily not is it real, is 21 21 ij f θ N i ) } x } e y cos sin ij T b ihrsett bases to respect with , − i 0 hna before as then , to f = i a = a θ a θ of , 12 { T θ e R f C ( ;fo (4.23) from l; i . ( a } n { 12 12 θ 12 e C ) h rnfrainlw(4.70) law transformation the → i cos sin sin n } T sbt ai for basis a both is , x y C and θ θ θ oamap a to rnfr ocomponents to transform m + − + . . { a a a f 22 22 22 i } cos sin cos eabssof basis a be { θ θ θ e )) i } , of C R (6.17) (6.18) n n C N e and and m is .

This is a supervisor’s copy of the notes. It is not to be distributed to students. ahmtclTio:I ler emty(atI 91 I) (Part Geometry & Algebra IA Tripos: Mathematical Remark. Example. if i.e. conjugate, matrices. real for there Deﬁnition. Similarly, matrices matrices. symmetric real to to matrices are complex matrices for orthogonal equivalent what an matrices is complex to are matrices Unitary if i.e. inverse, Deﬁnition. oevr rm(.) 61)ad(.9 tflosta,a required, as that, follows it (6.19) and (6.18) (6.3), from Moreover, ni h eodpr fti ore(n lehr) o h iebigw tt htHermitian that that such matrices. state matrices Hermitian we (i.e. as being matrices this well time Hermitian as about skew-symmetric the matrices, more includes For unitary learn that elsewhere). class will all (and You a can . course basis. . as . diagonalized, this of be change of always a part can by matrices second diagonalized the be in can in matrices of classes what Remark. that (6.16b) using from note we where choose to is basis) orthonormal an in results Hence hnefo 46)i olw that follows it (4.61) from Thence smnindaoe ignlmtie aedsrbepoete.I steeoeueu oknow to useful therefore is It properties. desirable have matrices diagonal above, mentioned As h metric The oee,i snttu that true not is it however, i.e. R e ope qaematrix square complex A ope qaematrix square complex A sadaoa arxif matrix diagonal a is eko rm(.9)ta elrtto arcsaeotooa,i.e. orthogonal, are matrices rotation real that (4.59b) from know We h ( e e A 1 − , 1 e e RA G 1 i sHriinsnefo 65) 69 n (6.12) and (6.9) (6.5a), from since Hermitian is ) 1 = 21 , = R e ( h G e e † 1 = = ) det ij sin , e e e e 1 a = R e 2 1 2 U = A θ 12 R e A i G e T 0 ssi obe to said is iθ 0 = √ = ji ∗ a ssi obe to said is 1 i 1 = 11 2 2 A = ± , e R e 1 = U + i ia − h 0 T − A † 1 iθ e a 22 R e √ h † R i e = j 21 2 1 e e R = e = 2 , cos sin and 2 R e U omlmatrices normal † e . , , A † I − . i = nta ent rm(.1 that (6.21) from note we Instead . e − e θ θ i unitary = 1 . 1 2 ∗ e e a i R e . Hermitian i 2 21 R = e †

− 1 = c R = e − cos − h 1 = 1 sin = ..olydmpcma.k ihems2006 Michaelmas [email protected], i e √ . i 1 θ , ± fisHriincnuaei qa oits to equal is conjugate Hermitian its if I 2 , θ , . ia e i.e. j 11 − i 1 fi seult t w Hermitian own its to equal is it if − ..mtie uhthat such matrices i.e. , ovnetnraiain(that normalisation convenient A . i ( = 1 i h , G e e − R e i ) 1 ij , i sdiagonal: is e e . j i = δ ij . RR A T † = A = † R A − T (6.20b) (6.22b) (6.20a) (6.22a) = A R (6.25) (6.21) (6.23) (6.19) (6.24) and ) A A = I † ;

This is a supervisor’s copy of the notes. It is not to be distributed to students.