Geometry 2014: Notes and Exercises
Sophie M. Fosson [email protected]
June 5, 2014 ii
These notes follow my exercise classes in chronological order; • are a support to student practice; • do not cover the whole course; • include some theory reminders; • include the exercises proposed during my exercise classes; • include further material and exercises; • are sure to contain several mistakes; • shall be updated every two-three weeks; • are expected to be enhanced by student feedback (that is, if you find any • mistakes please tell me! by email).
Thank you, s.m.f. Contents
1 Week 1 1 1.1 Introduction to matrices ...... 1 1.2 Homogeneous linear systems ...... 2 1.3 Determinant ...... 4 1.4 Cramer’s rule ...... 4
2 Week 2 7 2.1 Matrix reduction and rank ...... 7 2.2 Linear systems ...... 8 2.3 Matrix inversion ...... 9
3 Week 3 11 3.1 Rn as vector space ...... 11 3.2 Bases and dimension ...... 12 3.3 Products between vectors ...... 13
4 Week 4 15 4.1 Vector spaces ...... 15 4.2 Back to RN : straight lines and planes ...... 17 4.3 Lines and planes in R3: Cartesian Equations ...... 17 4.4 When a plane is a subspace? ...... 18 4.5 Geometric description of linear systems’ solutions ...... 18 4.6 Geometric description of linear systems’ solutions ...... 19
5 Week 5 21 5.1 Linear mappings ...... 21 5.2 Matrix associated with a linear mapping ...... 23
6 Week 6 25 6.1 Endomorphisms ...... 25 6.2 Eigenvectors, eigenvalues ...... 25 6.3 Multiple choice quizzes ...... 26
7 Week 7 29 7.1 Extra material ...... 29 7.2 Eigenvalues and multiplicities ...... 29 7.3 Change of basis ...... 31
iii iv CONTENTS
8 Week 8 33 8.1 Simple endomorphisms and diagonalizability ...... 33
9 Week 9 39 9.1 Circles and Spheres ...... 39 9.1.1 Circles in R2 ...... 39 9.1.2 Spheres ...... 40 9.1.3 Circles in R3 ...... 41
10 Week 10 43 10.1 Conics ...... 43 10.2 Quadrics ...... 43 10.3 Reduction of conics to canonical forms ...... 44
11 Week 11 47 11.1 Quadrics ...... 47 11.2 Critical points for functions of two variables ...... 48 11.3 Mixing quadrics and functions of two variables ...... 49
12 Week 12 51 Chapter 1
Week 1
1.1 Introduction to matrices
A matrix is a table of numbers, called the entries of the matrix; in these notes will mainly consider matrices whose entries are real numbers. We write A Rm×n or A Rm,n to name a real matrix A, with m rows and n columns: ∈ ∈
A11 A12 A1n ······ A21 A22 A2n ······ .. . A31 A32 . . A3n Aij R, i = 1, . . . , m; j = 1, . . . , n (1.1) ∈ . . . . . ...... Am1 Am2 Amn ······ The sum between two matrices in A, B Rn,m is the matrix in C = A+B n,m ∈ ∈ R such that Ci,j = Ai,j + Bi,j, i = 1, . . . , m, i = 1, . . . , n. Two matrices must have the same dimensions m, n to be summed. Given a scalar α R, the product C = αA is given by Ci,j = αAi,j, i = 1, . . . , m, i = 1, . . .∈ , n. The product between two matrices is a bit more complex. Given A Rm×n and B Rn×p, the entry (i, j) of the product AB is defined as ∈ ∈ n X (AB)ij = aihbhj (1.2) h=1
Notice that the number of columns of A must be equal to the number of rows of B. Moreover, the product is not commutative!
Exercise 1 Let
1/2 0 7 3 1 1 1 1 − A = 2 1 2 B = 0 2 1 C = 2 D = 1 1 4 3 0 0 1 3/4 −6 − Evaluate: (a)A + 3B;
1 2 CHAPTER 1. WEEK 1
(b) B2; (c) BT B. (d) A(2C + D);
Solution (a) 13/2 3 4 − A + 3B = 2 7 5 1 4 0 (b) 3 0 0 3 1 1 9 3 3 T B B = 1 2 0 0 2 1 = 3 5 3 1 1 1 0 0 1 3 3 3 (c) 3 1 1 3 1 1 9 5 5 2 B = 0 2 1 0 2 1 = 0 4 3 0 0 1 0 0 1 0 0 1 (d) 2 1 3 2C + D = 4 + 1 = 3 3/2 −6 15/2 3 51 − 3 = 24 . 15/2 15/2 − Exercise 2 Given xy x2 A = y2 −xy − where x, y R not both null, show that A2 = 0. ∈ Solution
xy x2 xy x2 A2 = = y2 −xy y2 −xy − − (xy)2 x2y2 x3y + x3y 0 0 = = . xy3 −xy3 −x2y2 + (xy)2 0 0 − − This exercise has shown that the power of a nonnull matrix may be null.
1.2 Homogeneous linear systems
A linear system Ax = b, A Rm,n, x Rn = Rn,1, b Rm = Rm,1 is said to be homogeneous when b = 0.∈ Homogeneous∈ systems have∈ at least one solution, namely x = 0. Performing the product, we get m equations for n unknowns. Given the equations, one could solve the system as learned in high school. The problem is 1.2. HOMOGENEOUS LINEAR SYSTEMS 3 that we aim at working even with large m, n, which makes such computations not feasible. This is why we work on the matrix in order to reduce the problem. More precisely, we are allowed to transform the matrix switching rows and performing any linear combinations between rows. In other terms, if rj is the generic jth row of the matrix, we can do:
rj ri ↔ rj crj, c R (1.3) → ∈ rj rj + cri, c R → ∈ These operations can be used to reduce the matrix. A rigorous definition of reduction will be given next week, however the intuition is that we can transform A into a matrix Ae which has many zeros, which makes the linear system easier.
Exercise 3 Solve the homogeneous system Ax = 0 where
1 2 0 A = 1 1 3 . (1.4) 1 2 1
Solution T 3 First, notice that, even if not remarked, it is clear that x = (x1, x2, x3) R and 0 = (0, 0, 0)T R3. ∈ ∈ Let us now perform some suitable operations from (1.3). But actually which operations? A more rigorous procedure will be explained next week, here just do operations that let you obtain some more zeros in the matrix! For example:
1. r3 r3 r1: → − 1 2 0 1 1 3 (1.5) 0 0 1
2. r2 r2 3r1: → − 1 2 0 1 1 0 (1.6) 0 0 1
3. r1 r1 r2: → − 1 2 0 Ae = 0 1 0 (1.7) 0 0 1
Given that Ax = 0 is equivalent to Axe = 0, we solve the second one. We have: 1 2 0 x1 0 x1 + 2x2 = 0 x1 0 Axe = 0 0 1 0 x2 = 0 x2 = 0 x2 = 0 ⇔ ⇔ ⇒ 0 0 1 x3 0 x3 = 0 x3 0 (1.8) For exercise, one can try to solve Ax = 0 with no reduction. This is not impossible, but a little bit more tricky. 4 CHAPTER 1. WEEK 1 1.3 Determinant
The determinant is a number associated with each square matrix. It is use- ful to infer many properties of a matrix, and also to solve systems (see next paragraph). The determinant can be computed as follows. In R2,2: a b a b det = = ad bc (1.9) c d c d − In R3,3:
a b c e f d f d e d e f = a b + c (1.10) i j − h j h i h i j and then extend by induction to Rn,n with n 3. ≥ 1.4 Cramer’s rule
The Cramer’s rule suggests a method to solve linear systems by using determi- nants. n,n T Given a square matrix A R and the system Ax = b, b = (b1, . . . , bn) , T ∈ x = (x1, . . . , xn) , then
det Ai xi = i = 1, . . . , n. (1.11) det A where Ai is the matrix obtained from A by replacing its i-th column by the column vector b. Notice that we must have det A = 0. PROOF 6 First, let us recall the properties of the determinant: given a square matrix, if we perform the column transformation ci ci + αcj the determinant remains → the same; otherwise, if we transform ci αci, α R, then the determinant of the new matrix is the determinant of the→ old one∈ multiplied by α. Now, consider b1 a12 a13 . . . a1n b2 a22 a23 . . . a2n A = (1.12) 1 . . . . . . . . . . bn an2 an3 . . . ann Pn and noting that bi = j=1 aijxj, i = 1, . . . , n, we obtain that det A1 = x1 det A which proves the thesis for x1. Iterating the procedure for all the xi, the Cramer’s rule is proved.
Exercise 4 Given A of previous exercise, solve the linear system Ax = (0, 0, 1)T using the Cramer’s rule.
Solution 1.4. CRAMER’S RULE 5
1 2 0 1 3 1 3 1 1 3 = 1 2 + 0 = (1 6) 2(1 3) = 1. (1.13) 2 1 − 1 1 − − − − 1 2 1
0 2 0 0 3 0 1 3 = 2 = 6. (1.14) − 1 1 1 2 1
1 0 0
1 0 3 = 3 (1.15) − 1 1 1
1 2 0 1 0 1 0 1 1 0 = 1 2 + 0 = 1 2 = 1. (1.16) 2 1 − 1 1 − − 1 2 1 Then, x1 = 6, x2 = 3, x3 = 1. − 6 CHAPTER 1. WEEK 1 Chapter 2
Week 2
2.1 Matrix reduction and rank
Marker of a row = first nonnull entry of a row (from left to right) • Step-reduced matrix (or reduced or row-echelon): a matrix in which • 1. each column has at most one marker 2. “scanning” the matrix in the direction , markers “move” in the direction ↓ −→ 3. if present, null rows are at the bottom Super-reduced matrix (or reduced row-echelon): a step-reduced matrix • in which in which each marker is equal to one and is the only nonzero entry of its column Rank of a matrix: number of nonnull rows in its step-reduced form. • N,1 vectors v1, v2,..., vm R are said linearly independent if a1v1 + • ∈ a2v2 + . . . amvm = 0, ai R, i = 1, . . . , m implies ai 0, i = 1, . . . , m ∈ ∈ rank maximum number of linearly independent rows of a matrix • maximum⇔ number of linearly independent columns of a matrix ⇔ Exercise 5 Compute the rank of 3 1 0 0 A = 3 7 2 1 . (2.1) 2 16 −4 −2 3 − 3 − 3 Can you find 3 linearly independent (l.i.) columns? Why? More in general: can you find 4 l.i. vectors in R3? Solution Reduction
1. r3 3r3 → 3 1 0 0 A = 3 7 2 1 . (2.2) 6 16 −4 −2 − −
7 8 CHAPTER 2. WEEK 2
2. r3 r3 2r2 → − 3 1 0 0 A = 3 7 2 1 . (2.3) 0 2− 0− 0
3. r2 r2 2r1 → − 3 1 0 0 A = 0 6 2 1 . (2.4) 0 2− 0− 0
4. r3 3r3 r2 → − 3 1 0 0 A = 0 6 2 1 . (2.5) − − 0 0 2 1
This is the reduced matrix (markers are boxed). It has 3 non-null rows, then rank(A) = 3.
2.2 Linear systems
given A Rm,n, b Rm,1, Ax = b has solution x Rn,1 if • ∈ ∈ ∈ r(A) = r(A b). | r= rank (A b) = augmented matrix (add the column b to A) | If r(A) = r(A b) = ρ, the system has n−ρ solutions, that is, n ρ free variables | ∞ −
Exercise 6 Solve the system Ax = b where
1 0 1 A = 0 1 0 (2.6) 3 0 1 − and b = (1, 0, 0)T .
Solution Let us consider b = (1, 0, 0)T and the enlarged matrix
1 0 1 1 0 1 0 0 (2.7) 3 0 1 0 − and let us reduce. It is sufficient to perform r3 r3 3r1 to have the row- echelon form: → − 1 0 1 1 0 1 0 0 (2.8) 0 0 4 3 − − 3 1 Then,x31 = 4 , x21 = 0, and x11 = 4 . 2.3. MATRIX INVERSION 9 2.3 Matrix inversion
inverse of a square matrix A: matrix X such that AX = I (I= identity • matrix). We write X = A−1. A square matrix A Rn,n admits inverse detA = 0 r(A) = n (is maximum) ∈ ⇔ 6 ⇔ Exercise 7 Consider the matrix 1 0 1 A = 0 1 0 . (2.9) 3 0 1 − Does A admit inverse (X)? If yes, compute it. Solution
A admits inverse as its determinant is not null. We compute the inverse. Method 1: solving a linear system for each column. Let us consider T T b = (1, 0, 0) and solve the system Ax = b where x = (x11, x21, x31) is the first column of the inverse. 3 1 Then,x31 = 4 , x21 = 0, and x11 = 4 (see previous exercise). Repeat the same for b = (0, 1, 0)T and b = (0, 0, 1)T . Method 2: super-reduction Given that A−1 exists, we have tried to −1 −1 compute it by solving AX = In. As AA = A A = In, let us multiply both −1 −1 −1 −1 −1 side of AX = In by A : A AX = A In, then InX = A , then X = A . −1 Therefore, if super-reduce (A In), that is, we obtain (In B), then B = A . In our case: | | 1 0 1 1 0 0 0 1 0 0 1 0 (2.10) 3 0 1 0 0 1 − By r3 r3 2r1 → − 1 0 1 1 0 0 0 1 0 0 1 0 (2.11) 0 0 4 3 0 1 − − In order to super-reduce, we need to nullify the entry in position (1, 3), and we can do it by r1 4r1 + r3: → 4 0 0 1 0 1 0 1 0 0 1 0 (2.12) 0 0 4 3 0 1. − − Finally, r1 r1/4, r1 r1/4: → → − 1 0 0 1/4 0 1/4 0 1 0 0 1 0 (2.13) 0 0 1 3/4 0 1/4. − In conclusion, 1/4 0 1/4 −1 A = 0 1 0 (2.14) 3/4 0 1/4. − 10 CHAPTER 2. WEEK 2
Method 3: Algebraic complements If you like determinants, there is a nice method to compute the inverse: it has been proved that 1 A−1 = CT det(A) where C is a N N matrix whose entries Cij are obtained through three steps: (a) we eliminate× row i and column j in A; (b) we compute the determinant of the so-obtained matrix; (c) we multiply the result by ( 1)i+j. − 2 1 0 We have, C11 = ( 1) det = 1, C12 = 0, C13 = 3, C21 = 0, − 0 1 − − − C22 = 4, C23 = 0, C31 = 1, C32 = 0, C33 = 1. Moreover,− det(A) = 4.− The reader can now write the final solution and check whether it matches− to ones obtained with the other methods. Remark: what about if det(A) = 0? This method seems to be not feasible, because we have to divide by 0 ... this is not really true: if det(A) = 0, A is not invertible so we simply do not compute the inverse!
Exercise 8 1 2 0 A = α β γ . (2.15) 1 2 0 For which α, β, γ R the solution of system Ax = 0 R3,1 has 2 free variables? ∈ ∈ Solution
Not using the rank: x1 + 2x2 = 0 αx1 + βx2 + γx3 = 0 ( 2α + β)x2 + γx3 = 0 −2α = β γ = 0 Using− the rank: we want r(A) = 1 then (α, β, γ) = µ(1, 2, 0), for some µ R. ∈ Chapter 3
Week 3
3.1 Rn as vector space
n A set U = u1, u2,... uk R is said to be linearly independent (l.i.) if • { } ∈ all its elements u1, u2,..., uk are l.i.
n Given any finite subset V = v1,..., vk R , (V ) = (v1,..., vk) is • { } ⊂ L L the set of all the possible linear combinations of v1,..., vk (we can say: set generated by v1,..., vk or span of v1,..., vk)
Subspace of Rn: a subset in V Rn which is closed under the operations • of ⊂
1. sum between vectors: v1, v2 V = v1 + v2 V ∈ ⇒ ∈ 2. product between a vector and a scalar: v1 V, c R = cv1 V ∈ ∈ ⇒ ∈ The null vector (0, 0,..., 0) always belongs to subspaces.
Exercise 9 For each of the following subsets of R4, determine whether the sub- set is a subspace: 4 (a) A = (x1, x2, x3, x4) R such that x1 = x2 . { ∈ 4 } (b) B = (x1, x2, x3, x4) R such that x1 = x2 and x1 + x2 + x3 + x4 = 0 . { ∈ R4 2 } (c) C = (x1, x2, x3, x4) such that x1 = x2 . { ∈ 4 } (d) D = (x1, x2, x3, x4) R such that x1 + x2 + x3 + x4 0 . { ∈ ≤ }
Solution (a) The set can be rewritten as A = x = (x1, x1, x3, x4), x1, x3, x4 R . Natu- { ∈ } rally, (0, 0, 0, 0) A. Then, if x, y A, then x+y = (x1+y1, x1+y1, x3+y3, x4+ ∈ ∈ y4) A since the first two components are still equal (which actually is the re- ∈ quired condition to be a vector of A). Finally, also cx = (cx1, cx1, cx3, cx4, ) A for any c R. Hence, A is a subspace. ∈ ∈ (b) B is a subset of A and we can write A = x = (x1, x1, x3, x3 2x1), x1, x3 { − − ∈ R . Given x, y B, then x + y = z B since z1 = z2 and z1 + z2 + z3 + z4 = } ∈ ∈ x1+y1+x2+y2+x3+y3+x4+y4 = 0; finally, cx B since cx1+cx2+cx3+cx4 = ∈ c(x1 + x2 + x3 + x4) = 0. B is a subspace.
11 12 CHAPTER 3. WEEK 3
(c) C is not a subspace since it is not closed with respect to both sum and 2 2 product. For instance, if x, y E and x + y = z, then z1 = x1 + y1 = x2 + y2 = 2 2 ∈ 6 z2 = (x2 + y2) . (d) D is not a subspace: for instance, consider ( 1, 0, 0, 0) D. Then, 1( 1, 0, 0, 0) = (1, 0, 0, 0) / D. − ∈ − − ∈
3.2 Bases and dimension
Let V be a subspace of Rn.A basis of V is a l.i. set B that generates V , • that is, V = (B). L n canonical basis of R is E = e1,..., en where ei is the vector with all • the entries equal to zero, except{ the i entry} which is 1. given a subspace, there are infinite possible bases. But all the bases have • the same number of elements. Such number is called the dimension of the subspace. Merging the two previous facts, we can say the the dimension of Rn is n. • If a subspace V has dimension d, any l.i. subset B V with d elements • is a basis for V ⊂
Exercise 10 Find a basis for subspace A of the previous exercise.
Solution Non-smart method: let us construct a basis. We choose arbitrarily a first vec- tor, for instance, let B = (1, 1, 0, 0) in A. This is not sufficient as a basis for A, as the vectors of A{are not all} multiple of (1, 1, 0, 0). Let us add on other vector such that B is linearly independent: for instance, (1, 1, 1, 1), that is B = (1, 1, 0, 0), (1, 1, 1, 1) . B is not yet a basis, since for instance (1, 1, 2, 3) cannot{ be written as a linear} combination of elements of B. Let us add an other element, for instance B = (1, 1, 0, 0), (1, 1, 1, 1), (1, 1, 1, 0) : now B still is linearly independent and it is also{ a basis since B = A. } L Smart method: first, find the dimension. The vectors of A have the first component equal to the second, then only three components are free. Hence, A has dimension 3. Now, it is sufficient to pick three vectors linearly independent of A to obtain a basis.
Exercise 11 Which of the following is a basis for R3? A = (1, 2, 0), (0, 0, 2) B = {A (1, 2, π) } C = A ∪ {(0, 1, 0),}(1, π, 0) D = A ∪ {(1, π, 0) } ∪ { } Solution R3 has dimension 3, so we immediately discard A and C which respectively have 2 and 4 elements. The elements of C are linearly dependent, then C is not a basis. The elements of D are linearly independent, then D is a basis. We report the properties of the four subsets in the following table. 3.3. PRODUCTS BETWEEN VECTORS 13
span l.i. basis A NO YES NO B NO NO NO C YES NO NO D YES YES YES
3.3 Products between vectors
Scalar product (= dot product) • n n X u = (u1, . . . , un), v = (v1, . . . , vn) R u v = u, v = uivi ∈ · h i i=1
u, v are said to be orthogonal if u v = 0 • ·
in R3 we define also the vector product (= cross product) •
i j k
u = (u1, u2, u3), v = (v1, v2, v3) u v = u1 u2 u3 × v1 v2 v3
(this is a “symbolic” expression!)
Exercise 12 Given u = ( 3, 1, 1) and v = (1, 1/3, 1/3) compute (a) u v − − − (b) u · v × (c) u u (v 1/3−→j ) · × − Solution (a) u v = 3 1/3 1/3 = 11/3. (b) u v = (0, 0, 0). Notice that u = 3v. (c) Non-smart· − − method:− u (v− 1/3j) = u ×(1, 0, 1/3) = (1/3, 0, 1) and finally− u (1/3, 0, 1) = 0. × − × − (c)· Smart method: no matter which is the w, u u w = 0 as u w is known to be orthogonal to u (and also to w). · × × 14 CHAPTER 3. WEEK 3 Chapter 4
Week 4
4.1 Vector spaces
We know what is a subspace of Rn. We extend this idea to other objects (outside Rn).
A vector space is any set which is closed with respect to two operations, that we call “sum” and “product”, that have the “usual” properties of sum and product.
From now on, we will call vectors the elements of a vector space (physical interpretation not required: in linear algebra, arrays of numbers, polynomials, functions may be vectors) Formally: a vector space is a set V associated with a field K such that we can define
1. an operation, that we call sum and we indicate by “+”, between two elements of V with the following properties:
(a) closure: for any u, v V , u + v V ∈ ∈ (b) commutative and associative properties; existence of an additive iden- tity (an element 0 V such that v + 0 = v); ∈ 2. an operation, that we call product, between an element of V and an ele- ment of K with the following properties:
(a) closure: for any u V , a K au V ∈ ∈ ∈ (b) commutative, associative, and distributive properties; existence of a product identity (1 K such that 1u = u for any u V ). ∈ ∈ Exercise 13 Determine whether of the following sets of functions is a vector space on the field R:
(a) R≤3[x] = all the polynomials in x of degree 3 . { ≤ }
(b) R=3[x] = all the polynomials in x of degree = 3 . { } 15 16 CHAPTER 4. WEEK 4
(c) V = all the functions f 2[0, 1] such that f 00(x) = x2f(x) . { ∈ C } (d) V = all the functions f 1[0, 1] such that f 0(x) = f 2(x) . { ∈ C } (e) V = all the polynomials in x having a zero at x = 5 . { } 2 3 (a) A generic element of R≤3[x] has the form a0 +a1x+a2x +a3x , ai R, i = 0, 1, 2, 3. ∈
Sum: we use the usual sum for polynomials. For any ai, bi R, i = • 2 3 2 3 ∈ 0, 1, 2, 3, a0 + a1x + a2x + a3x + b0 + b1x + b2x + b3x = (a0 + b0) + 2 3 (a1 + b1)x + (a2 + b2)x + (a3 + b3)x R≤3[x] closure w.r.t. sum; ∈ ⇒ Null element: P (x) 0 R≤3[x] is the additive identity for the sum; • ≡ ∈ 2 3 2 3 Product: λ R, λ(a0 +a1x+a2x +a3x ) = λa0 +λa1x+λa2x +λa3x • ∈ ∈ R≤3[x] closure w.r.t. product. ⇒ This is enough to state that R≤3[x] is a vector space. (b) P (x) 0 / R=3[x]: this is enough to conclude that R=3[x] is not a ≡ ∈ 3 3 2 vector space. If you want on other counterexample: x R=3[x], x + x ∈ − ∈ R=3[x], but their sum / R=3[x]. (c) ∈ 00 0 f0(x) 0 V since trivially f (x) = f (x) = f0(x) = 0 • ≡ ∈ 0 0 If f(x), g(x) V , then f(x) + g(x) = h(x) V by the linearity of the • derivative operation:∈ h00(x) = f 00(x) + g00(x) =∈ x2(f(x) + g(x)) = x2h(x). Analogously for the product: g(x) = cf(x), c R, then g00(x) = cf 00(x) = • cx2f(x) = x2g(x). ∈ Hence, V is a vector space. (d) V contains the null element f0(x) 0, but it is not closed with respect to the sum: given f(x), g(x) V , f(x) + g(x≡) = h(x) then h0(x) = f 0(x) + g0(x) = f 2(x) + g2(x) = h2(x) = f∈2(x) + g2(x) + 2f(x)g(x). 6
(e) P0(x) 0 V since trivially P0(0) = 0. Then, let us sum two vectors ≡ ∈ of C, say two polynomials P1(x) and P2(x) such that P1(5) = P2(5) = 0: we obtain P1(x) + P2(x) = P3(x) V as P3(5) = P1(5) + P2(5) = 0. Finally, ∈ let us check the product: given c R,P1(x) V , cP1(x) = Q(x) V , as ∈ ∈ ∈ Q(5) = cP1(5) = 0. V is a vector space.
Exercise 14 Given A = 1, x, x2, x3 x , B = x, x2, x3 x , C = 1, x, x2, x3 {− − } { − } {− − x, x + 3 , which of them span R≤3[x]? − } Method 1: use the definition of span A spans R≤3[x] if any vector R≤3[x] can be written as linear combination of 2 3 vectors in A. That is, given a generic a0 + a1x + a2x + a3x , there exist bi R, 2 3 ∈ i = 0, 1, 2, 3 such that I can write it as b0( 1) + b1x + b2x + b3(x x)? We recall that two polynomials are equal− when the coefficients of same− degree are equal, then
2 3 2 3 a0 + a1x + a2x + a3x = b0( 1) + b1x + b2x + b3(x x) − − 4.2. BACK TO RN : STRAIGHT LINES AND PLANES 17
corresponds to the linear system (in the unknowns bi, i = 0, 1, 2, 3) a0 = b0 − a1 = b1 b3 − a2 = b2 a3 = b3
which has solution b0 = a0, b1 = a1 + a3, b2 = a2, b3 = a3. − As the system has solution, A spans R≤3[x]. B does not span R≤3[x] as P (x) 1 cannot be written as linear combination of vectors of G. ≡
Method 2: use dimensions and bases The concepts of bases and dimensions defined for Rn can be extended to any vector spaces. In particular,
dim (R≤m[x]) = m + 1 2 m and the canonical basis for R≤m[x] is 1, x, x , . . . , x . { } Then, we can immediately conclude that B does not span R≤3[x] as it has only 3 elements. On the other hand, A is a basis since it has 4 elements and it is linearly 2 3 independent (a1( 1) + a2x + a3x + a4(x x) = 0 if and only if a1 = a2 = − − a3 = a4 = 0). Finally, C is not a basis (as it contains 5 elements), but as C = A x+3 ∪{− } it spans R≤3[x].
4.2 Back to RN : straight lines and planes
Straight line passing through a point P = (p1, . . . , pN ) and parallel to (u), L u = (u1, . . . , uN ): lP,u = P + tu, t R ∈
A line can be determined by a couple of points P0,P1: lP ,P = P0 + t(P1 0 1 { − P0)), t R ∈ } Plane passing through a point P RN and oriented according to (u, v), u, v RN : ∈ L ∈ πP,u,v = P + λu + µv, λ, µ R ∈
A plane can be determined by a triple of (non collinear) points P0,P1.P2: πP ,P ,P = P0 + λ(P1 P0)) + µ(P2 P0), λ, µ R . 0 1 2 { − − ∈ } 4.3 Lines and planes in R3: Cartesian Equations
3 Straight line in R :(x, y, z) = (p1, p2, p3) + (u1, u2, u3)t
x−p1 y = p2 + u2 u1 y−p2 z = p3 + u3 u2 18 CHAPTER 4. WEEK 4
a system of two linear equations in x, y, z (= intersection between two planes).⇒
Plane in R3:(x, y, z) = P + λu + µv ⇒ (x, y, z) P0 u v = 0 − · × i
⇒ ax + by + cx + d = 0
(a, b, c) u v || × 4.4 When a plane is a subspace?
Quiz 1 A plane ax + by + cz + d = 0 in R3 (a) is a subspace of dimension 2 if it contains (0, 0, 0) (b) is a subspace only if it is parallel to the coordinate plane z = 0 (c) never is a vector space (d) is a subspace of dimension 3 Solution (a) is the right sentence. The parametric equation of a plane passing through (0, 0, 0) can be written as λu + µv for suitable u, v R3 and any λ, µ R. But then the plane is given by (u, v), which is a subspace∈ of R3. The dimension∈ is then 2. It follows that allL the other sentences are false.
4.5 Geometric description of linear systems’ so- lutions
Quiz 2 Let us consider the system Ax = 0 where 1 3 1 A = 2 6 0 . 0 0 1 Its solution set is (a) a straight line (b) does not exist since the matrix is not reduced (c) the union of two parallel planes (d) is not a vector space 1 Solution Let us reduce by performing the operations r2 2 r2, r1 r1 r2 r3 (where means “changes into”) and then switching suitably→ the obtained→ − rows.− We obtain:→ 1 3 0 0 0 1 . 0 0 0 Therefore the solution set is v = ( 3a, a, 0) a R , and ( 3a, a, 0) = a( 3, 1, 0) is a straight line (passing{ through− (0, 0, 0)).∈ In particular,} − the solution set− is the subspace of dimension 1 ( 3, 1, 0) . (a) is the right sentence. L{ − } 4.6. GEOMETRIC DESCRIPTION OF LINEAR SYSTEMS’ SOLUTIONS19 4.6 Geometric description of linear systems’ so- lutions
Let us consider the system Ax = b where
1 3 1 A = 2 6 0 . 0 0 1
and b = (α, 0, α), α R, α = 0. Can you geometrically∈ describe6 its solution set? Which is the position of the solution set with respect to the solution of the homogeneous system Ax = 0? Solution
The reduced augmented matrix is
1 3 0 0 0 0 1 α . 0 0 0 0
Therefore the solution set is v = ( 3a, a, α) a R . This is a straight line, but not a subspace, as it does not{ contain− (0, 0, 0)∈ (assumed} that α = 0. This straight line is parallel to the solution set of Ax = 0, as ( 6 3a, a, α) = a( 3, 1, 0) + (0, 0, α). − − 20 CHAPTER 4. WEEK 4 Chapter 5
Week 5
5.1 Linear mappings
Given two vector spaces U and V on a field K, a map f : U V is a • linear mapping if for any α, β K and for any u, w U, → ∈ ∈ f(αu + βw) = αf(u) + βf(w).
Kernel of a l.m.: kerf = u U : f(u) = 0V where 0V is the null element • of V ; { ∈ } Image of a l.m.: Imf = v V such that there exists u U : f(u) = v . • { ∈ ∈ } Theorem: ker and Im are subspaces of U and V , respectively, and • dim(kerf) + dim(Imf) = dimU.
A linear mapping is determined when it is defined over the elements of a basis of U, that is, when we know the images through f of the elements of a basis.
Exercise 15 Find the linear mapping f : R3 R such that f(1, 2, 1) = f(0, 0, 1) = 0 and f(0, 1, 0) = 1. → Find kerf and Imf.
Solution First of all, we notice that B = b1, b2, b3 , where b1 = (1, 2, 1), b2 = (0, 0, 1), 3{ } b3 = (0, 1, 0), is a basis for R . Our goal is to write explicitly f(u) = ... for any u R3. As B is a basis, we know that there exist α, β, γ R such that ∈ ∈ u = αb1 + βb2 + γb3, and in particular it is easy to compute that α = u1, β = u3 u1, γ = u2 2u1. Now using the linearity: − − f(u) = f(αb1 + βb2 + γb3) = αf(b1) + βf(b2) + γf(b3)
= u1f(b1) + (u3 u1)f(b2) + (u2 2u1)f(b3) = u2 2u1. − − −
In conclusion, f(u) = u2 2u1. dim(kerf) + dim(Imf−) = dim(U) = 3. Now, Imf R, then its dimension can be 0 or 1. If it were equal to zero, Imf = 0 ,⊆ but this is not the case { } 21 22 CHAPTER 5. WEEK 5 since 1 belongs to it by the hypotheses of the exercise. Then dim(Imf) = 1, from which we conclude that dim(kerf) = 2. In this case the dimensions also determine the subspaces themselves. In fact, Imf = R (as for any U subspace of W , if dimU = dimW , then U = W ), and as b1, b2 kerf and b1 and b2 { } ⊂ are linearly independent, then kerf = b1, b2 . L{ } Quiz 3 Which of the following f : U V is not a linear mapping? → 3 3 (a) U = V = R , u = (u1, u2, u3) R , f(u) = (u3, u2, u1); ∀ ∈ (b) U = C[0, 1], V = R, g U, f(g) = g(0); ∀ ∈ 2 2 x 2x (c) U = R , V = C[0, 1], u R f(u) = u1e + u2e ; ∀ ∈ 2 2 x (d) U = R , V = C[0, 1], u = (u1, u2) R f(u) = u1u2e . ∀ ∈ Solution
(a) u, v R3, α, β R, ∀ ∈ ∈
f(αu + βv) = f(αu1 + βv1, αu2 + βv2, αu3 + βv3)
= (αu3 + βv3, αu2 + βv2, αu1 + βv1) = αf(u) + βf(v) l.m. ⇒ (b) g, h U, α, β R, ∀ ∈ ∈ f(αg + βh) = (αg + βh)(0) = αg(0) + βh(0) =αf(g) + βf(h) l.m. ⇒ (c) u, v R2, α, β R, ∀ ∈ ∈ x 2x f(αu + βv) = (αu1 + βv1)e + (αu2 + βv2)e x 2x x 2x = αu1e + αu2e + βv1e + βv2e = αf(u) + βf(v) l.m. ⇒ (d) u, v R2, α, β R, ∀ ∈ ∈ x f(αu + βv) = (αu1 + βv1)(αu2 + βv2)e x αf(u) + βf(v)(αu1u2βv1v2)e not l.m. ⇒ A linear mapping f : U V is said to be injective when f(u) = f(v), (u, v U), implies u = v, and is said→ to be surjective if Imf = V . ∈ It is easy to prove that a linear mapping is injective kerf = 0 dim(kerf) = 0. ⇔ { } ⇔
3 Quiz 4 Let f : R≤2[x] R a linear mapping such that f(1) = 0, f(19x) = 0, f(2x2 x) = 0. What is→ true? 6 6 (a)−f is6 necessarily injective; (b) f is necessarily surjective; (c) dim(Imf) 1 necessarily; (d) dim(kerf)≥ = 2 necessarily.
Solution Notice that f is not uniquely defined by the given properties. Then, “nec- essarily” can be read as “holds for any f with these properties”. 5.2. MATRIX ASSOCIATED WITH A LINEAR MAPPING 23
We know that dimR≤2(x) = 3, then dim(ker(f)) + dim(Im(f)) = 3. The possible combinations are: 0+3; 1+2; 2+1; 3+0. Since f(1) = 0 we know that ker(f) = R≤2(x), hence dim(kerf) < 3 (remind that if S is a subspace6 of V and dim(S6 ) = dim(V ), then S = V !). (c) is true since dim(ker(f)) 0, 1, 2 , then dim(Imf) 3, 2, 1 . We can easily∈ construct { } a counterexample∈ of { f with} ker of dimension 1, so that (a), (b), (d) turn out to be false (notice that if the kernel has dimension 1, the image has dimension 2, which means that the l.m. is not surjective since Imf = R3.) For6 example, let kerf = x2 , f(1) = (0, 0, 1), and f(x) = (1, 0, 0) defines a l.m. that fulfills the given propertiesL{ } and has kernel of dimension 1. Of course you can construct infinitely many other counterexamples.
5.2 Matrix associated with a linear mapping
If B = b1,..., bm is a basis and u = λ1b1 + + λmbm, we indicate • { } ··· by( u)B = (λ1, . . . , λm) the (unique) representation of u with respect to B. Order of the elements of the basis is now important
f : U V linear mapping, B = b1, b2,..., bm basis of U, C basis of V • → { } C,B Matrix Mf associated with f w. r. t. the bases B and C: . . . . . . . . C,B Mf = (f(b1))C (f(b2))C (f(bm))C . . ···. . . . . .
C,B Mf has dim(V ) rows and dim(U) = m columns.
Exercise 16 Let V = R≤2[x] be the set of all the polynomials in x, with coef- ficients in R, and with degree 2. ≤ 1. Prove that B = 1, x, x2 5 is a basis for V . { − } 2. Extract from (1, 0), (1, 1), (4, 1), (0, 1) a basis C of W = R2. { } 2 3. Prove that f : V W defined by f(a0 + a1x + a2x ) = (a0 + a1, a1 a2) is a linear mapping→ −
4. Find the matrix associated with f w. r. t. the bases B and C. 5. Discuss kernel, image, injectivity, surjectivity of f.
Solution
1. B is a basis as its cardinality (= number of elements) matches with the dimension of V . Moreover, B is linearly independent. 2. Any choice of two linearly independent vectors in the set is a basis. For example, C = (1, 0), (1, 1) { } 24 CHAPTER 5. WEEK 5
3. Intuitively: (a0 + a1, a1 a2) is a linear combination of the coefficients of the polynomial, so the mapping− is linear. 2 2 Formally: let P (x) = p0 + p1x + p2x , Q(x) = q0 + q1x + q2x two poly- nomials in V . Then, for any α, β R: ∈ 2 f(αP + βQ) = f(αp0 + βq0 + (αp1 + βq1)x + (αp2 + βq2)x )
= (αp0 + βq0 + αp1 + βq1, αp1 + βq1 αp2 βq2) − − = α(p0 + p1, p1 p2) + β(q0 + q1, q1 q2) − − = αf(P ) + βf(Q).
2 4. Let us call b1 = 1, b2 = x, b3 = x 5; c1 = (1, 0), c2 = (1, 1). −
f(b1) = (1, 0) = c1
then f(b1) = λ1c1 + λ2c2 if and only if (λ1, λ2) = (1, 0)
thus (f(b1))C = (λ1, λ2) = (1, 0)
f(b2) = (1, 1) = c2
then f(b1) = λ1c1 + λ2c2 if and only if (λ1, λ2) = (0, 1)
thus (f(b1))C = (λ1, λ2) = (0, 1)
f(b3) = ( 5, 1) − − then f(b1) = λ1c1 + λ2c2 = (λ1 + λ2, λ2) if and only if
λ2 = 1, λ1 = 5 λ1 = 4 − − − − thus (f(b1))C = (λ1, λ2) = ( 4, 1) − − In conclusion:
1 0 4 M C,B = . f 0 1 −1 − 2 5. From its definition, f(a0 + a1x + a2x ) = (0, 0) if and only if (a0 + a1, a1 − a2) = (0, 0), that is, (a0, a1, a2) = (α, α, α), α R. − − ∈ Then kerf = α(1 x x2), α R = 1 x x2 . Then, dimkerf = 1, and f is not injective.{ − − ∈ } L{ − − } As dim(kerf) + dim(Imf) = 3, dim(Imf) = 2, then we also conclude that Imf = R2, namely f is surjective. C,B Other method: use the fact that rank (Mf = dim(Imf), for any B, f, C (this follows from the fact Imf = f(b1), . . . , f(bm) ) L{ } Chapter 6
Week 6
6.1 Endomorphisms
1. A linear mapping f from a vector space V to itself (f : V V ) is also called endomorphism →
2. Given a v.s. V of dimension n and a basis B = (b1,..., bn), for any v V ∈ there exists a unique sequence of scalar coefficients α1, . . . , αn such that Pn v = i=1 αibi.
3.( v)B := (α1, . . . , αn) . . . . B,B 4. Mf = f(b1) B f(bn) B is a square matrix. . ··· . . .
endomorphisms square matrices ⇔
5. if V = Rn and B is the canonical basis, for any v V , ∈ B,B f(v) = Mf v
6.2 Eigenvectors, eigenvalues
6. Given A Rn×n, a nonnull v Rn is eigenvector of A if there exists λ R such∈ that Av = λv ∈ ∈ 7. Such λ is called eigenvalue of A.
8. How to compute the eigenvalues? By solving the equation
det(A λI) = 0 − (det(A λI) = characteristic polynomial). −
25 26 CHAPTER 6. WEEK 6
9. If you know the eigenvalues, you can find the eigenvectors by solving the system (A λI)v = 0 for each value of λ. −
10. Eigenspaces = vector spaces generated by eigenvectors.
6.3 Multiple choice quizzes
Quiz 5 The matrix
1 1 1 0 2 0 3− 0 1 − (a) has no eigenvectors since it is square (b) has only one eigenvalue λ = 2 (c) has two complex eigenvalues − (d) has only real eigenvectors
Solution of Quiz 5 (d) Eigenvalues:
1 λ 1 1 − 2 det 0 2 λ 0 = (2 + λ)(4 λ ) = 0 3− 0− 1 λ − − − λ1 = 2, λ2 = 2 ⇒ −
Eigenvectors for eigenvalue λ1 = 2:
1 1 1 − T T 0 4 0 v = 0 v = (α, 0, α) = α(1, 0, 1) , α R 3− 0 3 ⇒ ∈ − Eigenspace: (1, 0, 1)T . L{ } Eigenvectors for eigenvalue λ = 2: − 3 1 1 0 0 0 v = 0 v = (α, 0, 3α) = α(1, 0, 3), α R 3 0 1 ⇒ − − ∈ Eigenspace: (1, 0, 3)T . L{ − } Quiz 6 The matrix 1 0 0 4 k 0 0 0 1 (a) if k = 1 has null eigenvectors (b) if k = 1 has an eigenspace which is a plane (c) if k = 0 the associated endomorphism is surjective (d) if k = 0 the associated endomorphism is not linear 6.3. MULTIPLE CHOICE QUIZZES 27
Solution of Quiz 6 (b) If k = 1, the characteristic polynomial is (1 λ)3, then λ = 1 and µ = 3. The associated eigenvector is given by (0, α, β), −α, β R and since (0, α, β) = α(0, 1, 0)+β(0, 0, 1), then the eigenspace is (0, 1, 0),∈(0, 0, 1) . But a subspace of dimension 2 in R3 is a plane. In fact, theL{ parametric formula} of a plane is (x, y, z) = P + λQ + µR where P, Q, R R3 and λ, µ R and if it contains (0, 0, 0), it is a vector space (generated by∈Q and R). ∈ In order to check (c), we first remind that For any endomorphism f, the following are equivalent: 1. f is injective; 2. f is surjective; 3. f is bijective 4. kerf = 0 . { } One can verify that (c) is false observing that if k = 0 the matrix has rank 2 (non maximal), while a an endomorphism is bijective only if the associated matrix has maximal rank.
Quiz 7 Let f be a non surjective endomorphism of R3 having 2 and 4 among its eigenvalues. Then (a) f does not have other eigenvalues (b) there exist two linearly independent eigenvectors for the eigenvalue 2 (c) f has only one eigenspace (d) f has 3 distinct eigenvalues
Solution of Quiz 7 (d) If an endomorphism is non surjective, it always has the eigenvalue 0. In fact, we know that (see the solution of the previous quiz), if an endomor- phism f is not surjective, then its kernel contains something more than 0. Let us suppose that v = 0 is in the kernel of f. Then f(v) = 0, or also f(v) = 0v. Then v is an eigenvector6 for the eigenvalue 0! In conclusion we have 3 different eigenvalues λ1 = 0, λ2 = 2, λ3 = 4, (a) is then clearly false; to prove that (b) is false the best method is the analysis of the eigenvalues’ multiplicities (not yet introduced in these notes). To prove that (c) is false, let us construct a counterexample. For instance, suppose that f has as eigenvectors the vectors of the canonical basis: is this con- sistent? Yes, and the columns of matrix that defines f will be the eigenvectors multiplied by the eigenvalues (details are left to the reader).
Quiz 8 Let f be an endomorphism of R2 such that (1) u = (1, 2) is an eigen- vector of the eigenvalue 2 and (2) f(v) = u, where v = ( 1, 3) − 3 3 (a) M E,E = 1 is the matrix of f w. r. t. the canonical basis f 5 8 6 E = e1, e2 = (1, 0), (0, 1) { } { } 1 3 (b) M E,E = 1 is the matrix of f w. r. t. E f 5 8 6
2 1 (c) M B,B = is the matrix of f w. r. t. the basis B = u, v f 0 0 { } 28 CHAPTER 6. WEEK 6
2 1 (d) M B,B = is the matrix of f w. r. t. B = u, f 2(v) (f 2(u) = f 0 0 { } f(f(u))) Solution 8 (c) 1 0 Let us compute M E,E = f f . f 0 1 1 1 0 1 f = f + 2f 2 2 0 1 2 1 1 0 1 f − = f + 3f 3 − 0 1 2 Combining these equations we obtain 1 1 4 0 1 3 f = f = 0 5 8 1 5 6 and 1 4 3 M E,E = . f 5 8 6 Then (a) and (b) are false. (d) is false since f 2(u) = f(f(u)) = 2u, then u and f 2(u) are linearly dependent and do not form a basis. B,B To verify that (c) is true, let us compute Mf 1 1 1 f = 2 + 0 2 2 −3
1 1 1 f = + 0 −3 2 −3 Then the entries of the first column are 2 and 0, and the entries of the second column are 1 and 0. Quiz 9 Let f be the endomorphism of R3 defined as f(x, y, z) = (x + y + z, x + y + z, x + y + z): (a) ker(f) = (α, β, α β), α, β R (b) ker(f) = { (0, 0, −1), (0, 1, ∈1) } (c) f(x, y, z)L{ = (0, 0, 0)− if and only− } if x + y z = 0 (d)(1, 0, 1) and (0, 1, 1) are linearly independent− vectors belonging to ker(f) − − Quiz 10 Given f defined in Quiz 9 (a) (π, π, π) is an eigenvector associated with eigenvalue 3 (b) (π, π, π) is an eigenvector associated with eigenvalue 2 (c) (π, π, π) is not an eigenvector (d) a nontrivial subspace of ker(f) must have dimension 2 Solution 9 (d) The kernel contains all those vectors (x, y, z) s.t. x + y + z = 0, that is, of kind (a, b, a b) a, b R. Since (a, b, a b) = a(1, 0, 1) + b(0, 1, 1), it is clear that −(1,−0, 1), (0∈, 1, 1) is a basis− for− the kernel. − − Solution 10{ (a) − − } (d) is false because any nontrivial subspace (= a subspace different from 0 and from the space itself) must have dimension 1. { } Chapter 7
Week 7
7.1 Extra material
Carlini, 50 Multiple Choices in Geometry, CELID, http://www.celid. • it/node/194565
Few examples of past exams http://calvino.polito.it/~gatto/public/ • past_exams.htm
Extra material you can download to delve deeper into linear algebra
Beezer http://linear.ups.edu/version3/pdf/fcla-draft-solutions. • pdf
Denton and Waldron https://www.math.ucdavis.edu/~linear/linear. • pdf
Hefferson http://joshua.smcvt.edu/linearalgebra/ • Kaplan http://quod.lib.umich.edu/s/spobooks/5597602.0002.001 • 7.2 Eigenvalues and multiplicities
Let λ be eigenvalue of a matrix A Rn×n ∈ 1. We denote by µ the algebraic multiplicity of λ, which is the exponent of the corresponding factor of degree 1 in the characteristic polynomial
2. We denote by ν the geometric multiplicity of λ, which is the dimension of the eigenspace of λ.
3. It holds 1 ν µ for any eigenvalue. Moreover, the sum of the algebraic multiplicities≤ is≤ equal to n.
4. If for each eigenvalue, ν = µ, A is said to be diagonalizable (or equivalently we say that the endomorphism f associated with A is simple)
5. The eigenvectors of a diagonalizable matrix form a basis.
29 30 CHAPTER 7. WEEK 7
Exercise 17 (a) Construct a matrix 3 3 with eigenvalues λ1 = 0 and λ2 = 1 and respective eigenspaces × 1 0 1 Vλ1 = 0 , 1 Vλ2 = 0 L 1 0 L 1 − (b) Is such a matrix diagonalizable? (c) What about if 1 Vλ1 = 0 L 1
Solution
We can answer (b) before having constructed the matrix: since the eigenvec- tors form a basis, the matrix is diagonalizable. We can also notice that, since
νλ1 = 2 and νλ2 = 1 and the sum of all the geometric multiplicities cannot exceed the matrix dimension 3, there are no other eigevanlues and µλ1 = 2,
µλ2 = 1. (a) Let us construct the matrix by using the equations that define eigenvec- tors and eigenvalues. α R (i) th The generic vector of Vλ1 is β , α, β . Let A denote the i α ∈ column of the matrix A. We can write α α 0 (1) (3) (2) A β = 0 β α A + A + βA = 0 (7.1) α α ⇒ 0
For the arbitrarity of α and β, this equality hold if A(1) + A(3) = (0, 0, 0)T and A(2) = (0, 0, 0)T . Analogously, if we consider the second eigenvalue/eigenvector,
α α 1 (1) (3) A 0 = 1 0 A A = 0 (7.2) α α ⇒ − 1 − − − Combining equations (7.1) and (7.4), we obtain
1 (1) 1 A = 0 (7.3) 2 1 − In conclusion, 1 0 1 1 − A = 0 0 0 (7.4) 2 1 0 1 − (c) The construction procedure is the same. The so obtained equations lead to the same A(1) and A(3), and give no information about A(2). To fill A(2), let us choose arbitrarily a third eigenvalue, for instance, λ3 = 2 and let us 7.3. CHANGE OF BASIS 31 associated with it an eigenvector. Having 3 different eigenvalues, the matrix we will obtain is diagonalizable, then let us choose an eigenvector for λ3 that is linearly independent from (1, 0, 1)T and (1, 0, 1). For instance, (0, 1, 0)T . In conclusion − 1 0 1 1 − A = 0 2 0 . (7.5) 2 1 0 1 − What happens if you choose an eigenvector linearly dependent from the others? What’s wrong? What happens if you choose λ3 = λ1 or λ3 = λ2?
4×4 Quiz 11 Let A R have two distinct eigenvalues λ1 and λ2, and νλ = 3. ∈ 1 Let Vλ denote the eigenspace associated with the eigenvalue λ. Then
(a) dim(Vλ1 ) < dim(Vλ2 )
(b) dim(Vλ1 ) = dim(Vλ2 ) R4 (c) Vλ1 + Vλ2 = 3 6 (d) Vλ R as its dimension is 3. 1 ⊂ Solution of Quiz 11 (a) In general, we know that νλ µλ. Since µλ 3, µλ 1 and µλ +µλ = 4, ≤ 1 ≥ 2 ≥ 2 1 then µλ1 = 3, µλ2 = 1; moreover νλ2 = 1. Hence, (a) is true and (b) is false. (d) makes no sense. Notice that such an A is diagonalizable (the multiplicities coincide for each eigenvalue), hence the eigenvectors form a basis. As a consequence the eigenspaces R4 Vλ1 and Vλ2 intersect only in 0, and their sum is the whole space . Observe that if A was not diagonalizable, the intersection could be “larger” and the sum could be “smaller”.
7.3 Change of basis
Let v V , V vector space, dim(V ) = N, and let B = b1,..., bN ,C = ∈ { } c1,..., cN be bases of V . Let (v)B representation of v w.r.t. B. { } We have B,C (v)B = P (v)C
B,C B,C where P has (c1)B,... (cN )B on the columns. It is known that P is invertible, and (P B,C )−1 = P C,B; P B,C is called change-of-basis matrix.
Exercise 18 Let E be the canonical basis of R3 and C = (1, 2, 3), (1, 0, 0), (0, 0, 1) . { } Which is the change-of-basis matrix P E,C ? • And P C,E? • Check that P C,EP E,C = I • Let v = α(1, 2, 3), α R3. Which is its representation w.r.t. basis C? • Practical question: why∈ it could be advantageous to represent v w.r.t. C? 32 CHAPTER 7. WEEK 7
Solution
1 1 0 C,E P = 2 0 0 (7.6) 3 0 1
1 0 2 0 E,C 1 P = 1 2 0 (7.7) 0 − 3 1 − 2 T T If v = α(1, 2, 3) ,(v)C = (α, 0, 0) . This representation is sparse, that is, contains many zeros. A practical advantage is that sparse vectors are more compressible than others. For this motivation in different technologies one looks for bases in which vectors can be represented in a sparse way. For example, these sparsyfing changes of basis are the first step of the JPEG standard: imagine that the vector represents a digital image (each entry represents a pixel). In Image Processing, a change of basis that sparsifies is the so called Discrete Cosine Transform (DCT) An other practical problem of this kind is the following. Storage of big datasets is an issue when limited space is available. Suppose that you have to store a dataset of N = 1000 vectors RM , with M = 1000. Suppose also to know that all those vectors belong to a subspace∈ that has a basis B = b1,..., bH , where H = 10. Can you provide a smart way (= reduce the required{ space) to} store the dataset?
Non-smart way: store all the vectors, need N M = 106 memory slots. • × Smart way: store the representation of those vectors in the basis B We • need H N memory slots to store the vectors in the basis B and also M H ×memory slots to store the basis, which can be used to “decode” the× vectors, that is, to come back to the original representation. The total number of used memory slots is 2 104 << 106. × A nice lesson about change of basis and image compression http://www. youtube.com/watch?v=vGkn-3NFGck Chapter 8
Week 8
8.1 Simple endomorphisms and diagonalizabil- ity
A, B Rn×n are said to be similar if there exists P Rn×n such that • A = PBP∈ −1 ∈ A is said to be diagonalizable if it similar to a diagonal matrix • A is diagonalizable for each eigenvalue of A, algebraic multiplicity = • geometric multiplicity⇔ A is diagonalizable its eigenvectors form a basis of Rn • ⇔ An endomorphism associated with a diagonalizable matrix is said to be • simple A is symmetric A is diagonalizable • ⇒ A is symmetric the eigenvectors of A are orthogonal • ⇒ A is said to be orthogonal if A−1 = AT • If A is orthogonal, detA 1, 1 • ∈ {− } 2×2 Exercise 19 Let fα, α R, be the endomorphism of R defined as ∈ x y x + αy 2x + 2y f = (8.1) α z t z + 2t 2z + t
1. For which values of α fα is simple? 2. For α = 3, write a basis C of eigenvectors for R2×2 3. For α = 3, write the change-of-basis matrix P to change from the canonical basis to C 4. For α = 2, write a basis F of eigenvectors for R2×2 5. For α = 2, write the change-of-basis matrix P to change from the canonical basis to F
33 34 CHAPTER 8. WEEK 8
6. Can you deduce for which value of α the matrix associated with fα is symmetric just from the eigenvectors?
Solution
First observation: R2×2 is equivalent to R4 as a vector space, that is, once we have decided on ordering, any matrix 2 2 cna be written as a vector of length 4. You can easily prove that the operations× of sum and product by a scalar are equivalent. Let us then fix the ordering
x x y y (8.2) z t → z t and solve the exercise in R4. Thus
x x + αy 1 α 0 0 x y 2x + 2y 2 2 0 0 y fα = = . (8.3) z z + 2t 0 0 1 2 z t 2z + t 0 0 2 1 t Let us call 1 α 0 0 2 2 0 0 A = . (8.4) 0 0 1 2 0 0 2 1 Even if the exercise requires to do computations for 2 specific values of α, to some practice with parametric quantities, let proceed without fixing a numerical value for α. The first step would be the computation of eigenvalues and eigenvectors of A. Since A is a 4 4 matrix, the characteristic polynomial is of degree 4, then it might be difficult× to find its roots. For this motivation let us find eigenvalues and eigenvectors without computing the characteristic polynomial, but solving a parameteric linear system. The rest of the exercise has then to be intended as an exercise on parametric linear systems, where the major obstacle is the study the behavior as the parameters change. The standard solution via characteristic polynomial is left to the reader for further exercise (in this specific case, the characteristic polynomial can be reduced in a quite simple way). We know that eigenvalues λ and eigenvectors v satisfy the system: 1 λ α 0 0 v1 0 − 2 2 λ 0 0 v2 0 − = . (8.5) 0 0 1 λ 2 v3 0 − 0 0 2 1 λ v4 0 − Actually, this is a system with unknown v and two parameters α, λ to be studied. 8.1. SIMPLE ENDOMORPHISMS AND DIAGONALIZABILITY 35
For the structure of the matrix, we can decouple the problem into two sub- systems 1 λ α 0 0 v1 0 − 2 2 λ 0 0 v2 0 − = (8.6) 0 0 1 λ 2 v3 0 − 0 0 2 1 λ v4 0 − That is, (1 λ)v1 + αv2 = 0 − (8.7) 2v1 + (2 λ)v2 = 0 − (1 λ)v3 + 2v4 = 0 − (8.8) 2v3 + (1 λ)v4 = 0 − Let us start from (8.8): by sum and difference we obtain (3 λ)(v3 + v4) = 0 − (1 + λ)(v3 v4) = 0 − The following cases are then possible
If λ = 3, then v3 = v4 • If λ = 1, then v3 = v4 • − − If λ / 1, 3 , then v3 = v4 = 0 • ∈ {− } We immediately get that λ = 3 and λ1 = 1 are eigenvalues, because they − generate solutions with some degree of freedom (v3 = v3 or v3 = v4: in both cases one component is free!). − For what concerns (8.7) we have
λ−2 v1 = 2 v2 λ−2 (1 λ) v2 + αv2 = 0 − 2 2 2 The second equation is equal to (λ 3λ + 2 2α)v2 = 0 and (λ 3λ + 2 √ − − − − 2α) = 0 when λ = 3± 1+8α . √ 2 3± 1+8α If λ = the second equation is 0 = 0 and v1 or v2 is free; then, √ 2 3± 1+8α 1 2 are eigenvalues (one eigenvalues if α = 8 ). The following cases are then possible: −
1 3 If α = and λ = , then 4v1 = v2 • − 8 2 − √ √ 3± 1+8α 1 3± 1+8α 2 −2 If α = and λ = then v1 = v2 • 6 8 2 2 √ 3± 1+8α λ = , v1 = v2 = 0 • 6 2 Moreover, let us find the correspondences with λ = 1, 3 found through subsystem (8.8) − √ α = 1, then 3+ 1+8α = 3; • 2 √ α = 3, then 3− 1+8α = 1; • 2 − 36 CHAPTER 8. WEEK 8
Merging all the so-obtained cases for v1, v2, v2, v4, we have (γ, β R): ∈ α = 1 • − 8 3 Eigenvalues λ1 = 3 λ2 = 1 λ3 = − 2 Eigenvectors v(1) = γ(0, 0, 1, 1)T v(2) = γ(0, 0, 1, 1)T v(3) = γ(1, 4, 0, 0)T − − Geom. multiplicities ν1 = 1 ν2 = 1 ν3 = 1
Not simple, as at least for one eigenvalue the algebraic multiplicity must be greater than 1. α = 1 • Eigenvalues λ1 = 3 λ2 = 1 λ3 = 0 − Eigenvectors v(1) = (β, 2β, γ, γ)T v(2) = γ(0, 0, 1, 1)T v(3) = γ(1, 1, 0, 0)T − − Geom. multiplicities ν1 = 2 ν2 = 1 ν3 = 1
Simple! As µ1 + µ2 + µ3 = 4 and µi νi, i = 1, 2, 3,, imply µi = νi, i = 1, 2, 3. ≥ α = 3 • Eigenvalues λ1 = 3 λ2 = 1 λ3 = 4 − Eigenvectors v(1) = γ(0, 0, 1, 1)T v(2) = ( 3 β, β, γ, γ)T v(3) = γ(1, 1, 0, 0)T − 2 − Geom. multiplicities ν1 = 1 ν2 = 2 ν3 = 1
Simple! As µ1 + µ2 + µ3 = 4 and µi νi, i = 1, 2, 3,, imply µi = νi, i = 1, 2, 3. ≥ α / 1, 3, 1 • ∈ { − 8 } √ √ 3− 1+8α 3+ 1+8α λ1 = 3 λ2 = 1 λ3 = λ4 = − 2 2 v(1) = γ(0, 0, 1, 1)T v(2) = γ(0, 0, 1, 1)T v(3) = γ( λ3−2 , 1, 0, 0)T v(3) = γ( λ4−2 , 1, 0, 0)T − 2 2 ν1 = 1 ν2 = 1 ν3 = 1 ν4 = 1
Simple! As it has 4 distinct eigenvalues. At this point we have analysed all the possible cases and we can answer the questions:
1 1. fα is simple for any α = 6 − 8 2. (As explicitly required, let use come back to R2×2, that is, we reshape the eigenvectors): Basis: 0 0 3 1 0 0 1 1 , 2 , , 1 1 −0 0 1 1 0 0 − 3. 0 3/2 0 1 − 0 1 0 1 1 0 1 0 1 0 1 0 − 8.1. SIMPLE ENDOMORPHISMS AND DIAGONALIZABILITY 37
4. (As explicitly required, let use come back to R2×2, that is, we reshape the eigenvectors): Basis:
√ √ ( 3− 17 −2 ! 3+ 17 −2 !) 0 0 0 0 2 2 , , 2 1 , 2 1 1 1 1 1 − 0 0 0 0
5. √ √ 3− 17 3+ 17 2 −2 2 −2 0 0 2 2 0 0 1 1 1 1 0 0 1 1 0 0 − 6. To have symmetry we need diagonalizability and orthogonal eigenvectors (i.e., an orthonormal basis). For α = 1/8, we have orthogonality, but eigenvectors do not form a basis. For−α = 1, 3, in general we have no 1 orthogonal eigevenctors. Finally, if we consider α / 8 , 1, 3 , we have orthogonality only if α = 2. ∈ {− } Now substituting α = 2 in the matrix, you can easily check that this is the value that makes the matrix symmetric. 38 CHAPTER 8. WEEK 8 Chapter 9
Week 9
9.1 Circles and Spheres
Before introducing conics and quadrics, let us do some exercises on circles (in R2 and R3) and on spheres. Next week, we will see that a circle in R2 is a conic and a sphere is a quadric, but this week we can even ignore this.
9.1.1 Circles in R2 The canonical equation of a circle in R2 is
2 2 2 (x c1) + (y c2) = r (9.1) − − where r is the radius and C = (c1, c2) is the center of the circle. Any circle in R2 can be written in its canonical form by completion of the squares.
Exercise 20 Consider the family of circles kx2 + ky2 3x + 2ky = 0, k R, k > 0: − ∈ (a) Do they have the same radius? (b) Are they tangent to the same straight line at (0, 0)?
Solution
Let us first reduce the family of circles to the canonical form. Completion of the squares consists in the following procedure. Whenever we have an expression of kind a2 + 2ab, we can write is as (a + b)2 b2. In our context, we will always have a depending on a variable (x, y, . . . ) and− b constant.
3 9 3 kx2 3x = (√kx )2 (a = √kx, 2ab = 3x b = ) − − 2√k − 4k − ⇒ −2√k ky2 + 2ky = k(y2 + 2y) = k[(y + 1)2 1] − 3 9 kx2 + ky2 3x + 2ky = (√kx )2 + k(y + 1)2 k = ⇒ − − 2√k − 4k − 3 9 = k(x )2 + k(y + 1)2 k = 0 − 2k − 4k − (9.2)
39 40 CHAPTER 9. WEEK 9
-1
Figure 9.1: Two circles of Exercise 20: it is evident that they have different radii and different tangent lines at (0, 0).
Thus, the canonical form of the family of circles is obtained by dividing by k the last expression: 3 9 (x )2 + (y + 1)2 = 1 + (9.3) − 2k 4k2 3 (a) NO. In fact, for each k, the centre is Ck = , 1 and the radius is 2k − q 9 Rk = 1 + 4k2 . In particular, the radius changes.
(b) Let us call γk the circle obtained for a particular k. Any γk passes through the origin (substituting x = 0, y = 0, the given circle equation is always fulfilled), so each γk has tangent line at (0, 0). It is immediate to check that the line Nk orthogonal to the circle at (0, 0) is the line passing through (0, 0) and Ck, that is, 2kx Nk : = y. 3 − Then, Nk is different for each k, which implies also that the tangent lines (which are orthogonal to Nk) are different. More explicitly, as the direction of 3 Nk is given by 2k , 1 , the direction of the tangent lines may be expressed as 3 − 2k , 1 , and this is enough to show that the tangent lines are different for any k (in particular, they form a sheaf of lines passing through the origin).
9.1.2 Spheres The canonical equation of a sphere (in R3) is
2 2 2 (x c1) + (y c2) + (z c3) = r (9.4) − − − where r is the radius and C = (c1, c2, c3) is the center of the sphere. Any sphere can be written in its canonical form by completion of the squares.
Exercise 21 Let H : x + y + 1 = 0 S : x2 + y2 + z2 2x + 2y 4z + 3 = 0 − − 9.1. CIRCLES AND SPHERES 41
(a) Is H tangent to S? (b) Does H intersect S along a maximal circle? (c) Find a plane G parallel to H and tangent to S.
Solution By completion of the squares on x, y, z, we see that S is the sphere of centre C = (1, 1, 2) and radius √3. The canonical equation is − (x 1)2 + (y + 1)2 + (z 2)2 = 3. − − H is a plane (not a straight line!! as we are in R3). The intersection between a plane and a sphere may be (a) an empty set (no intersection); (b) a circle; (c) a point (in this case, we say that plane and sphere are tangent). (a) Substituting y = x 1 in the equation of S, we obtain x2 +( x 1)2 + (z 2)2 = 3, that is, 2x2−+− 1 + 2x + (z 2)2 = 3, and by simple computations− − − 1 2 2 1 − x 2 + (z 2) = 3 + 4 . This is the equation of an ellipse in xz plane, then it contains− many− points, so H and S are intersecting (not tangent). Pay attention! We are not saying that the intersection of S and H is an ellipse (we know that it must be a circle), but that the equation that we obtain would be an ellipse in the xz plane. But we are not in that plane. Such equation here has a pure algebraic role and we are just interested in the number of its solutions (the observation about the ellipse is useful just to be sure that is has many solutions). The situation is similar to x + y + 1 = 0, which would be a straight line in the xy plane, but in R3 is a plane... the same equation may represent different geometrical objects in different spaces! More details about circles in R3 will be given in next paragraph. (b) No, since the centre of S does not lie on H. (c) We must find a plane G : x + y = D which has distance √3 from the centre of the sphere. Given that L : (1, 1, 2) + t(1, 1, 0) is the line orthogonal to H and passing through the centre of− the sphere, we have that Q = L H is given by (1 + t) + (1 t) = D then t = D/2. Moreover, d(Q, C) = √3∩ when D = √6. − The± planes parallel to H and tangent to S are then x + y = √6. Given a straight line L and a curve γ, a cylindrical surface (or± cylinder) is given by all the lines parallel to L and passing through γ.
9.1.3 Circles in R3 Circles in R3 cannot be expressed by a single equation; they can be expressed as a system (= an intersection) between a sphere and a plane.
Exercise 22 Let
γ : z = x2 + y2 + z2 6z 16 = 0. − − (a) Verify that γ is a circle and calculate its centre and its radius. (b) Write the equation of the cylinder S that projects γ in the direction parallel to the z-axis. (c) Write the equation of the cylinder T that projects γ in the direction parallel to L : x = y z = 0. − 42 CHAPTER 9. WEEK 9
z
h = 8
x
h = 8 −
Figure 9.2: Illustration of Exercise 22 (projected on plane xz). The red segment is the projection of the circle, while the green and blue line determine the cylinders. The dashed orange lines indicate for which values of z we obtain two tangent circles by cutting the cylinders with planes parallel to xy .
(d) Determine h R such that the intersections of S and T with z = h are two mutually tangent∈ circles.
Solution (a) γ turns out to be the intersection between the plane z = 0 and the sphere x2 + y2 + z2 6z 16 = 0: − − z = 0 (9.5) x2 + y2 = 16. then C = (0, 0, 0) and R = 4.
(b) S : x2 + y2 = 16
(c) It is sufficient to think about the projection of the centre: moving (0, 0, 0) in the direction of L, when z = k, the centre is at (0, k, k). The dimensions of the circle do not change, then we have x2 + (y k)2 = 16. Hence, the equation of the cone is x2 + (y z)2−= 16. − (d) The intersections of S and T with any plane z = h are circles respectively of centres (0, 0, h) and (0, h, h). At which height are the circles determined by S tangent to the circles determined by T ? This occurs when the distance between the centres is equal to 8 (notice that all the circles have radius 4), that is, (0, 0, h) (0, h, h) = 8 h = 8. k ———-− k ⇒ ± Chapter 10
Week 10
Examples of Geometry tests (eng/ita) http://cantor.polito.it/didattica/index2.php?richiesta=esami&percorso= Geometria&orderby=lingua&orderhow=DESC&lingua=E
10.1 Conics
2 2 f(x, y) = a11x + a22y + 2a12xy + 2a13x + 2a23y + a33 = 0
a11 a12 a13 a11 a12 B = a12 a22 a23 A = a12 a22 a13 a23 a33
x x x f(x, y) = (x, y)A + 2(a , a ) + a = (x, y, 1)B y y 13 23 y 33 1
det(A) > 0 ellipse < 0 hyperbola = 0 parabola ⇒ ⇒ ⇒ det(B) = 0 degenerate (a point, a couple of lines) ⇒ x2 y2 x2 y2 2 2 Canonical forms: E: a2 + b2 = 1; H: a2 − b2 = ±1 P: y = ax , x = ay
10.2 Quadrics
“Extension” of the conics to R3
2 2 2 f(x, y, z) =a11x + a22y + a33z + 2a12xy + 2a13xz + 2a23yz
+ 2a14x + 2a24y + 2a34z + a44 = 0
B = (aij)i,j=1,2,3,4 A = (ai,j)i,j=1,2,3
x x (x, y, z)A y + 2(a14, a24, a34) y + a44 = 0 z z
43 44 CHAPTER 10. WEEK 10
Classification and shapes: http://en.wikipedia.org/wiki/Quadric#Euclidean_plane_and_space http://www.math.uiuc.edu/Courses/math241/quadrics/
10.3 Reduction of conics to canonical forms
Rotation + translation
Rotation: rotation matrix P = special (detP = 1) orthogonal matrix • (P −1 = P t) that diagonalizes A. A is symmetric such P exists; columns of P = (normalized) eigenvectors of A ⇒
Translation: completion of squares • Reduction steps:
T T 1. X AX + 2(a13, a23)X + a33 = 0 X = (x, y)
2. P −1AP = P T AP = D (D= diag matrix of eigenvalues)
3. X = PX0 where X0 = (x0, y0)T (rotated coordinate frame)
0 T 0 0 4.( PX ) A(PX ) + 2(a13, a23)PX + a33 = 0
0 T 0 0T 0 02 02 5.( PX ) A(PX ) = X DX = λ1x + λ2y
6. No mixed terms in x0y0 conclude by compl of squares ⇒ Exercise 23 Reduce 3x2 + 2xy + 3y2 + 2√2x = 0 to its canonical form.
Solution
3 1 √2 3 1 B = 1 3 0 A = (10.1) 1 3 √2 0 0 det A = 8 ellipse ⇒ Eigenvalues of A: λ1 = 2 and λ2 = 4. Eigenspaces Vλ = (1, 1) and Vλ = (1, 1) 1 L{ − } 2 L{ } Rotation matrix: 1 1 1 P = (10.2) √2 1 1 − (P corresponds to the rotation of angle θ = π/4)
2x02 + 4y02 + 2(√2, 0)P (x0, y0)T = 2x02 + 4y2 + (2, 4)(x0, y0)T = 2x02 + 4y02 + 2x0 + 4y0 = 0 x02 + 2y02 + x0 + 2y0 = 0 ⇒
002 002 Completion of squares: (x0+1/2)2+2(y0+1/2)2 1/4 1/2 = 0 x + y = 1 − − ⇒ 3/4 3/8 where x00 = x0 + 1/2 and y00 = y0 + 1/2. 10.3. REDUCTION OF CONICS TO CANONICAL FORMS 45
Quiz 12 Given f(x, y) = 4x2 + 12xy + 9y2 (a) There exists (x, y) such that f(x, y) < 0 (b) f(x, y) = 1 is a hyperbola (c) f(x, y) = 0 implies (x, y) = (0, 0) (d) The points that satisfy f(x, y) = 0 lie on a line
Quiz 13 tx2 + 2xy + ty2 + 2y = 0 (a) is an ellipse for t = 1 (b) is a parabola for some values of t (c) never is a hyperbola (d) a parabola for infinitely many values of t
Quiz 14 Q : z2 = 2x2 + 3y2 (a) Q πc, where πc : z = c are circles (b) There∩ exist planes π such that π Q are hyperbolas (c) x = y = z is contained in Q ∩ (d) (1, π, 9) Q ∈ Quiz 15 f(x, y, z) = x2 4xz + z2, then f(x, y, z) = 0 (a) is a parabola − (b) is a hyperboloid (c) xf(x, y, z) = 1 is a quadric (d) represents two incident planes
Solution 12 (d). f(x, y) = 4x2 + 12xy + 9y2 = (2x + 3y)2 then f(x, y) = 0 if and only if 2x + 3y = 0, which is a line t 1 0 Solution 13 (b). det 1 t 1 = t so the conic is non degenerate for 0 1 0 − t = 0. 6 t 1 det = t2 1 is equal to zero for t = 1. 1 t − ± Solution 14 (b). For instance, consider x = 1. 1 0 2 − Solution 15 (d). Let us reduce it: 0 0 0 has eigenvalues 0, 3, 1. 2 0 1 − − Let P be the orthonormal matrix of the eigenvectors (in this case, as it while be clear in a while, it is not necessary to compute it!), X = PX0 where X = (x, y, z)T and X0 = (x0, y0, z0), and in the variables X0 we have:
3y02 z02 + 2(0, 0, 0)PX0 = 0 z = y√3. − ⇒ ± 46 CHAPTER 10. WEEK 10 Chapter 11
Week 11
11.1 Quadrics
Exercise 24 Reduce and classify the quadric
y2 2xz √2(x + z) 2y 1 = 0. − − − − Remember that reduction for quadrics is analogous to reduction of conics... just one more variable!
T T 1. X AX + 2(a14, a24, a34)X + a44 = 0 X = (x, y, z)
2. P −1AP = P T AP = D (D= diag matrix of eigenvalues)
3. X = PX0 where X0 = (x0, y0)T (rotated coordinate frame)
4. ...
Solution
0 0 1 − A = 0 1 0 (11.1) 1 0 0 − Eigenvalues: λ1 = 1, µλ = 2; λ2 = 1, µλ = 1. 1 − 1 Eigenspaces: Vλ = (1, 0, 1), (0, 1, 0) and Vλ = (1, 0, 1) 1 L{ − } 2 L{ } Rotation matrix:
1/√2 0 1/√2 P = 0 1 0 . (11.2) 1/√2 0 1/√2 − Substituting X = PX0: x02 + y02 z02 2y0 2z0 = 1 − − − Compl. of squares, x02 + (y0 1)2 (z0 + 1)2 + 1 1 = 1. − − − Finally, x002 + y002 z002 = 1 where x00 = x0, y00 = y0 1, z00 = z0 + 1. − − (z constant circumference; x or y constant hyperbola) → → HYPERBOLOID OF ONE SHEET
47 48 CHAPTER 11. WEEK 11
Exercise 25 Consider the quadric of the previous exercise with no 2y:
y2 2xz √2(x + z) 1 = 0. − − − Reduce and classify it.
Solution Since we have just changed one term, we expect something similar to the pre- vious hyperboloid of one sheet...
0 0 1 − A = 0 1 0 (11.3) 1 0 0 − Eigenvalues: λ1 = 1, µλ = 2; λ2 = 1, µλ = 1. 1 − 1 Eigenspaces: Vλ1 = (1, 0, 1), (0, 1, 0) and Vλ2 = (1, 0, 1) Rotation matrix: L{ − } L{ } 1/√2 0 1/√2 P = 0 1 0 . (11.4) 1/√2 0 1/√2 − Substituting X = PX0: x02 + y02 z02 2z0 = 1 Compl. of squares, x02 + y02 −(z0 +− 1)2 = 0. Finally, x002 + y002 z002 = 0− where x00 = x0, y00 = y0, z00 = z0 + 1. (z constant circumference;− x or y constant two lines) CONE (can→ you why this cone is not ”so different“→ from the hyperboloid obtained in the previous exercise?)
11.2 Critical points for functions of two vari- ables
Given z = f(x, y), we define the partial derivatives ∂f ∂x = derivative of f with respect to x, that is, handle y as a constant ∂f ∂y = derivative of f with respect to y, that is, handle x as a constant We call gradient the vector
∂f ∂f f = , . ∇ ∂x ∂y We call Hessian matrix the matrix whose entries are all the second deriva- tives:
∂2f ∂2f ! ∂x2 ∂x∂y Hf = ∂2f ∂2f ∂y∂x ∂y2
∂2f Notice that ∂x2 just indicates ”derive twice w.r.t. x“, not a square... ∂2f ∂2f It is known that ∂x∂y = ∂y∂x The critical (or stationary) points are those whose gradient is null. To determine the nature of the critical points use the following test on the Hessian matrix: 11.3. MIXING QUADRICS AND FUNCTIONS OF TWO VARIABLES 49
∂2f - If det Hf (P ) > 0 and ∂x2 (P ) > 0, the critical point P is a minimum; ∂2f - If det Hf (P ) > 0 and ∂x2 (P ) < 0, the critical point P is a maximum; - If det Hf (P ) < 0, the critical point P is a saddle; - Otherwise: ?
11.3 Mixing quadrics and functions of two vari- ables
Exercise 26 Find maxima and minima p z = f(z, y) = 1 1 2x2 y2 − − − using (a) derivatives (b) quadrics
Solution
(a) First of all, notice that the domain of f is (x, y) R2 : 2x2 + y2 1 , that is, the region delimited by an ellipse (and the{ ellipse∈ itself). Let us≤ find} the critical points:
1 f = p ( 4x, 2y) = (0, 0) (x, y) = (0, 0). ∇ 2 1 2x2 y2 − − ⇔ − − (Notice that f is not defined on the boundaries, say the ellipse, then we will have to study∇ minima and maxima on the ellipse in a second time.) We have only one critical point. To study its nature (minimum, maximum, saddle) one should use the test on the Hessian matrix. But since in this case computing the second derivatives could be a bit long, I try to find another method taking information from the range of f. p We notice, in fact, that 1 2x2 y2 [0, 1] 1 2x2 y2 [0, 1] p − − ∈ ⇒ − − ∈ ⇒ 1 1 2x2 y2 [0, 1]. Now, since f(0, 0) = 0, P = (0, 0) certainly is a global− minimum.− − ∈ Let us study the boundaries (the ellipse). On the ellipse 1 2x2 y2 = 0, f(x, y) = 1. According to the studied range, 1 is the maximum that− the− function can reach, then all the points on the ellipse are (global) maxima points. For exercise, check this result with the test of the Hessian matrix. (b) This exercise could be solved also using geometry and in particular quadrics. In fact, we have
p z = f(z, y) = 1 1 2x2 y2 − − − that is
p 1 z = 1 2x2 y2. − − − Assuming
1 z 0 − ≥ 50 CHAPTER 11. WEEK 11
(otherwise the equation makes no sense), we can compute the squares both sides, without losing any information:
(1 z)2 = 1 2x2 y2. − − − This is a quadric! In particular it is an ellipsoid centered in (0, 0, 1). Well, not really the whole ellipsoid, since we are imposing z 1, that is, we cut the ellipsoid with the plane z = 1, and we then obtain half≤ ellipsoid. Given the length of the axes, it should be clear that this half ellipsoid has a minimum at (0, 0) and a ”plateau“ of maxima at z = 1. (Pay attention: an ellipsoid can not be the image of a function z = f(x, y), while half ellipsoid can be so. Why?)
Quiz 16 f(x, y) = 3xe2y (a) (0, 0) is a critical point (b) f is bounded (c) f is constant along the curve xy = 0 (d) the derivative of f at (0, 0) in the direction of (0, 4) is null. p Quiz 17 f(x, y) = x y (a) Domf = (x, y):| y| > 0 (b) f is differentiable{ at any} point s.t. x = 0 (c) x 2y 2√2z = 0 is the tangent plane to z = f(x, y) at Q = (2, 1, f(2, 1)) (d) There− − is no tangent plane at Q − −
Solution 16 (d). f(x, y) = 3xe2y has gradient f = e2y(3, 6x), which is never null. f is not bounded (compute the limits for∇x or y tending to infinity) and constant along x = 0, but not for y = 0 (then only one plane of the two forming the quadric xy = 0). Finally, the directional derivative at (0, 0) in the direction (0, 1) is f(0, 0) (0, 1) = (3, 0) (0, 1) = 0. Solution∇17 (c).·F (x, y, z) = f(···x, y) z. The tangent plane is normal to F . For y < 0, − ∇
√ y √x F (x, y, z) = − , , 1 ∇ 2√x −2√ y − √ − and F (Q) = ( √1 , 2 , 1), then the tangent plane is ∇ 2 2 − 2 − x √2 π : z + d = 0 2√2 − 2 − and given that Q must be in π, we get d = 0. p Further question: can you draw the graph of f(x, y) = x y ? p | | Notice that z = x y , so we fix z 0, x 0, and we can say z2 = x y . If y > 0, we have the| | quadric z2 =≥xy which≥ is a cone (reduction is| quite| easy). If y < 0, we have another cone z2 = xy. Actually, the graphs is the union of the differents parts of cones. Using− these hints, you can try to draw the graph. Chapter 12
Week 12
2 3 Quiz 18 Let M = 1 1 . 1/2 3−/4 (a) M 2 = MM has rank equal to 2 (b) M has determinant equal to 2 (c) M has rank equal to 2 (d) M T M has determinant equal to 2
2 0 0 Quiz 19 Let M = 1 1 1 . 2 2 2 (a) 3 is not an eigenavalue of M (b) 0 is not an eigenvalue of M (c) (0, 1, 1)T is an eigenvector of M (d) M is− not diagonalizable
Quiz 20 x = z = y + 2 (a) is a conic (b) corresponds to a subspace of R3 (c) is a line with direction given by (1, 1, 1) (d) is perpendicular to (7, 7, 7)
Quiz 21 The curve γ(t) = (t, t, t2) (a) is contained in the plane x y = 0 (b) is not plane − (c) is a circle (d) in P = (1, 1, 1), the tangent vector is parallel to (1, 1, 1)
x + y + z = 2 Quiz 22 The system z = 5 (a) has 2 solutions (b) has ∞ solutions ∞ 51 52 CHAPTER 12. WEEK 12
(c) (0, 0, 0) is a solution (d) has four solutions
Quiz 23 x2 y2 + 2x + 2y = 0 is (a) a parabola− (b) a non degenerate hyperbola (c) an ellipse (d) a couple of incident straight lines
Quiz 24 x2 y2 + z + 2x + 2y = 0 is (a) a paraboloid− (b) a hyperboloid (c) an ellipsoid (d) a sphere
Quiz 25 Given the plane π : 2y + 2z = 1 and the line l : x = z + y = 0 (a) l π − (b) l ⊂π is formed by two points (c) l ∩is normal to π (d) l π is empty ∩
Quiz 26 Given f(x, y, z) defined as the vector product between (1, 0, 3) and (x, y, z) (a) f is not a linear map (b) the image of f is a subset of R2 (c) is injective (d) dim(Im(f))=2
Quiz 27 f(x, y) = ex−y (a) is an endomorphism (b) has gradient (1, 1)f(x, y) (c) has gradient (1, −1) (d) has a minimum−
2 0 1 Exercise 27 Consider the matrix M = 1 1 0 . 0− 0 3 (i) Compute the eigenvalues of M (ii) Compute the eigenvectors and the eigenspaces (iii) Is M diagonalizable? Why? (iv) Is f defined by f(u) = Mu simple? (v) Write the matrix associated to f with respect to the basis = (1, 0, 0), (0, 1, 0), (1, 1, 1) B { }
2 Exercise 28 Consider f(x, y) = y2ex /(x−1) + 1. (i) Find the domain of f; is it an open or closed set? 53
(i) Compute the critical points and classify them (maximum, minimum, saddle) (ii) Find the tangent plane to the surface S : z = f(x, y) at P = (0, 0, 1)
Solution of Quiz 18: (c). The product MM does not exist, the determinant of T 21·169−432 2 M is null, the determinant of M M is 26 . Notice that 21 169 43 > 20 160 502 = 3200 2500 = 700 > 29, then the determinant is· larger− than 29−·6 = 8.− − Solution of Quiz 19: (c). (0, 1, 1)T is the eigenvector associated with eigen- value 0. To verify if x is an eigenvalue,− compute M xI: if the rank is not maximum, x is an eigenvalue. 0 and 3 turn out to− be eigenvalues. p(λ) = (2 λ)(λ2 3λ) hence the third eigenvalue is λ = 2. Since the three eigenvalues are− different,− M is diagonalizable. Solution of Quiz 20: (c). x = z and x = y + 2 then I fix x = t, t R and (x, y, z) = (0, 2, 0) + t(1, 1, 1). ∈ Solution of Quiz− 21: (a) It is sufficient to substitute (x, y, z) = (t, t, t2) in x = y: as x = t, y = t, we get t = t, which is true. Then, γ is contained in the plane x = y. Notice that γ is the intersection between the plane x = y and the parabolic cylinder z = x2, then it cannot be a circle. Solution of Quiz 22: (b) The solution is (α, 3 α, 5), α R. −1− 0 1 ∈ Solution of Quiz 23: (d). Notice that det 0 1 1 = 0 then the conic 1− 1 0 is degenerate. This is the first check that you have to do; then, if it is not de- 1 0 generate you compute the determinant of the submatrix to classify 0 1 − it (parabola, hyperbola or ellipse). Here you can obtain the solution even without computing the coefficients’ matrix: by completion of the squares, you get (x + 1)2 = (y 1)2, that is, x + 1 = (y 1) y = x + 2 or y = x. They are two perpendicular− lines. Solution±of− Quiz⇒ 24: (a). Again by completion− of the squares, we reduce the quadric: z = (x + 1)2 + (y 1)2 and we obtain a paraboloid (we recognize it by the fact that− one variable− is of degree 1). If we fix x equal to a constant, we obtain a parabola; if we fix z equal to a constant we obtain a hyperbola: then the quadric is hyperbolic paraboloid.) Solution of Quiz 25: (c). l : t(0, 1, 1) is parallel to (0, 2, 2). Solution of Quiz 26: (d) f(x, y, z)− = ( 3y, 3x z, y) =− (0, 0, 0) when y = 0 and 3x = z, one variable is free, then the− dimension− of the kernel is 1. Solution of Quiz 27: (b). Solution of Exercise 27. (i) The eigenvalues are the solutions of det(M λI) = 0. − 2 λ 0 1 − det 1 1 λ 0 = (2 λ)( 1 λ)(3 λ) = 0. 0− 0− 3 λ − − − − −
Then λ1 = 2, λ2 = 1, λ3 = 3. The three roots are single, say µλ = 1 for − i i = 1, 2, 3. As a consequence the geometric multiplicities νλi = 1 for i = 1, 2, 3, since it is known that for any eigenvalue 1 νλ µλ . ≤ i ≤ i (ii) Let us indicate by vi the eigenvector associated with λi. 54 CHAPTER 12. WEEK 12
λ1 = 2: 0 0 1 x 0 (M 2I)v1 = 1 3 0 y = 0 z = 0, x + y = 0 − 0− 0 1 z 0 ⇔
T therefore v1 = a(3, 1, 0) a R. The eigenspace is V1 = span (1, 1, 0) . ∈ { − } λ2 = 1: − 3 0 1 x 0 (M + I)v2 = 1 0 0 y = 0 x = 0, z = 0 0 0 4 z 0 ⇔
T therefore v2 = a(0, 1, 0) a R. The eigenspace is V1 = span (0, 1, 0) . ∈ { } λ3 = 3: 1 0 1 x 0 − (M 3I)v3 = 1 4 0 y = 0 x = z, x = 4y − 0− 0 0 z 0 ⇔
T therefore v2 = a(4, 1, 4) a R. The eigenspace is V1 = span (4, 1, 4) . (iii) Yes, M is diagonalizable∈ since algebraic and geometric{ multiplicities} coincide for each eigenvalues (this is the particularly easy case where all the algebraic multiplicites are equal to 1, which implies immediately that also the geometric multiplicites are equal to 1). (iv) Yes, f is simple since associated with the diagonalizable matrix M (notice also that the eigenvectors form a basis) B (v) Let us call Mf the matrix of f with respect to = b1, b2, b3 . As we B B { } studied some weeks ago, Mf is defined as the matrix that has on its columns (f(b1))B,(f(b2))B,(f(b3))B. First method: f(1, 0, 0)T = (2, 1, 0)T is the first column. f(0, 1, 0)T = (0, 1, 0)T is the second column. − f(1, 1, 1) = (3, 0, 3) = 3b3 3b2 then the third column is (0, 3, 3). − − 2 0 0 B Mf = 1 1 3 0− 0− 3 Second method: let B be the matrix that has the vectors of on the columns; B then the representation of a vector v in the basis is the vector (v)B given B by B(v)B = v. Therefore, given f(b1), f(b2), f(b3) (which can be easily computed), we can compute (f(b1))B,(f(b2))B,(f(b3))B Explicitly, for (f(b1))B: 1 0 1 x 2 0 1 1 y = 1 0 0 1 z 0
T has solution (2, 1, 0) ; for (f(b2))B 1 0 1 x 0 0 1 1 y = 1 0 0 1 z −0 12.1. NICE LINK... 55
T has solution (0, 1, 0) ; for (f(b3))B − 1 0 1 x 3 0 1 1 y = 0 0 0 1 z 3 has solution (0, 3, 3)T . − Solution of Exercise 28. (i) The domain of f is R2 (x, y): x = 1 . It is an open, not connectd, not bounded set. \{ } (ii) The gradient of f is
2 ex /(x−1) y2g(x), 2y
x2 where g(x) is the derivative of x−1 . The value of g(x) is not important in the evaluation of the critical points: it is immediate to see that the gradient is null if and only if y = 0. In other terms, all the points on the line y = 0 (excepts for the point (1, 0) which is not in the domain) are critical points. To classify the critical points we are used to compute the Hessian matrix, but in this case it is quite boring and even not conclusive (its determinant is null for y = 0). However, notice that f(x, y) 1, and f(x, 0) = 1: in conclusion all the points s.t. y = 0 are minima. ≥ (iii) In general, the tangent plane π to the surface z = f(x, y) is perpendicular to the gradient of F (x, y, z) = f(x, y) z, that is, ∂f , ∂f , 1 . In this case, − ∂x ∂y − in P the gradient is (0, 0, 1). Then the plane π must− satisfy z + d = 0. Given that P π, d = 1, hence π : z = 1. This makes sense since−P is a minimum and the∈ tangent plane is parallel to xy-plane.
12.1 Nice link...
... to visualize parametric curve in 3D: http://www.math.uri.edu/~bkaskosz/flashmo/parcur/