Contents 1 Normed Spaces. Banach Spaces. 2 1.1 Vector Space. 2 1.2 NormedSpace.BanachSpace.. 10 1.3 FurtherPropertiesofNormedSpaces.. 16 1.4 FiniteDimensionalNormedSpacesandSubspaces. 26 1.5 Linear Operators. 32 1.6 Bounded and Continuous Linear Operators. 38 1.7 LinearFunctionals. .............................. 46 1.8 Linear Operators and Functionals on Finite Dimensional Spaces. 55 1 Normed Spaces. Banach Spaces. 1.1 Vector Space. Definition 1.1. 1. An arbitrary subset M of a vector space X is said to be linearly independent if every non-empty finite subset of M is linearly independent. 2. A vector space X is said to be finite dimensional if there is a positive integer n such that X contains a linearly independent set of n vectors whereas any set of n +1 or more vectors of X is linearly dependent. n is called the dimension of X, written n = dim X. 3. If X is any vector space, not necessarily finite dimensional, and B is a linearly independent subset of X which spans X, then B is called a basis (or Hamel basis) of X. Hence if B is a basis for X, then every nonzero x X has a unique repre- • sentation as a linear combination of (finitely many!2 ) elements of B with nonzero scalars as coefficients. Theorem 1.2. Let X be an n-dimensional vector space. Then any proper subspace Y of X has dimension less than n. 1. Show that the set of all real numbers, with the usual addition and multiplication, constitutes a one-dimensional real vector space, and the set of all complex numbers constitutes a one-dimensional complex vector space. Solution: The usual addition on R and C are commutative and associative, while scalar multiplication on R and C are also associative and distributive. For R,the zero vector is 0R =0 R,theidentityscalaris1 =1 R,andtheadditive 2 R 2 inverse is x for any x R.ForC,thezerovectoris0 =0+0i C,the − 2 C 2 identity scalar is 1 =1+0i C and the additive inverse is z for all z C. C 2 − 2 2. Prove that 0x = 0, ↵0 = 0 and ( 1)x = x. − − Solution: 0x =(0+0)x =0x +0x = 0 =0x +( (0x)) ) − =0x +0x +( (0x)) − =0x + 0 =0x. ↵0 = ↵(0 + 0)=↵0 + ↵0 = 0 = ↵0 +( (↵0)) ) − = ↵0 + ↵0 +( (↵0)) − = ↵0 + 0 = ↵0. ( 1)x =( 1(1))x = 1(1x)= x. − − − − Page 2 3. Describe the span of M = (1, 1, 1), (0, 0, 2) in R3. { } Solution: The span of M is 1 0 span M = ↵ 1 + β 0 : ↵,β R 8 2 3 2 3 2 9 < 1 2 = 4 ↵5 4 5 : ; = ↵ : ↵,β R . 82 3 2 9 < ↵ +2β = 4 5 We see that span M corresponds: to the plane x = y on; R3. 4. Which of the following subsets of R3 constitute a subspace of R3? Here, x = (⇠1,⇠2,⇠3). (a) All x with ⇠1 = ⇠2 and ⇠3 =0. Solution: For any x, y W and any ↵,β R, 2 2 ⇠1 ⌘1 ↵⇠1 + ⌘1 ↵x + βy = ↵ ⇠ + β ⌘ = ↵⇠ + ⌘ W. 2 23 2 23 2 2 23 0 0 0 2 4 5 4 5 4 5 (b) All x with ⇠1 = ⇠2 +1. 2 3 Solution: Choose x = 1 W , x = 2 W ,then 1 2 3 2 2 3 0 2 0 2 4 5 4 5 2 3 5 x + x = 1 + 2 = 3 / W 1 2 2 3 2 3 2 3 0 0 0 2 4 5 4 5 4 5 since 5 =3+1. 6 (c) All x with positive ⇠1,⇠2,⇠3. Solution: Choose ↵ = 1, then for any x W , ↵x / W . − 2 2 (d) All x with ⇠ ⇠ + ⇠ = k. 1 − 2 3 Solution: For any x, y W , 2 ⇠1 + ⌘1 x + y = ⇠ + ⌘ . 2 2 23 ⇠3 + ⌘3 4 5 Page 3 Since ⇠ + ⌘ (⇠ + ⌘ )+(⇠ + ⌘ )=(⇠ ⇠ + ⇠ )+(⌘ ⌘ + ⌘ )=2k. 1 1 − 2 2 3 3 1 − 2 3 1 − 2 3 we see that W is a subspace of R3 if and only if k =0. j 5. Show that x1,...,xn ,wherexj(t)=t ,isalinearlyindependentsetinthespace C[a, b]. { } Solution: This is a simple consequence of Fundamental Theorem of Alge- bra. Fix a finite n>1. Suppose that for all t [a, b], we have 2 n n j λjxj(t)= λjt =0. j=1 j=1 X X Suppose λn = 0. Fundamental Theorem of Algebra states that any polynomials with degree 6n can have at most n real roots. Since the equation above is true for n all t [a, b], and the set of points in the interval [a, b]isuncountable, λ tj has 2 j j=1 to be the zero polynomial. Since n 1wasarbitrary(butfinite),thisshowsthatX ≥ any non-empty finite subset of xj j ⇤,⇤acountable/uncountableindexingset, 2 is linearly independent. { } 6. Show that in an n-dimensional vector space X, the representation of any x as a linear combination of a given basis vectors e1,...,en is unique. Solution: Let X is an n-dimensional vector space, with a basis e1,...,en . Suppose any x X has a representation as a linear combination{ of the basis} vectors, we claim2 that the representation is unique. Indeed, if x X has two representations 2 x = ↵1e1 + ...+ ↵nen = β1e1 + ...+ βnen. subtracting them gives n (↵ β )e + ...+(↵ β )e = (↵ β )e = 0. 1 − 1 1 n − n n j − j j j=1 X Since e1,...,en is a basis of X,bydefinitionitislinearlyindependent,which implies{ that ↵ }β =0forallj =1,...,n,i.e.therepresentationisunique. j − j 7. Let e1,...,en be a basis for a complex vector space X. Find a basis for X regarded as a{ real vector} space. What is the dimension of X in either case? Page 4 Solution: A basis for X regarded as a real vector space is e1,...,en, ie1, . , ien . The dimension of X is n as a complex vector space and 2n{ as a real vector space.} 8. If M is a linearly dependent set in a complex vector space X,isM linearly dependent in X,regardedasarealvectorspace? 1 i Solution: No. Let X = 2,withK = ,andconsiderx = and y = . C C i 1 − x, y is a linearly dependent set in X since ix = y. Now suppose K = R,and { } ↵ + βi 0+0i ↵x + βy = = 0 = . ↵i β 0+0i − Since ↵,β can only be real numbers, we see that (↵,β)=(0, 0) is the only solution to the equation. Hence x, y is a linearly independent set in X = C2 over R. { } 9. On a fixed interval [a, b] R,considerthesetX consisting of all polynomials with real coefficients and of degree⇢ not exceeding a given n,andthepolynomialx =0(for which a degree is not defined in the usual discussion of degree). (a) Show that X, with the usual addition and the usual multiplication by real num- bers, is a real vector space of dimension n + 1. Find a basis for X. Solution: Let X be the set given in the problem. It is clear that X is a real vector space. Indeed, for any P, Q X,withdeg(P ), deg(Q) n, 2 deg(P + Q) n and deg(↵P) n for any ↵ R.Asimilarargumentfrom Problem 5 shows that 1,t,t2,...,t n is a linearly2 independent set in X,and since 1,t,t2,...,tn spans{ X,itisabasisof} X and X has dimension n +1. { } (b) Show that we can obtain a complex vector space X˜ in a similar fashion if we let those coefficients be complex. Is X asubspaceofX˜? Solution: No. Consider P (t)=t X,choose↵ = i,then↵P(t)=it / X. 2 2 10. If Y and Z are subspaces of a vector space X,showthatY Z is a subspace of X, but Y Z need not be one. Give examples. \ [ Solution: Let Y and Z be subspaces of a vector space X.Takeanyx, y Y Z, note that x, y are both elements of Y and Z.Forany↵,β K, ↵x +2βy \ Y (since Y is a subspace of X)and↵x+βy Z (since Z is a subspace2 of X). Hence2 ↵x + βy Y Z. 2 2 \ Page 5 ↵ 0 For the second part, consider Y = : ↵ R and Z = : ↵ R .It 0 2 ↵ 2 ⇢ ⇢ can be (easily) deduced that Y and Z are subspaces of R2,butY Z is not a 1 0 [ 1 subspace of R2.Toseethis,choosex = and y = ,thenx + y = / 0 1 1 2 Y Z. [ 11. If M = ? is any subset of a vector space X,showthatspanM is a subspace of X. 6 Solution: This is immediate since a (scalar) field K is closed under addition and sums of two finite sums remain finite. 12. (a) Show that the set of all real two-rowed square matrices forms a vector space X. What is the zero vector in X? Solution: This follows from Problem 1 and the definition of matrix addition and matrix scalar multiplication: we prove that R is a vector space over R 00 or .ThezerovectorinX is 0 = . C 00 (b) Determine dim X. Find a basis for X. Solution: We claim that dim X = 4. To prove this, consider the following four vectors in X 10 01 00 00 e = e = e = e = . 1 00 2 00 3 10 4 01 ↵ ↵ Suppose ↵ e + ...+ ↵ e = 0 = 1 2 ,wehave↵ = ↵ = ↵ = ↵ =0, 1 1 4 4 ↵ ↵ 1 2 3 4 3 4 i.e. e1,e2,e3,e4 is a linearly independent set in X. However, any set of 5 { } ab or more vectors of X is linearly dependent, since any x = X can be cd2 written as a linear combination of e1,e2,e3,e4 ,i.e.