Chapter 6 Inner product spaces
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1 Chapter 6 Inner product spaces
2 6.1 Inner products and norms Definition 1 Let V be a vector space over F. An inner product on V is a function, : V V F such that the following conditions hold. x+z,y = x,y + z,y cx,y = c x,y x,y = x,y x,x > 0 if x 0 and 0,0 = 0.
3 Definition 2 The adjoint of an m n matrix X is the n m matrix A such that (A ) ij = A ji Theorem 1 (6.1) x,y +z = x,y + x,z x,cy = c x,y x,0, = 0,x = 0 x,x = 0 if and only if x = 0 If x,y = x,z for every x V, then y = z.
4 Definition 3 Let V be an inner product space. For x V, the norm of x is x = x,x. Theorem 2 (6.2) Let V be an inner product space over F. For x,y V and c F (a) cx = c x (b) x = 0 if and only if x = 0 and x 0 for any x. (c) (Cauchy-Schwarz Inequality) x, y x y (d) (Triangle Inequality) x + y x + y
5 Proof. (c) Expand x cy 2 and apply with c = x,y y,y (d) Note that x,y + y,x = 2Re x,y and Re x,y x,y. Definition 4 Vectors x,y V are called orthogonal if x,y = 0. A set of vectors S V is called orthogonal if any two distinct vectors are orthogonal.
6 S is called orthonormal if it is orthogonal and x = 1 for every x S. Odd Town Problem There are n people living in an odd town and they form clubs. A club must contain an odd number of members and for any two distinct clubs there must be an even (possibly zero) number of people in both of them. What is the maxiumum number of clubs that can be formed?
7 6.2 The Gram-Schmidt orthogonalization Definition 5 An ordered basis which is orthonormal is called an orthonormal basis. Theorem 3 (6.3) Suppose S = {v 1,...,v k } is an orthogonal subset of V such that v i 0. For y span(s), y = k i=1 y,v i v i 2v i. Corollary 4 Any orthogonal set of non-zero vectors is linearly independent.
8 Theorem 5 (6.4) (Gram-Schmidt algorithm) Let S = {w 1,...,w n } be linearly independent subset. Define S = {v 1,...,v n } as follows, v 1 := w 1 and for k 2 v k = w k k 1 j=1 w k,v j v j 2 v j. Then S is orthogonal and span(s ) = span(s) Proof. This is induction on S. For the inductive step, first check that v k 0 and then compute v k,v i for i < k. dim(span(s k )) = dim(span(s k )) because S k is linearly independent. Theorem 6 (6.5) Let V be a finite-dimensional inner product space and let β = {v 1,...,v n } be an orthonormal basis for V. Then for every x V, x = n i=1 x,v i v i.
9 Corollary 7 Let β = {v 1,...,v n } be orthonormal, let T : V V be linear, and let A = [T] β. Then A ij = T(v j ),v i. Fourier coefficient of x relative to β is x,y where y β. Definition 6 Let S be a non-empty subset of V. Define S = {x V x,y = 0 for every y S}. Note S is a subspace of V.
10 Theorem 8 (6.6) Let W be a finite-dimensional subspace of an inner product space V and let y V. Then there exist unique u W and z W such that y = u+z. Furthermore, if {v 1,...,v k } is an orthonormal basis for W, then u = y,v i v i. Proof. Let u = y,v i v i and z = y u. Check that z W. For the uniqueness u u W and z z W, Corollary 9 For any x W, y x y u and if y x = y u then x = u.
11 Theorem 10 (6.7) Suppose S = {v 1,...,v k } is an orthonormal set in an n-dimensional inner product space V. Then (a) S can be extended to an orthonormal basis {v 1,...,v n } for V; (b) {v k+1,...,v n } is an orthonormal basis for (span(s)). (c) If W is a subspace of V, then dim(w)+dim(w ) = dim(v).
12 6.3 The adjoint of a linear operator Theorem 11 (6.8) Let V be a finite-dimensional inner product space over F, and let g : V F be a linear transformation. Then there exists a unique vector y V such that g(x) = x,y for every x V. Proof. Take an orthonormal basis β = {v,...,v n } and define y = g(v i )v i. Theorem 12 (6.9) Let V be a finite-dimensional inner product space and let T : V V be linear. Then there exists a unique function T : V V such that T(x),y = x,t (y) for all x,y. Furthermore T is linear.
13 Proof. Fix y. Check that g(x) = T(x),y is linear. Theorem 6.8 gives unique y such that g(x) = x,y. Define T (y) = y. Note that T is a function and check that it is linear. Theorem 13 (6.10) Let V be a finite-dimensional inner product space and let β be an orthonormal ordered basis for V. If T is a linear operator on V, then [T ] β = [T] β.
14 Theorem 14 (6.11) Let V be an inner-product space, and let T,U be linear operators on V. Then (a) (T +U) = T +U (b) (ct) = ct (c) (TU) = U T (d) T = T (e) I = I
15 6.4 Normal and self-adjoint operators Lemma 15 Let T be a linear operator on a finite-dimensional inner product space V. If T has an eigenvector, then so does T. Theorem 16 (6.14) (Schur)Let T be a linear operator on a finite-dimensional inner product space V and suppose that the characteristic polynomial of T splits. Then there exists an orthonormal basis β for V such that [T] β is upper triangular. Proof. Induction on n := dim(v). T has a unit eigenvector z; W := span(z). Show that W is T-invariant and dim(w ) = n 1.
16 The characteristic polynomial of T W divided f T (t) (so it splits) and by IH for some γ [T W ] γ is upper triangular. Let β = γ {z}. Then [T] β is upper-triangular. Note:If V has an orthonormal basis of eigenvectors of T, then TT = T T. Definition 7 T : V V is normal if TT = T T. A M n n (F) is normal if AA = A A. Theorem 17 (6.15) Let T be a normal operator on V. Then
17 (a) T(x) = T (x) (b) T ci is normal for every c F. (c) If x is an eigenvector of T, then x is an eigenvector of T. (d) If λ 1,λ 2 are distinct eigenvalues of T with eigenvectors x 1,x 2, then x 1,x 2 = 0. Theorem 18 (6.16) Let T be a linear operator on a finitedimensional complex inner-product space V. Then T is normal if and only if there exists an orthonormal basis for V consisting of eigenvectors of T.
18 Proof. Suppose T is normal. T splits over C and so apply Schur s lemma to get an orthonormal basis β = {v 1,...,v n }. A := [T] β is upper triangular and so T(v 1 ) = A 11 v 1. Show that e k is an eigenvector by induction on k using the fact that A jk = T(v k ),v j. The converse is easy.
19 Definition 8 T : V V is self-adjoint (Hermitian) if T = T. A M n n (F) is Hermitian if A = A. Lemma 19 Let T be a Hermitian operator on a finite-dim inner product space V. Then (a) All eigenvalues of T are real. (b) If V is a real inner product space, then the characteristic polynomial splits. Theorem 20 (6.17) Let T be a linear operator on a finitedimensional real inner-product space V. Then T is Hermitian
20 if and only if there exists an orthonormal basis for V consisting of eigenvectors of T. Proof. The characteristic polynomial splits and so we may apply Schur s lemma. A := [T] β is upper triangular and so A. Thus it must be a diagonal matrix.
21 6.5 Unitary and orthogonal operators and their matrices Definition 9 Let V be a finite-dimensional inner product space over F and let T : V V be linear. If T(x) = x for every x V, then T is called unitary if F = C and orthogonal if F = R. Theorem 21 (6.18) Let T be a linear operator on a finitedimensional inner-product space V. Then the following statements are equivalent. (a) TT = T T = I (b) T(x),T(y) = x,y for all x,y V
22 (c) If β is an orthonormal basis, then so is T(β). (d) There exists an orthonormal basis β such that T(β) is orthonormal. (e) T(x) = x for every x. Proof. (d) (e) Let β = {v 1,...,v n } be orthonormal such that T(β) is orthonormal. Take x V. Then x = a i v i and x 2 = a i 2 = T(x) 2.
23 (e) (a) x,x = x,t T(x) by (e). Thus x,(i T T)(x) = 0 for every x. Set U := I T T. Then U is self-adjoint and so there is an orthonormal basis consisting of eigenvectors of U. Check that U(x) = λx implies λ = 0; U = T 0 ; T T = I; TT = I as well because [T] β is a square matrix. Definition 10 A square matrix A is called orthogonal if A t A = AA t = I and unitary if A A = AA = I. Note: AA = I iff rows of A are orthonormal.
24 A A = I iff columns of A are orthonormal. If λ is an eigenvalue of a unitary (orthogonal) matrix, then λ = 1. Definition 11 A,B M n n (C) (A,B M n n (R)) are unitarily (orthogonally) equivalent if there exists a unitary (orthogonal) matrix P such that A = P 1 BP. Theorem 22 (6.19) Let A M n n (C). Then A is normal if and only if A is unitarily equivalent to a diagonal matrix. Theorem 23 (6.20) Let A M n n (R). Then A is symmetric if and only if A is orthogonally equivalent to a diagonal matrix.
25 Theorem 24 (6.21) (Schur) Let A M n n (F) and suppose f A (t) splits over F. (a) If F = C, then A is unitarily equivalent to a complex uppertriangular matrix. (b) If F = R, then A is orthogonally equivalent to a real uppertriangular matrix.
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