Inner Product Spaces and Orthogonality

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1 Inner Product Spaces and Orthogonality week 3-4 Fall 2006 Dot product of R n The inner product or dot product of R n is a function, defined by u, v a b + a 2 b a n b n for u a, a 2,, a n T, v b, b 2,, b n T R n The inner product, satisfies the following properties: () Linearity: au + bv, w a u, w + b v, w (2) Symmetric Property: u, v v, u (3) Positive Definite Property: For any u V, u, u 0; and u, u 0 if and only if u 0 With the dot product we have geometric concepts such as the length of a vector, the angle between two vectors, orthogonality, etc We shall push these concepts to abstract vector spaces so that geometric concepts can be applied to describe abstract vectors 2 Inner product spaces Definition 2 An inner product of a real vector space V is an assignment that for any two vectors u, v V, there is a real number u, v, satisfying the following properties: () Linearity: au + bv, w a u, w + b v, w (2) Symmetric Property: u, v v, u (3) Positive Definite Property: For any u V, u, u 0; and u, u 0 if and only if u 0 The vector space V with an inner product is called a (real) inner product space Example 2 For x x x 2, y y y 2 R 2, define x, y 2x y x y 2 x 2 y + 5x 2 y 2 Then, is an inner product on R 2 It is easy to see the linearity and the symmetric property As for the positive definite property, note that Moreover, x, x 0 if and only if x, x 2x 2 2x x 2 + 5x 2 2 (x + x 2 ) 2 + (x 2x 2 ) 2 0 x + x 2 0, x 2x 2 0, which implies x x 2 0, ie, x 0 This inner product on R 2 is different from the dot product of R 2

2 For each vector u V, the norm (also called the length) of u is defined as the number u : u, u If u, we call u a unit vector and u is said to be normalized For any nonzero vector v V, we have the unit vector ˆv v v This process is called normalizing v Let B { u, u 2,, u n be a basis of an n-dimensional inner product space V For vectors u, v V, write u x u + x 2 u x n u n, The linearity implies We call the n n matrix v y u + y 2 u y n u n u, v n n x i u i, y j u j i n i j j n x i y j u i, u j u, u u, u 2 u, u n u 2, u u 2, u 2 u 2, u n A u n, u u n, u 2 u n, u n the matrix of the inner product, relative to the basis B Thus, using coordinate vectors u B x, x 2,, x n T, v B y, y 2,, y n T, we have u, v u T BAv B 3 Examples of inner product spaces Example 3 The vector space R n with the dot product u v a b + a 2 b a n b n, where u a, a 2,, a n T, v b, b 2,, b n T R n, is an inner product space The vector space R n with this special inner product (dot product) is called the Euclidean n-space, and the dot product is called the standard inner product on R n Example 32 The vector space Ca, b of all real-valued continuous functions on a closed interval a, b is an inner product space, whose inner product is defined by b f, g f(t)g(t)dt, a f, g Ca, b Example 33 The vector space M m,n of all m n real matrices can be made into an inner product space under the inner product A, B tr(b T A), where A, B M m,n 2

3 For instance, when m 3, n 2, and for we have Thus A a a 2 a 2 a 22 a 3 a 32, B b b 2 b 2 b 22 b 3 b 32 B T b a A + b 2 a 2 + b 3 a 3 b a 2 + b 2 a 22 + b 3 a 32 b 2 a + b 22 a 2 + b 32 a 3 b 2 a 2 + b 22 a 22 + b 32 a 32, A, B b a + b 2 a 2 + b 3 a 3 +b 2 a 2 + b 22 a 22 + b 32 a a ij b ij i j This means that the inner product space ( M 3,2,, ) is isomorphic to the Euclidean space ( R 3 2, ) 4 Representation of inner product Theorem 4 Let V be an n-dimensional vector space with an inner product,, and let A be the matrix of, relative to a basis B Then for any vectors u, v V, u, v x T Ay where x and y are the coordinate vectors of u and v, respectively, ie, x u B and y v B Example 4 For the inner product of R 3 defined by x, y 2x y x y 2 x 2 y + 5x 2 y 2, where x x x 2, y y y 2 R 2, its matrix relative to the standard basis E { e, e 2 is e, e A e, e 2 2 e 2, e e 2, e 2 5 The inner product can be written as x, y x T Ay x, x y y 2 We may change variables so the the inner product takes a simple form For instance, let { x (2/3)x + (/3)x 2 x 2 (/3)x (/3)x, 2 We have { y (2/3)y + (/3)y 2 y 2 (/3)y (/3)y 2 ( 2 x, y 2 3 x + )( 2 3 x 2 3 y + ) 3 y 2 ( 2 3 x + )( 3 x 2 3 y ) 3 y 2 ( 3 x )( 3 x 2 3 y ) 3 y 2 ( +5 3 x )( 3 x 2 3 y ) 3 y 2 x y + x 2y 2 x T y 3

4 This is equivalent to choosing a new basis so that the matrix of the inner product relative to the new basis is the identity matrix In fact, the matrix of the inner product relative to the basis B { u 2/3 /3 is the identity matrix, ie, u, u u 2, u u, u 2 u 2, u 2, u 2 /3 /3 0 0 Let P be the transition matrix from the standard basis {e, e 2 to the basis {u, u 2, ie, 2/3 /3 u, u 2 e, e 2 P e, e 2 /3 /3 Let x be the coordinate vector of the vector x relative the basis B (The coordinate vector of x relative to the standard basis is itself x) Then It follows that x e, e 2 x u, u 2 x e, e 2 P x x P x Similarly, let y be the coordinate vector of y relative to B Then y P y Note that x T x T P T Thus, on the one hand by Theorem, On the other hand, x, y x T I n y x T y x, y x T Ay x T P T AP y Theorem 42 Let V be a finite-dimensional inner product space Let A, B be matrices of the inner product relative to bases B, B of V, respectively If P is the transition matrix from B to B Then 5 Cauchy-Schwarz inequality B P T AP Theorem 5 (Cauchy-Schwarz Inequality) For any vectors u, v in an inner product space V, u, v 2 u, u v, v Equivalently, u, v u v Proof Consider the function y y(t) : u + tv, u + tv, t R Then y(t) 0 by the third property of inner product Note that y(t) is a quadratic function of t In fact, Thus the quadratic equation y(t) u, u + tv + tv, u + tv u, u + 2 u, v t + v, v t 2 u, u + 2 u, v t + v, v t 2 0 has at most one solution as y(t) 0 This implies that its discriminant must be less or equal to zero, ie, ( 2 u, v ) 2 4 u, u v, v 0 The Cauchy-Schwarz inequality follows 4

5 Theorem 52 The norm in an inner product space V satisfies the following properties: (N) v 0; and v 0 if and only if v 0 (N2) cv c v (N3) u + v u + v For nonzero vectors u, v V, the Cauchy-Schwarz inequality implies The angle θ between u and v is defined by The angle exists and is unique 6 Orthogonality cos θ u, v u v u, v u v Let V be an inner product space Two vectors u, v V are said to be orthogonal if u, v 0 Example 6 For inner product space C π, π, the functions sin t and cos t are orthogonal as sin t, cos t π π sin t cos t dt 2 sin2 t π π Example 62 Let u a, a 2,, a n T R n The set of all vector of the Euclidean n-space R n that are orthogonal to u is a subspace of R n In fact, it is the solution space of the single linear equation u, x a x + a 2 x a n x n 0 Example 63 Let u, 2, 3, 4, 5 T, v 2, 3, 4, 5, 6 T, and w, 2, 3, 3, 2 T R 5 The set of all vectors of R 5 that are orthogonal to u, v, w is a subspace of R 5 In fact, it is the solution space of the linear system x + 2x 2 + 3x 3 + 4x 4 + 5x 5 0 2x + 3x 2 + 4x 3 + 5x 4 + 6x 5 0 x + 2x 2 + 3x 3 + 3x 4 + 2x 5 0 Let S be a nonempty subset of an inner product space V We denote by S the set of all vectors of V that are orthogonal to every vector of S, called the orthogonal complement of S in V In notation, { S : v V v, u 0 for all u S If S contains only one vector u, we write u {v V v, u 0 Proposition 6 Let S be a nonempty subset of an inner product space V Then the orthogonal complement S is a subspace of V 5

6 Proof To show that S is a subspace We need to show that S is closed under addition and scalar multiplication Let u, v S and c R Since u, w 0 and v, w 0 for all w S, then u + v, w u, w + v, w 0, cu, w c u, w 0 for all w S So u + v, cu S Hence S is a subspace of R n Proposition 62 Let S be a subset of an inner product space V Then every vector of S is orthogonal to every vector of Span (S), ie, u, v 0, for all u Span (S), v S Proof For any u Span (S), the vector u must be a linear combination of some vectors in S, say, u a u + a 2 u a k u k Then for any v S, u, v a u, v + a 2 u 2, v + + a n u n, v 0 Example 64 Let A be an m n real matrix Then Nul A and Row A are orthogonal complements of each other in R n, ie, Nul A ( Row A ), ( Nul A ) Row A 7 Orthogonal sets and bases Let V be an inner product space A subset S { u, u 2,, u k of nonzero vectors of V is called an orthogonal set if every pair of vectors are orthogonal, ie, u i, u j 0, i < j k An orthogonal set S { u, u 2,, u k is called an orthonormal set if we further have u i, i k An orthonormal basis of V is a basis which is also an orthonormal set Theorem 7 (Pythagoras) Let v, v 2,, v k be mutually orthogonal vectors Then v + v v k 2 v 2 + v v k 2 Proof For simplicity, we assume k 2 If u and v are orthogonal, ie, u, v 0, then u + v 2 u + v, u + v u, u + v, v u 2 + v 2 Example 7 The three vectors v, 2, T, v 2 2,, 4 T, v 3 3, 2, T are mutually orthogonal Express the the vector v 7,, 9 T as a linear combination of v, v 2, v 3 Set x v + x 2 v 2 + x 3 v 3 v 6

7 There are two ways to find x, x 2, x 3 Method : Solving the linear system by performing row operations to its augmented matrix v, v 2, v 3 v, we obtain x 3, x 2, x 3 2 So v 3v v 2 + 2v 3 Method 2: Since v i v j for i j, we have where i, 2, 3 Then We then have v, v i x v + x 2 v 2 + x 3 v 3, v i x i v i, v i, x i v, v i, i, 2, 3 v i, v i x , x , x Theorem 72 Let v, v 2,, v k be an orthogonal basis of a subspace W Then for any w W, Proof Trivial 8 Orthogonal projection w v, w v, v v + v 2, w v 2, v 2 v v k, w v k, v k v k Let V be an inner product space Let v be a nonzero vector of V We want to decompose an arbitrary vector y into the form y αv + z, where z v Since z v, we have This implies that We define the vector v, y αv, v α v, v α Proj v (y) v, y v, v v, y v, v v, called the orthogonal projection of y along v The linear transformation Proj u : V V is called the orthogonal projection of V onto the direction v Proposition 8 Let v be a nonzero vector of the Euclidean n-space R n Then the orthogonal projection Proj u : R n R n is given by Proj v (y) v v vvt y; and the orthogonal projection Proj v : R n R n is given by ( Proj v (y) I ) v v vvt y 7

8 Write the vector v as v a, a 2,, a n T The for any scalar c, ca a c ca 2 cv a 2 c vc, ca n a n c where c is the matrix with the only entry c Note that v y v T y Then the orthogonal projection Proj v can be written as ( ) Proj v (y) (v y)v v v ( ) vv y v v ( ) vv T y v v This means that the standard matrix of Proj v is ( ) vv T v v Indeed, v is an n matrix and v T is a n matrix, the product vv T is an n n matrix The orthogonal projection Proj v : R n R n is given by ( Proj v (y) y Proj v (y) I ) v v vvt y This means that the standard matrix of Proj v is ( ) I vv T v v Example 8 Find the linear mapping from R 3 to R 3 that is a the orthogonal projection of R 3 ont the plane x + x 2 + x 3 0 To find the orthogonal projection of R 3 onto the subspace v, where v,, T, we find the following orthogonal projection Proj v (y) 3 ( v y v v ) v y + y 2 + y 3 3 Then the orthogonal projection of y onto v is given by Proj v y y Proj v (y) y y 2 y 3 ( I v v vvt y y 2 y 3 ) y 8

9 Let W be a subspace of V, and let v, v 2,, v k be an orthogonal basis of W We want to decompose an arbitrary vector y V into the form y w + z with w W and z W Then there exist scalars α, α 2,, α k such that Since z v, z v 2,, z v k, we have Then We thus define ŷ α v + α 2 v α k v k v i, y v i, α v + + α r v k + z α i v i, v i α i v i, y v i, v i, i k Proj W (y) v, y v, v v + v 2, y v 2, v 2 v v k, y v k, v k v k, called the orthogonal projection of v along W The linear transformation is called the orthogonal projection of V onto W Proj W : V V Theorem 82 Let V be an n-dimensional inner product space Let W be a subspace with an orthogonal basis B { v, v 2,, v k Then for any v V, Proj W (y) v, y v, v v + v 2, y v 2, v 2 v v k, y v k, v k v k, Proj W (y) y Proj W (y) In particular, if B is an orthonormal basis of W, then Proj W (y) v, y v + v 2, y v v k, y v k Proposition 83 Let W be a subspace of R n Let U u, u 2,, u k be an n k matrix, whose columns form an orthonormal basis of W Then the orthogonal projection Proj W : R n R n is given by Proof For any y R n, we have Proj W (y) UU T y Proj W (y) (u y)u + (u 2 y)u (u k y)u k Note that Then U T y u T u T 2 u T k y u T y u T 2 y u T k y UU T y u, u 2,, u k Proj W (y) u y u 2 y u k y u y u 2 y u k y 9

10 Example 82 Find the orthogonal projection where W is the plane x + x 2 + x 3 0 By inspection, the following two vectors form an orthogonal basis of W Then v Proj W (y) Proj W : R 3 R 3, 0 and v 2 ( ) v y v v y y ( v2 y v + v 2 v2 0 + y + y 2 2y Example 83 Find the matrix of the orthogonal projection where The following two vectors Proj W : R 3 R 3, W Span, 0 u / 3 / 3 / 3, u 2 / 2 / 2 0 ) v 2 form an orthonormal basis of W Then the standard matrix of Proj W is the product / 3 / 2 / 3 / 2 / / 3 / 3 / 3 / 2 /, which results the matrix 5/6 /6 /3 /6 5/6 /3 /3 /3 /3 Alternatively, the matrix can be found by computing the orthogonal projection: 0 Proj W (y) y + y 2 + y y y 2 + 2y 3 y + 5y 2 + 2y 3 2y + 2y 2 + 2y 3 5/6 /6 /3 /6 5/6 /3 /3 /3 /3 y y 2 y 3 + y y 2 2 y y 2 y 3 0

11 9 Gram-Schmidt process Let W be a subspace of an inner product space V Let B { v, v 2,, v k be a basis of W, not necessarily orthogonal An orthogonal basis B { w, w 2,, w k may be constructed from B as follows: More precisely, w v, W Span { w, w 2 v 2 Proj W (v 2 ), W 2 Span {w, w 2, w 3 v 3 Proj W2 (v 3 ), W 3 Span {w, w 2, w 3, w k v k Proj Wk (v k ), W k Span {w,, w k, w k v k Proj Wk (v k ) w v, w 2 v 2 w, v 2 w, w w, w 3 v 3 w, v 3 w, w w w 2, v 3 w 2, w 2 w 2, w k v k w, v k w, w w w 2, v k w 2, w 2 w 2 w k, v k w k, w k w k The method of constructing the orthogonal vector w, w 2,, w k is known as the Gram-Schmidt process Clearly, the vector w, w 2,, w k are linear combinations of v, v 2,, v k Conversely, the vectors v, v 2,, v k are also linear combinations of w, w 2,, w k : Hence v w, v 2 w, v 2 w, w w + w 2, v 3 w, v 3 w, w w + w 2, v 3 w 2, w 2 w 2 + w 3, v k w, v k w, w w + w 2, v k w 2, w 2 w w k, v k w k, w k w k + w k Span { v, v 2,, v k Span { w, w 2,, w k Since B {v, v 2,, v k is a basis for W, so is the set B { w, w 2,, w k Theorem 9 The basis { w, w 2,, w k constructed by the Gram-Schmidt process is an orthogonal basis of W Moreover, v, v 2, v 3,, v k w, w 2, w 3,, w k R, where R is the k k upper triangular matrix w,v 2 w,v 3 w,w w,w w 0 2,v 3 w 2,w w,v k w,w w 2,v k w 2,w 2 w 3,v k w 3,w 3

12 Example 9 Let W be the subspace of R 4 spanned by v, v 2 0 Construct an orthogonal basis for W, v 2 Set w v Let W Span {w To find a vector w 2 in W that is orthogonal to W, set 0 0 w 2 v 2 Proj W v 2 v 2 w, v 2 w, w w Let W 2 Span {w, w 2 To find a vector w 3 in W that is orthogonal to W 2, set w 3 v 3 Proj W2 v 3 v 3 w, v 3 w, w w w 2, v 3 w 2, w 2 w /3 /3 2/3 0 Then the set {w, w 2, w 3 is an orthogonal basis for W Theorem 92 Any m n real matrix A can be written as A QR, called a QR-decomposition, where Q is an m n matrix whose columns are mutually orthogonal, and R is an n n upper triangular matrix whose diagonal entries are Example 92 Find a QR-decomposition of the matrix A Let v, v 2, v 3, v 4 be the column vectors of A Set w v 0 2

13 Then v w Set w 2 v 2 w, v 2 w, w w /2 /2 Then v 2 (/2)w + w 2 Set w 3 v 3 w, v 3 w, w w w 2, v 3 w 2, w 2 w /2 /2 0 Then v 3 (3/2)w + w 2 + w 3 Set w 4 v 4 w, v 4 w, w w w 2, v 4 w 2, w 2 w /3 2/3 2/3 /2 /2 Then v 4 (/2)w + (/3)w 2 + w 4 Thus matrixes Q and R for QR-decomposition of A are as follows: /2 0 2/3 Q 0 0 2/3, /2 0 2/3 /2 3/2 /2 R 0 / Orthogonal matrix Let V be an n-dimensional inner product space A linear transformation T : V V is called an isometry if for any v V, T (v) v Example 0 For the Euclidean n-space R n with the dot product, rotations and reflections are isometries Theorem 0 A linear transformation T : V V is an isometry if and only if T preserving inner product, ie, for u, v V, T (u), T (v) u, v Proof Note that for vectors u, v V, T (u + v) 2 T (u + v), T (u + v) T (u), T (u) + T (v), T (v) + 2 T (u), T (v) T (u) 2 + T (v) T (u), T (v), u + v 2 u + v, u + v u, u + v, v + 2 u, v u 2 + v u, v It is clear that the length preserving is equivalent to the inner product preserving 3

14 An n n matrix Q is called orthogonal if QQ T I, ie, Q Q T Theorem 02 Let Q be an n n matrix The following are equivalent (a) Q is orthogonal (b) Q T is orthogonal (c) The column vectors of Q are orthonormal (d) The row vectors of Q are orthonormal Proof (a) (b) : If QQ T I, then Q Q T So Q T (Q T ) T Q T Q I This means that Q T is orthogonal (a) (c) : Let Q u, u 2,, u n Note that Q T Q u T u T 2 u T n u T u u T u 2 u T u n u u T 2 u u T 2 u 2 u T 2 u n, u 2,, u n u T n u u T n u 2 u T n u n Thus Q T Q I is equivalent to u T i u j for i j and u T i u j 0 for i j This means that Q is orthogonal if and only if {u, u 2,, u n is an orthonormal basis of V Theorem 03 Let V be an n-dimensional inner product space with an orthonormal basis B { u, u 2,, u n Let P be an n n real matrix, and v, v 2,, v n u, u 2,, u n P Then B { v, v 2,, v n is an orthonormal basis if and only if P is an orthogonal matrix Proof For simplicity, we assume n 3 Since v, v 2, v 3 u, u 2, u 3 P, ie, We then have v p u + p 2 u 2 + p 3 u 3, v 2 p 2 u + p 22 u 2 + p 32 u 3, v 3 p 3 u + p 23 u 2 + p 33 u 3 v i, v j p i u + p 2i u 2 + p 3i u 3, p j u + p 2j u 2 + p 3j u 3 p i p j + p 2i p 2j + p 3i p 3j Note that B is an orthonormal basis is equivalent to and P is an orthogonal matrix if and only if The proof is finished v i, v j δ ij, p i p j + p 2i p 2j + p 3i p 3j δ ij Theorem 04 Let V be an n-dimensional inner product space with an orthonormal basis B { u, u 2,, u n Let T : V V be a linear transformation Then T is an isometry if and only if the matrix of T relative to B is an orthogonal matrix 4

15 Proof Let A be the matrix of T relative to the basis B Then T (u ), T (u 2 ),, T (u n ) u, u 2,, u n A Note that T is an isometry if and only if {T (u ), T (u 2 ),, T (u n ) is an orthonormal basis of V, and that {T (u ), T (u 2 ),, T (u n ) is an orthonormal basis if and only the transition matrix A is an orthogonal matrix Example 02 The matrix A is not an orthogonal matrix The set 0 2 B,, 0 2 of the column vectors of A is an orthogonal basis of R 3 However, the set of the row vectors of A is not an orthogonal set The matrix / 3 / 2 / 6 U / 3 / 2 / 6 / 3 0 2/ 6 is an orthogonal matrix For the vector v 3, 0, 4 T, we have / 3 / 2 / 6 Uv / 3 / 2 / 6 / 3 0 2/ / / 6 3 8/ 6 The length of v is and the length of Uv is v ) 2 ( ) 2 Uv 2( 3 + 4/ / 6 5 Diagonalizing real symmetric matrices Let V be an n-dimensional real inner product space A linear mapping T : V V is said to be symmetric if T (u), v u, T (v) for all u, v V Example Let A be a real symmetric n n matrix Let T : R n R n be defined by T (x) Ax Then T is symmetric for the Euclidean n-space In fact, for u, v R n, we have T (u) v (Au) v (Au) T v u T A T v u T Av u Av u T (v) Proposition Let V be an n-dimensional real inner product space with an orthonormal basis B {u, u 2,, u n Let T : V V be a linear mapping whose matrix relative to B is A Then T is symmetric if and only the matrix A is symmetric Proof Note that T (u ), T (u 2 ),, T (u n ) u, u 2,, u n a a 2 a n a 2 a 22 a 2n a n a n2 a nn 5

16 Alternatively, If T is symmetric, then T (u j ) n a ij u i, j n i a ij u i, T (u j ) T (u i ), u j a ji So A is symmetric Conversely, if A is symmetric, then for vectors u n i a iu i, v n i b iu i, we have n n T (u), v a i b j T (u i ), u j a i b j a ji So T is symmetric i,j n a i b j a ij i,j u, T (v) i,j n a i b j u i, T (u j ) Theorem 2 The roots of characteristic polynomial of a real symmetric matrix A are all real numbers Proof Let λ be a (possible complex) root of the characteristic polynomial of A, and let v be a (possible complex) eigenvector for the eigenvalue λ Then Note that Thus Av λv i,j λ v 2 λv v (Av) v (Av) T v v T A T v v T A v v T Ā v v T Av v T λv v T ( λ v) λv T v λv v λ v 2 (λ λ) v 2 0 Since v 0, it follows that λ λ So λ is a real number Theorem 3 Let λ and µ be distinct eigenvalues of a symmetric linear transformation T : V V Then eigenvectors for λ are orthogonal to eigenvectors for µ Proof Let u be an eigenvector for λ, and let v be an eigenvectors for µ, ie, T (u) λu, T (v) µv Then Thus Since λ µ 0, it follows that u, v 0 λ u, v λu, v T (u), v u, T (v) u, µv µ u, v (λ µ) u, v 0 Theorem 4 Let V be an n-dimensional real inner product space Let T : V V be a symmetric linear mapping Then V has an orthonormal basis of eigenvectors of T Proof We proceed by induction on n For n, it is obviously true Let λ be an eigenvalue of T, and let u be a unit eigenvector of T for the eigenvalue λ Let W : u For any w W, T (w), u w, T (u ) w, λ u λ w, u 0 This means that T (w) W Thus the restriction T W : W W is a symmetric linear transformation Since dim W n, by induction hypothesis, W has an orthonormal basis {u 2,, u n of eigenvectors of T W Clearly, B {u, u 2,, u n is an orthonormal basis of V, and u, u 2,, u n are eigenvectors of T 6

17 Theorem 5 (Real Spectral Theorem) Any real symmetric matrix A can be diagonalized by an orthogonal matrix More specifically, there exists an orthonormal basis B {u, u 2,, u n of R n such that Au i λ i u i, i n, Q AQ Q T AQ Diag λ, λ 2,, λ n, where Q u, u 2,, u n ; and spectral decomposition A λ u u T + λ 2 u 2 u T λ n u n u T n Proof Let T : R n R n be defined by T (x) Ax For vectors u, v R n, T (u) v (Au) v (Au) T v u T A T v u T Av u (Av) u T (v) Then T is symmetric Thus R n has an orthonormal basis B {u, u 2,, u n of eigenvectors of T Let and Q u, u 2,, u n Then Alternatively, T (u i ) λ i u i, i n; Q AQ Q T AQ Diag λ, λ 2,, λ n D A QDQ QDQ T λ 0 0 u T 0 λ 2 0 u T 2 u, u 2,, u n 0 0 λ n u T n u T u T 2 λ u, λu 2,, λ n u n u T n λ u u T + λ 2 u 2 u T λ n u n u T n Note It is clear that if a real square matrix A is orthogonally diagonalizable, then A is symmetric Example 2 Is the matrix A orthogonally diagonalizable? 3 3 The characteristic polynomial of A is (t) (t + 2) 2 (t 7) There are eigenvalues λ 2 and λ 2 7 For λ 2, there are two independent eigenvectors v,, 0 T, v 2, 0, T Set w v, w 2 v 2 v 2 w w w w /2, /2, T 7

18 Then u / 2 / 2 0 form an orthonormal basis of E λ For λ 2 7, there is one independent eigenvector, u 2 v 3,, T / 6 / 6 2/ 6 The orthonormal basis of E λ2 is Then the orthogonal matrix Q diagonalizes the symmetric matrix A u 3 / 3 / 3 / 3 / 2 / 6 / 3 / 2 / 6 / 3 0 2/ 6 / 3 An n n real symmetric matrix A is called positive definite if, for any nonzero vector u R n, u, Au u T Au > 0 Theorem 6 Let A be an n n real symmetric matrix Let, be defined by u, v u T Av, u, v R n Then, is an inner product on R n if and only if the matrix A is positive definite Theorem 7 Let A be the matrix of an n-dimensional inner product space V relative to a basis B Then for u, v V, u, v u T BAv B Moreover, A is positive definite 2 Complex inner product spaces Definition 2 Let V be a complex vector space An inner product of V is a function, : V V C satisfying the following properties: () Linearity: au + bv, w a u, w + b v, w (2) Conjugate Symmetric Property: u, v v, u (3) Positive Definite Property: For any u V, u, u 0; and u, u 0 if and only if u 0 The complex vector space V with an inner product is called a complex inner product space Note that the Conjugate Symmetric Property implies that u, av + bw ā u, v + b u, w For any u V, since u, u is a nonnegative real number, we define the length of u to be the real number u u, u 8

19 Theorem 22 (Cauchy-Schwarz Inequality) Let V be a complex inner product space Then for any u, v V, u, v u v Like real inner product space, one can similarly define orthogonality, the angle between two vectors, orthogonal set, orthonormal basis, orthogonal projection, and Gram-Schmidt process, etc Two vectors u, v in a complex inner product space V are called orthogonal if u, v 0 When both u, v V are nonzero, the angle θ between u and v is defined by cos θ u, v u v A set { v, v 2,, v k of nonzero vectors of V is called an orthogonal set if the vectors v, v 2,, v k are mutually orthogonal A basis of V is called an orthogonal basis if their vectors are mutually orthogonal; a basis Bis called an orthonormal basis if B is an orthogonal basis and every vector of B has unit length Example 2 (Complex Euclidean space C n ) For vectors u z,, z n T,, v w,, w n T C n, define u, v u T v z w + z 2 w z n w n Then, is an inner product of C n, called the standard inner product on C n The vector space C n with the standard inner product is called the complex Euclidean n-space Theorem 23 Let B { v, v 2,, v n be an orthogonal basis of an inner product space V Then for any v V, v v, v v, v v + v, v 2 v 2, v 2 v v, v n v n, v n v n Let W be a subspace of V with an orthogonal basis {u, u 2,, u k Then the orthogonal projection Proj W : V V is given by Proj W (v) v, u u, u u + v, u 2 u 2, u 2 u v, u k u k, u k u k In particular, if V C n and {u, u 2,, u k is an orthonormal basis of W, then the standard matrix of the linear mapping Proj W is UŪT, ie, Proj W (x) UŪT x, where U u, u 2,, u k is an n k complex matrix Theorem 24 Let B { v, v 2,, v n be basis of an inner product space V Let A be the matrix of the inner product of V, ie, A a ij, where a ij v i, v j Then for u, v V, u, v u T A v, where u and v are the B-coordinate vectors of u and v, respectively where A complex square matrix A is called Hermitian if A A, A : ĀT A complex square matrix A is called positive definite if for any x C n, x T Ax 0 9

20 Theorem 25 Let A be an n n complex matrix Then A is the matrix of an n-dimensional inner product complex vector space relative to a basis if and only if A is Hermitian and positive definite A complex square matrix A is called unitary if A A, or equivalently, AA A A I Let A a ij Then A is unitary means that a i ā j + a 2i ā 2j + + a ni ā nj { if i j 0 if i j Theorem 26 Let A be a complex square matrix Then the following statements are equivalent: (a) A is unitary (b) The rows of A form an orthonormal set (c) The columns of A form an orthonormal set Theorem 27 Let V be an n-dimensional complex inner product space with an orthonormal basis B { u, u 2,, u n Let U be an n n real matrix, and v, v 2,, v n u, u 2,, u n A Then B { v, v 2,, v n is an orthonormal basis if and only if A is a unitary matrix Proof For simplicity, we assume n 3 Since v, v 2, v 3 u, u 2, u 3 A, ie, v a u + a 2 u 2 + a 3 u 3, v 2 a 2 u + a 22 u 2 + a 32 u 3, v 3 a 3 u + a 23 u 2 + a 33 u 3 We then have v i, v j a i u + a 2i u 2 + a 3i u 3, Note that B is an orthonormal basis is equivalent to and A is an unitary matrix if and only if The proof is finished a j u + a 2j u 2 + a 3j u 3 a i ā j + a 2i ā 2j + a 3i ā 3j v i, v j δ ij, a i ā j + a 2i ā 2j + a 3i ā 3j δ ij Let V be an n-dimensional complex inner product space A linear mapping T : V V is a called an isometry of V if T preserves length of vector, ie, for any v V, T (v) v 20

21 Theorem 28 Let V be an n-dimensional inner product space, and let T : V V be a linear transformation The following statements are equivalent: (a) T preserves length, ie, for any v V, (b) T preserves inner product, ie, for u, v V, (c) T preserves orthogonality, ie, for u, v V, T (v) v T (u), T (v) v, v T (u), T (v) 0 u, v 0 Proof note that if T preserves inner product, of course it preserves length Now for vectors u, v V, we have T (u + v) 2 T (u + v), T (u + v) T (u), T (u) + T (v), T (v) + T (u), T (v) + T (v), T (u) T (u) 2 + T (v) 2 + 2R T (u), T (v) ; T ( u + v ) 2 T (u) 2 + T (v) 2 + 2I T (u), T (v) On the other hand, If T preserves length, then It follows that Equivalently, u + v 2 u 2 + v 2 + 2R u, v ; u + v 2 u 2 + v 2 + 2I u, v T (u) 2 u 2, T (v) 2 v 2, T (u + v) 2 u + v 2, T ( u + v ) 2 u + v 2 R T (u), T (v) R u, v, T (u), T (v) u, v I T (u), T (v) I u, v 2

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