Module 6. be a sesquilinear form which is hermitian; that is, s(x, u) = s(u, x). Then s(x, u) + s(x, u) =

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1 Module 6 Orthogonal and Orthonormal Basis Let X be a vector space over K, K = R or C We note that R C and the conjugation Let X X s K be a sesquilinear form which is hermitian; that is, s(x, u) = s(u, x) Then s(x, u) + s(x, u) = s(x, u) + s(u, x) R for each x, u X If now s is positive ie Q(x) = s(x, x) 0 (note that s(x, x) = s(x, x) because s is hermitian so that s(x, x) R), let us write a = s(x, x), b = s(u, u), c = s(x, u) so that a 0, b 0, cc = c 2 0 For any r R, we have 0 s(x urc, x urc) = s(x, x) s(x, u)rc s(u, x)rc + s(u, u)r 2 cc = a 2ccr + br 2 cc = c 2 br 2 2 c 2 r + a The quadratic expression c 2 br 2 2 c 2 r + a positive iff its discriminant 4 c 4 4 c 2 ba 0 and thus we have, ifc 0, c 2 ba In case c = 0, this is true any way since a, b are positive real numbers Thus in all situations we have c 2 ab ie s(x, u) 2 s(x, x)s(u, u) When s is a positive hermitian form, which is positive-definite in the sense that s(x, x) = 0 forces x X to be o X, we shall denote s(x, u) by (x u) and call it an inner product on X In this notation, we have just proved (x u) 2 x 2 u 2 () This is known as the Cauchy-Schwarz inequality; note that we wrote x 2 for s(x, x) The number is called the norm of x X A vector space X over K (K = R, C) is called the inner product space if we have equipped it with a preferred inner product (From the theory of quadratic equations, we know that if At 2 + Bt + C = A(t α)(t β) Assuming without loss of generality α > β, we have (t α) > 0 (t β) > 0 and (t β) < 0 (t α) < 0 so that (t α)(t β) > 0 always holds and the expression At 2 + Bt + c has the same sign as A unless t has a value lying between α and β in which case the expression has the sign opposite to that of A If α, β are real with α = β then At 2 + Bt + C = A(t α) 2 has the [ ] same sign which A has If α and β are complex then At 2 + Bt + C = A (t + B 2A )2 + 4AC B2 4A with 2 4AC B 2 0 since the roots are complex and we conclude that again the expression At 2 + Bt + C has the same sign which A has To sum up: Suppose A, B, C real numbers Then

2 2 (i) At 2 + Bt + C 0 for all t R iff B 2 4AC 0, A > 0 we have used this here with A = c 2 b, B = 2 c 2, C = a (ii) At 2 + Bt + C 0 for all t R iff B 2 4AC 0, A < 0 ) We say X X s K is sesquilinear iff s(λx, u + wµ) = λs(x, u) + λs(x, w)µ and s(x + u, wµ) = s(x, w) + s(u, w)µ ie s is conjugate-linear in the first and linear in the second variable; sesqui means one and a half This is the physicists s convention ; the mathematician s convention is linear in the first and conjugate-linear in the second variable Clearly, if s is sesquilinear in the physicist convention, s given by s(x, u) := s(u, x) is sesquilinear in the mathematician s convention; one can adhere to either of the two Conjugate-linear is also called semi-linear whence the name sesqui-linear 2 Suppose now X is an inner-product space Then (x u) = (z u) for each u X means (x z x z) = (x x) (x z) (z x) + (z z) = (z x) (z z) (z x) + (z z) = 0 and since s(w, u) := (w u) is positive-definite (so that (w w) = 0 w = 0), we get x z = 0 ie x = z Similarly, (u x) = (u z) for each u X means (x z x z) = (x x) (x z) (z x) + (z z) = (x z) (x z) (z z) + (z z) = 0 and again, since s(w, u) := (w u) is positive-definite, we get x z = 0 ie x = z To sum up: If s is positive -definite (and in particular if s is an inner product) s(u, x) = s(u, z) for each u X forces x = z(in particular (u x) = (u z) for each u X forces x = z); similarly, s(x, u) = s(z, u) for each u X forces x = z) Further, (x u) = 0 for all u X means (x x) = 0 as well and then positive-definiteness ensures x = 0 3 Since the inner product is positive, x = + (x x) 0; further, since it is positive-definite, x = + (x x) = 0 forces x = 0 For λ K (K = C, or R) we have xλ = + (xλ xλ) = + λ(x x)λ = + λλ(x x) = λ x = x λ ( (x x) R) Further, x+u 2 = (x+u x+u) = (x x) + (x u) + (u x) + (u u) = x 2 + 2Re(x u) + u 2 x (x u) + u 2 x x u + u 2 = ( x + u ) 2 To sum up, we have proved (using the Cauchy-Schwarz inequality in the last argument) The norm function X x x [0, ) satisfies: (i) x = 0 x = 0 (of course x 0 and 0 = 0 hold)

3 (ii) xλ = x λ = λ x 3 (iii) x + u x + u for all x, u X, λ H (or λ C or λ R) The inequality x + u x + u is called the triangle inequality 4 In an inner product space, u ± x 2 = (u ± x u ± x) = u 2 ± 2Re(u x) + x 2 so that u + x 2 + u x 2 = 2 [ u 2 + x 2] for each u, x X We refer to this equation as the parallelogram law 5 We write u x (read: u is orthogonal to x) iff (u x) = 0; this is clearly a symmetric relation (u x iff x u) The Cauchy-Schwarz inequality (u x) u x ensures that if u, x are nonzero, we have (u x) 0 u x (u x) u x so that (u x)0 u x We define θ = cos (u x) 0 u x to be the angle between u and x This is called (i) obtuse if (u x) 0 < 0, and (ii) acute if (u x) 0 > 0; we see it is a right angle if (u x) 0 = 0 Note that now u±x 2 = u 2 ±2 u x cos θ+ x 2 Clearly, if cos θ = 0 we get u±x 2 = u 2 + x 2 Moreover, u x cos θ = 0 but cos θ = 0 u x unless K = R { A set {x α X} is called an orthogonal set iff x α x β whenever α β Then this property if each x α 0 and each vector in this set has norm x α x α } also has 6 (i) A set {x α X} is called an orthonormal set iff (x α x β ) = δβ α (δα β if α = β δβ α = 0 otherwise is the Kronecker delta, (ii) An orthogonal set of nonzero vectors must be linearly independent (If {x α } is orthogonal and x λ + + x n λ n = 0 then 0 = (x i 0) = (x i x λ + + x n λ n ) = (x i x λ + + x n λ n ) = (x i x )λ + + (x i x n )λ n = x i λ i, ( (x i x j ) = x i if i = j 0 otherwise ; thus λ i = 0 for i n because x i 0 for any i) (iii) Suppose {x α } is orthonormal and u x α for each α forces u = 0 Then we can surely find

4 4 no u X with u = such that u x α for each α which means that {x α } cannot be enlarged to a bigger orthonormal set and if {x α } {u} is orthonormal, u must be one of the {x α } Thus we decide to call a set total orthonormal set when it is an orthonormal set which cannot be enlarged to a bigger orthonormal set with the property that x X, x u α for each α forces x to be 0, {u α } is total A total orthonormal set will also be a linearly independent set (since every orthonormal set is linearly independent but it is not necessarily a maximal linearly independent set since it is not proved that it cannot be enlarged to a bigger linearly independent set; what is proved is that it cannot be enlarged to a bigger orthonormal set Thus a total orthonormal set is not necessarily a Hamal basis of the vector space under consideration Hamel basis=basis of a vector space defined earlier) However, we have the following Theorem 6 If X is a finite dimensional innerproduct space over K(C or R) then every total orthonromal set is a Hamel basis of X Proof Suppose {x 0,, x n } is an orthogonal set of nonzero vectors in an innerproduct space X having the property that (x i x) = 0 for 0 i n forces x = 0 Let x X and z = n x i (x i x) = x 0 (x 0 x) + + x n (x n x); then (x i z) = (x i x) so that (x i z x) = 0 n for 0 i n and we have x = z = x i (x i x) which shows that x is in the vector space generated by {x 0,, x n } On the other hand, {x 0,, x n } is linearly independent (since every orthogonal set of nonzero vectors is) Thus dim X = n and {x 0,, x n } is a Hamel basis of X 7 In any innerproduct space X, one says that a sequence {x n X} is a Cauchy sequence iff it is possible to find an integer k such that x k x n+k 0 as n and one says that the innerproduct space is a Hilbert space iff each Cauchy sequence is convergent in the sense that if {x n X} is a Cauchy sequence, one can find x X with x n x 0 as n ; the vector x X is then called the limit of {x n } and one writes x n x or x = lim x n Each sequence {x n } yields a new sequence n n s n := x j and one writes x = x j iff s n x saying that the series x j converges to x j= j= Theorem 8 When {x n } n=0 is a linearly independent set in an innerproduct space X, y n := x n n u j (u j x n ); u n = yn y, y n 0 = x 0 supply an orthonormal set {u n } n=0 such that span{u 0,, u n } = span{x 0, x n }

5 Proof If {x 0 } is linearly independent, y 0 = x 0 0 and u 0 = y0 y 0 5 is well defined and clearly span{u 0 } = span{x 0 } with {u 0 } orthonormal Assume now that y 0,, y n and u 0,, u n have been defined as above with {x 0,, x n } linearly independent and span{u 0,, u n } = n span{x 0,, x n }, u 0,, u n orthonormal Then y n = x n (u j x n ), u j 0 if we have {x 0,, x n, x n } linearly independent and u n = yn n (u m u n ) = (u m x n u j (u j x n )) y n y n is well defined Further, for 0 m n, = [(u m x n ) (u m u m )(u m x n )] y n ( (u m u j ) = δm j for 0 j n ) = [(u m x n ) (u m x n )] y n ( (u m u m ) = ) = 0 Thus u n u m for 0 m n and of course (u n u n ) = y n (y n y n ) n {u 0,, u n } is an orthonormal set Further, x n = u n y n + u j (u j x n ) = ( (u n x n ) = (u n y n ) = y n = so that n u j (u j x n ) j= n (u n u j )(u j x n ) = (u n u n y n ) = y n ) and thus x n span{u 0,, u n } proving that span{u 0,, u n } = span{x 0,, x n } since it is known that span{x 0,, x n } = span{u 0,, u n } Thus, in particular, if we have a finite dimensional Hilbert space, an orthonormal Hamel basis exists for it The process of construction an orthonormal set out of a linearly independent set outlined in the theorem is called Gram-Schmidt orthonormalization Definition 9 Say a function X T X (X an innerproduct space) is adjointable iff there is a function X T + X (read T + as T dagger) such that the adjointness condition (T + x u) = (x T u) for each x, u X is satisfied; we say T + is the Hilbert adjoint of T Proposition 92 If X T X is adjointable, the Hilbert adjoit T + is unique; further, both X T X and X T + X are linear Proof () We first note that if x = x, we have (T + x u) = (x T u) = (x T u) = (T + x u) at each u X so that(by positive- definiteness, para 2 page 3 above) we have T + x = T + x Thus the adjointness condition, if it holds, does define a function X T + X (2) If there are two Hilbert adjoints, say T + and for T, we have (T + x u) = (x T u) = x u) at each u X and hence (by positive-definiteness) T + x = x; since this holds at each x X, we have T + = Thus there is at most one Hilbert adjoint for T

6 6 (3) If the Hilbert adjoint T + exists, we have (w T (x + uλ)) = (T + w x + uλ) = (T + w x) + (T + w u)λ = (w T x) + (w T u)λ = (w T x + (T u)λ) at each w X so that (by positive-definiteness again) we have T (x + uλ) = T x + (T u)λ; this being true at each x, u X, λ K, we see that T is (right)linear Further, (T + (x + wλ) u) = (x + wλ T u) = (x T u) + (wλ T u) = (x T u) + λ(w T u) = (T + x u) + λ(t + w u) = (T + x u) + ((T + w)λ u) = ((T + x + (T + w)λ) u) for each u X Therefore(by positive-definiteness) we get T + (x + wλ) = T + (x) + (T + (W ))λ, and this being true at each x, w X, λ K, we conclude that X T + X is (right)linear 0 The Next question is to determine which X T X are adjointable The question makes sense clearly for linear operators X T X only Suppose X is finite dimensional, say dim X = n < coordinatizing by say (e, ɛ), ɛ is the dual basis of e, we already know that X X s K is a hermitian sesquilinear form iff [s j i ] n n (s j i = s(e j, e i )) is a hermitian matrix ie s j i = si j (verify this) Thus s i i must be real (for K = C) For an inner product on X (thus X is the finite dimensional Hilbert space) we choose an orthonormal basis e (which is possible by Gram-Schmidt) and note n n n n n (x u) = ( e i x i e j u j ) = x i (e i e j )u j = x i u i ( (e i e j ) = 0 for i j (e i e i ) = ) i= Thus (x u) = x u where x is the conjugate-transpose of the column vector x = and u is the column vector u 0 u n (being calculated with respect to (e, ɛ) ) n Therefore (u T x) = u T x = u (x T ) (T being the matrix of T with respect to (e, ɛ) and T its congate-transpose) x 0 x n n

7 = (x T u) = (T u) x ( x = x) 7 = u x) where X X is the operator having the matrix T with respect to (e, ɛ) and sending u to T u But since (u T x) = (T + u x) and the Hilbert adjoint T + is unique, we conclude that = T + and thus the matrix of T + with respect to (e, ɛ) T To sum up: Every linear operator X T X is adjointable when X is a finite dimensional Hilbert space This result does not hold in infinite dimensional inner product space in general (when X is a Hilbert space and T := considerations in this course) sup T x <, it is however true We are not concerned with these x Now take K = C such that X is a finite dimensional complex Hilbert space, say C n Let L(X, X) be the space of all linear operators X X Then we know that L(X, X) has dimension n 2 and if (e, ɛ) is a coordinatization of C 2, e j ɛ i 0 i n is a basis of L(X, X) 0 j n For X A n X, we define trace (A) := ɛ i Ae i ; (Thus if we denote the associated matrix of n A by a = [a j i ], we have trace (A) = a i i; we then define the trace of a square matrix to be the sum of its diagonal elements and in view of the bijectivity K n n A a Lim(X, X) (just define A by Ax := ax for x X) the two definitions record the same mathematical concept) Clearly, L(X, X) trace C is a linear form (verify it) If (d, δ) δ is the dual basis of d, any other coordinatization for X we use the decomposition of identity X id P ei ɛ X = X i P dj δ X = X j X to find n n trace (A) = ɛ i Ae i = ɛ i A( de i ) n n = ɛ i Ad j δ j e i n n = δ j e i ɛ i Ad j n n = δ j d(ad j ) = δ j Ad j

8 8 which means that the trace of A is coordinate-free (trace (A) is the same whatever coordintization is chosen) 2 Considering (X, (e, ɛ)) A, B a, b (X, (e, ɛ)) we define (A B) := trace (A + B) = trace (a b) C (since a is the matrix of A + as noted in para (0) above) Then (i) (λa + B C + Dµ) = trace ((λa + b) (c + dµ)) = trace ((b + a λ)(c + dµ)) = trace (λa c + b c + λa dµ + b dµ) = λtrace (a c) + trace (b c) + λtrace (a + d)µ + trace (b d)µ = λ(a C) + (B C) + λ(a D)µ + (B D)µ which says that (A B) as defined is a sesquilinear form on L(X, X) (ii) (A B) = trace (a b) = trace (b (a ) ) = trace (b a) = (B A) which says it is hermitian, and n n n (iii) (A A) = trace (a a) = (a a) i i = (a ) i ja j i n n n = a j i aj i = n a j i 2 which says (A A) 0, (A A) = 0 iff each a j i = 0, 0 i n, 0 i j ie a = 0 ie A = 0 ie this is positive definite Thus (A, B) (A B) = trace (A B) = trace (a b) defines an inner product on L(X, X) into a complex Hilbert space of dimension n 2 This inner product is called the Hilbert-Schmidt inner product on L(X, X) and equipped with this inner product, L(X, X) is called the Liouville space of the Hilbert space X (the name Liouville space is used mostly in quantum mechanics; see Quantum Information: An Overview by Gregg Jaeger; spinger 2007 page 248) (iv) If dim V = n, we have essentially V = C n and V, the dual, also has dimension n when V is an inner product space, let us write, given z V, V ϕz C supplied by ϕ z (x) := (z x) Then ϕ z (x+uλ) = (z x+uλ) = (z x)+(z u)λ = ϕ z (x)+(ϕ z (u))λ so that ϕ z V Further, V ϕ V tr defined by ϕ(z) := ϕ z obeys (ϕ(λz + w))(x) = (λz + w x) = λ(z x) + (w x) = λϕ z (x) + ϕ w (x) at each x X which means ϕ(λz + w) = λϕ(z) + ϕ(w) showing that V ϕ V is conjugatelinear If ϕ z = ϕ w the at each x V we have ϕ z (x) = (z x) = ϕ w (x) = (w x) which means(by positive definiteness) z = w Thus this ϕ is a bijective(injectivity would ensure this because dim V = dim V ) Explicitly, Suppose f V and choose an orthonormal basis {e 0,, e n } of V ; with respect to this basis of V (and the basis {} of C the linear form V f C will be given by a row vector n n [a 0,, a n ] Mat n (C) so that for any x = e i x i we have f(x) = f(e i )x i, f(e i ) = a i so that if a = a j e j, we have ϕ a (x) = (a x) = ( a j e j e i x i ) = a i x i = f(x) at each x V

9 and thus f = ϕ a 9 To sum up: Riesz Representation Theorem: If dimv =, there exists a conjugate-linear bijection V T V given by T (z) V computing as (T (z))(x) = (z x) for z, x V Given f V, it is z = a 0 a n V which is such that T (z) = f; we say z is the representor for f and this column vector is with reference to a chosen orthonormal basis e; a i = f(e i )

Elementary linear algebra

Chapter 1 Elementary linear algebra 1.1 Vector spaces Vector spaces owe their importance to the fact that so many models arising in the solutions of specific problems turn out to be vector spaces. The