Dual vector space Aim lecture: We generalise the notion of transposes of matrices to arbitrary linear maps by introducing dual vector spaces. In most of this lecture, we allow F to be a general field. Defn Let V = F-space. The dual of V is the F-space V ∗ = L(V ; F). The elements of V ∗ are called linear functionals. n ∗ n E.g.1 (F ) = L(F ; F) = M1n(F), the set of length n row vectors in F. Note that n n n the transpose map (·)T : F −! (F )∗ : v 7! vT & F is an isomorphism! E.g. 2 If X is a set & V = Fun(X ; F), then for every x 2 X we obtain an element ∗ evx 2 V . E.g. 3 Given an inner product (·|·) on V & v 2 V we have (v|·) 2 V ∗. Daniel Chan (UNSW) Lecture 35: Dual vector spaces. Transpose Semester 2 2012 1 / 8 Transpose We can now generalise the transpose map for matrices. Prop-Defn Let S : V −! W be linear. The transpose of S is the fn S T : W ∗ −! V ∗ defined S f by S T f = f ◦ S : V −! W −! F for f 2 W ∗. S T is linear. Proof. Linearity follows from the distributive law: for f ; g 2 W ∗ we have (f + g) ◦ S = f ◦ S + g ◦ S & the fact that (βf ) ◦ S = β(f ◦ S) for β 2 F. d T ∗ ∗ E.g. Let S = dx : R[x] −! R[x] so S : R[x] −! R[x] . What's T ∗ S ev0 2 R[x] ? A Daniel Chan (UNSW) Lecture 35: Dual vector spaces. Transpose Semester 2 2012 2 / 8 Connection with matrix transpose n m Q Let A 2 Mmn(F)& TA : F −! F be the assoc lin map. What is T m ∗ n ∗ TA :(F ) = M1m(F) −! (F ) = M1n(F)? T Simplest answer TA is pre-composition (= right composition) with TA. Since composition corresponds with matrix multn, this is right multn by A. Note, TA is left multn by A. T m ∗ n More formally, let v 2 M1m(F) = (F ) ; w 2 F . Then T T T T (TA v )w = (v ◦ TA)w = (v A)w T T T so TA v = v A. T Q What's the matrix representing TA wrt natural co-ordinate systems m m n n (·)T : F −! (F )∗; (·)T : F −! (F )∗. Answer T T The matrix representing TA wrt the natural co-ord systems is A . Why? Daniel Chan (UNSW) Lecture 35: Dual vector spaces. Transpose Semester 2 2012 3 / 8 Functoriality Prop T ∗ ∗ 1 (id V ) = id V ∗ : V −! V . T S 2 Consider a composite of linear maps S ◦ T : U −! V −! W . Then (S ◦ T )T = T T ◦ S T : W ∗ −! U∗. 3 In particular, if S : V −! W is an isomorphism, so is S T . T Proof. For 1) note (id V ) : f 7! f ◦ id V = f . 2) For f 2 W ∗, we have 3) Suffice show S −T = (S −1)T is inverse to S T . But by 1) & 2) S T ◦ (S −1)T = (S −1 ◦ S)T = id T = id & sim (S −1)T ◦ S T = id so we are done. Daniel Chan (UNSW) Lecture 35: Dual vector spaces. Transpose Semester 2 2012 4 / 8 Dimension of the dual vector space Prop Let V = fin dim F-space. Then dim V ∗ = dim V . d Proof. Let C : F −! V be a co-ord system so dim = d. Then d C T : V ∗ −! (F )∗ is also an isomorphism so ∗ d ∗ dim V = dim(F ) = d: ∗ E.g. fev0; ev1g is a basis for C[x]≤1 since it is lin indep (ex) & ∗ dim C[x]≤1 = dim C[x]≤1 = 2. Daniel Chan (UNSW) Lecture 35: Dual vector spaces. Transpose Semester 2 2012 5 / 8 Linearity of the transpose operator T T The matrix transpose operator (·) : Mmn(F) −! Mnm(F): A 7! A is linear. This suggests Prop Let β 2 F & R; S : V −! W be linear. Then 1 (R + S)T = RT + S T 2 (βS)T = βS T In particular the map (·)T : L(V ; W ) −! L(W ∗; V ∗): S 7! S T is linear. Proof. For f 2 W ∗, this follows from Daniel Chan (UNSW) Lecture 35: Dual vector spaces. Transpose Semester 2 2012 6 / 8 Inner products & dual vector spaces Let now V = F-space equipped with an inner product (·|·) so F = R or C. Prop 1 The canonical map D : V −! V ∗ : v 7! (v|·) is an injective conjugate linear map. 2 If V is fin dim then D is a conjugate linear isomorphism. Proof. 1) (Ex) Conjugate linearity of (·|w) shows D is conjugate linear. To check injectivity suppose Dv = Dv0 so that by conjugate linearity (v − v0jv − v0) = (vjv − v0) − (v0jv − v0) = (Dv)(v − v0) − (Dv0)(v − v0) = 0 so v − v0 = 0 by inner product axioms. 2) We now check D is onto & show l 2 V ∗ is in the image of D. Now D0 = 0 so we can assume l 6= 0 so has image F. Our isomorphism thm applied to l : V −! F =) for any vector space complement W to ker l is isomorphic to F ? via the restricted map ljW : W −! F. This holds in particular for W = (ker l) so we can find w 2 W with l(w) = 1. Daniel Chan (UNSW) Lecture 35: Dual vector spaces. Transpose Semester 2 2012 7 / 8 Proof completed w It suffices now to show D kwk2 = l. Now any vector in V = W ⊕ (ker l) can be written in the form βw + v with β 2 F; v 2 ker l. Then w w w D (βw + v) = ( jβw + v) = β( jw) = β = l(βw + v) kwk2 kwk2 kwk2 & we are done. R 1 E.g. Let V = R[x]≤1 with inner product (f jg) = 0 f (t)g(t)dt. Find f 2 V such that (f |·) = ev0. Daniel Chan (UNSW) Lecture 35: Dual vector spaces. Transpose Semester 2 2012 8 / 8.