(III.D) Linear Functionals II: The Dual Space

Transcription

1 IIID Linear Functionals II: The Dual Space First I remind you that a linear functional on a vector space V over R is any linear transformation f : V R In IIIC we looked at a finite subspace [=derivations] of the infinitedimensional space of linear functionals on C M Now let s take a finite-dimensional vector space V and consider V := {vector space consisting of all linear functionals on V}, read V-al Two functionals f, f V are equal if they give the same value on all v V : f v = f v Let B = { v,, v n } be a basis for V, and let f i : V R be some special linear functionals de f ined by { 0, i = j f i v j = δ ij =, i = j By linearity, on any v = a i v i, this makes f i v = a i PROPOSITION The { f,, f n } are a basis for V the al basis, and dimv = dimv[= n] PROOF that the { f i } span V : let f : V R be any linear functional, and let β i = f v i Then on any v i = a i v i, f v = a i f v i = a i β i = β i f i v Therefore f = β i f i as a functional

2 IIID LINEAR FUNCTIONALS II: THE DUAL SPACE that the { f i } are linearly independent: suppose α j f j = 0 as a f unctional ; that is, on all v V, α j f j v = 0 In particular, applying this to v = v i for each i, So all the α i are zero 0 = j α j f j v i = α j δ ij = α i j Change of basis Take another B = { v,, v n } for V, and the corresponding al B = { f,, f n } for V : that is, f i v j = δ ij Regardless of the basis B, f v = i α i f i j In terms of vectors, with [ f ] B = β j v j = i,j α α i β j f i v j = i,j, [ v] B = β α i β j δ ij = α i β i i, α n this just says f v = t [ f ] B [ v] B = β n α α n β β n So for the different bases, f v = t [ f ] B [ v] B = t [ f ] B P B B[ v] B = = t t P B B[ f ] B [ v]b for all v, which is to say [ f ] B = t P B B[ f ] B But by definition of P B B, [ f ] B = P B B [ f ] B

3 IIID LINEAR FUNCTIONALS II: THE DUAL SPACE 3 and comparing equalities gives the formula P B B = t P B B describing change of coordinates for covectors If you stretch a basis, say B = { v,, v n } B = { v,, v n }, then vectors appear to shrink by half that is, their coordinates with respect to the basis shrink, since the vectors must stay the same This formula, in the same sense, says that covectors would appear to double in size Brief aside on calculus In R, let x be the coordinate with respect to B = {ê} and u the coordinate with respect to B = {ê} Then x ê = u ê u = x expresses the shrinking of coordinates of a fixed vector as the basis stretches In the last lecture we suggested that coefficients of first-order linear differential operators transform infinitesimally { } like coordinates { } of vectors Here this looks as follows: B = d, B = d, d = d the coefficient of d d is half that of ; intuitively, d d should be bigger than because d f is the rise in f per unit run of ê, while d f is the rise in f per unit run of ê The covectors in this situation ie, the functionals on the space of linear differential operators are differential forms and defined by d =, d = ; thus =, which is exactly what you know from calculus: u = x = = So the change of coefficients goes instead of More generally, on R n you may define a basis for the differential forms by i = δ ij ; these give at each point the al xj to the space of first-order linear differential operators They are what appears under the integral sign in calculus simply because that is, functionals In this context, the prefix co is essentially a synonym for al When the latter are thought of as tangent vectors, the differential -forms are cotangent vectors

4 4 IIID LINEAR FUNCTIONALS II: THE DUAL SPACE they transform in exectly the right way under change of variable u -substitution Dual of a linear transformation Every linear transformation T : V W inces a al linear transformation T : W V by pullback of functionals: given g W, define 3 T g := g T The picture is: T g g R V T W As usual take B, C bases for V, W resp; B, C their al bases for V, W We would like to relate the matrices C[T] B and B [T ] C it turns out they are just transposes of one another Before doing this in full generality, let s step back and look at a simple case Consider S : R m R m with matrix [S]ê = A with respect to the standard basis We can think of R m as its own al space, as follows Any l R m gives a linear functional 4 on R m by matrix multiplication of a m matrix by an m matrix: l v := t l v 3 also written T g, as al spaces/bases are often written V /B I m avoiding this so as to minimize confusion when the Hermitian transpose is introced later on 4 We are really mapping R m R m by taking l to this functional This map also takes ê to the al basis ê, so ê gives its own al basis in the same sense as l gives functionals Identifying a vector space with its al is equivalent to giving an inner proct, and what s going on here is again an ad hoc use of the dot proct

5 IIID LINEAR FUNCTIONALS II: THE DUAL SPACE 5 The al transformation S : R m R m, with matrix B, is defined by S l v = ls v In terms of matrices this translates to t B l v = t l A v t l tb v = t l A v If this is true for all l, v then t B = A or B = t A So in this context, the fact that the matrix of S -al is the transpose of the matrix of S, is very simple 5 Now to prove the corresponding fact for T and T above, start with the definition of T g, for any g W : T g v = g T v = gt v for all v V Now recall that for any 6 v V, f V, and basis B for V, f v = t [ f ] B [ v] B Applying this fact to the end terms of, we have t [T g] B [ v] B = t [g] C [T v] C Now using the defintion of the matrices of T resp T with respect to B and C resp B and C, for example C [T] B [ v] B = [T v] C, becomes 3 t B [T ] C [g] C [ v]b = t [g] C C [T] B [ v] B 5 Remark for physicists: this argument may remind anyone exposed to quantum mechanics of adjoint operators, perhaps more so if we write l v = t l v as l, v : then the definition of S reads S l, v = l, S v 6 the same of course goes for any w W, g W, and basis C for W

6 6 IIID LINEAR FUNCTIONALS II: THE DUAL SPACE Applying the rule t A B = t B ta for multiplying transposes of matrices, to the left hand-side of 3 gives t [g] C tb [T ] C [ v] B = t [g] C C[T] B [ v] B, and since this holds for any g W and v V, we conclude that t B [T ] C = C [T] B or B [T ] C = t C [T] B The double al Associated to any subspace U V there is a subspace U V of complimentary dimension, called the annihilator of U It consists simply of all linear functionals f V such that f x = 0 for all x U For example, suppose U is the plane in R 3 consisting of solutions to x + x + 3x 3 = 0 Then in terms of the al standard basis ê see the supplement, U is just the line in R 3 spanned by since then for any x U, f x = t [ f ]ê [ x]ê = [ f ]ê = 3 3, x x x 3 = x + x + 3x 3 = 0 Now suppose you take one more al, and consider the annihilator of U, U V! Well, it turns out that you just get back what you started with: V = V, and moreover U is again just U We will prove only the first statement, by checking that the linear transformation given by V V α L α defined, on any f V, by L α f := f α

7 IIID LINEAR FUNCTIONALS II: THE DUAL SPACE 7 is -to- and onto In fact we only have to show it is -to- since dimv = dimv = dimv, and the image of a -to- transformation has the same dimension as its domain α = 0 LEMMA Let α = α i v i V If f α = 0 for all f V, then PROOF For all f, 0 = f α = α i f v i ; in particular, this holds for f = f j { f j } the al basis to { v i } and so 0 = α i f j v i = α i δ ij = α j for each j Therefore α = α j v j = 0 Now let α, α be two vectors V, and suppose L α = L α ; that is, for all f V, L α f = L α f Using the definition of L this says f α = f α or f α α = 0 for all f V ; the Lemma = α α = 0, ie α = α We have shown L α = L α = α = α, and so the correspondence α L α is -to- and as discussed above therefore an isomorphism of V with its double-al A note on finding al bases B, given a basis B = { v,, v n } of R n where the v i s are written in terms of the standard basis The al ê to the standard basis consists by definition of functionals ê, ê, ê 3 satisfying Notice that this means ê i x = ê i x j ê j = x i = ê i ê j = δ ij 0 0 i x x n Now we want to find functionals { v,, v n } or { f,, f n } use either notation satisfying v i v j = δ ij

8 8 IIID LINEAR FUNCTIONALS II: THE DUAL SPACE We need some way of writing them down Take n = 3 for concreteness If you write for the linear functional f defined by 3 x f x = 3 = x + x + 3x 3, you are writing f in the basis ê ; that is, [ f ]ê =, and f x = t [ f ]ê [ x]ê 3 So you need to find 3 vectors [ f ]ê, [ f ]ê, [ f 3 ]ê whose transposes obey t [ f ]ê t [ f ]ê t [ f 3 ]ê x x 3 t [ f i ]ê v j = δ ij, ie v v v 3 = But this just means finding the inverse matrix otherwise known as P B and interpreting the rows as [transposes of] the al basis vectors functionals written in the al standard basis ê Exercises Let B = { v, v, v 3 } be the basis of R 3 defined by v = 0, v =, v 3 = 0 Find the al basis of B Let V be the vector space of all polynomial functions over the field of real numbers Let a and b be fixed real numbers and let f

9 be the linear functional on V defined by f p = EXERCISES 9 b a px If D is the differentiation operator on V, what is D f? 3 One interesting example of a linear functional is the trace map tr : M n F F, where we are regarding the n n matrices as a vector space of dimension n over F i Show that trab = trba ii Dece that similar matrices have the same trace iii Do there exist A, B M n F with AB BA = I n? 4 Show that the annihilator of the image of a transformation T : V W is the kernel of its al transformation: ImT = kert This translates for T : R n R n with matrix A to ImA = ker t A, in which form it is sometimes called the Fredholm alternative by functional analysts 5 Let W be the subspace of R 4 spanned by 0 v = and v 3 = Which linear functionals f x, x, x 3, x 4 = c x + c x + c 3 x 3 + c 4 x 4 are in the annihilator of W? 6 Let V be a finite-dimensional vector space over R, and v,, v N a finite set of nonzero vectors in V Does there exist a linear functional f V which has f v i = 0 for every i?