hw7

Math 113 Homework 7 Due Friday, May 24, 2013 by 4 pm Please remember to write down your name and Stanford ID number, and to staple your solutions. Solutions are due to the Course Assistant, Graham White, in his oce, 380-380R (either hand your solutions directly to him or leave the solutions under his door). As usual, please justify all of your solutions and/or answers with carefully written proofs. Book problems : Solve Axler Chapter 6 problems 10, 12, 17, 20, 24, 25, 26 (page 122-125). 1.Let (V;h
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i) be a nite dimensional inner product space over F(either RorC). If we are given a basis v1;:::;vn, letg1;:::;gn2Vbe the functions gi:V!Fdened by gi(v) =hv;vii: (a) Prove that giis a basis for V. (b) Recall that the dual basis v 1;:::;v nis given by (1) v j(vi) =( 1i=j 0i6=j Prove that the basis ( v 1;:::;v n) is equal to the basis ( g1;:::;gn) if and only if v1;:::;vnis an orthonormal basis forV. 2.Orthonormal lists in innite dimensions . LetV=C0([;];R) denote the vector space of continuous functions from the interval [ ;] toR. EquipVwith the inner product hp;qi:=1 Z p(x)q(x)dx Show that the innite set f1p 2;cosx;sinx;cos 2x;sin 2x;cos 3x;sin 3x;:::;g is an orthonormal list with respect to this inner product. (The identities sinkxsinmx=1 2(cos(km)xcos(k+m)x) coskxcosmx=1 2(cos(km)x+ cos(k+m)x) sinkxcosmx=1 2(sin(km)x+ sin(k+m)x) may be helpful). Remark : Of course, this set is not an orthonormal basis , as nite linear combinations of this set do not span V. However, it is a deep theorem in Fourier analysis that in 1 fact, certain convergent innite linear combinations do spanV! Namely, any continuous functionf2Vcan be written as an innite convergent sum f(x) =a0p 2+X mamcosmx+X nbnsinnx: Moreover, in this sum, the coecient of a given orthonormal list element, is given by the usual projection formula, e.g., the coecient amof cosmxis am=hf;cosmxi: 3.Abstract operations on vector spaces II: Tensor products . Given a pair of vector spaces V andW, last week we dened their formal direct sum VW; This week we will have dened a formal product of vector spaces, known as the tensor product and denoted by V W: One heuristic property rst, to motivate the denition: in the same manner that di- rect sums are additive in dimension, tensor products will be multiplicative in dimension. Onto the denition: V Wis dened to be the set of elements of the formX kakvk wk; whereakis a scalar, vk2V, and wk2W, and the sum is a nite sum ( is just a formal symbol used to denote the concatenation of vkandwkin this context). Moreover inV W, there are relations, which give us rules for simplifying expressions. That is, the following expressions are equal: a(v w) = (av) w=v (aw); (v+v0) w=v w+v0 w v (w+w0) =v w+v w0:(2) The zero element is 0V 0W, addition is just given by formally adding two nite sums together (and simplifying if possible by using the relations above), and scalar multiplica- tion is as one might expect: a
X kakvk wk:=X ka(akvk wk) Elements of the form v ware called pure tensors . A general element of V W will not be a pure tensor, but rather a sum of such terms. There are a few dierences to note from formal direct sums: note rst that v w+v0 w06= (v+v0) (w+w0); in fact (v+v0) (w+w0) =v w+v w0+v0 w+v0 w0; 2 which is to say that the formal symbol \ " acts much like a product, in its distributivity properties. In a similar vein, note that, using the relations above, 0V w= (0
0V) w=0V 0w=0V 0W=0V W; another property formally similar to multiplication. Similarly, v 0W=0V W. (a) There is a natural map :VW!V W (v;w)7!v w: Prove that this is a bilinear map, where bilinear maps were dened on HW6. (b) In fact is the universal bilinear map , as we will see in this exercise. Namely, prove the following: if T:VW!X is abilinear map , in the sense of homework 6, prove that there exists a unique linear map T:V W!X such thatT=T. That is,Tfactors uniquely as VW!V WT!X whereTis a linear map. Hint : What does it mean for a map T:V W!Xto be a linear? Firstly, it means that on a formal sum X kakvk wk Tis additive and homogenous, so T(X kakvk wk) =X kakT(vk wk): It also means that Tshould not be aected by applying the relations (2); that is, T((v+v0) w) should be equal to T(v w+v0 w) and so on. (c) Prove that dim( V W) = (dimV)
(dimW) (Hint: given a basis ( v1;:::;vk) ofV and a basis w1;:::;wlofW, prove that the collection of pure tensors fvi wjj1 ik;1jlgis a basis of V W). Crucial Hint : Proving that these elements span is relatively straightforward. How- ever, proving that this collection is linearly independent directly may be a bit more tricky (given how many ways one can try to simplify and expand expressions using the relations (2)). Here is a suggested shortcut: suppose there is a linear relationship among thefvi wjg, so some sum of them with coecients aijis 0. Recall on HW6 you found bilinear maps fij:VW!Fthat were 1 on the input vi;wjand zero 3 on other pairs vs;wt. Apply part (b) to obtain a linear map fij:V W!F. What happens if you apply this fijto the linear relationship? (d)An example. LetF[x] denote the vector space of polynomials in a variable x(which we normally callP(F)), and let F[y] denote the vector space of polynomials in a variable y(this is the same vector space, where we've relabled the variable). Construct an isomorphism F[x] F[y]!F[x;y] where F[x;y] is a new vector space: the vector space of polynomials in two variables ; i.e. nite sums of the form X i0;j0aijxiyj (e) The tensor product functions much like a product on vector spaces (minus the exis- tence of multiplicative inverses!) That is, up to canonical isomorphism it distributes with formal direct sum, (3) V (WX)=(V W)(V X) there is a multiplicative identity (4) V F=F V=V and multiplying by 0 (the additive identity) results in 0: (5) V f0g=f0g V=f0g: Prove most of these facts. More precisely, construct and verify the canonical isomor- phisms (4) and (5), and construct a canonical map for (3) (no need to verify it's an isomorphism). Remark : For those interested in applications, a brief motivation: Tensor products have many incarnations in applied elds, especially all over physics. If one likes thinking in terms of coordinates, the tensor product Rm Rncan be identied with mnmatrices, and higher tensor products Rm1


Rmkcan be identied with m1


mkmulti- dimensional arrays of scalars. 4