hw7_solutions

Math 113 — Homework 7 Graham White May 28, 2013 Book problems 10. The first element of our basis is the function 1. The second element is proportional to xhx;1i1 =x1 2. Normalising, we getp 12(x1 2). The third element is proportional to x2hx2;1i1hx2;p 12(x1 2)i(x1 2) =x21 3p 12(1 41 6)p 12(x1 2) =x21 3(x1 2) =x2x+1 6 Normalising, we getp 5(6x26x+ 1) . 12. Ife1is a vector with Span(e1) = Span(v1), thene1is a scalar multiple of v1. Lete1=cv1. Thenhe1;e1i= jcj2hv1;v1i, which is equal to 1exactly whenjcj=1p hv1;v1i. Hence there are two choices for e1. For the remainder of the argument, we will only be concerned with the span of e1and whether vectors are orthogonal to it, so it doesn’t matter which choice we make. Now, orthogonal complement to Span(e1)inside Span(v1;v2)is one dimensional. As before, there are two choices of vector e2which are contained in this subspace with he2;e2i= 1. We now repeat this argument. There are two choices for e3in the orthogonal complement of Span(e1;e2)inside Span(v1;v2;v3), etc. For eachei, there were two possibilities, and these choices could be made independently. Therefore there are 2npossible listsfe1;e2;:::;engsatisfying the desired condition. 17. Any element of Vcan be written as v= (vP(v)) +P(v). We have that (vP(v))2ker(P), because P2=P, and thatP(v)2im(P)by definition. Therefore ker(P)andim(P)spanV. Every element of ker(P) is orthogonal to every element of im(P)and a vector is orthogonal to itself only if it is the zero vector, so ker(P)\im(P) =f0g. ThereforeV= ker(P)im(P). Using the above decomposition, the component of vinim(P)isP(v), soPis the orthogonal projection onto im(P). 20. IfUandU?are both invariant under T, then letvbe an arbitrary element of V. We may write v=u+u0, u2Uandu02U?. Then we have that PUT(v) =PUT(u+u0) =PU(T(u) +T(u0)) =T(u)(T(u)2U;T(u0)2U?) =T(PU(u+u0)) = TPU(v) ThereforePUT=TPU. 1 Now, assume that PUT=TPU. Letube any element of U. We have that PUT(u) =TPU(u), soPUT(u) = T(u). Therefore T(u)is fixed byPU, so is an element of U. HenceUis invariant under T. We have that IdV=PU+PU?, and IdVandPUboth commute with T, soPU?commutes with T. Therefore the same argument shows that U?is invariant under T. We have shown that UandU?are invariant under Tif and only if PUT=TPU. 24. The required condition is linear in p, so it suffices to give a qthat works for p= 1;p=xandp=x2. That is, we need the following three conditions: Z1 0q(x)dx= 1 Z1 0xq(x)dx=1 2 Z1 0x2q(x)dx=1 4 Letq(x) =ax2+bx+c. Our equations reduce to a 3+b 2+c= 1 a 4+b 3+c 2=1 2 a 5+b 4+c 3=1 4 Solving these gives a=15;b= 15;c=3 2. Therefore the polynomial q(x)isq(x) =15x2+ 15x3 2. 25. Proceeding as in the previous part, we have the equations Z1 0q(x)dx=Z1 0cos(x)dx Z1 0xq(x)dx=Z1 0xcos(x)dx Z1 0x2q(x)dx=Z1 0x2cos(x)dx Using integration by parts, we evaluate these as Z1 0q(x)dx= 0 Z1 0xq(x)dx=2 2 Z1 0x2q(x)dx=2 2 As before, let q(x) =ax2+bx+c. We get that a 3+b 2+c= 0 a 4+b 3+c 2=2 2 a 5+b 4+c 3=2 2 Solving these gives a= 0;b=24 2;c=12 2. Therefore the polynomial q(x)isq(x) =24 2x+12 2. 2 26. Let an orthonormal basis for Vbefe1;:::;eng. We have that for each v2V,hTv;ai=hv;Tai. (Note that the first inner product is in Vand the second is in F). Takingvto beei, we get that for each i,hTei;ai=hei;Tai. But now we have an expression for hei;Taifor eachei, which we can use to reconstitute Ta. We get that Ta=ha;Te 1ie1+:::+ha;Tenien: (The order of the elements ha;bivshb;aiis important, but reversing it just results in complex conjugation). Other problems: 1. (a) Assume that some linear combination a1g1+:::+angn= 0. Then for any v2V, we have that a1hv;v1i+:::+anhv;vni= 0 Therefore hv;a1v1+:::+anvni= 0: This equation is true for all v2V, includingv=a1v1+:::+anvn. But the dot product of a vector with itself is zero only if that vector is zero. Therefore a1v1+:::+anvn= 0. Butv1;:::;vnis linearly independent, so each aiis zero. Therefore each aiis zero, and we have shown that g1;:::;gnare linearly independent. We know that the dimension of Visn, so any set of nlinearly independent vectors is a basis. Therefore g1;:::;gnis a basis of V. (b) The basis (v 1;:::;v n)being equal to the basis (g1;:::;gn)is equivalent to the statement that hvi;vji= ijfor eachiandj, by the definition of v i. But this is the definition of orthonormality of a basis. 2. In order to show that this list is orthonormal, we need to show that the inner product of any element with itself is one and that any two elements are orthogonal. We check these as follows, where kandlare any positive integers,k6=l. 3 h1p (2);1p (2)i=1 Z 1 2dx =1 ( 2 2) = 1 hcos(kx);cos(kx)i=1 Z cos2(kx)dx =1 Z cos(2kx) + 1 2dx =1 sin(2kx) 2k+x 2  =1 ( 2 2) = 1 hsin(kx);sin(kx)i=1 Z sin2(kx)dx =1 Z 1cos(2kx) 2dx =1 x 2sin(2kx) 2k  =1 ( 2 2) = 1 4 h1p 2;cos(kx)i=1p 21 Z cos(kx)dx =1 2sin(kx) k  = 0 h1p 2;sin(kx)i=1p 21 Z sin(kx)dx =1 2cos(kx) k  = 0 hcos(kx);cos(lx)i=1 Z cos(kx) cos(lx)dx =1 Z cos(kx+lx) + cos(kxlx) 2dx = 0 (as calculated above, as k+landklare nonzero) hsin(kx);sin(lx)i=1 Z sin(kx) sin(lx)dx =1 Z cos(kx+lx)cos(kxlx) 2dx = 0 (as calculated above, as k+landklare nonzero) hcos(kx);sin(lx)i=1 Z cos(kx) sin(lx)dx =1 Z sin(kx+lx)sin(kxlx) 2dx = 0 (as calculated above, as k+landklare nonzero) Therefore the given set of functions is orthonormal. 3. (a) We check that is bilinear by using the relations used in the definition of the tensor product, as follows (av+bv0;w) = (av+bv0) w =av w+bv0 w =a(v w) +b(v0 w) =a(v;w) +b(v0;w) Linearity in the second variable is checked via a similar calculation. (b) We define T(v w) =T(v;w)for eachv2Vand eachw2W, and extend this definition to all of V W linearly. The function Tis well-defined because Tis bilinear. (Could you write down the calculations that would verify this?). By construction, we have that T=T. Therefore there exists at least one map T with the required properties. Assume there were two linear maps T;T0withT=TandT=T0. ThenTandT0would be equal on the image of . But this means that TandT0are equal on any pure tensor, and any element of V Wis a linear combination of pure tensors. Therefore T=T0, so there is a unique linear map with this property. (c) Any element of V Wis a linear combination of pure tensors. Using the bilinearity of the tensor product, we may write any pure tensor v was a linear combination of putative basis elements vi wjby expanding vandwin terms of the bases of VandW. Therefore the elements vi wjspanV W. 5 As in the hint, assume that there is some linear dependence relation between the vi wj, with coefficients aij. The function fijis equal to 1onvi wjand is zero on all other elements of our supposed basis (Why? Check this using part b) ). Applying fijto our linear dependence relation gives that aij= 0. We repeat this for each pair (i;j), showing that each aij= 0. Therefore the vi wjare linearly independent. We have shown that the set of vi wjspansV Wand is linearly independent, so it is a basis. Therefore the dimension of the V Wis the product of the dimensions of Vand ofW, by counting the elements of this basis. (d) Consider the map defined by(p(x);q(y)) =p(x)q(y)and extended linearly to all of F[x] F[y]. The function is linear because polynomial multiplication is bilinear. Define another map 1by 1(xayb) =xa yb, extended linearly to F[x;y]. These maps are inverses of one another, which is easily verified on monomials. (Why is it enough to check this on monomials?). Therefore is an isomor- phism from F[x] F[y]toF[x;y], as required. (e) For (3), the canonical isomorphism is the linear map which takes v (w;x)to(v w;v x). For(4), note that for any v2Vanda2F, we have that v a=a(v 1). Therefore, let be the linear map defined by (v a) =avand its inverse be 1(v) =v 1. It is easily checked that these maps linear and are inverses of one another (on pure tensors in V Fand on any element of V). A similar argument works for F V. For(5), note that for any v2V, we have that v 0 = 0 . Therefore, the vector space V f0gis the unique vector space with only one element. There is only one map between vector spaces which each have one element. This map is trivially linear and bijective, thus is an isomorphism. Likewise for 0 V. 6