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LINEAR ALGEBRA FALL 2007/08 PROBLEM SET 9 SOLUTIONS In MATH 110: LINEAR ALGEBRA FALL 2007/08 PROBLEM SET 9 SOLUTIONS In the following V will denote a finite-dimensional vector space over R that is also an inner product space with inner product denoted by h ·; · i. The norm induced by h ·; · i will be denoted k · k. 1. Let S be a subset (not necessarily a subspace) of V . We define the orthogonal annihilator of S, denoted S?, to be the set S? = fv 2 V j hv; wi = 0 for all w 2 Sg: (a) Show that S? is always a subspace of V . ? Solution. Let v1; v2 2 S . Then hv1; wi = 0 and hv2; wi = 0 for all w 2 S. So for any α; β 2 R, hαv1 + βv2; wi = αhv1; wi + βhv2; wi = 0 ? for all w 2 S. Hence αv1 + βv2 2 S . (b) Show that S ⊆ (S?)?. Solution. Let w 2 S. For any v 2 S?, we have hv; wi = 0 by definition of S?. Since this is true for all v 2 S? and hw; vi = hv; wi, we see that hw; vi = 0 for all v 2 S?, ie. w 2 S?. (c) Show that span(S) ⊆ (S?)?. Solution. Since (S?)? is a subspace by (a) (it is the orthogonal annihilator of S?) and S ⊆ (S?)? by (b), we have span(S) ⊆ (S?)?. ? ? (d) Show that if S1 and S2 are subsets of V and S1 ⊆ S2, then S2 ⊆ S1 . ? Solution. Let v 2 S2 . Then hv; wi = 0 for all w 2 S2 and so for all w 2 S1 (since ? S1 ⊆ S2). Hence v 2 S1 . (e) Show that ((S?)?)? = S?. ? ? Solution. Applying (d) to (b) with S1 = S and S2 = (S ) , we get ((S?)?)? ⊆ S?: Apply (c) to S?, we get span(S?) ⊆ ((S?)?)? but by (a), span(S?) = S?. Hence we get equality. ? (f) Show that either S \ S must be either the empty set ? or the zero subspace f0V g. ? ? ? Solution. If S \ S 6= ?, then let v 2 S \ S . Since v 2 S , we have hv; wi = 0 2 for all w 2 S. Since v 2 S, in particular, kvk = hv; vi = 0, implying that v = 0V . In ? ? other words, the only vector in S \ S is 0V . So S \ S = f0V g. On the other hand, if ? ? 0V 2= S, then 0V 2= S \ S and so S \ S = ? (if not, then the previous argument gives a contradiction). 2. Let W be a subspace of V . The subspace W ? is called the orthogonal complement of W . (a) Show that V = W ⊕ W ?. ? Solution. Since W and W are both subspaces (the latter by Problem 1(a)), 0V 2 W ? ? and 0V 2 W . So by Problem 1(f), we have that W \W = f0V g. It remains to show that Date: December 9, 2007 (Version 1.0). 1 ? ? W + W = V . Clearly W + W ⊆ V . For the converse, let v 2 V and let fw1;:::; wrg be an orthonormal basis of W . Consider x := hv; w1iw1 + hv; w2iw2 + ··· + hv; wriwr and y := v − x: Clearly x 2 W . We claim that y 2 W ?. For any w 2 W , we could write w = hw; w1iw1 + hw; w2iw2 + ··· + hw; wriwr since fw1;:::; wrg is an orthonormal basis of W . So hy; wi = hv − x; wi = hv; wi − hx; wi D Xr E DXr E = v; hw; wiiwi − hv; wiiwi; w i=1 i=1 Xr Xr = hv; wiihw; wii − hv; wiihwi; wi i=1 i=1 = 0 ? since hw; wii = hwi; wi. Hence v = x + y 2 W + W . (b) Show that W = (W ?)?. Solution. By Problem 1(b), we have W ⊆ (W ?)?. But by (a), we have W ⊕ W ? = V = W ? ⊕ (W ?)? and so dim(W ) + dim(W ?) = dim(W ?) + dim((W ?)?) and so dim(W ) = dim((W ?)?): Hence W = (W ?)?. m×n n m 3. Let A 2 R and regard A : R ! R as a linear transformation in the usual way. Show that nullsp(A>) = (colsp(A))? and rowsp(A) = (nullsp(A))?: Hence deduce that m > R = nullsp(A ) ⊕ colsp(A) and n R = rowsp(A) ⊕ nullsp(A): m n ? Solution. Clearly colsp(A) = fAx 2 R j x 2 R g. So if v 2 (colsp(A)) , then hv;Axi = 0 n for all x 2 R . So hA>v; xi = (A>v)>x = v>Ax = hv;Axi = 0 n > for all x 2 R . In particular, we may pick x = A v to get kA>vk2 = hA>v;A>vi = 0; giving us A>v = 0: Hence v 2 nullsp(A>). Conversely, if v 2 nullsp(A>), then hv;Axi = v>Ax = (A>v)>x = 0>x = 0 2 n ? for all x 2 R . So v 2 (colsp(A)) . This proves the first equality. For the second equality, just note that rowsp(A) = colsp(A>) and apply the first equality to get (colsp(A>))? = nullsp(A); and then apply Problem 2(b) to get rowsp(A) = ((rowsp(A))?)? = (nullsp(A))?: The next two inequalities are immediate consequences of Problem 2(a). n×n n 4. Let A; B 2 R . Let h ·; · i denote the standard inner product on R . n (a) Show that if hAx; yi = 0 for all x; y 2 R , then A = O, the zero matrix. n n Solution. Let fe1;:::; eng be the standard basis of R . Observe that if A = [aij]i;j=1, then > aij = ei Aej = hei;Aeji = hAej; eii = 0 by letting x and y be ei and ej respectively. Since this is true for all i and j, we have that A = O. n (b) Is it true that if hAx; xi = 0 for all x 2 R , then A = O, the zero matrix? 2 n > Solution. Recall notation from Homework 4, Problem 1(d). Let A 2 ^ (R ), ie. A = −A. Then hAx; xi = x>A>x = −x>Ax = −hx;Axi = −hAx; xi; implying that hAx; xi = 0. So any non-zero skew-symmetric matrix would be a counter example. An explicit 2 × 2 example is given by 0 −1 A = : 1 0 n (c) Is it true that if hAx; xi = hBx; xi for all x 2 R , then A = B? Solution. Any two distinct skew-symmetric matrices would provide a counter example. An explicit 2 × 2 example is given by 0 −1 0 0 A = ;B = : 1 0 0 0 n 5. Let B1 = fx1;:::; xng and B2 = fy1;:::; yng be two bases of R . B1 and B2 are not necessarily orthonormal bases. n×n (a) Show that there exists an orthogonal matrix Q 2 R such that Qxi = yi; i = 1; : : : ; n; if and only if hxi; xji = hyi; yji; i; j = 1; : : : ; n: Solution. By a theorem we have proved in the lectures, an orthogonal matrix Q preserves inner product and so satisfies hyi; yji = hQxi;Qxji = hxi; xji: It remains to show that if the latter condition is satisfied, then such an orthogonal matrix may be defined. First, observe that since B1 and B2 are bases, the equations Qxi = yi; i = 1; : : : ; n; 3 1 n defines the matrix Q uniquely . We only need to verify that Q is orthogonal. Let x 2 R and let Xn x = αixi i=1 be the representation of x in terms of B1. Then kQxk2 = hQx;Qxi DXn Xn E = αiQxi; αjQxj i=1 j=1 DXn Xn E = αiyi; αjyj i=1 j=1 Xn Xn = αiαjhyi; yji i=1 j=1 Xn Xn = αiαjhxi; xji i=1 j=1 DXn Xn E = αixi; αjxj i=1 j=1 = hx; xi = kxk2: n Since this is true for all x 2 R , ie. Q preserves norm, the aforementioned theorem in the lectures imply that Q is orthogonal. n×n (b) Show that if B1 is an orthonormal basis and if Q 2 R is an orthogonal matrix, then fQx1;:::;Qxng is also an orthonormal basis. Solution. Clearly Qx1;:::;Qxn are linearly independent since if α1Qx1 + ··· + αnQxn = 0; then Q(α1x1 + ··· + αnxn) = 0; and so > α1x1 + ··· + αnxn = Q 0 = 0 and so α1 = ··· = αn = 0 by the linear independence of x1;:::; xn. Hence fQx1;:::;Qxng n is a basis (since the set has size n = dim(R )). The orthonormality follows from the fact that Q preserves inner product and that x1;:::; xn form an orthonormal basis: ( 0 if i 6= j; hQxi;Qxji = hxi; xji = 1 if i = j: 1 In fact, Q is just the matrix whose coordinate representation with respect to B1 and B2 is the identity matrix, ie. [Q]B1;B2 = I. Explicitly, if we let X and Y be the matrices whose columns are x1;:::; xn and y1;:::; yn respectively, then Q = YX−1. 4.
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