# 5 since dim r 2 2 the third answer must be wrong for

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(5) Since dim R 2 = 2 the third answer must be wrong. For the first we do have ((1, -1), (-1, -1)) = 0, but 11(1, -1)11 = 11(-1, -1)11 = J2 =I- l. Compare to the last of the four definitions on page 139. (6) Apart from physicists, who have already worked this out as an exercise in Chapter 4, you will not find it so easy to see the correctness of the third answer. It follows from (x,y) = (1/4)(llx + yW -llx - yW). The second answer, although not correct, is not so far off the mark. For such maps we always have f( x) = At.p( x) for all x E V and some A E Rand a suitable orthogonal map t.p. (7) Orthogonal maps are always injective (see the fact on page 143). There- fore, we must have Ker Pu = U.L = 0, if Pu is to be orthogonal. By the corollary on page 143 we then have U = V, and Pv = Idv is certainly orthogonal. (8) See the fact on page 144. (9) The second answer is not as wrong as the first, since (Il, +) would indeed be a group, but the formulation is not acceptable. N is not unfit to have some group structure N X N -7 N, but since" +" here denotes the ordinary addition of numbers, axiom (3) is not satisfied. (10) A 2k X 2k matrix is not a k X k matrix, so the first answer must be wrong. The statement (_l?k = 1 in the third is only a decoy, quite irrelevant to the question. SO(2k) =I- O(2k) because, for example, ( -1 1 1) E O(2k) " SO(2k).
196 CHAPTER 12: ANSWERS TO THE TESTS CHAPTER 9 TEST 1 2 3 4 5 6 7 8 9 10 x x x x x x x x x x x (1) If f(x) = AX is to mean anything, x and f(x) must belong to the same vector space. Look at the definition on page 151. (2) fe-x) = AX:::} fe-x) = (-A)(-X) and f(x) = (-A)X, so x and -x are eigenvectors for - A . (3) E>. = Ker(f - AId) contains zero as well as the eigenvectors for A. eigenvalue A = 1. (5) See the corollary on page 153. (6) f(x) = AX :::} x = f-l(AX) = Af-l(x) :::} f-l(X) = lx. Note that A is in fact nonzero, otherwise f would fail to be injective and could not be an automorphism. (7) All three conditions imply diagonalizability (see page 153, the corollary on page 153, and, of course, the definition of diagonalizability on page 152), but the first two are not equivalent to diagonalizability, as the example ( 1 0 0) f = 0 1 0 : lR 3 --+ lR 3 000 shows. (8) We have only restricted IF to be lR or C on page 159, so as to avoid certain technicalities with polynomials over arbitrary fields. (10) An easy calculation shows that v is an eigenvector of f for the eigenvalue A, if and only if <p -1 (v) is an eigenvector of 9 for the eigenvalue A.
Answers to Chapter 10 Test 197 CHAPTER 10 TEST 1 2 3 4 5 6 7 8 9 10 x X X X X X X X X X X X (1) Endomorhpisms having the first property are called orthogonal, while the third property imposes no conditions on f (symmetry of the inner product). (2) The Ai are real numbers, not elements of V, therefore the first state- ment is meaningless. For the correctness of the two others, see Fact 1 on page 162. (3) The same argument as used in Fact 2 on page 162 shows that the first two answers are both correct: from u E U and w E U.L it follows that (f(w),u) = (w,f(u)) = 0 for a self-adjoint f, and (f(w),u) = (f-If(w),f-l(u)) = (w,f-l(u)) = 0 for an orthogonal f. (Note that in the latter case f : V -+ V and flU: U -+ U are even isomorphisms. ) (4) For the first matrix al4 =F a41 , for the third al2 =F a21 .