Linear Algebra

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Unformatted text preview: 1 Math 110 Homework 11 Partial Solutions If you have any questions about these solutions, or about any problem not solved, please ask via email or in office hours, etc. 5.2.12 (a) Let be the eigenspace of T corresponding to , and let −1 be the eigenspace of T−1 corresponding to −1 . If x ∈ ≥ then , −1 −1 −1 T(x ) = x . Thus x = T (x ) = T (x ) and so T (x ) = −1 . Thus x ∈ −1 . By the same argument, we also get −1 ⊆ ≥ and thus that they are equal. (b) If T is diagonalizable, then there exists a basis fi for such that [T]fi is diagonal. By (a), [T−1 ]fi must then also be diagonal. 5.2.20 Suppose that = W1 ⊕ · · · ⊕ . Then, by Theorem 5.10, we can find ordered bases fi1 k for W1 k , respectively, so that fi = fi1 ∪ · · · ∪ is a basis for . This then gives that dim( ) = dim(W1 ) + · · · + dimWk by counting basis vectors. On the other hand, suppose that the dimension equality holds. Picking a basis fii for each Wi, we have that fi = fi1 ∪· · ·∪ has dim( ) vectors. Since W1 + · · · + Wk = , we also have that fi spans . Thus fi is a basis for , and so = W1 ⊕ · · · ⊕ . 5.4.13 Suppose that W is the T-cyclic subspace of generated by v and that w ∈ . Since W = span({ (v) T 2 (v) }), this means that there exists an integer n and a0 1 ∈ such that w = a0 + n(v) = (a + a + · · · + a Tn)(v). Setting g(t) = a1 (v) + · · · + anT 0 1 n a0 + a1 + · · · + antn, we have that w = g(T)(v). 5.4.17 Let f (t) = ± + cn−1 −1 + c 1 + c0 . By the Cayley-Hamilton Theorem, f ( ) = 0. In other words, n = ±(cn−1 −1 + · · · + c1 + c0 ). −1 +k Thus n ∈ span({ }). Inductively, if n n+1 ∈ −1 n+k+1 = ±(c +k k+2 span({ }), then + · · · + c1 + n−1 −1 2 c + 0 k+1 ) ∈ span({ }). Thus span({ }) = −1 span({ }) has dimension ≤ . 5.4.18 (a) Note that det( ) = det( − 0I ) = f (0) = a0 . Thus if and only if det( ) = 0 if and only if a0 = 0. is invertible 2 (b) By the Cayley-Hamilton Theorem, we have (−1)n An + an−1 An−1 + a1 A + a0 I = 0. This can be rearranged to give −a0 I = (−1)n An + an−1 An−1 + a1 A. Since A is invertible (and thus by part (a), a0 = 0), we can multiply this entire equation by −a−1 A−1 to give 0 the desired result. 5.4.20 If there exists a g such that U = g (T ), then it is clear that U T = T U . On the other hand, suppose that U T = T U . Suppose that V is the T cyclic subspace of itself generated by v . By exercise 5.4.13 (using U (v ) in place of w), we can find a polynomial g so that g (T )(v ) = U (v ). We then claim that U = g (T ). Notice first that U (T k (v )) = T k (U (v )) = T k (g (T )(v )) = g (T )(T k (v )) for every k ≥ 0. Then, if x ∈ V is any vector, we can write x as some linear combination of finitely many of the vectors T k (v ). By linearity, we see (calculation suppressed) that U (x) = g (T )(x). Since this is true for any x ∈ V , we have that U = g (T ). ...
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5.2+5.4 - 1 Math 110 Homework 11 Partial Solutions If you...

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