hmw11 - Solutions to Homework 11. Math 110, Fall 2006. Prob...

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Solutions to Homework 11. Math 110, Fall 2006. Prob 5.4.1. (a) False, { 0 } and the whole space are T -invariant for any T . (b) True (Theorem 5.21). (c) False: just take w to be a nonzero multiple of v , then the corresponding T -cyclic subspaces are the same, but the vectors are not. (d) False, span { v,Tv,T 2 v,. .. } is in general different from span { Tv,T 2 v,T 3 v,. .. } . (e) True (by the Cayley-Hamilton theorem). (f) True – it is the characteristic polynomial of its companion matrix introduced on p. 316. (g) True (Theorem 5.25). Prob 5.4.4. If TW W , then T 2 W = T ( TW ) TW W , and analogously T k W W for all powers k = 0 , 1 , 2 ,... . Also, since W is a subspace, we get the following chain of inclusions: ( a 0 I + a 1 T + ··· + a n T n ) W a 0 W + a 1 ( TW ) + ··· + a n ( T n W ) a 0 W + a 1 W + ··· + a n W W + W + ··· + W = W. Prob 5.4.5. Suppose each W j is a T -invariant subspace of V . Take x ∈ ∩ j W j . Since x W j for each fixed j , the T -invariance of W j implies Tx W j . So, Tx is in all subspaces W j , hence Tx ∈ ∩ j W j . Since x was arbitrary, this shows that j W j is T -invariant. Prob 5.4.17. First let us prove that A k span { I,A,A 2 ,...,A n - 1 } for all k IN . Indeed, this holds for k < n trivially and for k = n since A satisfies its characteristic polynomial A n + a n - 1 A n - 1 + ··· + a 0 I = 0 , (1) and hence A n span { I,A,. ..,A n - 1 } . Now argue by induction: assume that all the powers of A up to A n + m are already shown to be in the span { I,A,. ..,A n - 1 } . Multiply (1) by A m +1 to obtain A n + m +1 + a n - 1 A n + m + ··· + a 0 A m +1 = 0 . This shows that A n + m +1 is a linear combination of the preceding powers of A . Since those are already proved to be in span { I,A,. ..,A n - 1 } , so must be A n + m +1 . This proves that span { A j : j ZZ + } = span { I,A,. ..,A n - 1 } . Now, since the space span { I,A,. ..,A n - 1 } is spanned by n matrices, its dimension is at most n , QED. Prob 5.4.18. (a) Since f ( t ) = det( A - tI ), by evaluating both sides at 0, we get det A = f (0) = a 0 . So, A is invertible if and only if det A 6 = 0 if and only if a 0 6 = 0. (b) If a 0 6 = 0, then the characteristic equation can be re-written as - 1 a 0 ( ( - 1) n A n - 1 + a n - 1 A n - 2 + ··· + a 1 I ) · A = I.
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This note was uploaded on 10/02/2009 for the course MATH 54554 taught by Professor Holtz during the Spring '06 term at University of California, Berkeley.

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hmw11 - Solutions to Homework 11. Math 110, Fall 2006. Prob...

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