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MIT18_06S10_pset10_s10_sol

MIT18_06S10_pset10_s10_sol - 18.06 Problem Set 10 Solution...

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18.06 Problem Set 10 Solution Total: 100 points Section 6.6. Problem 12. These Jordan matrices have eigenvalues 0 , 0 , 0 , 0. They have two eigenvectors (one from each block). But the block sizes don’t match and they are not similar : 0 1 0 0 0 1 0 0 J = 0 0 0 0 0 0 0 1 and K = 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 For any matrix M , compare JM with MK . If they are equal show that M is not invertible. Then M 1 JM = K is impossible; J is not similar to K . Solution (4 points) Let M = ( m ij ). Then m 21 m 22 m 23 m 24 0 m 11 m 12 0 JM = 0 0 0 0 m 41 m 42 m 43 m 44 and MK = 0 m 21 m 22 0 0 m 31 m 32 0 . 0 0 0 0 0 m 41 m 42 0 If JM = MK then m 21 = m 22 = m 24 = m 41 = m 42 = m 44 = 0 , which in particular means that the second row is either a multiple of the fourth row, or the fourth row is all 0’s. In either of these cases M is not invertible. Suppose that J were similar to K . Then there would be some invertible matrix M such that K = M 1 JM , which would mean that MK = JM . But we just showed that in this case M is never invertible! Contradiction. Thus J is not similar to K . Section 6.6. Problem 14. Prove that A T is always similar to A (we know that the λ ’s are the same): 1. For one Jordan block J i : find M i so that M i 1 J i M i = J i T (see example 3). 2. For any J with blocks J i : build M 0 from blocks so that M 0 1 JM 0 = J T . 1
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3. For any A = MJM 1 : Show that A T is similar to J T and so to J and so to A . Solution (4 points) 1. Suppose that we have one Jordan block J i . Then 1 λ 1 0 0 1 λ · · · . . 1 . . 1 = 1 λ 0 1 λ . . . λ 1 0 · · · λ 0 · · · . . . 1 λ 1 0 0 0 λ · · · so J is similar to J T . 2. Suppose that each J i satisfies J i T = M i 1 J i M i . Let M 0 be the block-diagonal matrix consisting of the M i ’s along the diagonal. Then = = M 1 1 J 1 M 1 M 1 2 J 2 M 2 M 1 JM 0 0 . . . . . . . . . M n 1 J n M n M 1 1 J 1 M 1 M 1 J 2 M 2 2 . . . M n 1 J n M n J T 1 J T 2 = 3. A T = ( MJM 1 ) T = ( M 1 ) T J T M T = ( M T ) 1 J T ( M T ) . So A T is similar to J T , which is similar to J , which is similar to A . Thus any matrix is similar to its transpose. Section 6.6. Problem 20. Why are these statements all true? 2 = J T . . . J T n
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(a) If A is similar to B then A 2 is similar to B 2 . (b) A 2 and B 2 can be similar when A and B are not similar. (c) 3 0 0 4 is similar to 3 0 1 4 .
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