MIT18_06S10_pset10_s10_sol

MIT18_06S10_pset10_s10_sol - 18.06 Problem Set 10 Solution...

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Unformatted text preview: 18.06 Problem Set 10 Solution Total: 100 points Section 6.6. Problem 12. These Jordan matrices have eigenvalues , , , 0. They have two eigenvectors (one from each block). But the block sizes don’t match and they are not similar : ⎞ ⎛ ⎞ ⎛ 1 1 J = ⎜ ⎜ ⎝ 1 ⎟ ⎟ ⎠ and K = ⎜ ⎜ ⎝ 1 ⎟ ⎟ ⎠ 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 m 11 m 12 JM = ⎜ ⎜ ⎝ m 41 m 42 m 43 m 44 ⎟ ⎟ ⎠ and MK = ⎜ ⎜ ⎝ m 21 m 22 m 31 m 32 ⎟ ⎟ ⎠ . m 41 m 42 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 from blocks so that M − 1 JM = J T . 1 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 1 λ ··· ⎜ ⎜ ⎜ ⎜ ⎝ . . 1 ⎟ ⎟ ⎟ ⎟ ⎠ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ ⎜ ⎜ ⎜ ⎜ ⎝ . . 1 ⎟ ⎟ ⎟ ⎟ ⎠ = ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ 1 λ 1 λ . . . ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ λ 1 ··· λ ··· . . . 1 λ 1 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 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 . . . . . . . . . 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....
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This note was uploaded on 08/22/2011 for the course MATH 1806 taught by Professor Strang during the Fall '10 term at MIT.

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MIT18_06S10_pset10_s10_sol - 18.06 Problem Set 10 Solution...

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