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challenge8 - 18.06 Spring 2005 Problem Set 8 Solution to...

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18.06 - Spring 2005 - Problem Set 8 Solution to the Challenge Problem Challenge Problem: Consider the 3 × 3 matrix a b c A = 1 d e 0 1 f Determine the entries a, b, c, d, e, f so that: the top left 1 × 1 block is a matrix with eigenvalue 2; the top left 2 × 2 block is a matrix with eigenvalues 3 and -3; the top left 3 × 3 block is a matrix with eigenvalues 0, 1 and -2. Solution. Let A i denote the top left i × i block of A . The matrix A 1 is the matrix ( a ). Since a is the only eigenvalue of this matrix, we conclude that a = 2. We now move on to determining the entries of the matrix A 2 , the top left 2 × 2 block of A : 2 b A 2 = 1 d Since the sum of the eigenvalues of A 2 is 0 by hypothesis, and it is also equal to the trace of A 2 , we obtain that 2 + d = 0, or d = 2. Moreover, the product of the eigenvalues of A 2 is -9 by hypothesis, and it is equal to the determinant of A 2 . Thus we have 9 = 2 d b = 4 b and we deduce that b = 5 and therefore 2 5 A 2 = 1 2 Finally, consider A = A 3 . Again, the sum of the eigenvalues of A is -1 and it is
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