Diagonalizing a MatrixDate: 04/12/2001 at 02:14:11 From: john cholewa Subject: Regarding a class of linear algebra problem. Hi Dr. Math - In Herstein's _Topics in Algebra_ there is a problem in section 6.10 on page 349 stating that: If Q is a real symmetric matrix satisfying Q^k = I for some integer k >= 1, prove that Q^2 = I. Also in an online problem archive there is a problem stating: If A is a 2x2 matrix with integer entries, and A^k = I for some k >= 1, then A^12 = I. The obvious thing that jumps to mind is that the Det() of the matrix is +-1, and its invertible. With A^(k-1) = A^(-1). However, to go any farther I need something on the other direction of the Caley-Hamilton to force something, like if one knows TQT^(-1) = I for some T, then it's the unique diagonalizing matrix for Q. (Actually looking for a generalized theorem that given a certain condition on A [normal?], and A^k = I for some k >= 1, then A^(n) = I for some n.) Hope you can help me here. Thanks. Date: 04/12/2001 at 13:16:38 From: Doctor Rob Subject: Re: Regarding a class of linear algebra problem. Thanks for writing to Ask Dr. Math, John. Any real symmetric matrix can be diagonalized. If Q is that matrix, there is a real orthogonal matrix P such that P'*Q*P = D where D is diagonal and P' means the transpose of P, and P*P' = P'*P = I. Then Q = P*D*P'. Now taking the kth power, I = Q^k = P*D^k*P', I = P'*I*P = D^k. Since D is diagonal and real, that means all the diagonal entries of D must satisfy d^k = 1, so each must be 1 or -1. That implies that d^2 = 1, and so D^2 = I, and so Q^2 = I. Two-by-two integer matrices are a different story. Usually they cannot be diagonalized. On the other hand, they do satisfy quadratic characteristic equations, by the Cayley-Hamilton Theorem. Thus there are integers a and b such that a = trace(Q), b = det(Q), Q^2 - a*Q + b*I = 0. On the other hand, you know that Q^k = I for some k, so det(Q^k) = det(Q)^k = 1, det(Q) = 1 or -1. Thus b = 1 or -1. The task at hand is to show that the only possibilities for the minimal polynomial of Q are x - 1, x + 1, x^2 - 1, x^2 + 1, x^2 - x + 1, x^2 + x + 1, and all of these are divisors of x^12 - 1. That I leave to you. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/ |
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