Cayley-Hamilton TheoremDate: 04/19/99 at 13:03:32 From: Sarah Mastura Subject: The Cayley-Hamilton Theorem - a proof I was wondering if you could possibly send me a proof of the Cayley- Hamilton Theorem, or if not point me in the right sort of direction of one. The proof in my notes is quite incomplete and I can't, right now, get hold of a book that it might be in. Thanks very much. Date: 04/19/99 at 16:10:08 From: Doctor Rob Subject: Re: The Cayley-Hamilton Theorem - a proof Thanks for writing to Ask Dr. Math. Cayley-Hamilton Theorem: If A is an n-by-n matrix and f(x) its characteristic polynomial, then f(A) = 0. Proof: Let B = x*I - A, where I is the identity matrix of the same size as A. Let C be the classical adjoint of B, so that BC = (det B)I. Recall f(x) = det(B), by definition. We wish to show that f(A) = 0. It suffices to show that f(A)*e(k) = 0 for 1 <= k <= n, where e(k) is the k-th unit vector, so fix a value of k in that range. Now x = A implies that for all j, x*e(j) = A*e(j), n = SUM A(i,j)*e(i), i=1 n n x*SUM delta(i,j)*e(i) = SUM A(i,j)*e(i), i=1 i=1 n SUM [x*delta(i,j)-A(i,j)]*e(i) = 0, i=1 n SUM B(i,j)*e(i) = 0, i=1 n C(j,k)*SUM B(i,j)*e(i) = 0, i=1 n SUM B(i,j)*C(j,k)*e(i) = 0. i=1 Then, summing over all j, reversing the order of summation, and using the definition of the classical adjoint, n n SUM SUM B(i,j)*C(j,k)*e(i) = 0, j=1 i=1 n n SUM [SUM B(i,j)*C(j,k)]*e(i) = 0, i=1 j=1 n SUM [det(B)*delta(i,k)]*e(i) = 0, i=1 n SUM [f(x)*delta(i,k)]*e(i) = 0, i=1 n SUM [f(A)*delta(i,k)]*e(i) = 0, i=1 n f(A)*SUM [delta(i,k)]*e(i) = 0, i=1 f(A)*e(k) = 0. This proves that f(A) annihilates all the unit vectors, so must be the zero matrix, and f(A) = 0. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/ |
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