Matrix Inversion by the Cayley-Hamilton Theorem
Date: 02/27/98 at 18:34:43 From: DuBois Ford Subject: Matrix Inversion by way of the Cayley-Hamilton Theorem I would like to know what the Cayley-Hamilton Theorem is and how it is used to find the inverse of a matrix. I have checked many math encyclopedias but could only find information on Cayley and Hamilton, not the theorem.
Date: 02/28/98 at 10:47:28 From: Doctor Anthony Subject: Re: Matrix Inversion by way of the Cayley-Hamilton Theorem The Cayley-Hamilton theorem states that every square matrix A satisfies its own characteristic equation. The characteristic equation is the equation whose roots are the eigenvalues of the matrix. If these terms are unfamiliar, I will illustrate with a 2x2 matrix, but the ideas can be generalized to an nxn matrix. A = [a b| |c d] If k is an eigenvalue of the matrix then k is found by solving the equation |a-k b | = 0 | c d-k| (The lefthand side is a determinant) (a-k)(d-k) - cb = 0 ad - ak - dk + k^2 - cb = 0 k^2 - (a+d)k + ad-bc = 0 This is the characteristic equation of A. Then Cayley-Hamilton states that this equation is satisfied by A. A^2 - (a+d)A + (ad-bc).I = 0 where I is the identity matrix. Example: If A =[1 -1| |2 3] The characteristic equation of A is k^2 - 4k + 5 = 0 ...........(1) so A^2 - 4A + 5I = 0 ...........(2) If you are familiar with the ideas of eigenvalues and eigenvectors, then if k is an eigenvalue and u an eigenvector, we have A.u = k.u multiply through by A A^2.u = A(k.u) A^2.u = k.A.u A^2.u = k^2.u Apply these ideas to equations (1) and (2) Multiply (1) by the vector u, and we have k^2.u - 4k.u + 5.u = 0 so replacing k^n.u by A^n.u we have A^2.u - 4A.u + 5I.u = 0 (A^2 - 4A + 5I)u = 0 and so A^2 - 4A + 5I = 0 and this proves Cayley-Hamilton. Cayley-Hamilton can be used to find powers of matrices or the inverse of a matrix. For example, if A is the matrix given above, we can write A^2 = 4A - 5I = [4 -4| - [5 0| |8 12] |0 5] = [-1 -4| | 8 7] To find the inverse of A we write the equation in the form: 5I = 4A - A^2 now multiply by A^(-1) 5A^(-1) = 4I - A 5A^(-1) = [4 0| - [1 -1| |0 4] |2 3] = [3 1| |-2 1] A^(-1) = (1/5)[3 1| |-2 1] Although I have demonstrated the methods on a 2x2 matrix, the methods are clearly valid for any nxn matrix. -Doctor Anthony, The Math Forum http://mathforum.org/dr.math/
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