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Matrix Inversion by the Cayley-Hamilton TheoremDate: 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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