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### Powers of Matrices

```
Date: 6/27/96 at 14:54:56
From: Jose Miguel
Subject: Matrices Exponential

I am in the 2nd year of the "Mathmatics teaching" course, and I'll be
taking a test next Monday, and I can't get anywhere information about
the chapter concerning Matrices Exponential.

Thanks.

Sofia
```

```
Date: 6/29/96 at 18:1:8
From: Doctor Anthony
Subject: Re: Matrices Exponential

Powers of Matrices.

There are two common methods for finding powers of square matrices.
These are:

(1) Find the eigenvalues and eigenvectors of the matrix.  If k1, k2,
k3 are the eigenvalues and s1, s2, s3 the corresponding
eigenvectors,  then we have for the matrix A

A*s1 = k1*s1    A*s2 = k2*s2   and     A*s3 = k3*s3

Now make up a matrix P with s1 as the first column, s2 the second
column and s3 the third column, then

A*P = P*D  where D is the diagonal matrix with elements k1, k2, k3

Post multiply by P^(-1) and we get

A = P*D*P^(-1)

A^2 = P*D*P^(-1)*P*D*P^(-1)   Now P^(-1)*P = I
= P*D^2*P^(-1)     continuing in this style

A^3 = P*D^3*P^(-1)

and in general  A^n = P*D^n*P^(-1)

Now this a simple result to evaluate.  P was simply written down
knowing s1, s2 and s3, P^(-1) is the inverse matrix - not unduly
difficult to find.  D^n is a diagonal matrix with elements k1^n,
k2^n and k3^n, results which can be read off a calculator.  We
note too that using this method it is as easy to find A^500 as
A^3.

Any standard book on algebra will show how to find the eigenvalues
and eigenvectors.  Briefly we solve det(A-kI) = 0 to get the
eigenvalues, and having got the k values, we get the eigenvectors
from A*s1 = k1*s1 etc.

(2) The Cayley-Hamilton theorem and powers of matrices.

If det(A-kI) = a0 + a1*k + a2*k^2 + a3*k^3 = 0

This is called the characteristic equation of the matrix, and the
Cayley-Hamilton theorem states that every square matrix, A,
satisfies its own characteristic equation.  That is

a0*I + a1*A + a2*A^2 + a3*A^3 = 0

Example;  If A = |1 -1|
|2  3|  find A^8

The characteristic equation is k^2 - 4k + 5 = 0

So A^2 - 4A + 5I = 0   and:

A^2 = 4A - 5I
A^4 = 16A^2 - 40A + 25I
= 16(4A-5I) - 40A + 25I
= 24A - 55I
A^8 = 576A^2 - 2640A + 3025I
= 576(4a-5I) - 2640A + 3025I
= -336A + 145I

This method is clearly not as good as the previous one for finding
large powers of A.

-Doctor Anthony,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```
Associated Topics:
College Linear Algebra
High School Linear Algebra

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