# Computation of Eigenvalues

For a square matrix A of order n, the number is an eigenvalue if and only if there exists a non-zero vector C such that

Using the matrix multiplication properties, we obtain

This is a linear system for which the matrix coefficient is . We also know that this system has one solution if and only if the matrix coefficient is invertible, i.e. . Since the zero-vector is a solution and C is not the zero vector, then we must have

Example. Consider the matrix

The equation translates into

which is equivalent to the quadratic equation

In other words, the matrix A has only two eigenvalues.

In general, for a square matrix A of order n, the equation

will give the eigenvalues of A. This equation is called the characteristic equation or characteristic polynomial of A. It is a polynomial function in of degree n. So we know that this equation will not have more than n roots or solutions. So a square matrix A of order n will not have more than n eigenvalues.

Example. Consider the diagonal matrix

Its characteristic polynomial is

So the eigenvalues of D are a, b, c, and d, i.e. the entries on the diagonal.

This result is valid for any diagonal matrix of any size. So depending on the values you have on the diagonal, you may have one eigenvalue, two eigenvalues, or more. Anything is possible.

Remark. It is quite amazing to see that any square matrix A has the same eigenvalues as its transpose AT because

For any square matrix of order 2, A, where

the characteristic polynomial is given by the equation

The number (a+d) is called the trace of A (denoted tr(A)), and clearly the number (ad-bc) is the determinant of A. So the characteristic polynomial of A can be rewritten as

Let us evaluate the matrix

B = A2 - tr(A) A + det(A) I2.

We have

We leave the details to the reader to check that

In other word, we have

This equation is known as the Cayley-Hamilton theorem. It is true for any square matrix A of any order, i.e.

where is the characteristic polynomial of A.

We have some properties of the eigenvalues of a matrix.

Theorem. Let A be a square matrix of order n. If is an eigenvalue of A, then:

1.
is an eigenvalue of Am, for
2.
If A is invertible, then is an eigenvalue of A-1.
3.
A is not invertible if and only if is an eigenvalue of A.
4.
If is any number, then is an eigenvalue of .
5.
If A and B are similar, then they have the same characteristic polynomial (which implies they also have the same eigenvalues).

The next natural question to answer deals with the eigenvectors. In the next page, we will discuss the problem of finding eigenvectors..

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Author: M.A. Khamsi