# Eigenvalues of a 3*3 Matrix

• Nov 6th 2008, 09:51 PM
Jimmy_W
Eigenvalues of a 3*3 Matrix
I'm having trouble finding the eigenvalues of a 3*3 matrix.

I know how to find them for a 2*2 matrix, but when its a 3*3 I get completely lost.

If I have the following matrix

$\begin{bmatrix}a&b&c\\d&e&f\\g&h&i \end{bmatrix}$

so

$\begin{bmatrix}a - \lambda &b&c\\d&e-\lambda &f\\g&h&i - \lambda \end{bmatrix}$

how then is the quickest and easiest method to find the eigenvalues?

• Nov 6th 2008, 11:58 PM
mr fantastic
Quote:

Originally Posted by Jimmy_W
I'm having trouble finding the eigenvalues of a 3*3 matrix.

I know how to find them for a 2*2 matrix, but when its a 3*3 I get completely lost.

If I have the following matrix

$\begin{bmatrix}a&b&c\\d&e&f\\g&h&i \end{bmatrix}$

so

$\begin{bmatrix}a - \lambda &b&c\\d&e-\lambda &f\\g&h&i - \lambda \end{bmatrix}$ *

how then is the quickest and easiest method to find the eigenvalues?

Get the determinant of the matrix marked * (read Matrices and determinants) and equate it to zero. Solve the resulting equation for $\lambda$.
• Nov 7th 2008, 12:50 AM
Jimmy_W
Thanks for the link. It contained exactly what I was looking for.
• Nov 7th 2008, 03:10 AM
Herbststurm
Hi,

what you have with the lamdas in the main diagonal is called the characteristic matix. Now you have to calculate the determinant. If you have 3x3 matrices the fastest way is the rule of sarrus, which is nearly the same like a cross product of two vectors.

Than you don't get a concrete number of the determinant, because of the lamdas, but you will get a polynom which is called characteristic polynom.

Now find the roots of this polynom by setting it to zero and solve. Mostly with linear factors and Polynom Divison. Horner's theorem

The roots are the eigenvalues. At this point you are ready searching the eigenvalues.

Greetings