# Finding eigenvectors of a matrix

JaysFan31

## Homework Statement

I need to find the eigenvectors for the matrix shown below.

## Homework Equations

Nothing relevant.

## The Attempt at a Solution

I have the matrix
2 1 -1 0
0 4 -2 0
0 3 -1 0
0 3 -2 1

The eigenvectors are 1,1,2,2.

To get the eigenvectors for 1,1, I use the matrix
-1 1 -1 0
0 -3 -2 0
0 3 2 0
0 3 -2 0

When I bring this to RREF, I have
1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 0

So I know that the vector
0
0
0
1
is an eigenvector.

But the book also says that
1
2
3
0
is an eigenvector for the eigenvalue 1.

If the matrix only has one solution, how do I get this other eigenvector?

## Homework Equations

Nothing relevant.

See if this equations help:

$$det(A-{\lambda}I) = 0$$

$$A{\cdot}x = {\lambda}x$$

where $$A$$ is the matrix, $$x$$ is an eigenvector and $$\lambda$$ is it's corresponding eigenvalue

D H
Staff Emeritus
When an eigenvalue has a multiplicity > 1 it means that there multiple linearly independent eigenvectors corresponding to that eigenvalue. In particular, the eigenvectors span an N-dimensional subspace, where N is the multiplicity of the eigenvalue.

As a trivial example, the n-dimensional identity matrix $I_n$ has eigenvalue 1 with multiplicity n. The eigenvalues span the original space. (In other words, any vector in $\mathbb R^n$ is an eigenvector of $I_n$).

HallsofIvy
Homework Helper

## Homework Statement

I need to find the eigenvectors for the matrix shown below.

## Homework Equations

Nothing relevant.

## The Attempt at a Solution

I have the matrix
2 1 -1 0
0 4 -2 0
0 3 -1 0
0 3 -2 1

The eigenvectors are 1,1,2,2.

To get the eigenvectors for 1,1, I use the matrix
-1 1 -1 0
0 -3 -2 0
0 3 2 0
0 3 -2 0
This is, I take it, your original matrix with 1 subtracted from the diagonal values? Or, at least it's supposed to be: check the second row: 4-1= 3, not -3.

When I bring this to RREF, I have
1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 0

So I know that the vector
0
0
0
1
is an eigenvector.

But the book also says that
1
2
3
0
is an eigenvector for the eigenvalue 1.

If the matrix only has one solution, how do I get this other eigenvector?

1 1 -1 0
0 3 -2 0
0 3 -2 0
0 3 -2 0

row reduces to
1 0 -1/3 0
0 1 -2/2 0
0 0 0 0
0 0 0 0

That gives both (1, 2, 3, 0) as eigen value.