1. ## eigenvector

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?

2. Originally Posted by SwedishMan
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.
How can the eigenvectors be 1, 1, 2, 2? You mean the eigenvalues are 1, 1, 2, 2 right?

You know that one of the eigenvectors (call this one x) is
0
0
0
1

Since you have two equal eigenvalues you will necessarily have another eigenvector with eigenvalue 1. We may assume that these two eigenvectors are orthogonal, so assume another eigenvector with eigenvalue equal to 1 of the form (call this one y):
a
b
c
d

such that
x (inner product) y = 0

Thus d = 0.

So y is of the form:
a
b
c
0.

So calling your original matrix A, the eigenvector equation becomes:
Ay = (1)y

2a + b - c = (1)a
4b - 2c = (1)b
3b - c = (1)c
3b - 2c = (1)0

The last three conditions are identical and say that c = (3/2)b. Putting this into the first equation gives:
2a - (1/2)b = a ==> a = (1/2)b

So the eigenvector takes the form:
(1/2)b
b
(3/2)b
0

We may set any value of b for this, and choosing b = 2 gives you your book's eigenvector.

-Dan