# Trying to understand Eigenvectors

1. May 4, 2014

### noelo2014

1. The problem statement, all variables and given/known data

Find the Eigenvalues of A=

4 0 1
-2 1 0
-2 0 1

Then find the eigenvectors corresponding to each of the eigenvalues.

2. Relevant equations

3. The attempt at a solution

I found the Characteristic Polynomial of the matrix, computed the Eigenvalues which are 1,2,3.

What I'm trying to get my head around is the concept of the eigenvectors.

First of all I attempted to find the eigenvector(s) for λ=1. So I constructed the matrix (A-Iλ), row-reduced and got the matrix:

1 0 0
0 0 1
0 0 0

This matrix corresponds to the set of linear eqns (A-Iλ)x, and x must be non-zero. So normally I'd just read the solutions from this matrix and tell myself

x1=0
and
x3=0

I did this in maple and it gave me the value (0,1,0) as the eigenvector corresponding to λ=1, but x2 doesn't equal zero in any of these rows. Can someone explain this to me?

2. May 4, 2014

### CAF123

There is no constraint on x2, so it is arbitrary. 1 is chosen so that the eigenvector is properly normalized. The eigenspace corresponding to eigenvalue $\lambda =1$ is then all vectors of the form $\langle 0, k, 0\rangle$.

3. May 4, 2014

### noelo2014

Ok, so what you're saying is that x1 has to be zero and x3 has to be zero but x2 can be anything?

I think I get it, I think I'll just need to practice more questions like this. It's a bit subtle.

4. May 4, 2014

### Staff: Mentor

It might be helpful to put some words with the work you did.

When you were finding the eigenvectors for λ = 1, you ended up with this matrix:
$$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}$$

Your work started with the matrix equation (A - 1I)x = 0. By row-reduction, you got to the matrix I show above.

This matrix represents the linear system
x1 + 0x2 + 0x3 = 0
0x1 + 0x2 + x3 = 0
0x1 + 0x2 + 0x3 = 0

Or more simply,
x1 = 0
x3 = 0

Since there are no conditions or restrictions on x2, you can substitute any real value for x2. This means that any vector of the form <0, k, 0> is an eigenvector for λ = 1. For convenience's sake, k = 1 is chosen, making the eigenvector <0, 1, 0>. Any multiple of that vector would also be an eigenvector for this eigenvalue.