Homework Help: 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.