Eigenvectors from eigenvalues doesn't add up

I having some trouble understanding how to find eigenvectors of a matrix. I have the following matrix:

$A=a\left[ \begin{matrix} -1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \\ \end{matrix} \right]$

And then I'm told that the state $\psi$ is an eigenvector of $A$, which is given by:

$\psi =\frac{1}{\sqrt{3}}\left[ \begin{matrix} 1 \\ -1 \\ 1 \\ \end{matrix} \right]$

Then I'm told to find the eigenvalue of that, which turns out to be $-a$, and then I have to find the last eigenvalues and their respective eigenvectors.

The last to eigenvalues is $a$ and another $-a$. Now, if I just use a computer to find the eigenvectors I end up with:

$\left[ \begin{matrix} 0 \\ -1 \\ 1 \\ \end{matrix} \right],\left[ \begin{matrix} 1 \\ 0 \\ 0 \\ \end{matrix} \right],\left[ \begin{matrix} 0 \\ 1 \\ 1 \\ \end{matrix} \right]$,

where the first two is from $-a$ and the last is $a$.

Now, my question is. Where does the eigenvector, $\psi$, I have been told come from ? According to the solution of this problem the remaining eigenvectors, besides $\psi$, should be the first and the second vector given above, but not the third.

So basically, I'm a bit confused. So I was hoping anyone could point in the right direction to why this is so ?

• $\psi$ is a linear combination of the 1st and 2nd eigenvector found by computer. Since these two eigenvectors correspond to the same eigenvalue (equal to $-a$), any their linear combination is again an eigenvector with the same eigenvalue. – Start wearing purple Jun 7 '13 at 18:12
• But why is the solutions given, NOT the eigenvector given by the eigenvalue $a$, but only the two $-a$ ? That is what makes no sense to me :/ – Denver Dang Jun 7 '13 at 18:19
• If the book claims a basis of eigenvectors is as you describe, then it is in error. – David Mitra Jun 7 '13 at 18:20
• @DavidMitra There is no error here. The computer found one particular pair, but there are many other acceptable pairs of eigenvectors corresponding to the eigenvalue $-a$ – SheetJS Jun 7 '13 at 18:24
• You said (I think) that, from the solution, $\psi$ and the first two vectors in your last display form a maximal, independent set of eigenvectors of $A$. This is not the case for the reasons O.L. wrote in his comment. – David Mitra Jun 7 '13 at 18:24

Matrix-vector multiplication is linear (in particular, that means $A(\phi+\rho) = A\phi + A\rho$ for any two vectors $\phi$ and $\rho$)
Thus, if two vectors $\phi$ and $\rho$ have the same eigen value $\lambda$, then
$A(\phi+\rho) = A\phi + A\rho = \lambda\phi + \lambda\rho = \lambda(\phi+\rho)$
and $\phi + \rho$ is an eigenvector of $A$ with eigenvalue $\lambda$ as well. This applies to all linear combinations of $\phi$ and $\rho$.
The original eigenvector you were given is a linear combination of the first two eigenvectors (with coefficients $1/\sqrt3$ for both vectors), both of which have the same eigenvalue.