Finding eigenvectors of a 3x3 matrix with a root of multiplicity 3 I have the matrix \begin{bmatrix}1&0&0\\2&2&-1\\0&1&0\end{bmatrix}
I know that the only eigenvalue is 1 with multiplicity 3
I solved for the first eigenvalue and got \begin{bmatrix}0\\1\\1\end{bmatrix}
How do I find the other two?  I know they are \begin{bmatrix}0\\1\\0\end{bmatrix} and \begin{bmatrix}1/2\\0\\0\end{bmatrix}
but when I do $(A-\lambda I)v_2 = v_1$, I get the system of equations $2x + y -z = 1$, 
$y -z =1$.  I don't see how that gives the second eigenvector.
Thanks

If you use a Linear Algebra tool like in MATLAB, or a libraries for programming languages, like NumPy with Python or MathNet Numerics with C#, you can find out that the eigenvectors calculated by these tools are:
[0, 0, 0]
[(√2)/2, (√2)/2, (√2)/2]
[(√2)/2, (√2)/2, (√2)/2]
The eigenvalues are as stated:
(1,0)
(1,0)
(1,0)
1 of multiplicity 3.
We should bear in mind that eigenvalues could be complex numbers.
