How to calculate eigenvectors for equal eigenvalues? I've had this matrix on a test some time ago:
$$ \begin{bmatrix}
 0 & 1 & 0\\ 
 -4& 4 & 0\\ 
 -2& 1 & 2
\end{bmatrix} $$
It's eigenvalues are $\ \lambda _1= \lambda _2= \lambda _3=2$.
In my case, that got me the eigenvectors of: $\ v_1(0,0,0), v_2 (0,0,1), v_3 (1,2,0)$.
I got the $v_3$  by using standard $(A- \lambda I)X=0$ system, but I had to guess the other two. In his case, it was easy, since in the system of equations, the $x_3$ never shows up, but I want to know if there is some formal way to determine the other vectors in a situation such as this one. 
 A: The $(A-\lambda I)X=0$ equation should give all the solutions. You simply obtain a solution space of dimension greater than 1. So your question is basically how to find solution spaces to linear equation systems.
To do this, you need to obtain the Canonical Echelon form of the matrix. In our case it has only zeroes except in the upper row which is $(1, \frac{1}{2}, 0)$. Now, the variables corresponding to the first "1" in every row of the Echelon matrix is a "bound" variable; the remaining variables are free. To obtain solutions which are not dependent, simply substitute 1 in one free variable and 0 in the others and see what you get for each such substitution.
In our case, if the variables are $x,y,z$ we have $x+\frac{1}{2}y+0\cdot z=0$, so $x$ is bound and $y,z$ are free. For $y=1, z=0$ we get $x=-\frac{1}{2}$, (so the solution we get is $(\frac{1}{2},1,0)$ which is the same as $(1,2,0)$ and for $y=0, z=1$ we get $x=0$ so the solution is $(0,0,1)$.
Note that $(0,0,0)$ is a trivial solution and not considered eigenvector; in our case the eigenspace of $\lambda = 2$ is of dimension 2, so we are done (the matrix is not diagonizable).
