Finding two eigenvectors from one eigenvalue My assignment tells me to Find a diagonal matrix similar to the matrix:  $$\begin{bmatrix}0 &-4 &-6\\-1 &0 &-3\\1 &2 &5\end{bmatrix}$$  It's eigenvalues are 1, 2 and 2. And the eigenvector for lambda = 1 is $$\begin{bmatrix}-2\\-1\\1\end{bmatrix}$$ I have noe problem finding the eigenvalues or the eigenvectors for 1 and 2, but since I'm going to diagoanlize it, I need 3 linaerly independent eigenvectors. And I have no clue how to do that.
The matrix for lamba=2 (after subtracting lambda, and making it in RREF) is 
$$\begin{bmatrix}1 &0 &5\\0 &1 &-1\\0 &0 &0\end{bmatrix}$$ 
Meaning z is free. y=z, and x=-5z  Any help on how to find two eigenvectors from this, such that all three eigenvectors are linearly indep. would be awesome. I don't need any further help (I think) on how to diagonalize it, as that is shown in great detail on youtube. Also; I'm new to the MathJax layout/setup, so sorry if it could be made more easy to read. 

Your RREF is wrong. It should be $1,2,3$ on the first line, and zero elsewhere. Thus giving the eigenbasis $(-3,0,1)$, $(-2,1,0)$ for the eigenvalue $2$.
