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.

• the corresponding equations are a + 5c = 0 and b -c = 0 from which you can choose two lin independent vectors <a,b,c> for example <5,0,-1> and you think about another one. Now you have three lin independent eigenvectors. With your notes/boiok you should be able to set up the diagonal matrix – imranfat Apr 25 '13 at 17:22
• Something must have gone wrong to obtain the RREF of the matrix for $\lambda =2$. The original matrix is indeed diagonalizable, so the rank of the RREF matrix should be $1$, so that there are two linearly independent vectors in the kernel. – Jared Apr 25 '13 at 17:24

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 $2$.