Diagonalization of the matrix $\begin{pmatrix}1&-3&3\\3&-5&3\\6&-6&4\end{pmatrix}$ We need to find whether the matrix $$\begin{pmatrix}1&-3&3\\3&-5&3\\6&-6&4\end{pmatrix}$$
is diagonalizable. If so, we have to find the diagonal matrix and also the matrix that will diagonalize it.
I have found out the eigen values: $-2,-2,4.$
And using the eigen values, the supposed diagonal matrix is$$\begin{pmatrix}-2&0&0\\0&-2&0\\0&0&4\end{pmatrix}.$$
But I need to prove that $D=P^{-1}AP.$ That's where the problem arises:

*

*I have a problem in finding the eigen vectors. How can I find the eigen vector when the eigen value is $-2?$ I am getting only one equation!


*Also, how can I find the "$P$" so that I can do $P^{-1}AP?$ (If I had 3 different eigen values and for that if we had got 3 different eigen vectors, finding $P$ was pretty straight forward. But here we have 2 repetitions of the same eigen value and I can't even find out the eigen vector.)

You were nearly done, since you already found two independent eigenvectors for $-2,$ namely $(1,1,0)^T$ and $(-1,0,1)^T.$
An eigenvector for $4$ is $(1,1,2)^T.$
These 3 columns will do for $P.$
