Diagonalization of the matrix $\begin{pmatrix}1&-3&3\\3&-5&3\\6&-6&4\end{pmatrix}$

Question

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:

1. 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!

2. 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.)

Answer 1

You only need continue, when the eigenvalue is $-2$, the eigenspace can be find with the definition directly, as in the other eigenvalue $$E_{-2}=\{(x,y,z)\in {\bf R}^3: Av=-2v\}$$ That is, $$\left[\begin{array}{ccc|c} 3&-3&3&0\\ 3&-3&3&0\\ 6&-6&6&0 \end{array}\right]\sim \left[\begin{array}{ccc|c} 1&-1&1&0\\ 0&0&0&0\\ 0&0&0&0 \end{array}\right]\implies x-y+z=0 $$ Then, $$E_{-2}=\{(x,y,z)\in{\bf R}^3: x-y+z=0\}={\rm span}\{(1,1,0),(-1,0,1)\}$$ Similar with the other eigenvalue. Thus, $$P=\begin{bmatrix}-1&1&1\\0&1&1\\1&0&2\end{bmatrix}, \quad D=\begin{bmatrix}-2&0&0\\0&-2&0\\ 0&0&4\end{bmatrix}$$

Answer 2

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 $P.$