Find a non-diagonizable matrix that commutes with $\begin{bmatrix}0 & 0& -1\\1 & 1 &1\\0 & 0 & 1\end{bmatrix}$ I'm really lost on this one, with no clue on where to begin. Any help would be appreciated.
 A: HINT:
A general $3 \times 3$ matrix can be written as 
$${\bf G} = \begin{bmatrix} a & b & c \\ d & e & f \\ g  & h & i \end{bmatrix}$$
Let your matrix be ${\bf Y}$. You need to multiply out ${\bf GY}$ and ${\bf YG}$ and then set them equal to one another. 
The matrices ${\bf GY}$ and ${\bf YG}$ are equal if, and only if, each of the entries are equal. Putting ${\bf GY} = {\bf YG}$ will give you nine (linear) simultaneous equations in $a,b,c,d,e,f,g,h$ and $i$. Solve these and then substitute them back into ${\bf G}$ will give you your answer.
I have worked through these equations and they are quite simple to solve. For example, you will have $b+g=0$ and $b+h=0$ as the first two of these equations. Once you have the $a,b,c,d,e,f,g,h$ and $i$ you can find the conditions for ${\bf G}$ to be diagnoalisable.
A: Not a general approach, but one solution is easy to see. You have for example
$$\begin{pmatrix} 0&0&0 \\ 0&0&1\\0&0&0\end{pmatrix}\cdot
\begin{pmatrix}0&0&-1\\1&1&1\\0&0&1\end{pmatrix} = \begin{pmatrix} 0&0&0 \\ 0&0&1\\0&0&0\end{pmatrix}=\begin{pmatrix}0&0&-1\\1&1&1\\0&0&1\end{pmatrix}\cdot\begin{pmatrix} 0&0&0 \\ 0&0&1\\0&0&0\end{pmatrix}$$
A: The given matrix has eigenvectors $v_1=\begin{pmatrix}1\\-1\\0\end{pmatrix}$ with $Av_1=0$, 
$v_2=\begin{pmatrix}0\\1\\0\end{pmatrix}$ with $Av_2=v_2$, and 
$v_3=\begin{pmatrix}-1\\0\\1\end{pmatrix}$ with $Av_3=v_3$. (Espsecially, it is diagonalizable - contrary to popular belief).
With respect to the basis $v_1,v_2,v_3$ the given matirx is just 
$$\begin{bmatrix}0&0&0\\0&1&0\\0&0&1\end{bmatrix} $$
and a simple example of a commuting nondiagonalizable matrix would be 
$$\begin{bmatrix}0&0&0\\0&0&1\\0&0&0\end{bmatrix}. $$
Incidentally, this matrix remains unchanged when we go back to expressing it with respect to the standard basis. So the final answer is comprised in 
$$\begin{bmatrix}0 & 0& -1\\1 & 1 &1\\0 & 0 & 1\end{bmatrix}\cdot \begin{bmatrix}0&0&0\\0&0&1\\0&0&0\end{bmatrix}=\begin{bmatrix}0&0&0\\0&0&1\\0&0&0\end{bmatrix}\cdot \begin{bmatrix}0 & 0& -1\\1 & 1 &1\\0 & 0 & 1\end{bmatrix}=\begin{bmatrix}0&0&0\\0&0&1\\0&0&0\end{bmatrix}. $$
