We define $A:\mathbb{R}^3\rightarrow\mathbb{R}^3$ as a linear operator such that $A\left(\begin{bmatrix}x\\y\\z\end{bmatrix}\right) = \begin{bmatrix}3x-2y+2z\\4x-4y+6z\\2x-3y+5z\end{bmatrix}$ where, $B$ is a basis for $\mathbb{R}^3$.
$$B=\left\{ \begin{bmatrix}1\\3\\6\end{bmatrix}, \begin{bmatrix}-3\\-5\\6\end{bmatrix}, \begin{bmatrix}3\\3\\4\end{bmatrix} \right\}$$
I am guessing that we find $[A]_B^B$ and prove that it is not diagonalizable. How does one prove such claims?