Build up a not diagonalizable linear map

Question

I need an hint for this problem.

Let be $M = \begin{bmatrix}2 & 1 \\ -2 & 0\end{bmatrix} \in M_2(\mathbb{K})$ and $H=\{A \in M_2(\mathbb{K}) : AM=MA \} $

Build up a linear map $f: M_2(\mathbb{K}) \rightarrow M_2(\mathbb{K})$ such that:

1. $rk(f) = 2$
2. $H$ is an eigenspace for $f$
3. $f \left ( \begin{matrix}0 & 2 \\ -4 & -2\end{matrix} \right ) = \begin{bmatrix}2 & 1 \\ -6 & 2\end{bmatrix}$
4. $f$ is not diagonalizable

My doubt: I found that a basis for $H$ is $B_H=\left\{ \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}, \begin{bmatrix}0 & 2 \\ -4 & -2\end{bmatrix} \right\}$, so how it can be possible that H is an eigenspace if $\begin{bmatrix}0 & 2 \\ -4 & -2\end{bmatrix}$ in not an eigenvector?