Finding kernel and image of endomorphism. I have been trying to solve this, but I can't. I am given the endorphism: $$f(X) = A\cdot X - X\cdot A$$
Where $X$ is any real-valued $2 \times 2$ matrix and $$A = \begin{pmatrix}1 & 2 \\ 2 & -1\end{pmatrix}$$
And I have to find a basis of the kernel and a basis of the image of $f$. Finding a basis for the kernel was easy. By definition, the basis is the set of matrices $X$ that verify $f(X) = (0)$, therefore:
$$A\cdot X - X \cdot A = 0$$
$$A\cdot X = X \cdot A$$
$$X = A^{-1}\cdot X \cdot A$$
So, since X must be a diagonal matrix with its two elements equal to its eigenvalues, we know it must be a matrix of the form: $$X =\begin{pmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{pmatrix}$$
And since both $\lambda_1$ and $\lambda_2$ must verify $A-\lambda I_2 = 0$. Writing out the just mentioned equation in matrix form, one gets a matrix with only one non-zero entry for each eigenvalue, being the result in both cases that $\lambda_1= \lambda_2$. Therefore, a basis for the kernel could be: $$B_{ker(f)} = \Big{\{} \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\Big{\}}$$
But I am stuck here. Since this is an endomorphism on the space of square matrices of order $2$, the dimension equation tells us that $$\dim(Im(f)) + \dim(ker(f)) = \dim(Im(f)) + 1 = 4 \Rightarrow \dim(Im(f)) = 4$$
I am not sure I am right and I don't know how to find a basis for the image of $f$, though.

Hint: Observe
\begin{align}
AX-XA =&\
\begin{pmatrix}
1 & 2\\
2 & -1
\end{pmatrix}
\begin{pmatrix}
x & y\\
z & w
\end{pmatrix}
-
\begin{pmatrix}
x & y\\
z & w
\end{pmatrix}
\begin{pmatrix}
1 & 2\\
2 & -1
\end{pmatrix}\\
=&
\begin{pmatrix}
x+2z & y+2w\\
2x-z & 2y-w
\end{pmatrix}
-
\begin{pmatrix}
x+2y & 2x-y\\
z+2w & 2z-w
\end{pmatrix}\\
=&\ 
\begin{pmatrix}
2(z-y) & 2(w-x)+2y\\
2(x-w)-2z & 2(y-z)
\end{pmatrix}.
\end{align}
