Finding if a linear transformation is diagonalisable Hi i am having some trouble tackling this question for my exam revision.
Let $M_{(2,2)}(\mathbb{R})$ denote the vector spce of 2x2 matrices over the real numbers. Also, let A denote the matrix
$$\begin{bmatrix}2&\lambda\\1&0\end{bmatrix}$$
where $\lambda$ is a real number. Consider the map
$$\psi : V\rightarrow V, \psi(X)=AX-XA. $$
Compute the eigenvalues for $\psi$. For which values of $\lambda$ is $\psi$ diagonalisable?(check the dimension of the kernel of $\psi$.)
It is clear to see that $\psi$ is a linear transformation and hence it is diagonalisable if it has two distinct real eigenvalues.
The problem i am having is how do I find the eigen values of this linear transformation. I have tried finding the eigen values for the matrix $A$ and obtain $\lambda > -1$ but i do not think this is correct as I haven't considered the kernel or the actual function itself at all.
Secondly i tried using the hint and firstly i obtained the kernel
$$AV - VA=0$$
$$\begin{bmatrix}2&\lambda\\1&0\end{bmatrix} \begin{bmatrix}v_1&v_2\\v_3&v_4\end{bmatrix}-\begin{bmatrix}v_1&v_2\\v_3&v_4\end{bmatrix}\begin{bmatrix}2&\lambda\\1&0\end{bmatrix}=\begin{bmatrix}0&0\\0&0\end{bmatrix}$$
rearranging I obtained
$$\begin{bmatrix}-v_2+\lambda v_3&2v_2+\lambda v_4-\lambda v_1\\v_1-2v_3-v_4&v_2-\lambda v_3\end{bmatrix}=\begin{bmatrix}0&0\\0&0\end{bmatrix}$$
from this i obtained $\lambda=\frac{v_2}{v_3}$
so the ker($\psi$)=$c\begin{bmatrix}2&\frac{v_2}{v_3}\\1&0\end{bmatrix}$ for any $c,v_2,v_3 \epsilon\mathbb{R}$ i am stuck on how am i supposed to compute the eigen values ? and i can see that the $dim(ker)=1$ using the rank nullity theorem this means the $dim(im(\psi))$=1

$\psi$ is a linear transformation between $M_{2\times 2} \rightarrow M_{2\times 2},$ so you need to find a matrix representation $[\psi]$ of $\psi$ and compute the eigenvalues and eigenvectors of $[\psi]$. Since $M_{2\times 2}$ is 4-dimensional, $[\psi]$ will be a $4 \times 4$ matrix. It will be diagonalizable  if and only if it has 4 linearly independent eigenvectors. It is not necessary that it has 4 distinct eigenvalues. It may have only 1 eigenvalue with an eigenspace of dimension 4, for example. 
