Given a linear transformation $T$ find a basis $B$ such that $[T]_B$ is a diagonal matrix Given

$ A = \begin{bmatrix} 1 & 3 \\ 0 & 6 \end{bmatrix} $

and the linear transformation $T:M_2(\mathbb{R}) \rightarrow M_2(\mathbb{R})$ which is defined as

$T(v) = Av$

find a basis $B$ of $M_2(\mathbb{R})$ such that $[T]_B$ is a diagonal matrix.
I have no idea how to approach this problem, any help would be appreciated.

A matrix $M = \begin{pmatrix} x & y \\ z & t \end{pmatrix}\neq 0$ is an eigen vector of $T$ for eigenvalue $\lambda$ if and only if
\begin{align}
\begin{pmatrix} x + 3z & y + 3t \\ 6z & 6t \end{pmatrix} = \begin{pmatrix} \lambda x & \lambda y \\ \lambda z & \lambda t \end{pmatrix}
\end{align}
and then, if and only if
\begin{align}
(1-\lambda)x + z &= 0 \\
(1-\lambda )y+  t&= 0 \\
(6-\lambda)z &=0\\
(6-\lambda)t&=0
\end{align}
So now, the problem reduces to determine for which value of $\lambda$ the above system have a non-zero solution. If you can determine $4$ independant matrices $M_1,M_2,M_3$ and $M_4$ that are solution (maybe they are eigenvectors with the same eigenvalue $\lambda$!), you can diagonalize $T$ in the basis $(M_1,M_2,M_3,M_4)$.
A trick is to notice that $\begin{pmatrix}1 \\ 0 \end{pmatrix}$ is an eigenvector of $M$ with eigenvalue $1$, so $M_1=\begin{pmatrix} 1 & 0\\ 0 & 0  \end{pmatrix}$ and $M_2 = \begin{pmatrix}0 & 1 \\ 0 & 0 \end{pmatrix}$ are eigenvectors of $T$ with eigenvalue $1$. You already have two independant eigenvectors for $T$, so just focus on finding two others.
In fact, if $V_1$ and $V_2$ are two linearly independant eigenvectors of $M$, then the $2\times 2$ matrices $(V_1,0)$, $(0,V_1)$, $(V_2,0)$ and $(0,V_2)$ will be independant eigenvectors of $T$. One can see this by looking at the above system: the $x$ and $z$ coordinates are coupled, independantly of the $y$ and $t$ coordinates (in fact, they are not so independant because they have to have the same eigenvalue!). So $(x,z)$ and $(y,t)$ have to be eigenvectors of $M$ with the same eigenvalue $\lambda$.
