How to show that the linear operator is diagonalizable 
Let $\mathbf{A}=\begin{pmatrix}a&b\\c&d\end{pmatrix}\in\mathcal{M}_{2\times2}(\mathbb{R})$. Define a linear operator $\mathit{T}$ on $\mathcal{M}_{2\times2}(\mathbb{R})$ such that
$\qquad\qquad\qquad\qquad\qquad\qquad\mathit{T}\left(\mathbf{X}\right)=\mathbf{AX}\quad$ for $\mathbf{X}\in\mathcal{M}_{2\times2}(\mathbb{R})$
Show that if $\mathbf{A}$ is diagonalizable, then $\mathit{T}$ is diagonalizable.

My intended approach is to show that there exists a diagonal matrix $\mathbf{D}$ such that $\mathbf{D}=\mathbf{P}^{-1}\left[\mathit{T}\right]_C\mathbf{P}$, where $C$ is the standard basis for $\mathcal{M}_{2\times2}(\mathbb{R})$, then I can conclude $\mathbf{D}=\left[T\right]_B$ for an ordered basis $B$ for $\mathcal{M}_{2\times2}(\mathbb{R})$, and thus $\mathit{T}$ is diagonalizable.
Can anyone provide some advice on this?
 A: If $A$ is diagonalizable, then there is some matrix $P$ such that 
$$
P^{-1}AP=\begin{pmatrix} a & 0\\0 &b \end{pmatrix}.
$$
Notice that 
$$
A\begin{pmatrix} 1 & 0 \\ 0& 0 \end{pmatrix}= \begin{pmatrix} a & 0 \\ 0& 0 \end{pmatrix},
$$
$$
A\begin{pmatrix} 0 & 1 \\ 0& 0 \end{pmatrix}= \begin{pmatrix} 0 & a \\ 0& 0 \end{pmatrix},
$$
and so on for the other two standard basis vectors for $M_{2\times 2}(\mathbb{R})$.
A: Let $V$ be the vector space of $n\times n$ matrices over the field $F$. 
Let $A$ be  a fixed $n\times n$ matrix over $F$.
let $T$ be a linear operator on $V$ defined as  $T(B) = AB$.
Question is to check whether :
$A$ and $T$ have same eigen values??? (Linear algebra: Hoffman kunze 6.2.15)
let $\lambda \in F$ be an eigenvalue of $T$ i.e., $T(B)=\lambda B$ for some $B\neq 0$.
But, $T(B)=AB$ so, we have $AB=\lambda B$ i.e., $(A-\lambda I)B=0$
Now, $\det ((A-\lambda I).B)=0$ i.e., $\det (A-\lambda I).\det(B)=0$
Suppose $\det (A-\lambda I)\neq0$ this would imply $A-\lambda I$ is invertible and so $(A-\lambda I)B=0$ imply $(A-\lambda I)^{-1}(A-\lambda I)B=0$ i.e., $B=0$ 
So, $\det(A-\lambda I)=0$ which imply that $\lambda$ is an eigenvalue of $A$ 
Conversely, Suppose $\lambda \in F$ is an eigenvalue of $A$.
I need to find some $B\neq 0$ such that $T(B)=\lambda B$
$T(B)=AB$ So, i have $AB=\lambda B$ i.e., $(A-\lambda I)B=0$
As $\lambda$ is an eigenvalue of $A$, $A-\lambda I$ is not invertible.
Now, Thanking an equivalent condition of if $AB\neq 0$ for any non zero matrix $B$ then $A$ is invertible , we see that there does exist $B\neq 0$ such that $(A-\lambda I)B=0$ and thus we are done.
Thus, we see that $A$ and $T$ have same eigen values 
Thus, $A$ is diagonalizable iff $T$ is diagonalizable.
A: If the field is not algebrically closed then the fact that $A$ is diagonalizable does not imply that T is diagonalizable.. suppose minimal polynomial of $A$ is $(x-c1)(x-c2)\dots(x-cn)$
and minimal polynomial of $T$ is $(x-c1)\dots(x-cn)(x^2+c)$ Then what?
