Q:

Properties of a linear transformation $T : M_2(\mathbb{R}) \rightarrow M_2(\mathbb{R})$

Let $A \in M_2(\mathbb{R})$ be of trace $2$ and determinant $-3$. Consider the linear transformation $T: M_2(\mathbb{R}) \rightarrow M_2(\mathbb{R})$ by $$T(B)=AB$$ Then
1) $T$ is diagonalizable
2) $2$ is an eigenvalue of $T$
3) $T$ is invertible
4) $T(B)=B$, for some non zero $B$ in $M_2(\mathbb{R})$

My try:
3):
$Ker(T)=\{B : T(B)=0\}=\{B : AB=0\}=\{0\}$, since $A$ is invertible and $AB=0 \Rightarrow B=0.$ Hence $T$ is one-one and so invertible. Therefore,
3) is true 
4): Since there are so many matrices having trace $2$ and det $-3$, so i take $A$ as $A=\begin{bmatrix}1&2\\2&1\\\end{bmatrix}$ and
take $B=\begin{bmatrix}a&b\\c&d\\\end{bmatrix}$ and so $T(B)=B$ implies $AB=B$. That is, 
$$\begin{bmatrix}1&2\\2&1\\\end{bmatrix} \begin{bmatrix}a&b\\c&d\\\end{bmatrix}=\begin{bmatrix}a&b\\c&d\\\end{bmatrix}$$
which implies $$a+2c=a$$
$$b+2d=b$$
$$2a+c=c$$
$$2b+d=d$$
Hence $a=b=c=d=0$ and so $B=0$. Therefore, 4) is false
2): Since the matrix of $T$ in the standard basis is
$$[T]=\begin{bmatrix}1&0&2&0\\0&1&0&2\\2&0&1&0\\0&2&0&1\end{bmatrix}$$
  The eigenvalues of the above matrices are $-1$ and $3$, each with multiplicity $2$. Form this, 2) is false
1): $$\text{multiplicity of 3}=2=4-rank([T]-3I)$$ which is true
Similarly for the eigenvalue $-1$. so 1) is true
My Question is:
Is my all answers valid ? 
Is there any short cut method to find each one ? 

A:

You solved 2 for a particular matrix. In general, the matrix of $T$ in the standard basis is\begin{bmatrix}\alpha&0&\beta&0\\0&\alpha&0&\beta\\\gamma&0&\delta&0\\0&\gamma&0&\delta\end{bmatrix}if $A=\left[\begin{smallmatrix}\alpha&\beta\\\gamma&\delta\end{smallmatrix}\right]$. So, the characteristic polynomial of $T$ is\begin{multline*}x^4-2 \alpha  x^3-2 \delta  x^3+\alpha^2 x^2+\delta^2 x^2-2 \beta  \gamma  x^2+4 \alpha  \delta  x^2-2 \alpha  \delta^2 x+2 \alpha  \beta  \gamma  x-2 \alpha^2 \delta  x+\\+2 \beta  \gamma  \delta  x+\beta^2 \gamma^2+\alpha^2 \delta^2-2 \alpha  \beta  \gamma  \delta =\bigl((x-\alpha )(x-\delta )-\beta \gamma \bigr)^2=\\=\bigl(x^2-(\alpha +\delta )x+\alpha \delta -\beta \gamma \bigr)^2=\\=(x^2-\operatorname{tr}(A)x+\det(A))^2=(x^2-2x-3)^2\text,\end{multline*}whose roots are $-1$ and $3$.

