Characteristic polynomial of a mapping from matrices space to matrices space 
Let $T$ be the linear map from $M_n \to M_n$ given by TX=AX, while A is as well a matrix $n \times n$
(a) Write out the characteristic polynomials for $T$ 
(b) Show that if A is diagonalzible then T is also diagonalizble

I know that the characteristic polynomial of B, some  $n \times n$ matrix is the expansion of $$\det(B - I \lambda ).$$
The problem is with finding the matrix for $T$.
On one hand it seems so natural to say that A itself is the matrix that represents T by the natural base. 
On the other hand, if I do take the natural base of matrixes $n \times n$ (Where every vector in the base has a 1 somewhere and the rest of the matrix is zeros), the dimension of the base is $n^2$
so my matrix $T_A(x)$ is the size of $n^2$? 
I'm not sure if I explained myself correctly, but I don't even know where to start.. if A is simply the matrix that represent T, the answer to (b) is quite obvious, it's the declaration for T to be digonizable. (or at least a sentence that we already proved in lesson)
Moreover, what can I write as the characteristic polynomial? just $det(A-\lambda I)$
I would really appreciate any help understanding these things. 
Thanks.
 A: Let $E_{ij}$ the matrix which all its entries are $0$ except the entry located at the $i^{th}$ row and $j^{th}$ column is equal $1$ then we know that $$\left(E_{ij}\right)_{1\le i,j\le n}=(E_{11},\ldots,E_{n1},E_{12},\ldots,E_{n2},\ldots,E_{1n},\ldots,E_{nn})$$ is a basis of $\mathcal M_n(\Bbb R)$
and we see that the matrix of $T$ relative to this basis is the diagonal block matrix:
$$ M=\begin{pmatrix}
 \large A &  &&       &  \\
         & \ddots & &{\huge 0}&& \\
     &   &  \ddots     &  & \\
&  {\huge 0}  &      & \ddots & \\
&   &       &  &A \\
 \end{pmatrix} $$
so 
$$\chi_T(x)=\det(M-xI_{n^2})=\prod_{k=1}^n\det(A-xI_n)=\left(\chi_A(x)\right)^n$$
Now we have
$$T^2X=T(TX)=T(AX)=A^2X$$
hence we see easily that (we denote $\pi_A$ the minimal polynomial of $A$)
$$\pi_A(T)=0$$
so $\pi_T|\pi_A$ hence if $A$ is diagonalizable then $\pi_A$ has a simple roots so $T$ is also diagonalizable and
$$\pi_T=\pi_A$$
A: For b):
Consider a basis $(x_i)$ of eigenvectors of $A$:
$$
Ax_i = a_ix_i
$$
Then consider the matrices
$$
X_{i, j} = (0, \ldots, e_{i}\ldots, 0)
\\
TX_{i,j} = a_iX_{i, j} 
$$ (with $e_i$ in the $j^{th}$ position).
and then check that $(X_{i, j})$ is a basis (rather easy).
You conclude that the eigenvalues are (a_i), each with multiplicity multiplied by $n$.
For a):
Use the density of the diagonalizable matrices and the continuity of the 
application giving the characteristic polynomial.
You conclude that the following relationship between characteristic polynomials remains in general:
$$
\Pi_T(X) = \left(\Pi_A(X)\right)^n 
$$ 
