Characteristic polynomial of $T:M_n(\mathbb{F}) \rightarrow M_n(\mathbb{F}) ,\ \ TX = AX \ \ (A\in M_n(\mathbb{F}))$ Question - how would I proceed to find the characteristic polynomial of $T:M_n(\mathbb{F}) \rightarrow M_n(\mathbb{F}) ,\ \ TX = AX  \ \ (A\in M_n(\mathbb{F}))$ ?
What I've been trying:
Given the the standard base $\{E_{11}, E_{12}, \dots, E_{nn}\}$ of $M_n(\mathbb{F})$ in which ($E_{ij})_{kl} =\left\{\begin{matrix}
 1,& k=i \ \ and \ \ l=j \\ 0, &otherwise
\end{matrix}\right.$
$T$ can be represented by the following $n^2\times n^2$ matrix:
$$[T] = \begin{pmatrix}
(A)_{11}I_n&(A)_{12}I_n&\cdots&(A)_{1n}I_n\\
(A)_{21}I_n&(A)_{22}I_n&\cdots&(A)_{2n}I_n\\
\vdots&\vdots&\ddots&\vdots\\
(A)_{n1}I_n&(A)_{n2}I_n&\cdots&(A)_{nn}I_n
\end{pmatrix} $$
Now, from from here I'd like to calculate $det([T]-tI_{n^2})$, and this is the point where I got stuck.
I'd be glad for ideas on how to proceed from here, or ideas for other ways to tackle the problem.
 A: You have $(AE_{i,j})_{k,l}= 0$ if $l\neq j$  and $a_{ki}$ if $l=j$.
Thus, if you consider the basis $(E_{11}, E_{21}, ..., E_{n1}, E_{12}, ... , E_{n2}, ... , E_{1n}, ... , E_{nn})$ of $M_n(\mathbb{F})$, then you're lead to computing the determinant of a bloc diagonal matrix of size $n^2\times n^2$, whose $n$ blocks are all equal to $A-XI_n$.
This gives you that $\chi_T(X)=\chi_A(X)^n$.
A: You can view $M_n(\Bbb{F})$ as the direct sum of $n$ copies of $\Bbb{F}^n$, say $V_1,V_2,\ldots,V_n$, where $V_i$ represents the $i$th column.  So, basically, $T:\bigoplus_{i=1}^nV_i\to \bigoplus_{i=1}^n V_i$ by sending $$(v_1,v_2,\ldots,v_n)\mapsto (Av_1,Av_2,\ldots,Av_n).$$  So, each $V_i$ is a $T$-invariant subspace, and $T|_{V_i}:V_i\to V_i$ is the same as the linear operator $A$.  We can use a more general result below to prove that $\chi_T(t)=\big(\chi_A(t)\big)^n$.
Let $U_1,U_2,\ldots,U_k$ be finite dimensional vector spaces and let $S_i:U_i\to U_i$ be linear maps.  Then, the direct sum  $S=\bigoplus_{i=1}^kS_i$ of the linear maps $S_1,S_2,\ldots,S_k$ is the linear transformation $S:\bigoplus_{i=1}^kU_i\to \bigoplus_{i=1}^k U_i$ such that
$$S(u_1,u_2,\ldots, u_k)=(S_1u_1,S_2u_2,\ldots,S_ku_k).$$
Then, the characteristic polynomial $\chi_S(t)$ of $S$ is the product of the characteristic polynomials $\chi_{S_i}(t)$ of each $S_i$.  Id est,
$$\chi_{\bigoplus_{i=1}^kS_i}(t)=\prod_{i=1}^k\chi_{S_i}(t).$$  (A proof can be done by choosing a good basis so that the matrix of $S$ is a block matrix with zero non-diagonal blocks.)
