Relation of characteristic polynomials of a matrix $A$ and $\Phi(B) = AB$ where $B$ is a matrix Let $A \in M \left(n\times n, K\right)$ be a matrix and
\begin{align*}
\Phi: M\left(n\times n, K\right) &\rightarrow M\left(n\times n, K\right) \\
X &\mapsto AX
\end{align*}
a linear map.
The characteristic polynomials of $\Phi$ and $A$ have the property that
$\deg P_A = n$ and $\deg P_\Phi = n^2$. I was also able to show that the minimal polynomials of $\Phi$ and $A$ are equal.
But what do I need to do when I want to show that $\left( P_A \right)^n = P_\Phi$?
 A: If $V$ is an eigenvector of $A$ associated to an eigenvalue $\lambda$ then define the matrix $X=(V\cdots V)$ where the columns of $X$ are $V$ and we get
$$\Phi(X)=\lambda X$$
and so $X$ is an eigenvector of $\Phi$.
Conversely, if $X=(V_1\cdots V_n)$ is an eigenvector of $\Phi$ associated to $\lambda$ then
$$\Phi(X)=\lambda X\implies AV_i=\lambda V_i$$
so $V_i$ is an eigenvector of $A$. Hence
$$E_\lambda(\Phi)=\left\{X=(V_1\cdots V_n)\in \mathcal M_n(\Bbb K)\mid V_i\in E_\lambda(A)\right\}$$
and hence if $A$ is diagonalizable, the algebraic multiplicity of $\lambda$ in $P_\Phi$ is $\dim E_\lambda(\Phi)=(\dim E_\lambda(A))^n$. Now we can deduce the desired equality.
A: This is true in any field so there needs to be a proof which does not assume the existence of eigenvalues or diagonalisability of the matrix $A$. So ... 
If one uses the standard $E_{ij}$ basis for $M_{n\times n}(K)$, and orders them first by $j$ then by $i$, then the matrix of $\Phi$ is just a block diagonal matrix with $A$ repeated $n$ times on the diagonal, and zeros elsewhere. 
It is then clear that $\chi_{\Phi}(t)=\chi_{A}(t)^n$.
