Characteristic values of linear operator. Let $V$ be the space of $n\times n$ matrices over $\mathbb F$. Let $A$ be a fixed $n\times n$ matrix. Let $T$ be a linear operator on $V$ defined by $T(B)=AB$ for all $B$. Is it true that $A$ and $T$ have the same characteristic values?
 A: Suppose  $\lambda \in \mathbb F$ (wlog algebraically closed) is a characteristic value of $T$. Then $AB =\lambda B$, for some $B\not=0$, i.e. 
$$ (A-\lambda I ) B = 0 \tag{*}.$$
If $\det (A -\lambda I) =0$, then $\lambda$ is a characteristic value of $A$. If not, the matrix $C =(A-\lambda I)$ is invertible, so we can left-multiply $(*)$ by $C^{-1}$ , and conclude that $B= 0$, so $B$ was not an eigen-vector - contradiction. So all characteristic values  of $T$ are characteristic values of $A$. 
Going the other way, consider the action of $T$ on the matrix $B= (v_1,\cdots, v_n)$ ($v_k$ are [column] vectors of $V$):
 $$ T (v_1,\cdots, v_n) = ( Av_1,\cdots, Av_n). \tag{**}$$
Taking $v_1$ as an eigen-vector of $A$, all other $v_k =0$, we see that the characteristic values of $A$ are characteristic values of $T$. 
Note: this could be said in terms of modules over the ring $\mathbb F [x]$ and the module isomorphism implied in $(**)$.
Edit: After the fact, the equation $(**)$ is obviously sufficient to see that that characteristic values of $T$ and $A$ are identical - one doesn't need the first part of the argument, since if $\lambda$ is a characteristic value for $T$ the equation $(**)$ also exhibits it as an characteristic values for $A$, as at least one of  the $v_k$ must be non-zero.  
