Is it true that $[T_A]=[L_A]$? Given two linear transformations, $T_A: M_n(F)\rightarrow M_n(F), T_A(B)=AB$ and $L_A:F^n\rightarrow F^n, L_A(v)=Av$, I am trying to show that their eigenvalues are the same. 
I would like to say that their matrix representations are equal, i.e. $[T_A]=[L_A]$ with respect to some basis. If this is true, then $\det(T_A-\lambda id_{M_n(F)})=\det([T_A-\lambda id_{M_n(F)}])=\det([T_A]-\lambda[id_{M_n(F)}])=\det([L_A]-\lambda I_n)=\det(L_A-\lambda id_{F^n})$.
Am I on the right track? Is my initial assumption true?

No, this can't be true, as $\dim F^n=n$, so $[L_A]$ is an $n\times n$ matrix, but $\dim M_n(F)=n^2$, so $[T_A]$ will be much bigger, an $n^2\times n^2$ matrix.
To prove they share the eigenvalues, show that


*

*any eigenvalue of $L_A$ is an eigenvalue of $T_A$,

*any eigenvalue of $T_A$ is an eigenvalue of $L_A$.



 Consider an element of $M_n(F)$ as an $n$-tuple of the column vectors: $B=(b_1|\dots|b_n)$. 
 Then $T_A(B)=AB=A\cdot(b_1|\dots|b_n)=(Ab_1|\dots|Ab_n)$.

