Let $V$ be the space of $n\times n$ matrices, $B$ a fixed matrix, prove that $L_B:A\mapsto BA$ and $B$ have the same eigenvalues.

Question

I know that $\operatorname{trace}(L_B) = n\times \operatorname{trace}(B)$ and $\det(L_B)= (\det(B))^n$. Moreover, I know that the matrix of $L_B$ in the standard basis is

\begin{pmatrix} B_{11}&B_{12}&\cdots &B_{1n}\\ B_{21}&B_{22}&\cdots &B_{2n}\\ \cdots\\ B_{n1}&B_{n2}&\cdots &B_{nn}\\ \end{pmatrix}

where $B_{ij}$ is a diagonal matrix with as diagonal value the element $b_{ij}$ of the matrix $B$.

Now, what about eigenvalues and eigenvectors of $B$ and $L_B$?

Answer 1

Hint: Prove by induction that $L^k_B(A)=B^kA$ for all $k\in\Bbb N$ and deduce that for all polynomial $P$

$$P(L_B)(A)=P(B)A$$ and deduce the desired result.