Characteristic polynomial for an operator.

Question

For a fix matrix $A\in M_{n\times n}(\mathbb{K})$. Let the operator $L\colon M_{n\times n}(\mathbb{K}) \to M_{n\times n}(\mathbb{K})$ defined by $L(X)=AX$.

I want to find the characteristic polynomial for $L$ in terms of characteristic polynomial of $A$.

How do I proceed?

Answer 1

Hints. Let $V_j$ be the subspace of all matrices of which all columns except perhaps the $j$-th ones are zero.

1. Since each $V_j$ is an invariant subspace of $L$ and $V$ is the direct sum of all $V_j$s, we have $p_L=p_{L|_{V_1}}p_{L|_{V_2}}\cdots p_{L|_{V_n}}$, where $p_L$ denotes the characteristic polynomial of $L$ and likewise for each $p_{L|_{V_n}}$.
2. Observe that the extraction $\theta_j:V_j\to \mathbb K^n$ defined by $\pmatrix{0&\cdots&0&x&0&\cdots&0}\mapsto x$ is an isomorphism. Therefore the two linear operators $L|_{V_j}:V_j\to V_j$ and $(\theta_j\circ L|_{V_j}\circ\theta_j^{-1}):\mathbb K^n\to\mathbb K^n$ have the same characteristic polynomials. What is the relationship between $\theta_j\circ L|_{V_j}\circ\theta_j^{-1}$ and $A$?