Set $R = M_n(\mathbb{C})$ and let $f_A : R \to R$ be a $\mathbb{C}$-linear map such that $$ f_A(X) = [X,A] = XA - AX. $$
Obvious fact: Note that there are $a_{ij} \in \mathbb{C}$ such that $$ f_A^n (X) = \sum_{i+j=n} a_{ij} A^i X A^j. $$ Thus if $A$ is nilpotent then $f_A$ is nilpotent, obviously. But the assumption "$A$ is nilpotent" is too strong. I want to extend this fact.
My prediction: Assume that $f_A$ has a non-zero eigenvalue $\lambda \in \mathbb{C}$. Then there are $\beta,\gamma \in \mathbb{C}$ which are eigenvalues of $A$ such that $\beta - \gamma = \lambda$
My effort Assume that $n=2$. Let $X$ be a non-zero element of $E(\lambda,f_A)$. Then $X^k \in E(k\lambda, f_A)$. So we get $X$ is nilpotent. So there is a $P \in GL_2(\mathbb{C})$ such that $$ \Lambda := PXP^{-1} = \begin{pmatrix} 0 &1 \\ 0& 0 \end{pmatrix}. $$ Set $$ B=PAP^{-1} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}. $$ Then $[\Lambda, B] = \lambda \Lambda$ implies that $c=0$ and $d-a = \lambda$. So we obtain $$ \Phi_A(x) = \Phi_B(x) = (x-a)(x-a-\lambda). $$
My question: What is a proper extension of the "obvious fact"? Is my prediction true? If so, how to prove ?
This is a direct consequence of the fact that the spectrum of $f_A$ is $\{\mu_i-\mu_j:\ 1\le i,j\le n\}$ when the spectrum of $A$ is $\{\mu_1,\ldots,\mu_n\}$.
This fact can be proved as follows. By a continuity argument, we may assume that $A$ is diagonalisable. Let $u_i$ and $v_i$ be respectively a right eigenvector and a left eigenvector of $A$ corresponding to the eigenvalue $\mu_i$. Then $[u_jv_i^T,A]=u_jv_i^TA-Au_jv_i^T=(\mu_i-\mu_j)u_jv_i^T$. Hence $\{u_jv_i^T:\ 1\le i,j\le n\}$ is an eigenbasis and $\{\mu_i-\mu_j:\ 1\le i,j\le n\}$ is the spectrum of $f_A$.
Alternatively, the matrix representation of $f_A$ is $A^T\otimes I-I\otimes A$. By transforming $A$ into its Jordan form, it is obvious that the eigenvalues of $f_A$ are $\mu_i-\mu_j$ for all $i,j\in\{1,2,\ldots,n\}$.
Another consequence of the aforementioned fact is that $f_A$ is nilpotent if and only if all eigenvalues of $A$ are the same.