$\DeclareMathOperator{\SEnd}{End}$Let $X$ be a finite-dimensional vector space over some algebraically complete field $F$. Let $\Phi,\Psi \in \SEnd[X]$ be endomorphisms on $X$. Define the operator $L \in \SEnd[\SEnd[X]]$ by $$L(T) := \Phi T - T \Psi $$ What are the eigenvalues of $L$?
If $(\lambda,u)$ and $(\nu,v)$ are eigenvalue-eigenvector pairs for $\Phi,\Psi^\intercal$, then we can define $T := uv^\intercal$ (where $u,v$ are viewed as vectors in $F^n$). Then $$\begin{align*} L(uv^\intercal) &= \Phi uv^\intercal - uv^\intercal \Psi \\ &= \lambda uv^\intercal - u(\Psi^\intercal v)^\intercal \\ &= \lambda uv^\intercal - u(\mu v)^\intercal \\ &= (\lambda - \mu) uv^\intercal \end{align*}$$ Since $\lambda - \mu$ where $\lambda,\mu$ ranges over the eigenvalues of $\Phi,\Psi$ have the potential to take $n^2$ different values, I suspect that all eigenvalues of $L$ can be written in this form. Is this true?
Addendum: The answer seems to be affirmative using minimal polynomials. If no one answers I will fill in an answer soon.