$AB - BA = \lambda B$ has a nontrivial solution, iff $\lambda = \lambda_i - \lambda_j$?

Question

I have the following question in hand.

If $\lambda_1,\cdots,\lambda_n$ are the eigenvalues of a given matrix $A \in M_n$, then prove that the matrix equation $AB - BA = \lambda B$ has a nontrivial solution $B \neq 0 \in M_n$, if and only if $\lambda = \lambda_i - \lambda_j$ for some $i,j$.

Answer 1

Consider the operator $B\mapsto [A,B]$. What are its eigenvalues?

Answer 2

Here is my attempt. Does this make sense to you experts?

If we vectorize such that \begin{align} AB - BA &= \lambda B \\ &\Downarrow \\ \mbox{vec}\left(AB - BA \right) &= \mbox{vec}(\lambda B) \\ \mbox{vec}\left(ABI - IBA \right) &= \mbox{vec}(\lambda B) \\ \left(\left(I \otimes A\right) - \left(A^{\rm T} \otimes I\right)\right)\mbox{vec}(B) &= \lambda \mbox{vec}(B) \\ \end{align}

So, according to Theorem 13.16, the eigenvalues of the Kronecker sum $\left(\left(I \otimes A\right) - \left(A^{\rm T} \otimes I\right)\right)$ would be $\lambda_i - \lambda_j$. Hence, the solution should be non-trivial if and only if $\lambda = \lambda_i - \lambda_j$.