TITLE: Problem from the shortlist of the Romanian Mathematical olympiad
QUESTION [9 upvotes]: Let $A,B\in\mathit{M_{n}\left(\mathbb{C}\right)}$ and $c\in\mathbb{C}^{*}$ such that $AB-BA=c\left(A-B\right)$. Prove that $A$ and $B$ have the same eigenvalues. 

My idea was to prove that $\operatorname{Tr}(A^k)=\operatorname{Tr}(B^k)$, $\forall k\in \mathbb{N}$.
For $k=1$ this is obvious since $\operatorname{Tr}(AB-BA)=0$.
I could prove this for $k=2$ by multiplying the given relation by $A$ to the left and to the right respectively and then doing the same thing for $B$. However, I was not able to use the same technique for higher powers and mathematical induction didn't work either.
I think that my idea works because this was shortlisted for Grade 11 students from Romania, who only learn linear algebra, so a solution without abstract algebra should be possible, and the result I mentioned seems suitable for such a question. Furthermore, since it works even for $k=2$ it is just a matter of finding a generalisation.

REPLY [14 votes]: The given equation is equivalent to the following equation : $$ \forall\lambda\in\mathbb{C} \qquad (B-(\lambda-c)I_n)(A-\lambda I_n)=(A-(\lambda-c)I_n)(B-\lambda I_n)$$ Thus , we have : $$\frac{\chi_A(\lambda)}{\chi_B(\lambda)}=\frac{\chi_A(\lambda-c)}{\chi_B(\lambda-c)}$$
Hence  : 
$$
\forall m\in\mathbb{N}, \  \forall \lambda \in \Bbb C \qquad \frac{\chi_A(\lambda)}{\chi_B(\lambda)}=\frac{\chi_A(\lambda-mc)}{\chi_B(\lambda-mc)}
$$ $m\rightarrow \infty\quad$  will give :   $$\forall \lambda \in \Bbb C \qquad \frac{\chi_A(\lambda)}{\chi_B(\lambda)}=1$$
Which means that $A$ and $B$ have the same characteristic polynomial, then they have the same eigenvalues.