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.

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    $\begingroup$ Set $X = A-B$. Then, $AB - BA = c\left(A-B\right)$ becomes $XB - BX = cX$. Setting $Y = c^{-1}B$, this further becomes $XY - YX = X$. But this is a well-known condition known to imply that $X$ is nilpotent. Not sure how useful this is. $\endgroup$ – darij grinberg Nov 8 at 21:41
  • $\begingroup$ Note that the condition is equivalent to the same condition replacing $A$ by $A-r$ and $B$ by $B-r$. Therefore, it is enough to prove that if $B$ is invertible then $A$ is also invertible, since this and the symmetry of the expression imply that the complement of their spectra coinside. But $A=B+(A-B)$ the sum of an invertible matrix plus a nilpotent. $\endgroup$ – conditionalMethod Nov 8 at 21:50
  • $\begingroup$ Oops. An invertible plus a nilpotent is not always invertible. But in our case $XB=cX+BX$, which implies it by the same proof as $1$ plus nilpotent. $\endgroup$ – conditionalMethod Nov 8 at 22:12
  • $\begingroup$ @darijgrinberg Thank you ! If the matrices commuted, I know for sure that this would imply that $A$ and $B$ have the same eigenvalues, but I don't know if this somehow still holds here (they obviously don't commute since the initial condition would imply $A=B$ and this is just a trivial case). $\endgroup$ – JustAnAmateur Nov 8 at 23:19
  • $\begingroup$ An observation: we have $$ \begin{align} \operatorname{trace}(A^2) &= \operatorname{trace}[A(A - B) + AB] \\ &= \operatorname{trace}[c^{-1}A(AB - BA) + AB] \\ &= \operatorname{trace}[c^{-1}A^{2}B - c^{-1}ABA + AB] \\ &= \operatorname{trace}[c^{-1}A^{2}B - c^{-1}A^2B + AB] = \operatorname{trace}(AB). \end{align} $$ This lets us prove the $k=2$ case, and I suspect that some logic along these lines can be leveraged for a more general proof. $\endgroup$ – Omnomnomnom Nov 8 at 23:33

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.

  • $\begingroup$ I hope you don't mind my edit. Excellent answer! $\endgroup$ – Omnomnomnom Nov 9 at 0:29
  • $\begingroup$ It's okay , thanks ! $\endgroup$ – lessili Nov 9 at 0:31
  • $\begingroup$ Or mine (the letter $n$ was serving double duty). $\endgroup$ – darij grinberg Nov 9 at 2:06

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