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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.

user26857
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JustAnAmateur
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    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. – darij grinberg Nov 08 '19 at 21:41
  • 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. – conditionalMethod Nov 08 '19 at 21:50
  • 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. – conditionalMethod Nov 08 '19 at 22:12
  • @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). – JustAnAmateur Nov 08 '19 at 23:19
  • 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. – Ben Grossmann Nov 08 '19 at 23:33
  • My attempt at generalizing: we have

    $$ \begin{align} \operatorname{trace}[A^jB^k] &= \operatorname{trace}[A^{j-1}(A - B)B^k + A^{j-1}B^{k+1}] \ &= \operatorname{trace}[c^{-1}A^{j-1}(AB - BA)B^k + A^{j-1}B^{k+1}] \ &= \operatorname{trace}[c^{-1}A^{j}B^{k+1} - c^{-1}A^{j-1}BAB^k + A^{k-1}B] \ &= \operatorname{trace}[c^{-1}A^{j}B^{k+1} - c^{-1}A^{j-1}BAB^{k-1} + A^{k-1}B] \end{align} $$ this tells us something when $j = 1$ or $k=1$. However, for other positive integers I see no way to deal with the second term using only cyclic permutations of products.

     – Ben Grossmann Nov 08 '19 at 23:55

1 Answers1

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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.

darij grinberg
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Laassila souhayl
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