Given $X, Y \in \Bbb C^{n \times n}$ and $XY+2YX=3I$, show $[X,Y]$ is nilpotent

Question

Given $X, Y \in \Bbb C^{n \times n}$ and $XY+2YX=3I$, show that $[X,Y]$ is nilpotent.

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I tried with Jacobson Theorem but it does not work. Trying to calculate traces of powers by induction to see if we get all eigenvalues by Newton system equal to $0$ I don't think it is a right idea. Maybe we can get to the characteristic polynomial since $XY$ and $YX$ have the same characteristic polynomial and same eigenvalues. If someone can help me I would like to know if there exists any generalization for this type of problems.

Answer 1

In general, suppose that $I-XY=k(I-YX)$ for some complex number $k$ that is not a root of unity. (In your case, $k=-2$.) Since $I-XY$ and $I-YX$ have the same spectrum $S$, we have $S=kS$. As $k$ is not a root of unity, we must have $S=\{0\}$, or else $S$ would be infinite. Hence $I-XY$ and $I-YX$ are nilpotent. In turn, $[X,Y]=(I-YX)-(I-XY)=(1-k)(I-YX)$ is nilpotent too.

Answer 2

another approach is to use commutativity:
$XY\big(XY+2YX\big)=3XY = \big(XY+2YX\big)XY$
$\implies XY \text{ and }\big(XY+2YX\big)$ are simultaneously upper triangularizable
$R= S^{-1}\big(XY\big)S$ and $T = S^{-1}\big(XY+2YX\big)S \implies S^{-1}(YX)S = \frac{1}{2}\big(T-R\big) =U$
i.e. $(XY)$ and $(YX)$ are simultaneously upper triangularizable

Thus $S^{-1}\big[X,Y\big]S = R - U$ which is upper triangular with eigenvalues $\lambda_k^{(R)}-\lambda_k^{(U)}$. Notice for some permutation matrix $P$ the original equation $\frac{1}{3}\big(XY+2YX\big)=I$ tells us

$\frac{1}{3}\begin{bmatrix} I+2P &\mathbf 0\\ I &2I\end{bmatrix}\begin{bmatrix}\lambda_1^{(R)} \\ \vdots \\ \lambda_n^{(R)}\\\lambda_1^{(U)}\\ \vdots \\ \lambda_n^{(U)}\\\end{bmatrix}=\mathbf 1$

where $\frac{1}{3}\begin{bmatrix} I+2P &\mathbf 0\\ I &2I\end{bmatrix}\mathbf 1 = \mathbf 1$ gives a satisfying solution, and in fact this is unique since $\det\left(\begin{bmatrix} I+2P &\mathbf 0\\ I &2I\end{bmatrix}\right) = \det\big(I+2P\big)\det\big(2I\big)=2^n\cdot \det\big(I+2P\big)\neq 0$.