MHB Is the commutator of two complex matrices nilpotent?

[Ackbach]

Here is this week's POTW:

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If the commutator, $Z$, of two complex $n\times n$ matrices commutes with one of those matrices, must $Z$ be nilpotent?

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[Ackbach]

No one answered this week's POTW. Source credit: this problem, and quite possibly a few more in the coming weeks, is due to Euge. Here is the solution:

[sp]Yes. Let $Z = [X,Y]$ and suppose $ZX = XZ$. Inductively, $Z^kX = XZ^k$ for all natural numbers $k$. For every $k \ge 1$, $$\operatorname{trace}(Z^k) = \operatorname{trace}(Z^{k-1}XY) - \operatorname{trace}(Z^{k-1}YX) = \operatorname{trace}(XZ^{k-1}Y) - \operatorname{trace}(XZ^{k-1}Y) = 0$$ using the properties $\operatorname{trace}(MN) = \operatorname{trace}(NM)$ and $Z^{k-1}X = XZ^{k-1}$. This implies $Z$ is nilpotent. Indeed, since $Z$ is complex, it has $n$ eigenvalues. If its eigenvalues are not identical, suppose $\lambda_1,\ldots, \lambda_d$ are the distinct eigenvalues of $Z$ with multiplicities $m_1,\ldots, m_d$, respectively. Since $\operatorname{trace}(Z^k) = 0$ for all $k$, then $\sum_{j = 1}^d m_j \lambda_j^k = 0$ for $k = 1,2,\ldots, d$. The Vandermonde determinant of $\lambda_1,\ldots, \lambda_d$ is nonzero since the $\lambda_i$ are distinct, whence $m_1 = \cdots = m_d = 0$. This contradicts the equation $\sum_{j = 1}^d m_j = n$. So let $\lambda$ be the unique eigenvalue of $Z$. The condition $\operatorname{trace}(Z) = 0$ forces $\lambda = 0$. Finally, the characteristic polynomial of $Z$ is $p(t) = t^n$, implying the $Z$ is nilpotent.[/sp]