Let $V$ be a vector space finitely-generated over $\mathbb{C}$. Do there exist endomorphisms $α$ and $β$ of $V$ satisfying the condition that $σ_1 + αβ − βα$ is nilpotent? (Exercise 771 from Golan, The Linear Algebra a Beginning Graduate Student Ought to Know)

I just know that if $V=\mathbb{C}$ the assumption is not true because $α$ and $β$ commute and $σ_1$ is not nilpotent, but I dont know how to star if $\dim(V)>1$.

$σ_1$ is the identity function ($σ_1:x\mapsto x$).


  • $\begingroup$ What is $\sigma_1$? $\endgroup$ – Alex Fok May 20 '15 at 2:56
  • $\begingroup$ the identity fuction $f(x)=x$ $\endgroup$ – José May 20 '15 at 2:57
  • $\begingroup$ And $\beta \sigma$ is composition? $\endgroup$ – Gary. May 20 '15 at 3:02
  • $\begingroup$ yes, the product of endomorphism is composition $\endgroup$ – José May 20 '15 at 3:05

It is impossible to find such endomorphisms because the eigenvalues of $\alpha\beta-\beta\alpha$ would be all $-1$ in order for $\sigma_1+\alpha\beta-\beta\alpha$ to be nilpotent, a contradiction as the trace of $\alpha\beta-\beta\alpha$, which is the sum of eigenvalues, should be $0$.

  • $\begingroup$ just one question, how do you know that the trace must be zero? there is a proposition that establish that? $\endgroup$ – José May 20 '15 at 3:17
  • $\begingroup$ Yes. It is known that $\text{tr}(\alpha\beta)=\text{tr}(\beta\alpha)$, which can shown by direct computation. $\endgroup$ – Alex Fok May 20 '15 at 3:21

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