let the set of all n by n matrices with complex entries be denoted by $M_n(\mathbb{C}).$ Let $A,B\in M_n(\mathbb{C})$ be such that $AB-BA $ commutes with $A$. Show that $AB-BA$ is nilpotent.
I need to show that there exists a positive integer $k$ so that $(AB-BA)^k = 0. $ No nilpotent matrix can be invertible since otherwise we'd get the contradiction that it equals zero. There's a theorem that an $n\times n$ matrix with entries from a field is nilpotent iff its characteristic polynomial is $t^n$. Thus we just need to show that the characteristic polynomial of $AB-BA$ is $t^n. $ Let $C = AB-BA.$ I think I can show something like $trace(C^m) = 0$ for all positive integers m. Note that the trace is the sum of the eigenvalues of any matrix (by Vieta's formulas).
