Let $ n\geq 2 $ and $ A,B,C \in M_{n}(\mathbb{C}) $ be three matrices so that $$ A^{2}B+BA^{2}=2ABA $$ and $ C=AB-BA $.
Prove that $ \mbox{tr}(C^{k})=0,\forall k\in \mathbb{N}. $ I tried solving it by mathematical induction, but it didn't work. The only thing I know is that $ \mbox{tr}(C)=0. $
The given equality is equivalent to $AC=CA$. According to the Jacobson lemma, $C$ is nilpotent and we are done.
cf. page 1 of https://jankobracic.files.wordpress.com/2011/02/on-the-jacobsons-lemma.pdf
EDIT. to @ George R. Assume that we replace the underlying field $\mathbb{C}$ with a commutative ring $R$. Then the previous reference shows that there is $k$ s.t. $n!A^k=0_n$; if $n!$ is not a zero divisor in $R$, then $A^k=0$ and $A$ is nilpotent. Unfortunately, we know little about the traces of $(A^j)_{j\leq k}$ and even $A^n$ is not necessarily $0$ ! For instance, consider $A=2I_2$ over $R=\mathbb{Z}/16\mathbb{Z}$; $A$ is nilpotent because $A^4=0$ but $tr(A)=4\not= 0,tr(A^2)=8\not=0$ and $A^3\not=0$.
Now, we replace $\mathbb{C}$ with a field $K$ s.t. $charac(K)>n$; then $n!$ is invertible and $A^k=0$; if $spectrum(A)=(\lambda_i)_i$ (the eigenvalues), then $spectrum(A^k)=(\lambda_i^k)_i=(0,\cdots,0)$. Thus, for every $i$, $\lambda_i=0$, and moreover, for every $i,p\geq 1$, $\lambda_i^p=0$. Finally, for every $p\geq 1$, $tr(A^p)=\sum_i\lambda_i^p=0$.
Conversely, we can prove that if $tr(A)=\cdots=tr(A^n)=0$, then $spectrum(A)=\{0,\cdots,0\}$ and $A$ is nilpotent.