TITLE: $A^2=AB+BA$. Prove that $\det(AB-BA)=0$
QUESTION [5 upvotes]: Let $A,B$ be two $3\times 3$ matrices with complex entries, such that $A^2=AB+BA$. Prove that $\det(AB-BA)=0$
Nice problem, and I want to find a solution.
$AB-BA=A^2-2BA=(A-2B)A$ so if $|A|=0$ we have done, if $|A| \not=0$ I can't prove.

REPLY [16 votes]: $$\begin{align}
\det(AB-BA)&= \det(A^2-2BA)\\
&= \det(A-2B)\det(A) \\
&= \det(A)\det(A-2B) \\
&= \det(A^2-2AB) \\
&= \det(BA-AB) \\
&= (-1)^3 \det(AB-BA) \\
&= -\det(AB-BA)
\end{align}$$
So $$\det(AB-BA)=0$$