How to show that $A^2=AB+BA$ implies $\det(AB-BA)=0$ for $3\times3$ matrices? 
Let $A$ and $B$ be two $3\times 3$ matrices with complex entries such that $A^2=AB+BA$. Prove that $\det(AB-BA)=0$.

(Is the above result true for matrices with real entries?)
 A: This was easier than I originally thought.  Note that
$$
AB - BA = AB + BA - 2BA = A^2 - 2BA = (A-2B)A\\
-(AB - BA) =
AB + BA - 2AB=
A^2 - 2AB = A(A - 2B)
$$
and that we have both $\det(AB - BA) = \det(-(AB - BA))$ and $\det(AB - BA) = -\det(-(AB - BA))$.

My original approach: note that $AB - BA$ has trace zero.  Then, we note that
$$
AB - BA = AB + BA - 2BA = A^2 - 2BA = (A-2B)A\\
-(AB - BA) =
AB + BA - 2AB=
A^2 - 2AB = A(A - 2B)
$$
From there, we note by Sylvester's determinant theorem that
$$
\det(AB - BA - \lambda I)= \\
\det((A - 2B)A - \lambda I) =\\
\det(A(A - 2B) - \lambda I) =\\
\det(-(AB - BA) - \lambda I) =\\
(-1)^3 \det(AB - BA + \lambda I)
$$
So, if $\lambda$ is an eigenvalue of $AB - BA$, then so is $-\lambda$.  (Alternatively, directly note that $(A - 2B)A$ and $A(A - 2B)$ must have the same eigenvalues).
Thus, we know that $AB - BA$ has $3$ eigenvalues whose sum is zero and such that $\lambda$ is an eigenvalue iff $-\lambda$ is an eigenvalue.  We conclude that $AB - BA$ must have $0$ among its eigenvalues.  That is, $\det(AB - BA - 0I) = \det(AB - BA) = 0$.
A: Let $A,B\in F^{n,n}$, $F$ any field of characteristic $\ne 2$, $n$ odd. Assume they satisfy $A^2 = AB+BA$.
From $A^2=AB+BA$ it follows
$$
AB-BA= A^2-2BA= (A-2B)A.
$$
Then it holds
$$
\det(AB-BA) = \det((A-2B)A) = \det(A(A-2B)).
$$
Now we have
$$
A(A-2B) = A^2-2AB = BA-AB.
$$
This proves
$$
\det(AB-BA) = \det(BA-AB) = \det( -(AB-BA)) = (-1)^n \det(AB-BA),
$$
and since $n$ is odd (and $1+1\ne0$), $\det(AB-BA)=0$ follows.

If the field has characteristic 2 then the result is true as well: In this case $AB-BA=AB+BA=A^2$. Suppose $A$ is invertible. Then the assumption $A^2=AB+BA$ implies $I_3=BA^{-1}+A^{-1}B$. The trace of the matrix on the left is $1$, the trace of the matrix on the right is $tr(BA^{-1}+A^{-1}B)=tr(BA^{-1})+tr(BA^{-1})=0$, contradiction. So $A$ cannot be invertible, and $\det(AB-BA)=\det(A^2)=0$.
