Show determinant is non-negative Let $A,B \in M_{2}(\mathbb R)$ . Show that $\det((AB+BA)^4 + (AB-BA)^4)\geq 0$
My attempt: expression becomes $\det(2(M-N)^2+16MN)$ where $M=(AB)^2$ and $N=(BA)^2$.
Not sure how to continue from here.
Any hints appreciated.

Let $d=\det(AB-BA)$ and $\lambda_1,\lambda_2$ be the two eigenvalues of $AB+BA$. Since $X^2=-\det(X)I_2$ and in turn $X^4=\det(X)^2I_2$ for any traceless $2\times2$ matrix $X$, we get
$$
\det\left[(AB+BA)^4 + (AB-BA)^4\right]
=\det\left[(AB+BA)^4 + d^2I_2\right]
=(\lambda_1^4+d^2)(\lambda_2^4+d^2).
$$
As $AB+BA$ is real, either $\lambda_1$ and $\lambda_2$ are both real or they are complex conjugates to each other. In either case, the assertion follows immediately.
