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.
 A: 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.
A: I believe that approaching the problem is easier by showing the Positive Semi-Definiteness of M and N. This way you can generalize it to $A,B \in R^{n \times n}$.
Expanding the square of matrix $AB$ and $BA$ according to this source, $M = (AB)^2 = ABB^TA^T, N = (BA)^2 = BAA^TB^T$ are symmetric. As the result their eigenvalues are real.
Moreover, M and N are PSD (shown at the end of the answer). The determinant of a PSD matrix is greater or equal to zero. Thus, it is enough to show that $2(M - N)^2 + 16MN = 2M^2 + 12MN + 2N^2$ is PSD. And this is trivial since it is a summation and product of symmetric PSD matrices.
M is PSD if $x^TMx \geq 0$ $\forall x \in R^n$ $\implies x^TABB^TA^Tx = \sum_{i=1}^n\langle x,(AB)_i\rangle^2 \geq 0$ where $(AB)_i$ is $i$th column of matrix $AB$.
