# 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.

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 $AB$.

• Hi, thanks for your response. When you say 'eigenvalues of T4T4 and T′4T′4 are non-negative or in other words they are Positive Semi Definite since they are raised by the power of 4', what if the eigenvalues of T and T' are complex? How can you guarantee that their fourth powers will be non-negative? Oct 1, 2017 at 15:46
• But in which condition do they become complex while $A,B \in R^{n \times n}$? Oct 1, 2017 at 16:03
• For example look at the matrix $\pmatrix{0&1\\-1&0}$. Oct 1, 2017 at 16:07
• Can you explain how you have got $\det(2(M-N)^2+16MN$? Oct 1, 2017 at 21:01
• The expansion of $(AB)^2$ is not how I would usually interpret things. I would usually put $(AB)^2=ABAB$ instead. Oct 2, 2017 at 3:26