<s>[INST] Prove that $ \det(A^4 + A^2 B^2 + 2A^2 + I) \geq 0 $ Problem:
Let $ A $ and $ B $ be an $ n \times n $ matrices with real entries. If $ AB = -BA $, prove that
$$ {\det}{\left(A^4 + A^2B^2 + 2A^2 + I\right)} \geq 0. $$
My Approach:
If $ A $ invertible, then
$$ AB = -BA \implies AA^{-1}B = -BA^{-1}A \implies B = -B \implies B = O. $$
So,
$$ {\det}{\left(A^4 + A^2B^2 + 2A^2 + I\right)} = {\det}{\left(A^4 + 2A^2 + I\right)} = {\det}{\left(\left(A^2 + I\right)^{2}\right)} = \left({\det}{\left(A^{2} + I\right)}\right)^2 \geq 0. $$
If $ B $ invertible, then
$$ AB = -BA \implies AB^{-1}B = -BB^{-1}A \implies A = -A \implies A = O. $$
So,
$$ {\det}{\left(A^4 + A^2B^2 + 2A^2 + I\right)} = {\det}{(I)} = 1 \geq 0. $$
My Questions:

*

*Can we 'center' multiply both sides of the equation $ AB = -BA $ with a matrix? For example, in my solutions, $ AB = -BA \implies AA^{-1}B = -BA^{-1}A $.

*How do we prove it when $ A $ and $ B $ not invertible?

Thanks
 [/INST] This answer is not really different from Anurag A's, but I think it is clearer to present the answer in the way below.
Let $A=\pmatrix{0&A_2\\ 0&A_4}$ where $A_4$ is invertible. The condition $AB=-BA$ implies that $B=\pmatrix{B_1&B_2\\ 0&B_4}$ where $A_4$ and $B_4$ anti-commute. Hence it suffices to prove the inequality for the matrix pairs $(0,B_1)$ and $(A_4,B_4)$, i.e., for the cases where $A$ is either zero or invertible.
When $A$ is zero, the determinant is $1$ and the inequality is true. When $A$ is invertible, we have
$$
\det(A^4+A^2B^2+2A^2+I)
=\det(A)^2\det\left(B^2+(A+A^{-1})^2\right).\tag{1}
$$
Since $A$ anti-commutes with $B$, so do $A^{-1}$ and $A+A^{-1}$. In general, if $BC+CB=0$, then $B^2+C^2=(B+C)^2$. Therefore the RHS of $(1)$ is equal to $\left[\det(A)\det(B+A+A^{-1})\right]^2$, which is nonnegative because $A$ and $B$ are real.
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