Show that $\det(AB-BA) × \det(AC-CA) \geq 0$ if $A^2 = -BC$ We have $A,B,C$ three $n×n$ matrices with real entries. We know that
$$
  A^2 = -BC
$$
and we want to show that
$$
  \det(AB-BA) × \det(AC-CA) \geq 0 \,.
$$
We can easily show that for $n=2k$ we have $\det(B) × \det(C) \geq 0$, and for $n = 2k + 1$ we have $\det(B) × \det(C) \leq0$. Then for $n = 2k$ we can expand $\det(AB-BA)$ like
$$
  2 × \det(A) × \det(B) - c_1 + c_2 - c_3 + c_4 - \dotsb +c_{n-2} - c_{n-1}
$$
and expand $\det(AC-CA)$ like
$$
  2 × \det(A) × \det(C) - d_1 + d_2 - d_3 + d_4 - \dotsb + d_{n-2} - d_{n-1} \,.
$$
For $n = 2k + 1$, we can expand $\det(AB - BA)$ like
$$
  c_1 - c_2 + c_3 - c_4 + \dotsb + c_{n-2} - c_{n-1}
$$
and $\det(AC - CA)$ like
$$
  d_1 - d_2 + d_3 - d_4 + \dotsb + d_{n-2} - d_{n-1} \,.
$$
If we multiply these two expanded forms in every case $n = 2k$ and $n = 2k+1$, then we get an ugly answer and I don’t know we can check the sign like that. I am now stuck. Maybe we need to use eigenvalues? Please someone help me solve this problem.
 A: Since $-A^3 = ABC = BCA,$ I think a cute solution is as follows:
$$\det\left((AB-BA)CB(AC-CA)\right) = \det\left((ABC-BAC)(BAC-BCA)\right) = \det((-A^3-BAC)(BAC+A^3)) = (-1)^n\det(A^3+BAC)^2.$$
But the LHS is just:
$$\det\left((AB-BA)CB(AC-CA)\right) = (-1)^n\det(AB-BA)\det(AC-CA)\det(A^2) = $$
so the potential minus sign will cancel and we are done.
EDIT: Just so people know, this answer is incomplete as the case $\det A = 0$ is not covered. It was simply meant to be an alternative to the corresponding part of answer below, which is complete with the limiting argument.
Furthermore, the now-deleted solution by user1551 can be modified - even though $A^2+CB$ is not zero necessarily, it does not alter the block matrix manipulation since we still have:
$$\det\begin{pmatrix} A & 0 \\ B & D \end{pmatrix} = \det A\det D$$
assuming the dimensions conform. But since the answer is deleted, I have no way of adding this modification.
A: Since $A^2+BC=0$,
$$
\begin{aligned}
&\quad\det(AB-BA)\det(AC-CA)\\
&=\det\pmatrix{AB-BA&0\\ A^2+CB&AC-CA}\\
&=\det\left[\pmatrix{A&-B\\ C&A}\pmatrix{B&-A\\ A&C}\right]\\
&=\det\left[\pmatrix{A&-B\\ C&A}\pmatrix{-I&0\\ 0&I}\pmatrix{A&-B\\ C&A}\pmatrix{0&I\\ I&0}\right]\\
&=\det\pmatrix{A&-B\\ C&A}\times(-1)^n\times\det\pmatrix{A&-B\\ C&A}\times(-1)^n\\
&=\det\pmatrix{A&-B\\ C&A}^2\ge0.
\end{aligned}
$$
A: Disclaimer: this is most likely not the intended / easiest solution. I wrote it here because it is a very general trick that can be applied to a variety of similar matrix questions.
Step 1: Solve the problem when the matrices involved are invertible.
Suppose $A, B, C$ are invertible. In this case, we have $B = -A^2C^{-1}$, so
$$AB - BA = - A^3C^{-1} + A^2C^{-1}A = A^2C^{-1}(AC - CA)C^{-1}.$$
So we have
$$\text{det}(AB - BA) = \text{det}(A)^2 \text{det}(C)^{-2}\text{det}(AC - CA)$$
Thus $\text{det}(AB - BA)$ and $\text{det}(AC - CA)$ have the same sign.
Step 2: Approximate non-invertible matrices by invertible matrices.
We need the following general lemma.
Lemma: If $A = BC$, and a sequence of invertible matrices $\{A_i\}$ converges to $A$, then there exists sequences of invertible matrices $\{B_i\}$, $\{C_i\}$ converging to $B$, $C$ respectively such that $A_i = B_iC_i$.
To prove the lemma, one can reduce to the case when $B$ is diagonal by say LR-decomposition, in which case one can let $B_i$ be diagonal as well. Anyway the lemma should intuitively be true.
To apply this lemma to this problem, let $\{A_i\}$ be a sequence of invertible matrices converging to $A$. Then there exists sequences of invertible matrices $\{B_i\}$ and $\{C_i\}$ converging to $B, C$ respectively, such that $A_i^2 = -B_iC_i$. By step 1, we have
$$\text{det}(A_i B_i - B_i A_i)\text{det}(A_i C_i - C_i A_i) \geq 0.$$
Taking the limit $i \to \infty$ gives the desired result
$$\text{det}(A B - B A)\text{det}(A C - C A) \geq 0.$$
