Prove: $(\det(A-B)+\det(A+B) )^2 \ge 4\det(A^2-B^2 )$

Question

Let $A,B \in \mathcal{M}_n (\mathbb{R})$ two matrices so that:

a) $AB^2=B^2 A$ and $BA^2=A^2 B$

b) $\text{rank}(AB-BA)=1$.

Prove: $$(\det(A-B)+\det(A+B) )^2≥4\det(A^2-B^2 )$$

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This is a solution:

Denote $C=AB-BA$. Then $rank(C)=1$ so $C=p\cdot q^T$ where $p,q$ are column vectors and $tr(C)=0$. This proves that $C^2=p\cdot q^T\cdot p\cdot q^T=tr(C)C=0.$

If $D=A^2-B^2$ then a) implies that $CD=DC$ and $D$ commutes with $A,B$.

As a consequence $tr(CD^{-1})=0$.

If $\det(D)=0$ we have nothing to prove. Else $D$ is invertible and $(CD^{-1})^2=0$.

We have $(A-B)(A+B)=A^2-B^2+AB-BA=D+C$. We would like to prove that $\det(A-B)(B-A)=\det(A^2-B^2)$.

For this define $ f(t)=\det(AB-BA+t(A^2-B^2))$

and see that $f(t)=\det(D)\det(CD^{-1}+tI)=\det(A^2-B^2)\cdot t^n$.

Replace $t=1$ in the previous relation to get $\det(AB-BA+A^2-B^2)=\det(A^2-B^2)$ and we are done.

But i don't understand why $tr(CD^{-1})=0$ ??

Answer 1

Because $(CD^{-1})^2=0$, then $CD^{-1}$ only has zero eigenvalue. There should be a mistake in the order of proof.

Answer 2

Note $D^{-1}$ also commutes with $C$, so

$CD^{-1}CD^{-1}=C^2D^{-1}D^{-1}=0$

Let $\lambda_i$ be any eigenvalue of $CD^{-1}$. Then $\lambda_i^2$ is eigenvalue of $(CD^{-1})^2$. So $\lambda_i^2=0$ and $\lambda_i=0$.

Thus $Tr(CD^{-1})=\sum \limits_{k=1}^n\lambda_k=0$.