Equality involving ranks of matrices

Question

Let $A,B\in M_n(\mathbb{C})$ such that $A^2=B^2=I_n$. Prove that $$\hbox{rank}~((A+I_n)(B-I_n))+\hbox{rank}~((A-I_n)(B+I_n))=\hbox{rank}~(A-B).$$ I have managed to prove the (easier) inequality (using the subadditivity property of the rank): $$\hbox{rank}~((A+I_n)(B-I_n))+\hbox{rank}~((A-I_n)(B+I_n))=\hbox{rank}~((A+I_n)(B-I_n))+\hbox{rank}~(-(A-I_n)(B+I_n))=$$ $$=\hbox{rank}~(AB-A+B-I_n)+\hbox{rank}~(-AB-A+B+I_n))\geq \hbox{rank}~((AB-A+B-I_n)+(-AB-A+B+I_n))=$$ $$=\hbox{rank}~(-2(A-B))=\hbox{rank}~(A-B).$$

Answer 1

$A$ and $B$ are diagonalisable and their eigenvalues are $\pm1$. Hence $$ A=P\pmatrix{I_r&0\\ 0&-I_{n-r}}P^{-1} \ \text{ and }\ B=Q\pmatrix{I_s&0\\ 0&-I_{n-s}}Q^{-1} $$ for some invertible matrices $P$ and $Q$ and for some integers $r$ and $s$. Let $$ P^{-1}Q=\pmatrix{X&Y\\ Z&W} $$ where $X$ is $r\times s$. Then \begin{align} (A+I)(B-I)&=P\pmatrix{0&-4Y\\ 0&0}Q^{-1},\\ (A-I)(B+I)&=P\pmatrix{0&0\\ -4Z&0}Q^{-1}. \end{align} It is now clear that \begin{align} &\operatorname{rank}\bigl((A+I)(B-I)\bigr) +\operatorname{rank}\bigl((A-I)(B+I)\bigr)\\ =\ &\operatorname{rank}\bigl((A+I)(B-I)+(A-I)(B+I)\bigr). \end{align}