TITLE: A problem about a rank's inequality of a complex matrix
QUESTION [8 upvotes]: Let $X,Y \in M_{n\times n}(\mathbb{C})$ such that $X^{2}=Y^{2}=I_{n}$. Prove that
$$\operatorname{rank}((X+I_n)(Y-I_n))+\operatorname{rank}((X-I_n)(Y+I_n))=\operatorname{rank}(X-Y)$$
My approach: Using the well-known inequality: $\operatorname{rank}(X)+\operatorname{rank}(Y)\geq \operatorname{rank}(X+Y)$, we can see that $$\operatorname{rank}(X-Y)=\operatorname{rank}(X-XY+XY-Y)\leq \operatorname{rank}(X-XY)+\operatorname{rank}(XY-Y)$$
but, how can I continue from here?

REPLY [5 votes]: As $X$ and $Y$ are involutions we have that
$$
P_x=\frac12(X+I)\qquad\text{and}\qquad P_y=\frac12(Y+I)
$$
are idempotent (projections). It is straightforward to see that the rank identity is equivalent to
$$
\operatorname{rank} P_x(I-P_y)+\operatorname{rank}(I-P_x)P_y=\operatorname{rank}(P_x-P_y)
$$
which follows easily from the fact that
$$
P_x-P_y=P_x(I-P_y)-(I-P_x)P_y
$$
together with the standard result that the images of $P_x$ and $I-P_x$ are complementary.