TITLE: Rank of idempotent matrices
QUESTION [5 upvotes]: Let $B_1, B_2, \dots, B_k$ be idempotent matrices, i.e., $B_i^2=B_i$. Can we prove that 
$$\mbox{rank}(I-B_1\cdots B_k)\leq \sum\limits_{i=1}^k \mathrm{rank}(I-B_i)$$
where $I$ is the identity matrix?

REPLY [6 votes]: Edit: The inequality is true for general $B_i$s. Idempotence is not needed (thanks to robinson for his/her comment.)
\begin{align*}
\mathrm{rank}(I-AB)
&= \mathrm{rank}(I-A+A-AB)\\
&\le \mathrm{rank}(I-A)+\mathrm{rank}(A-AB)\tag{1}\\
&= \mathrm{rank}(I-A)+\mathrm{rank}(A(I-B))\\
&\le \mathrm{rank}(I-A)+\mathrm{rank}(I-B)\tag{2},
\end{align*}
where $(1)$ is due to the fact that $\mathrm{rank}(X+Y)\le\mathrm{rank}(X)+\mathrm{rank}(Y)$ and $(2)$ is due to the inequality $\mathrm{rank}(XY)\le\min\left(\mathrm{rank}(X), \mathrm{rank}(Y)\right)$.
Apply the displayed inequality recursively, we get the result.