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Let $B_i, i=1,2,\cdots,k$ be idempotent matrices, i.e. $B_i^2=B_i$. Can we prove that $\mathrm{rank}(I-B_1\cdots B_k)\leq \sum\limits_{i=1}^k \mathrm{rank}(I-B_i)$, where $I$ is the identity matrix?

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up vote 4 down vote accepted

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.

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Seems like you don't need idempotence for step (2) – robinson Jun 1 '13 at 8:25
@user1551 Year, as robinson said, the idempotence is not used in (2). since we have $rank(AB)\leq \min(rank(A),rank(B))$. – XLDD Jun 1 '13 at 8:40
@robinson Thanks. Edited. – user1551 Jun 1 '13 at 8:43
@XLDD Thanks. Actually I noticed that too when I wrote my answer, but the inequality rank(AB) <= rank(A) is more obvious when A is idempotent than in the general case. So, I used idempotence as the reason then. – user1551 Jun 1 '13 at 8:46

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