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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?

Rodrigo de Azevedo
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XLDD
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1 Answers1

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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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user1551
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  • Seems like you don't need idempotence for step (2) – robinson Jun 01 '13 at 08:25
  • @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 01 '13 at 08:40
  • @robinson Thanks. Edited. – user1551 Jun 01 '13 at 08:43
  • @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 01 '13 at 08:46