If $X$ is an $n \times p$ matrix of rank $r$ and $C = AX$ for some $q \times n$ matrix $A$ with rank$(A) = q$, how do I show that rank $(X(I-C^{-}C))=$ rank$(X)-$ rank$(C)$? I can show that rank $(X(I-C^{-}C))\geq $ rank$(X)-$ rank$(C)$, but how do I get the reverse inequality?

Note : $C^{-}$ is a generalized inverse of $C$.

Any help would be appreciated.


$C^-$ is denoted $C^+$. The fact that $A$ has maximal rank is useless.

1) We show that, for every $X,C$, one has $rank(X(I_p-C^+C))=dim(\ker(C))-dim(\ker(X)\cap \ker(C))$.

Proof: $C^+C$ is the orthogonal projection onto $im(C^*)$, that is, onto the orthogonal of $\ker(C)$. Then $I_p-C^+C$ is the orthogonal projection onto $\ker(C)$. Thus $im(I-C^+C)=\ker(C)$ and $im(X(I-C^+C))=X(\ker(C))$. Let $\ker(C)=(\ker(C)\cap\ker(X))\oplus Z$ ; then $X(\ker(C))=X(Z)$ and $X$ is an isomorphism from $Z$ onto $X(\ker(C))$ and we are done.

2) Here $\ker(X)\subset \ker(C)$ ; we deduce that $rank(X(I_p-C^+C))=dim(\ker(C))-dim(\ker(X))=rank(X)-rank(C)$.

EDIT: @Dunn,@Gigili, I see that you are unable to send a simple "thank you". Incredible rudeness.

  • $\begingroup$ I'm sorry, but what should I thank you for? $\endgroup$
    – Gigili
    Jan 15 '14 at 13:46
  • $\begingroup$ @Gigili, you offered a bounty for a possible answer. I am not interested by your bounty, but, at least, I think that you can indicate that you have read the answer $\endgroup$
    – user91684
    Jan 16 '14 at 9:45

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