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I know this is an old question and there are several answers for this using eigenvalues and matrix factorization but they have not taught in my matrix analysis course yet. Therefore, my question would be a different proof for the following:

$$\text{rank}(p)=\text{trace}(p)$$

where $p$ is an $n \times n$ matrix. The only thing that we are allowed to use is $p^2=p$ and the fact that $\text{trace}(AB)=\text{trace}(BA)=$ where $A,B$ are conformable matrices.

Please notice what I have asked before answering the question.

  • For most, $p$ and $P$ will not denote the same thing. If you mean them to refer to the same thing, use the exact same symbol. – Arturo Magidin Apr 22 '19 at 04:23
  • I did. Thank you. –  Apr 22 '19 at 04:24
  • well it appears that this very problem has spawned a paper http://www-personal.umich.edu/~dsbaero/library/BaBeTre1.pdf – Andres Mejia Apr 22 '19 at 04:28
  • Definitely, we have not studied Moore–Penrose inverse. –  Apr 22 '19 at 04:30
  • @Saeed is this a homework question, then? What tools do you have that you might expect to be useful here? – Ben Grossmann Apr 22 '19 at 04:32
  • Just those two in the statement. –  Apr 22 '19 at 04:33
  • NB: If you had entered "projection" instead of "projector", the related-questions tab would have immediately told you about the old thread(s). – darij grinberg Apr 22 '19 at 04:34
  • Also, FFS, using Hermitian adjoints and Moore-Penrose inverses is hell of an overkill for this question. If the article is using the existence of a Moore-Penrose inverse, I am really wondering how it can lay a claim on an intrinsic proof. – darij grinberg Apr 22 '19 at 04:35
  • Please read the statement, I am not looking for fancy proof using the one that you quoted, I want to show it using those properties that I mentioned. –  Apr 22 '19 at 04:39

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Consider a rank-factorization $p = cf$ where $f:\Bbb R^n \to \Bbb R^r$ and $c:\Bbb R^r \to \Bbb R^n$, where $r = \operatorname{rank}(p)$, $c$ is injective, and $f$ is surjective. We note that $$ p^2 = p \implies cfcf = cf. $$ That is, we have $c(fc)f = c(\operatorname{id}_{\Bbb R^r})f$. Because $f$ has a right-inverse and $c$ has a left-inverse, we may conclude from the above that $fc = \operatorname{id}_{\Bbb R^r}$. It follows that $$ \operatorname{trace}(p) = \operatorname{trace}(cf) = \operatorname{trace}(fc) = \operatorname{trace}(\operatorname{id}_{\Bbb R^r}) = r $$ which was the desired result.

Ben Grossmann
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