Prove that $X'X$ is rank-preserving of X

Asked Aug 27 '19 at 15:39

Active Aug 27 '19 at 15:45

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Given $X_{nxp}$ of rank p, we know that $(X'X)^{-1}$ exists because $X'X$ also has rank $p$. But how do you prove that $X'X$ has rank $p$ (it is rank-preserving of $X$)? What properties can you use to prove this?

Clarification: $X'$ is the transpose of $X$, and the values of $X$ are real.

edited Aug 27 '19 at 15:45

asked Aug 27 '19 at 15:39

bob

Is $X'$ the transpose of $X$? – Angina Seng Aug 27 '19 at 15:42
@LordSharktheUnknown yes – bob Aug 27 '19 at 15:43
And does $X$ have real entries? – Angina Seng Aug 27 '19 at 15:44
@LordSharktheUnknown yes. just updated the question – bob Aug 27 '19 at 15:44
Then prove that $X'Xv=0$ iff $Xv=0$. – Angina Seng Aug 27 '19 at 15:45
@LordSharktheUnknown could you explain how you would show that those two null spaces are equal? – bob Aug 27 '19 at 16:03

Prove that $X'X$ is rank-preserving of X

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