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Why does the rank of the design matrix $\boldsymbol X$ equal the rank of $\boldsymbol{X'X}$? Is this true in all circumstances?

If X is not linearly independent, what would the rank of X'X be?

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  • $\begingroup$ Welcome to the site, @kurt. I tried to edit your Q for greater clarity. Please make sure it's still asking what you want to know. In addition, I couldn't make sense of your 2nd paragraph; did you mean "If X is not linearly independent, what would the rank of X'X be"? $\endgroup$ – gung - Reinstate Monica Feb 13 '13 at 4:02
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For any matrix $X$, $R(X'X) =R(X)$. Where R() is the rank function.

You could prove this using null space. If $Xz=0$ for some $z$, then clearly $X'Xz =0$. Conversely, if $X'Xz=0$, then $z'X'Xz=0$, and it follows that $Xz=0$. This implies $X$ and $X'X$ have the same null space. Hence the result.

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