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Let $X$ be a $T \times K$ matrix. $X'X$ positive definite means that for all $c \not = 0$ $c'X'X c >0$, so then $(Xc)' Xc > 0$ which implies $(X\cdot c)\cdot (X\cdot c)$ (I'm not sure what proper notation is here, sorry), and therefore $X\cdot c \not = 0$ for any $c$., so $X$ has full column rank, and $X'X$ has full rank?

In other words, $X'X$ positive definite is a necessary condition for $X'X$ to have full rank?

I am using $'$ to denote the transpose. So $X' = X^T$.

Thanks.

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    $\begingroup$ It does imply all the eigenvalues are real, and positive. What does that imply about the determinant? $\endgroup$ – Clement C. Feb 8 '16 at 3:51
  • $\begingroup$ That the determinant is real and positive, since the determinant is the product of the eigenvalues. $\endgroup$ – majmun Feb 8 '16 at 3:52
  • $\begingroup$ This answers the question in the title... however, it looks like the one in the body is not quite the same. (Title asks for sufficient, body asks for necessary.) Which one are you asking? $\endgroup$ – Clement C. Feb 8 '16 at 3:52
  • $\begingroup$ Sorry, I will edit, the title, as I am more curious about that. $\endgroup$ – majmun Feb 8 '16 at 3:53
  • $\begingroup$ It's necessary and sufficient. $\endgroup$ – Hans Engler Feb 8 '16 at 3:55
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You already answer the question. In summary $X'X$ is a positive-semidefinite matrix so eigenvalues are either $0$ or positive. When you say $X'X$ is positive-definite you are saying that eigenvalues are strictly positive so the matrix is full-rank.

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