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.
Best Answer
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.