I read that If $X^T$ is full column rank ($X$ is not necessarily square), then $X^TX$ is positive definite, and I'm trying to see why this is the case.
I know that if $X^T$ is full column rank, then $A=X^TX$ is also full rank. We also know that $A$ has only positive entries, and that $A$ is symmetric. But knowing this doesn't seem to guarantee positive definiteness.
What is an intuitive way to arrive at that $A$ is also positive definite?
Best Answer
For $v \ne 0$, we know that $Xv \ne 0$ by things you already know.
Now look at $$ v^t A v = v^t X^t X v = \langle Xv, Xv \rangle > 0. $$