TITLE: If $A$ is full column rank, then $A^TA$ is always invertible
QUESTION [28 upvotes]: I need to prove that 

If $A$ is full column rank, then $A^TA$ is always invertible.

I know when an $m \times n$ matrix is full column rank, then its columns are linearly independent. But nothing more to use to prove the above theorem. I'd appreciate if you could give me some hints.

REPLY [23 votes]: It suffices to show that if $A^T A x = 0$ for some vector $x$, then $x = 0$. If $A^T A x = 0$, then
$$0 = x^T A^T A x = (Ax)^T(Ax) = \langle Ax, Ax \rangle = \lVert Ax \rVert^2,$$
which on the other hand implies that $Ax = 0$, so since $A$ has full rank, $x = 0$.