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.
Best Answer
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, $x = 0$.