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

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.

• suppose $A^TAx=0$ for some non zero $x$ then??? – user87543 Jan 9 '14 at 13:27
• @praphulla: I'm not sure. Then either $A$ or $x$ must be zero. Should we use the fact that determinant of $A^TA$ must be non-zero? – Gigili Jan 9 '14 at 14:05
• why do you think $A$ or $x$ must be zero? – user87543 Jan 9 '14 at 14:07
• @Praphulla: Um, because otherwise how is their multiplication equal to zero? – Gigili Jan 9 '14 at 14:11
• I (kind of) lost interest in this problem as the whole excitement is ruined by that full answer... please have a look at that answer.... I am sorry for not being helpful to you! – user87543 Jan 9 '14 at 14:14

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, $x = 0$.
• Thank you for your answer. Did you multiply the equation by $x^T$? – Gigili Jan 9 '14 at 14:30