# 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

## 1 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, $x = 0$.

• what is the point of giving full answer when someone has given a hint just before ten minutes and OP has not responded to that..... – user87543 Jan 9 '14 at 13:45
• Hmm, the same as would be the point of giving it with no hint given; did I break an unwritten rule in doing so? OP is as free to use the hint as the answer. Moreover, one could always discuss what constitutes a full answer -- depending on OP's knowledge about linear algebra, I'd say that fuller answers exist, seeing as for example there is a 12 minute video on YouTube explaining the same three lines. – fuglede Jan 9 '14 at 14:03
• it is not the question that full answer exist or not... for that matter almost half of the problems that are asked here in basic courses linear algebra/calculus there would be definitely a solution in Google... But the point is that OP may wish to "interact" and giving full answer does not help much in that case... It would be up to the person giving the answer though... – user87543 Jan 9 '14 at 14:08
• Thank you for your answer. Did you multiply the equation by $x^T$? – Gigili Jan 9 '14 at 14:30
• @Gigili: Yep, that's right. – fuglede Jan 9 '14 at 16:10