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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.

Gigili
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    suppose $A^TAx=0$ for some non zero $x$ then??? –  Jan 09 '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 09 '14 at 14:05
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    why do you think $A$ or $x$ must be zero? –  Jan 09 '14 at 14:07
  • @Praphulla: Um, because otherwise how is their multiplication equal to zero? – Gigili Jan 09 '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! –  Jan 09 '14 at 14:14

1 Answers1

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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, $x = 0$.

fuglede
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    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..... –  Jan 09 '14 at 13:45
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    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 09 '14 at 14:03
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    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... –  Jan 09 '14 at 14:08
  • Thank you for your answer. Did you multiply the equation by $x^T$? – Gigili Jan 09 '14 at 14:30
  • @Gigili: Yep, that's right. – fuglede Jan 09 '14 at 16:10
  • So you supposed that $x$ is non-zero to be able to do that, right? – Gigili Jan 09 '14 at 17:11
  • That's where you get contradiction? Thanks again. – Gigili Jan 09 '14 at 18:02
  • @Gigili: You could do that if you like, but it's just as good to make no assumptions like that and simply conclude that $x = 0$. (For various reasons, proofs by contradiction are considered somewhat less preferable than those without; at least when they're easily avoidable like here.) – fuglede Jan 09 '14 at 18:57
  • Your proof also shows that the two are equivalent: $A$ has full column rank if and only if $A^T A$ is invertible. – Sungjin Kim Oct 18 '16 at 21:43