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I read that If $X^T$ is full column rank ($X$ is not necessarily square), then $X^TX$ is positive definite, and I'm trying to see why this is the case.

I know that if $X^T$ is full column rank, then $A=X^TX$ is also full rank. We also know that $A$ has only positive entries, and that $A$ is symmetric. But knowing this doesn't seem to guarantee positive definiteness.

What is an intuitive way to arrive at that $A$ is also positive definite?

24n8
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1 Answers1

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For $v \ne 0$, we know that $Xv \ne 0$ by things you already know.

Now look at $$ v^t A v = v^t X^t X v = \langle Xv, Xv \rangle > 0. $$

John Hughes
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  • Ah that makes sense. This doesn't make use of $X^TX$ being full rank right? – 24n8 May 17 '20 at 20:20
  • Also, my linear algebra is a bit rusty. We know that $Xv \neq 0$ because $X$ is full rank, which means that the nullspace of $X$ is empty, hence $Xv = 0 \iff v = 0$? – 24n8 May 17 '20 at 20:24
  • Yep...that's right. (Almost: the nullspace isn't EMPTY, but it does consist solely of the zero vector!) – John Hughes May 17 '20 at 21:04