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$A$ is an invertible matrix over $\mathbb{R}$ (nxn). Show that $A^{T}A$ is positive definite.

I looked up for it and found this two relevent posts but still need help. positive definite and transpose help me understand a line in an “$A^TA$ is positive, semi-definite” proof

Any suggestions? thanks

(The whole excercise is about bilinear forms)

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Jenni201
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2 Answers2

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I like to think of positive definite being positive semi-definite with the added condition that $Mx = 0$ implies $x=0$. Since $A$ is invertible, so is $A^T$. Thus $A^TA$ is invertible and so $A^TAx = 0$ implies $x=0$. Thus all we are left with is to prove that $A^TA$ is positive semi-definite.

Well to see this, all we need to do is inspect what happens to $\langle x,A^TAx\rangle$. We can however rewrite this in a nicer form: $\langle x,A^TAx\rangle = \langle Ax,Ax\rangle$ by definition. Can you see how to proceed?

If you're not familiar with inner product notation (the $\langle\cdot,\cdot\rangle$ notation), you can instead consider $x^TA^TAx$ since this is the same expression in this case but by the property of transposes, $x^TA^T = (Ax)^T$ so we are left with $(Ax)^T(Ax)$.

Cameron Williams
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  • What's a positive semi-definite ? Does it mean that A>0 rather than A>=0 ? – Jenni201 Jun 14 '14 at 23:52
  • A matrix (linear transformation) $M$ is positive semi-definite if $\langle x,Mx\rangle \ge 0$ or equivalently $x^TMx \ge 0$ for any vector $x$. – Cameron Williams Jun 14 '14 at 23:52
  • Oh ok. thanks :) – Jenni201 Jun 14 '14 at 23:53
  • Umm can you proceed abit farther ? How does writing (Ax)^t*(Ax) helps ? thanks – Jenni201 Jun 15 '14 at 00:00
  • I guess I have to show that (Ax)^t(Ax)=(Ax)(Ax) and that proves. But how can I show that (Ax)^t = Ax ? – Jenni201 Jun 15 '14 at 00:03
  • That's not quite the argument you want to use. The way it's written can be a little bit confusing so let's let $y=Ax$. Then what we have is $y^Ty$. What do you know about this quantity? – Cameron Williams Jun 15 '14 at 03:02
  • Maybe it's the norm of y=Ax ? (because say A is nxn and x is nx1 then Ax is nx1 so (Ax)^T is 1xn and if you multiply it by Ax (which is nx1) you get the norm of y=Ax ? Am I right ? (norm satisfys >=0 and =0 if and only if y=Ax=0 => x=0) – Jenni201 Jun 15 '14 at 10:03
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Hint: A matrix $B$ is positive definite when $\langle Bv, v \rangle \geq 0$ with $= 0$ if and only if $v = 0$; it is common to also assume that $B$ is symmetric to apply the definition of positive definite.

So, for you, $\langle A^t Av, v \rangle = \langle Av, Av \rangle$...

Tom
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