Prove that $A^{t}A$ is positive definite

Question

$A$ 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)

Answer 1

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 $(Ax)^T(Ax)$.

Answer 2

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