Prove that if $A$ is invertible then $AA^\top$ is positive definite

Question

I need to prove that if $A$ is a square invertible matrix then $AA^\top$ ($A$ multiply $A$ transpose) is positive definite.

I tried to prove that all the eigenvalues are positive. I know that $AA^\top$ is symmetric, and its determinant is positive, and trace is positive.

Therefore: product of eigenvalues is positive , and sum of eigenvalues is positive.

In addition $0$ is not an eigenvalue. That's what I know so far but I still cannot conclude that every eigenvalue for itself is positive. What have I missed? Thanks!

Answer 1

I assume that $A$ is a real matrix (otherwise, the statement is not true).

Prove positive definiteness by definition:

$B$ is positive definite if $\langle x,Bx\rangle> 0$ for all $x\neq 0$. For $B=BB^T$, you have $$\langle x, BB^T x\rangle.$$

Now, prove that this is $\geq 0$ by taking into account that $\langle x, Ax\rangle = \langle A^T x, x\rangle$ for any matrix $A$.

Answer 2

Let $x$ be a non-zero vector from $\mathbb{R}^n$ (here $x$ is a row vector). Since $A$ is invertible, $xA$ is also non-zero. Then $$ 0 < \langle xA,xA \rangle = xA(xA)^\top = x AA^\top x^\top = x (AA^\top) x^\top. $$ The latter implies that $AA^\top$ is positive definite.

Answer 3

If you set $B=A^T$ and $C=AA^T=B^TB$, then it should be easier: the definition of positive definite is that $$ x^TCx>0 $$ for all $x\in\mathbb{R}^n$, $x\ne0$.

Since $$ x^TCx=x^TB^TBx=(Bx)^T(Bx) $$ you can set $y=Bx$; then you know that $y^Ty>0$ if $y\ne0$; can $y$ be zero when $x\ne0$?