Rank of $AA^T$ if $A$ is a full rank matrix

Question

Assume we have $A \in \mathbb{R}^{m \times n}$, $rg(A) = m$. Will $\det(AA^T)$ be non-zero? Same question is for $\det(\Phi^T\Phi)$, where $\Phi \in \mathbb{R}^{n \times m}, rg(\Phi) = m$.

Answer 1

Yes, it is true: if $A$ is $m \times n$ with rank $m$, then $\det(AA^T)$ will be non-zero. Similarly, if $A$ is $m \times n$ with rank $n$, then $\det(A^TA)$ will be non-zero.

There are a few ways you could go about proving this. One is to use the fact that $$ \operatorname{rank}(A) = \operatorname{rank}(A^TA) = \operatorname{rank}(AA^T), $$ and a square matrix will have non-zero determinant if and only if its rank is equal to its size. Alternatively, you could use the Cauchy Binet formula to express either determinant as a sum of squares.