How to show that $\operatorname{det}(I + 2 AA^T) \neq 0$?

Question

How to show that $|I + 2 AA^T| \neq 0$, where $A$ is real $m \times n$ matrix?

My attempt: Suppose that $|I + 2AA^T| = 0$. Then $2AA^T$ has $-1$ as an eigenvalue. Now recall that $AA^T$ is symmetric positive semidefinite, i.e. cannot have negative eigenvalues. Hence, we have a contradiction.

Am I correct?

Answer 1

As pointed out in the comments, your argument is correct.