Prove that $\det(AA^T+I)\ge 1$ 
If $A$ is a matrix with real entries, prove that $$\det(AA^T+I)\ge 1.$$

I tried using the eigenvalues. One thing came into my mind: maybe $AA^T$ is positive definite (I don't know whether this is true or not). However, I prefer a solution that does not use properties of positive definite matrix.
So I have 2 questions here:


*

*Is the statement "$AA^T$ is positive definite" true?

*Could you help me with a solution that does not use properties of positive definite matrix?
Thanks a lot.
 A: $\require{begingroup} \begingroup$Note$\let\geq\geqslant\newcommand\norm[1]{\|#1\|}$ that $AA^T$ is symmetric (hence diagonalisable) and thus has real nonnegative eigenvalues. Indeed, if $v$ is a (non-zero) eigenvector corresponding to $\lambda$, then
$$\lambda\norm v^2=\lambda v^Tv=v^TAA^Tv=\norm{A^Tv}^2\geq0\implies\lambda\geq0.$$
$AA^T+I$ is also symmetric and thus has real eigenvalues.
Furthermore, $\lambda$ is an eigenvalue of $AA^T$ iff $\lambda+1$ is an eigenvalue of $AA^T+I$:
$$AA^Tv=\lambda v\iff(AA^T+I)\cdot v=(\lambda+1)\cdot v.$$
Let $\lambda_1,\ldots,\lambda_n$ be the eigenvalues of $AA^T$ (with multiplicity, so they need not be different). Then
$$\det(AA^T+I)=(\lambda_1+1)(\lambda_2+1)\cdots(\lambda_n+1)\geq1.\endgroup$$
A: Your idea that $AA^T$ is positive definite is a good idea. More precisely, you can show that $AA^T$ is positive semidefinite. This is easiest to show using definitions, because $M$ is positive semidefinite if for all $x$, you have 
$$x^TMx \geq 0,$$ and in your case, this inequality is fairly simple to show.
A: Note that $AA^T$ is positive-semi definite. Indeed, $<x,AA^Tx>=<A^Tx,A^Tx>=\|A^Tx\|^2\geq 0.$ Next it is easy to prove.
