If $A$ is a real matrix, then $det(AA^T)\geq0$, where $A$ is $m\times n$ matrix without using eigenvalues? I searched the http://math.stackexchange.com but most of the answer for this question use eigenvalues etc, but I need more elementary ways to show that why $det(AA^T)\geq0$ for any real matrix, and $A^T$ is the transpose of $A$. I could easily prove that when $m=n$. I need to show that when $m&lt;n$ or $m&gt;n$. 

I guess also when $m&gt;n$, then $det(AA^T)=0$. Is it right? If yes, how?

Thanks!
Without eigenvalues:

If $A^T$ has rank deficit, so does $AA^T$. Thus, in that case we have $\det(AA^T) = 0$. Now, assume $m \le n$ and rank of $A$ equals $m$. Let 
$$ A^T = Q \begin{bmatrix} R \\ 0 \end{bmatrix} $$
be a QR decomposition of $A^T$ (for example given by gram schmidt), that is, $Q$ is square and orthongal and $R$ is invertible and upper triangular. Then, weh have
$$ \det(AA^T) = \det(R^T R) = \det(R)^2 &gt; 0. $$
Why without using eigenvalues? For $A\in \mathbb{C}^{n\times m}$, $AA^H$ is a Hermitian $n\times n$ matrix. Thus, its eigenvectors form an orthonormal basis for $\mathbb{C}^n$ and their corresponding eigenvalues are real. Let $(\lambda,x)$ be an eigen-pair of $AA^H$, then
$$AA^Hx=\lambda x \Longrightarrow 0\leq\|Ax\|_2^2=\lambda$$
where last equation by taking inner product with $x$. Since $\lambda$ was arbitrary eigenvalue, det$(A^HA)\geq 0$.