If $A$ is a real matrix, then $\det(A A^T) \geq 0$ without using eigenvalues?

Question

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 $\det(A A^T) \geq 0$ 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 $m < n$ or $m > n$.

I guess also when $m > n$, then $\det(A A^T) = 0$. Is it right? If yes, how?

Thanks!

Answer 1

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 orthogonal and $R$ is invertible and upper triangular. Then, we have $$ \det(AA^T) = \det(R^T R) = \det(R)^2 > 0. $$

Answer 2

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$(A^HA)\geq 0$.