How to prove the inequality $\det (AA^T) \ge 0$?

Question

How to prove for any matrix $A \in \Bbb R^{n \times n}$, that the inequality $\det(AA^T) \ge 0$ is true?

Answer 1

You need to know two properties of determinant:

1. It is multiplicative. In other words, $\det(AB)=\det(A) \cdot \det(B)$
2. $\det(A^T)=\det(A)$

Then the proof becomes easy.

$\det(AA^T)=\det(A)\det(A^T)=\det(A)^2 \geq 0$

Answer 2

$\det(A)=\det(A^T)$ and determinants preserve products.

Answer 3

As has been remarked in other answers, the inequality becomes clear once the fact that ${\rm det}(A^{T}) = {\rm det}(A)$ is established (assuming also that ${\rm det}$ is known to be multiplicative). This is not immediately obvious ( at least by any proof known to me). Here is the (standard) proof using the definition of determinant in terms of signed products, with one term from each row and each column, in case you haven't seen it. So, if $A = [a_{ij}]$ is an $n \times n$ matrix, we have ${\rm det}(A) = \sum_{\tau \in S_{n}} {\rm sign}(\tau)\prod_{i=1}^{n}a_{i \tau(i)} = \sum_{\tau \in S_{n}} {\rm sign}(\tau ^{-1})\prod_{i=1}^{n}a_{\tau^{-1}(i)i}$ since $\tau$ is odd if and only if $\tau^{-1}$ is odd. Now $\tau^{-1}$ runs through $S_{n}$ exactly once as $\tau$ runs over $S_{n},$ so the last sum is $\sum_{\tau \in S_{n}} {\rm sign}(\tau)\prod_{i=1}^{n}a_{\tau(i) i}$,which is ${\rm det}(A^{T}).$

Answer 4

The result is true for every $(n\times m)$ matrix $A$. Indeed $AA^T$ is a symmetric real matrix and it is sufficient to prove that $AA^T$ is non-negative. Let $x\in\mathbb{R}^n$. Then $x^T(AA^T)x=||A^Tx||^2\geq 0$ and we are done.