Is the Gramian determinant always nonnegative? Is Gramian determinant  $\det (A^TA)$ always nonnegative (or at least when $A$ has no more columns than rows)? It's used to compute a volume element as in this article
https://en.wikipedia.org/wiki/Volume_element#Volume_element_of_a_linear_subspace
and I usually see square root taken directly of this determinant and not of its absolute value. This and the fact that it is nonnegative for a square matrix makes me suspect it is always nonnegative.
 A: $A^TA$ is positive semidefinite, hence $\det(A^TA)\geq 0.$ 
Proof of $A^TA$ is positive semidefinite:
$x^TA^TAx=\left\|Ax\right\|^2 \geq 0.$
A: It is, of course, possible to note that $A^TA$ is positive semidefinite as in the other answer.
Another method is to apply the Cauchy-Binet formula, which allows us to see that (when $A$ is $m \times n$ with $m  \leq n$)
$$
\det(A^TA) = \left(\sum_{S \in \binom{[m]}{n}} \det A_{S,[n]}\right)^2 \geq 0
$$
A: Let the SVD of $A \in \mathbb R^{m \times n}$ be
$$A = U \Sigma V^T = \begin{bmatrix} U_1 & U_2\end{bmatrix} \begin{bmatrix} \hat\Sigma & O\\ O & O\end{bmatrix} \begin{bmatrix} V_1^T\\ V_2^T\end{bmatrix}$$
where the zero matrices may be empty. The eigendecomposition of $A^T A$ is, thus,
$$A^T A = V \Sigma^T U^T U \Sigma V^T = V \Sigma^T \Sigma V^T = \begin{bmatrix} V_1 & V_2\end{bmatrix} \begin{bmatrix} \hat\Sigma^2 & O\\ O & O\end{bmatrix} \begin{bmatrix} V_1^T\\ V_2^T\end{bmatrix}$$
and
$$\det (A^T A) = \det \begin{bmatrix} \hat\Sigma^2 & O\\ O & O\end{bmatrix} \geq 0$$
As the singular values are real, their squares are nonnegative. Thus, the eigenvalues of $A^T A$ are nonnegative, which implies that $A^T A$ is either positive semidefinite or positive definite.
