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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.

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    $\begingroup$ @Hmm: Here $A$ is not necessarily rectangular. $\endgroup$ – anomaly Jun 22 '16 at 20:29
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$A^TA$ is positive semidefinite, hence $\det(A^TA)\geq 0.$

Proof of $A^TA$ is positive semidefinite:

$x^TA^TAx=\left\|Ax\right\|^2 \geq 0.$

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    $\begingroup$ If $A$ has full column rank, then $A^T A$ is positive definite. $\endgroup$ – Rodrigo de Azevedo Jun 22 '16 at 20:48
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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 $m \leq n$) $$ \det(A^TA) = \left(\sum_{S \in \binom{[m]}{n}} \det A_{S,[n]}\right)^2 \geq 0 $$

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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.

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  • $\begingroup$ The other answer has a much quicker way to see that $A^TA$ is positive semidefinite, if you haven't seen it already. $\endgroup$ – Omnomnomnom Jun 22 '16 at 20:46
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    $\begingroup$ @Omnomnomnom I saw it. This was just a way of TEX'ing SVDs so I can use them in future answers. $\endgroup$ – Rodrigo de Azevedo Jun 22 '16 at 20:47

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