A question about $\det(A^TA)$ Assume that $A\in\mathbb{R}^{m\times n}$, where $m\leq n$. The aim is to find a lower and an upper bound for $\mathrm{det}(A^TA)$. What can be said about this issue?
 A: PS: I think the singular value decomposition of $A$ can help us to find an answer to this question. In fact, let us assume that $A=U\Sigma V^T$ in which $U\in\mathbb{R}^{m\times m}$ and $V\in\mathbb{R}^{n\times n}$ are orthonormal matrices, i.e., $U^TU=I_m$ and $V^TV=I_n$. Moreover, the matrix $\Sigma$ is defined as
$$\Sigma=diag(\sigma_1,\sigma_2,\ldots,\sigma_r,0,\ldots,0)_{m\times n}\quad\text{s.t.}\quad \sigma_1\geq\sigma_2\geq\cdots\geq\sigma_r,$$
in which $\sigma_1$ is the largest singular value of $A$. Following this discussion, we have
$$\mathrm{det}(A^TA)=\mathrm{det}(V\Sigma^TU^TU\Sigma V^T)=\mathrm{det}(V\Sigma^T\Sigma V^T)=\mathrm{det}(\Sigma^T\Sigma)=\prod_{i=1}^r \sigma_i^2\leq\sigma_1^{2r}.$$
On the other hand, since $0\leq\mathrm{det}(A^TA)$, we can say that
$$0\leq\mathrm{det}(A^TA)\leq \sigma_1^{2r}.$$
For this reason, $0$ and $\sigma_1^{2r}$ can be considered respectively as a lower bound and an upper bound for $\mathrm{det}(A^TA)$. Is it possible to find other upper bounds for $\mathrm{det}(A^TA)$?
A: As described in the other answer, $\det(A^TA)=\sigma^2_1\cdots\sigma^2_n$.
Using the GM-AM inequality $$\det{A^TA}=\sigma^2_1\cdots\sigma^2_n\le\left(\frac{\sigma_1^2+\cdots+\sigma_n^2}{n}\right)^{n}=\left(\frac{\|A\|^2_2}{n}\right)^{n}=\left(\tfrac{1}{n}\sum_{ij}a_{ij}^2\right)^{n}$$
