Following up on this question: Let $A \in \mathbb{R}^{m \times n}$ be a matrix such that $m \ll n$ and let $Q \in \mathbb{R}^{n \times r}$ with $r<m$ such that $Q^TQ = I$.
By Courant–Fischer Max-Min theorem, we have that the spectrum of $Q^T A^T A Q $ is contained in that of $A^T A$.
However, the answer to the linked question states that for all $i=1,\dots,r$ we have that $\sigma_{\min} (A) \leq \sigma_i (AQ) \leq \sigma_{\max}(A)$, and I guess that this answer takes into account the broadest definition of singular values, in which $\sigma_{\min}(A) = 0 $, and in this case, the lower bound on the smallest non zero singular value of $AQ$ is trivial (0).
When considering only non zero singular values, then the claim that $\sigma_{\min} (A) \leq \sigma_i (AQ)$ does not hold anymore. For example, consider
$$A = \left[ \begin{matrix} -2 & 1 & 3 & 4 \\ -4 & 3 & 2 & 0 \\ 1 & 2 & 3& -4 \\ \end{matrix}\right] \text{ and } Q = \left[ \begin{matrix} 1/\sqrt{2} & 0 \\ 1/\sqrt{2} & 0 \\ 0 & 1/\sqrt{2} \\ 0 & 1/\sqrt{2} \end{matrix}\right] $$ then the singular values of $A$ are : $\sigma_1 (A) = 6.82, ~\sigma_2 (A) = 5.92,~ \sigma_3 (A) = 2.74 $, and for $AQ$ we have $\sigma_1 (AQ) = 5.34, ~\sigma_2 (AQ) = 1.98$.
When only non zero singular values are considered, we have that $\sigma_{\min} (AQ) = \sigma_2 (AQ) = 1.98 < \sigma_{\min}(A) = \sigma_3 (A) = 2.74$.
Is there a way to bound the smallest non-zero singular value of $AQ$ from below?