The smallest singular value is $1$-Lipschitz for the Frobenius norm: $|\sigma_{\min}(A)-\sigma_{\min}(B)|\le\|A-B\|_F$

Let $A, B \in \mathbb{R}^{N\times n}$ be real matrices, and let

denote the smallest singular value of $A$ (with the convention $\inf\emptyset = 0$, so $\sigma_{\min} = 0$ when $n = 0$), and let $\|M\|_F = \big(\sum_{i,j} M_{ij}^2\big)^{1/2}$ denote the Frobenius norm. Then

This is the Lipschitz estimate that makes Gaussian concentration applicable to the smallest singular value of a Gaussian matrix: combined with Gordon's bound $\mathbb{E}\,\sigma_{\min}(G) \ge \sqrt N - \sqrt n$, it yields the lower tail $\mathbb{P}\{\sigma_{\min}(G) \le \sqrt N - \sqrt n - u\} \le e^{-u^2/2}$.

Formalization Note. sMin A is the infimum of $\sqrt{(Ax)\cdot(Ax)}$ over the subtype $\{x \mid x\cdot x = 1\}$ and frobNorm is $\sqrt{\sum_{i,j} A_{ij}^2}$. No hypothesis on $n$ is needed: for $n = 0$ both sides of the inequality reduce to $0 \le \|A-B\|_F$.