Consider a perturbed matrix $\widetilde{H}=H+\delta H$ such that $\|\delta H\|_\text{F}\leq \varepsilon$,
where $\widetilde{H}$ and $\varepsilon$ are known.
I am wondering how I can find the smallest singular value $\sigma_{\text{min}}$ of $H$? Are there any bounds for $\sigma_{\text{min}}$? I do appreciate your insights.
This is what I have done so far:
To find $\sigma_{\text{min}}$, I did a perturbation analysis of the singular value problem $H= U \Sigma V^T$, i.e.,
$$\Sigma + \delta\Sigma = U^T(H+\delta H)V.$$
Since $U$ and $V$ are orthogonal or unitary, they preserve norms. Accordingly,
$$\|\delta\Sigma\|_\text{F} = \|\delta H\|_\text{F}.$$
Also, from Mirsky's theorem, the inequality
$$\sqrt{\sum_{i=1}^{n} (\tilde{\sigma_i}-\sigma_i)^2} \leq \|\delta H\|_\text{F}$$
holds, where $\tilde{\sigma_i}$ is the approximate of $\sigma_i$ due to the perturbation matrix $\delta H$. I am not sure how to proceed from here.
By Weyl's inequality $$ |\sigma_{\min}(\widetilde H)-\sigma_{\min}(H)| \leq \|\widetilde H-H\|_2 \leq \|\delta H\|_2 \leq \|\delta H\|_F \leq \varepsilon. $$