I was reading an article of matrix completion and met the following lemma The concentration inequality for $\sigma_{\max}$ part is a standard result. However, I didn't find any results like the $\sigma_{\min}$ part. The most similar expression was found in Theorem 6.1 of Wainwright's book as follows \begin{equation} P\left[\frac{\sigma_{\min}(X)}{\sqrt{n}}\leq\sigma_{\min}(\sqrt{\Sigma})(1-\delta)-\sqrt{\frac{\operatorname{tr}(\Sigma)}{n}}\right]\leq \exp\{-n\delta^2/2\} \end{equation} but it cannot be directly relaxed to the form like \begin{equation} \sigma_{\min}\left(\frac{1}{n}X^TX\right)\geq c\sigma_{\min}(\Sigma) \end{equation} Since it involves a trace term. Moreover, the author refer its proof from another paper. But seems there are no explicit statement in it.

So my question is, how to prove that result? And can we obtain similar result for subgaussian random vectors?