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Jan 17, 2023 at 21:48 comment added tony I see. Thank you!
Jan 16, 2023 at 22:41 comment added Carlo Beenakker it just means your statement about $\lambda_2(A)>0$ holds if $\|A-B\|<\lambda_2(B)$ for any PSD $B$; the fact that $A$ is random and $B=\mathbb{E}[A]$ is irrelevant, it does not constrain $B$ in any way.
Jan 16, 2023 at 21:11 comment added tony Hi @Carlo Beenakker. I don't understand what does this mean "Since the weight of 𝐴 in the statistical average can be arbitrarily small, we might as well replace $\mathbb{E}(A)$ by an arbitrary PSD matrix $B$ with $\lambda_2(B)>0$." Could you clarify more on it? I understand the answer based on: if $B$ is PSD and $\lambda_2(B)>0$, $\|A-B\|<\lambda_2(B)$, then $\lambda_2(A)>0$.
Jan 16, 2023 at 12:29 vote accept tony
Jan 15, 2023 at 14:13 history edited Carlo Beenakker CC BY-SA 4.0
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Jan 15, 2023 at 13:02 history edited Carlo Beenakker CC BY-SA 4.0
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Jan 15, 2023 at 12:53 history edited Carlo Beenakker CC BY-SA 4.0
added 120 characters in body
Jan 15, 2023 at 12:14 history answered Carlo Beenakker CC BY-SA 4.0