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Let $A$ be a PSD random matrix, which has $0$ as one of its eigenvalues. The second smallest eigenvalue of the expectation of $A$ writes as $\lambda_2(\mathbb{E}(A))>0$.

Why the following statement holds?

If $\|A-\mathbb{E}(A)\|<\lambda_2(\mathbb{E}(A))$, then $\lambda_2(A)>0.$

Note that $\|\cdot\|$ denotes the operator norm.

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    $\begingroup$ @CarloBeenakker I want to prove $\lambda_2(A)>0$ instead of $\lambda_2(\mathbb{E}(A))>0$ $\endgroup$
    – tony
    Commented Jan 14, 2023 at 16:53
  • $\begingroup$ clear, thank you for the clarifying edit. $\endgroup$
    – Carlo Beenakker
    Commented Jan 14, 2023 at 19:40

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The random-matrix connection is a bit of a red herring: 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 $\lambda_2(B)>0$.
The statement in the OP

If $\|A-B\|<\lambda_2(B)$, then $\lambda_2(A)>0,$

is a consequence of the "eigenvalue stability" inequality $$|\lambda_i(A)-\lambda_i(B)|\leq \|A-B\|,$$ which is proven, for example, in Tao's notes (equation 13).

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  • $\begingroup$ 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 $\lambda_2(A)>0$. $\endgroup$
    – tony
    Commented Jan 16, 2023 at 21:11
  • $\begingroup$ 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. $\endgroup$
    – Carlo Beenakker
    Commented Jan 16, 2023 at 22:41
  • $\begingroup$ I see. Thank you! $\endgroup$
    – tony
    Commented Jan 17, 2023 at 21:48

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