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tony
tony
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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>0.$$\lambda_2(A)>0.$

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

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 $\lambda_2>0.$

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

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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tony
tony
  • 405
  • 2
  • 8

Deviation of random matrix from its expectation informs the positiveness of its second smallest eigenvalue

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 $\lambda_2>0.$

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