For an abstract matrix $A$ of dimension $p\times p$ and $p$ can approach infinity, we known that $\|A\|_2 \leq \sqrt{p}\|A\|_\infty$. However, in some papers, e.g., the sentences below A.18 in page 13, and the sentences above Lemma 3 in page 13, I see that when $A$ is symmetric, $\|A\|_2 \leq \|A\|_\infty$, how to prove it?
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What is your definition of the norms $\Vert A \Vert_2, \Vert A \Vert_\infty$? Are those the Schatten-norms? Or the operator norm from $\mathbb{R}^p$ with the $\Vert \cdot \Vert_p$ norms? – Severin Schraven Oct 12 '23 at 02:11
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@SeverinSchraven $|A|2=\sqrt{\lambda\max(A'A)}$ and $|A|\infty=\max{1\leq i\leq p}\sum_{j=1}^p|a_{ij}|$ – Patrick Star Oct 12 '23 at 02:15
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Hmmm, ok. So for a a symmetric matrix, the first norm is just the maximum of the absolute values of the eigenvalues. Now I don't see how to relate this to the second norm :( – Severin Schraven Oct 12 '23 at 02:24
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@SeverinSchraven I have listed two papers where the result was used. – Patrick Star Oct 12 '23 at 02:32
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First, the 2-norm of any symmetric matrix is equal to its maximum absolute eigenvalue. This was proved in this question. Note that a real symmetric matrix has real eigenvalues. So, the maximum eigenvalue can be bounded by the maximum absolute row sum, by the Gershgorin circle theorem. The maximum absolute row sum is in turn the infinity norm.
Not sure if this is the simplest way to prove this, just putting it out there)
