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