Given that $A$ is symmetric, for any norm $\|\cdot\|_*$: $$\|A\|_2 \le \|A\|_*$$
Why is it so? I think it has something to do with the fact that $\rho(A) \le \|A\|$ (Where $\rho (A)$ is the spectral radius of $A$)
2026-04-25 04:09:35.1777090175
Why $\|A\|_2 \le \|A\|_*$?
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Your intuition is correct. If $A$ is symmetric, it is orthogonally diagonalisable and hence its singular values are just the absolute values of its eigenvalues. Hence $\|A\|_2=\rho(A)$.