How can I prove this inequality?
For any even number $k$ and symmetric matrix $A$,
$||A||^{k}\leq Tr(A^{k})$
where ||.|| stands form the spectral norm and $Tr$ for the trace of the matrix.
2026-03-29 20:46:17.1774817177
Spectral norm is smaller than trace
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This is true because for every real symmetric $A$, we have $\|A\|^{2k}\le\operatorname{tr}(A^{2k})$, or equivalently, $\left(|\lambda|_\max(A)\right)^{2k}\le\sum_i\lambda_i(A)^{2k}$.