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Let $\{u_i\}_{i=1}^n$ be a set of orthonormal basis and $M$ a symmetric matrix. Suppose \begin{equation} \left|\langle u_iu_i^T,M\rangle\right|\leq \tau\ \ \forall i=1,\dots,n. \end{equation} Can I conclude that the operator norm (i.e. the largest singular value) $\lVert M\rVert_{\infty}\leq \tau$ ? Or for some $c>0$, $\lVert M\rVert_{\infty}\leq c\tau$?

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No, your bound gives the largest possible diagonal entry of $M$ if you choose the standard basis, and the largest eigenvalue of $M$ could be larger. Consider for example the matrix $M$ of all ones. The largest eigenvalue of this matrix is $n$. In general the situation is even worse; all diagonal entries could be zero and the off-diagonal entries could all be $1$. Then you would get $\tau = 0$ even though the largest eigenvalue is $n-1$. If you have a positive definite symmetric matrix, I believe the situation is better. Maybe you want to rephrase your question that way? Or maybe you want to say that the bound $\tau$ is the universal least upper bound that holds for all orthonormal bases? Then I think you can say something.

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  • $\begingroup$ Thanks for prompt answer. So we cannot extract any information for the norms (nuclear, Frobenius, operator) of $M$? $\endgroup$
    – yue
    Oct 15, 2013 at 4:41

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