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Let $A\in M_n$. How can I show that $$\left|{\textrm{Tr}(A)\over\sqrt{n}}\right|\leq \Vert A\Vert_F$$ I tried it using the Cauchy-Schwarz inequality.

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  • $\begingroup$ Frobenius norm was $\sqrt{\sum a_{ij}^2}$? $\endgroup$ – Daniel Fischer Nov 5 '13 at 21:50
  • $\begingroup$ The trace only depends on the diagonal entries of $A$. What does this suggest about an approach? $\endgroup$ – hardmath Nov 5 '13 at 21:56
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Hint: Given $A\in\mathbb{C}^{t \times t}$ with entries, $a_{ij}$, we have $\mathrm{tr}(A) = \sum_{i=1}^n a_{ii}$, and $\|A\|_F = \sqrt{\sum_{i,j=1}^n |a_{ij}|^2}$. We immediately have $\|A\|_F \geq \sqrt{\sum_{i=1}^n |a_{ii}|^2}$. Hence, to prove your result it is sufficient to prove that $$\left|\sum_{i=1}^n a_{ii} \right| \leq \sqrt{n}\sqrt{\sum_{i=1}^n |a_{ii}|^2},$$ which now has nothing to do with matrices.

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Let $\operatorname{vec}(A)$ denotes the vector obtained by stacking the columns of $A$ on top of one another. Then your inequality can be rewritten as $\left|\langle\operatorname{vec}(A),\operatorname{vec}(I)\rangle\right|\le\|\operatorname{vec}(I)\|_2\|\operatorname{vec}(A)\|_2$, which is Cauchy-Schwarz inequaltiy.

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