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I'd like some help showing the equivalence of these two norms when $p = \log n$.

Recall the $p$-th Schatten norm of a linear operator $A$ acting on $\mathbb{R}^{n}$. In the particular case of $p = \log(n)$, Schatten norm should be equivalent to the spectral norm, see last line on p.18. Moreover, $\| A\|_{C^{n}_{\log n}} \leq \| A \|$.

On the other hand, consider an example of an operator on $\mathbb{R}^{3}$:

$ A = \begin{pmatrix} 1 & 0 & 0\\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{pmatrix} $

Obviously, $\| A \| = 3$. However when $p = \log(3)$, the corresponding Schatten's norm is

$ \| A \|_{C^{3}_{\log 3}} = \left( 1^{\log 3} + 2^{\log 3} + 3^{\log 3}\right)^{\frac{1}{\log 3}} \approx 5.48 $

implying that $\| A\|_{C^{n}_{\log n}} > \| A \|$.

Am I doing something wrong without realizing it? I'm confused here. I'd like to show the equivalence of two norms but I'm stuck...

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In finite dimensional vector space all norms are equivalent, and you can use a general theorem to prove your assertion, otherwise $ \lVert A \rVert_{C^n_p} := \left(\sum_1^n s_i(A)^p\right)^{1/p} \leq \\ \left(n \cdot\sup_i\{s_i(A)^p\}\right)^{1/p} = n^{1/p} \cdot \sup_i\{s_i(A)\} = n^{1/p} \cdot \lVert A \rVert $ and $ \lVert A \rVert := \sup_i \{s_i(A)\} = \left(\sup_i\{s_i(A)^p\}\right)^{1/p} \leq \left(\sum_1^n s_i(A)^p\right)^{1/p} = \lVert A \rVert_{C^n_p} $

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    The article "Random Vectors in the Isotropic Position" by M. Rudelson also has the inequality that has the same 'wrong' form as that you mentioned.2013-11-20