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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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up vote 1 down vote accepted

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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Thanks for the answer, it cleared some things up. However, could you confirm that the inequality stated in the paper I refered to, i.e. $\| A\|_{C^{n}_{\log n}} \leq \| A \| \leq e\|A\|_{C^{n}_{\log n}}$ is actually false? It certainly seems so given my "counter-example". Adapting it from the proof, the correct one should look like $$\|A\|_{C^{n}_{\log n}} \leq e\|A\| \leq n\|A\|_{C^{n}_{\log n}}$$ which is obtained setting $p = \log n$. – johnny Oct 8 '12 at 7:56
To add to my comment, I now guess that the inequalities can be sharpened. It seems that in all cases $\|A\| \leq \| A \|_{C^{n}_{p}}$ for $p \geq 1$ so that we have $\| A \|_{C^{n}_{p}} \leq e\|A\| \leq e\| A \|_{C^{n}_{p}}$. I guess that the paper missed an $e$ somewhere. – johnny Oct 8 '12 at 8:23
@johnny I confirm that the inequality stated in the paper is false and, as you suggested, I sharpened the second inequality in my answer. – AlbertH Oct 8 '12 at 14:51
Thanks. Your help is greatly appreciated. – johnny Oct 8 '12 at 15:11
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. – user110203 Nov 20 '13 at 14:10

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