It is an easy fact that for a matrix $A \in M_n(\mathbb C)$, the matrix $A' = (|A(i,j)|)_{i,j \leq n}$ has a larger operator norm than $A$. By operator norm I mean the norm as an operator on $\ell^2_n$, or equivalently its largest singular value.

My question is:

What happens when the operator norm is replaced by the Schatten $p$-norm?

Let me recall that for $1 \leq p \leq \infty$ and a matrix $A \in M_n(\mathbb C)$, its Schatten $p$-norm $\|A\|_p$ is the $\ell^p$-norm of its singular values, or equivalently $\|A\|_p= Tr((A^*A)^{p/2})^{1/p}$ if $p<\infty$. If $p=\infty$, this is just the operator norm.

Some remarks:

- If $p=2$ the equality $\|A\|_2 =\|A'\|_2$ is obvious.
- If $p$ is an even integer, the inequality $\|A\|_p \leq \|A'\|_p$ holds (just expand $\|A\|_p^p = Tr( (A^*A)^{p/2})$ and use the triangle inequality to prove that $\|A\|_p^p \leq \|A'\|_p^p$).
- When $A$ is a complex Hadamard matrix (a unitary with all entries having absolute value $1/\sqrt n$), $\|A\|_p < \|A'\|_p$ for $p>2$ and $\|A\|_p> \|A'\|_p$ for $p<2$: indeed, $p$-norm of $A$ is $n^{1/p}$ and the $p$-norm of $A'$ is $\sqrt n$. A naive guess would be that these inequalities always hold, BUT:
- When $p<6$ and $p \neq 2,4$, the quantities $\|A\|_p$ and $\|A'\|_p$ are not comparable (with constants independant of $n$). For $p<4$, take $A=\begin{pmatrix} 1 & -1 & 0\\ 0 & 1 & 1 \\ 1 & 0 & 1\end{pmatrix}$, and for$4<p<6$ take $A = \begin{pmatrix} 0 & 1&1 & 0\\ 1&0 & 1 & 0 \\ 1 & 0&0 & -1\\0&1&0&1\end{pmatrix}$.

So the question is really for $p>6$. For example, I could not find any matrix $A$ such that $\|A\|_7 > \|A'\|_7$.

**Update** I was informed by Gilles Pisier that this question has already been considered. This question has been first answered by V. Peller, and the answer was refined successively by B. Simon, M. Dechamps-Gondim, F. Lust-Piquard and H. Queffélec and L. Rosen. The counterexamples given in DLQ are the same as the ones I give in my answer, with a slightly different proof.

anyunitarily invariant norm.... $\endgroup$ – Suvrit Oct 3 '12 at 8:163more comments