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?
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> 
> 


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](http://www.ams.org/mathscinet-getitem?mr=MR83g%3A47030), and the answer was refined successively by [B. Simon](http://www.ams.org/mathscinet-getitem?mr=MR83b%3A15017), [M. Dechamps-Gondim, F. Lust-Piquard and H. Queffélec](http://www.ams.org/mathscinet-getitem?mr=MR86h%3A47024) and [L. Rosen](http://www.ams.org/mathscinet-getitem?mr=MR97b%3A15028). The counterexamples given in [DLQ](http://www.ams.org/mathscinet-getitem?mr=MR86h%3A47024) are the same as the ones I give in my answer, with a slightly different proof.



Comparison of the norm of a matrix and its entry-wise absolute value.
Let me answer my own question by contructing, for $p>2$ not an even
integer (say $2k<p<2k+2$), a matrix $A$ such that $\|A\|\_p> \|A'\|\_p$. In fact I construct a family of matrices $A\_n \in M\_n(\mathbb C)$ such that $\|A\_n\|\_p > \|A\_n'\|\_p$ whenever $n-k$ is an even positive integer, and $\|A\_n\|\_p < \|A'\_n\|\_p$ if $n-k$ is an odd positive integer$.


This will imply that $\|A\|\_p \leq \|A'\|\_p$ for all $n$ and all $A \in M\_n$ if and only if $p$ is an even integer or $p=\infty$. For other values of $p$, these two quantities are not comparable, and if we go the the Schatten classes (ie we allow infinite matrices), there is no implication between the properties $\|A\|\_p<\infty$ and $\|A'\|\_p<\infty$.


Here is the construction. For every integer $n$, consider $S\_n \in M\_n(\mathbb C)$ to be the matrix of a cyclic permutation of $\{{1,\dots,n\}}$ in which one of the $1$'s is replaced by a $-1$. Take $A\_n=Id+S\_n$, so that $A\_n'=Id+S\_n'$. I only sketch the proof that $A\_n$ works, since I might be the only one interested ;).


There are two independent claims:



> 
> *Claim 1*: the function $p \mapsto \|A\_n\|\_p^p - \|A\_n'\|\_p^p$ has at most $n-1$ zeros (counting multiplicities).
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> 
> *Claim 2*: $\|A\_n\|\_p = \|A\_n'\|\_p$ for all even integers $2 \leq p \leq 2n-2$.
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> 


These two claims together imply that $\|A\_n\|\_p^p -\|A\_n'\|\_p^p$ is non zero outside of $\{{2,4,\dots,2n-2\}}$ and changes signs at each of these values of $p$. Since it is negative for $p=2n$, we have the announced properties.


The second claim is easier. A first observation is that $Tr(S\_n^k) = Tr({S'\_n}^k)$ for every $k$ with $-n+1 \leq k \leq n-1$: $k=0$ is obvious, and if $k \neq 0$ both matrices have a zero diagonal. Hence, since $A\_n^\* A\_n = 2+S\_n +S\_n^\*$ and ${A'\_n}^\* A'\_n = 2+ S'\_n +{S'\_n}^{\*}$ we have $\|A\_n\|\_p = \|A\_n'\|\_p$ for all even integers $p \leq 2n-2$.


To prove the first claim, first notice that the eigenvalues of $S'\_n$ (resp. $S\_n$) are $\lambda\_k=\exp(2ki\pi/n)$, $k=1\dots n$ (resp. $ \mu\_k=\exp( (2k+1) i \pi/n)$, $k=1\dots n$). Thus the singular values of $A'\_n$ (resp. $A\_n$) are $|1+\lambda\_k|$ (resp. $|1+\mu\_k|$). In particular if $N(B)$ denotes the number of distinct non-zero singular values of a matrix $B$, we have $N(A\_n)+N(A'\_n)=n$. Hence $\|A\_n\|\_p^p - \|A\_n'\|\_p^p$ can be written in the form $\sum\_{j=1}^n \alpha\_j e^{\beta\_j p}$, and such a function cannot have more than $n-1$ zeros unless it is identically zero.


