Comparison of the norm of a matrix and its entry-wise absolute value.

Question

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.

Answer 1

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).

Claim 2: $\|A_n\|_p = \|A_n'\|_p$ for all even integers $2 \leq p \leq 2n-2$.

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.

Answer 2

Mikael's explicit construction is extremely nice. Here's an ugly numerical example for those who like matlab based counterexamples.

This problem turned out to be quite hard to defeat numerically by brute force, which is a common characteristic of many singular value inequalities. I tried slightly more structured matrices, and am tempted to believe that I have found computational counterexamples.

Here's the brute-force construction.

a=randn;b=2*randn;c=rand;
m=gallery('tridiag', b*ones(n,1), a*ones(n+1,1), c*ones(n,1));
m=full(m);
m(n,1)=-2;   % this additional entry is important.
lhs = norm(svd(m),p);
rhs = norm(svd(abs(m)),p);

Now, if you loop over the above random "almost" tridiagonal matrices, Matlab quickly reports a counterexample where "lhs" is bigger than "rhs" (I tried $n=9$ and $p=7$).