A pinching over $M_n({\mathbb C})$ is an endomorphism $T$ where the $(i,j)$-entry of $T(M)$ is given either by $0$ or by $m_{ij}$, depending on the pair $(i,j)$. Let us say that a pinching is symmetric if the rule is the same for $(i,j)$ and $(j,i)$ whenever $j\ne i$.

R. Bhatia has shown that the pinching $M\mapsto D(M):={\rm diag}(m_{11},\ldots,m_{nn})$ is a contraction for every unitarily invariant norm $\|\cdot\|$. In other words, $$\|D(M)\|\le\|M\|,\qquad\forall M\in M_n({\mathbb C}).$$ Remark that the map $D$ sends the cone of positive definite Hermitian matrices $HPD_n$ into itself.

It is not difficult to extend Bhatia's result to block pinching $\Delta$, in which $\Delta(M)$ is block diagonal, made from diagonal blocks of $M$. Again $\Delta$ sends $HPD_n$ into itself. On an other hand, it is known that some non-block diagonal pinching are not contracting and do not preserve $HPD_n$. For instance, the linear map $$M=\left( \begin{array}{ccc} a & b & c \\\\ d & e & f \\\\ g & h & k \end{array} \right)\mapsto B(M)=\left( \begin{array}{ccc} a & b & 0 \\\\ d & e & f \\\\ 0 & h & k \end{array} \right).$$ The operator norm of $B$ (when $M_n({\mathbb C})$ is endowed with a unitarily invariant norm) is larger than $1$ (Bhatia), and there exists $H\in HPD_n$ such that $B(M)$ is even not semi-positive definite.

Question. Are the following three properties equivalent ?

  1. The pinching $T$ is symmetric and contracting for every unitarily invariant norm of $M_n({\mathbb C})$.
  2. The pinching $T$ sends $HPD_n$ into itself.
  3. The pinching $T$ is block diagonal.
  • 1
    $\begingroup$ I don't understand what "block diagonal" means; why is $M\mapsto B(M)$ not block diagonal? Also, I think I understand that a "pinching" is defined purely by giving some subset $P\subseteq n\times n$ where $T(M)_{ij} = m_{ij}$ if $(i,j)\in P$, and $0$ otherwise. Is that right? $\endgroup$ – Matthew Daws Oct 11 '10 at 12:23
  • $\begingroup$ @Matthew. Your description of pinching'' is perfect. Using your notation, a pinching is block diagonal'' if $P$ is the union of sets $Q_\ell\times Q_\ell$, where $Q_1\cup\cdots\cup Q_r$ is a partition of $[1,\ldots,n]$. $\endgroup$ – Denis Serre Oct 11 '10 at 12:28
  • $\begingroup$ So in other words, if $E_{i,j}\ (1\leq i,j \leq n)$ is the canonical basis of $M_n(\mathbb{C})$, a pinching is a map of the form $\sum_{i\in I} E^∗_{i,j}$ for some $I \subseteq \{1, \dots, n\}^2$,and it is a block−diagonal pinching if the subset I is of the form $\cup_{\ell} Q_{\ell}^2$ as you described. (And to preserve definiteness, we need to disallow blocks of zeroes on the diagonal, hence the partition.) Correct? (Damn! We really could use a preview function for comments too!) $\endgroup$ – Thierry Zell Oct 11 '10 at 12:56
  • 1
    $\begingroup$ Do you want to restrict yourself to pinchings preserving the diagonal ? (otherwise you have to take care of the trivial pinching mapping everything to zero, for example) $\endgroup$ – Guillaume Aubrun Oct 11 '10 at 14:54
  • $\begingroup$ @Guillaume. Yes, that would be better. $\endgroup$ – Denis Serre Oct 11 '10 at 15:20

R. Bhatia proved (in Amer. Math. Monthly 107) that the operator $D_k$ taking $M$ to its $k$-th diagonal ($M_{ij}$, $j-i=k$) contracts any unitarily invariant norm, so 1 implies neither of 2, 3. Still, it is an interesting question to characterize those "contractive pinchings".

EDIT (partly answering the modified question): Consider a symmetric pinching, $M\mapsto P(M)=(p_{ij}M_{ij})_{ij}$, $p_{ij}=p_{ji}\in\{O,1\}$. Property (2) implies $p_{ii}=1$ for all $i$, since a positive definite matrix has positive diagonal elements. Property (1) doesn't imply the same thing, since all $p_{ij}$ might be zero (as by Guillaume's comment) [EDIT: a slightly less trivial example is $p_{11}=p_{22}=0$, $p_{12}=p_{21}=1$, and $p_{ij}=\delta_{i,j}$ if $i>2$ or $j>2$ ]. So let us assume $p_{ii}=1$ as part of the hypothesis. Then both (1) and (2) imply that the relation between indices $\{(i,j): p_{ij}=1\}$ is reflexive and symmetric. If it is not transitive (i.e. if it is not an equivalence relation, or equivalently (3) doesn't hold), then there are three distinct indices $i,j,k$ with $p_{ij}=p_{jk}=1$ but $p_{ik}=0$. Let $I=\{i,j,k\}$. Then, considering your $B$ example, $P$ doesn't preserve positive definiteness, since the $I \times I$ principal minor of $P(M)$ can be negative for a positive definite $M$. Hence (2) implies (3), and they are equivalent. Similiarly, restricting $P$ to matrices supported on $I\times I$, if $B$ doesn't contract some "natural" unitarily invariant norm on $3\times 3$ matrices,the same must be true of $P$. But it is easily seen that $B$ doesn't contract the trace norm (sum of singular values), since the all ones $3\times3$ matrix $E$ has trace norm $3$ and $B(E)$ has trace norm $1+2\sqrt{2}$ (cf Bhatia). Hence (1),(2),(3) are equivalent, if in (1) one assumes $p_{ii}=1$ for all $i$ (i.e. the symmetric pinching also preserves the diagonal).

  • $\begingroup$ I think you need to be careful. On page 1 of this article, the term "pinching" is described: this is equivalent to finding some orthonormal basis with respect to which we take a "block diagonal pinching" as Denis defined it above. So it seems that Denis's meaning of "pinching" is more general than this articles... $\endgroup$ – Matthew Daws Oct 11 '10 at 14:28
  • $\begingroup$ Sorry, but that doesn't affect your main point: indeed, this article does show that (1) does not imply (2) or (3). Sorry... $\endgroup$ – Matthew Daws Oct 11 '10 at 14:31
  • $\begingroup$ @BS. Of course, you're right, and I should have been more careful, since I had this paper in mind. I was really thinking about pinching that are symmetric wrt the diagonal. I'll edit and make it precise. $\endgroup$ – Denis Serre Oct 11 '10 at 14:37

A pinching has the form $M \mapsto T * M$, where $*$ is the entrywise product and $T$ is a $0/1$-matrix. I have the impression that (1)-(3) are all equivalent to

(4) T is positive,

and that it can be proved via the following: if a $0/1$-matrix $T$, with $1$ on the diagonal, avoids the pattern $\left( \begin{array}{ccc} 1 & 1 & 0 \\ 1 & 1 & 1 \\ 0 & 1 & 1 \end{array} \right)$, then $T$ must be block-diagonal.

Either (1) or (2) imply that $T$ avoids this pattern: for (1) via your example, for (2) because it implies (4) (apply the pitching to the matrix with all entries equal to 1).

  • $\begingroup$ In what sense do you mean $T$ is positive? It is certainly not necessarily positive definite (the 3x3 matrix with a 2x2 block of 1s and a 1 in the last position on the diagonal is only positive semi-definite, but it maps HPD to HPD). $\endgroup$ – Willie Wong Oct 11 '10 at 16:05
  • $\begingroup$ I see ... actually I used the word positive to mean "positive semi-definite" (the French terminology), and I also assumed erroneously that Denis considered pinchings preserving positive semi-definiteness. $\endgroup$ – Guillaume Aubrun Oct 11 '10 at 16:43
  • $\begingroup$ Actually my answer is more or less the same as BS's edited answer $\endgroup$ – Guillaume Aubrun Oct 11 '10 at 17:12

If you restrict $T$ to be symmetric, than don't you already have the answer stated in your question? You've claimed that

(3) $\implies$ (1) by generalizing Bhatia.

Now (3) $\implies$ (2) trivially. It suffices to show that "not (3)" $\implies$ "not (1)/(2)". But this follows from your claim about the operation $B$, after noting that:

if $T$ is a symmetric matrix with entries either 0 or 1, with only 1s on the diagonal, and $T$ is not block diagonal, then there exists $i,j,k$ distinct indices such that $T_{ij} = T_{jk} = 1$ and $T_{ki} = 0$.

Proof: By re-indexing the basis, you can bring the matrix of $T$ to a form where if $j > i$, $T_{ij} = 0$, then $T_{ik} = T_{lj} = 0$ for all $l < i < j < k$. Take $a < b$ such that $T_{ab} = 0$, $T_{a+1,b} = T_{a,b-1} = 1$. Now, if all such $a,b$ differ only by 1 ($b-1 = a$), then clearly the matrix of $T$ is block diagonal. If $a,b$ differs by at least two, chose $a, a+1, b$ to be the triplet.


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