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 ?

- The pinching $T$ is symmetric and contracting for every unitarily invariant norm of $M_n({\mathbb C})$.
- The pinching $T$ sends $HPD_n$ into itself.
- The pinching $T$ is block diagonal.

`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`$I \subseteq \{1, \dots, n\}^2$`

,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