Pinching and positive definite matrices - MathOverflow

Pinching and positive definite matrices

2010-10-11T07:55:58Z

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.

Answer by BS for Pinching and positive definite matrices

2010-10-11T13:11:17Z

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).

Answer by Guillaume Aubrun for Pinching and positive definite matrices

2010-10-11T15:57:52Z

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).

Answer by Willie Wong for Pinching and positive definite matrices

2010-10-11T17:16:24Z

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.