Pinching and positive definite matrices - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T04:48:09Z http://mathoverflow.net/feeds/question/41758 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/41758/pinching-and-positive-definite-matrices Pinching and positive definite matrices Denis Serre 2010-10-11T07:55:58Z 2010-10-11T21:47:32Z <p>A <em>pinching</em> 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 $j\ne i$. </p> <p>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.</p> <p>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 $$M=\left( \begin{array}{ccc} a &amp; b &amp; c \\ d &amp; e &amp; f \\ g &amp; h &amp; k \end{array} \right)\mapsto B(M)=\left( \begin{array}{ccc} a &amp; b &amp; 0 \\ d &amp; e &amp; f \\ 0 &amp; h &amp; k \end{array} \right).$$ 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.</p> <p><strong>Question</strong>. Are the following three properties equivalent ?</p> <ol> <li>The pinching $T$ is symmetric and contracting for every unitarily invariant norm of $M_n({\mathbb C})$.</li> <li>The pinching $T$ sends $HPD_n$ into itself.</li> <li>The pinching $T$ is block diagonal.</li> </ol> http://mathoverflow.net/questions/41758/pinching-and-positive-definite-matrices/41782#41782 Answer by BS for Pinching and positive definite matrices BS 2010-10-11T13:11:17Z 2010-10-11T21:47:32Z <p>R. Bhatia proved (in <a href="http://www.ams.org/mathscinet-getitem?mr=1786234" rel="nofollow">Amer. Math. Monthly 107</a>) 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".</p> <p>EDIT (partly answering the modified question): Consider a symmetric pinching, <code>$M\mapsto P(M)=(p_{ij}M_{ij})_{ij}$, $p_{ij}=p_{ji}\in\{O,1\}$</code>. 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 <code>$\{(i,j): p_{ij}=1\}$</code> 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 <code>$I=\{i,j,k\}$</code>. Then, considering your $B$ example, $P$ doesn't preserve positive definiteness, since the <code>$I \times I$</code> 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 <code>$I\times I$</code>, 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 <em>trace norm</em> (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).</p> http://mathoverflow.net/questions/41758/pinching-and-positive-definite-matrices/41799#41799 Answer by Guillaume Aubrun for Pinching and positive definite matrices Guillaume Aubrun 2010-10-11T15:57:52Z 2010-10-11T20:13:42Z <p>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 </p> <p>(4) T is positive,</p> <p>and that it can be proved via the following: if a $0/1$-matrix $T$, with $1$ on the diagonal, avoids the pattern <code>$\left( \begin{array}{ccc} 1 &amp; 1 &amp; 0 \\ 1 &amp; 1 &amp; 1 \\ 0 &amp; 1 &amp; 1 \end{array} \right)$</code>, then $T$ must be block-diagonal.</p> <p>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).</p> http://mathoverflow.net/questions/41758/pinching-and-positive-definite-matrices/41811#41811 Answer by Willie Wong for Pinching and positive definite matrices Willie Wong 2010-10-11T17:16:24Z 2010-10-11T17:16:24Z <p>If you restrict $T$ to be symmetric, than don't you already have the answer stated in your question? You've claimed that</p> <p>(3) $\implies$ (1) by generalizing Bhatia. </p> <p>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:</p> <blockquote> <p>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 $T_{ki} = 0$. </p> </blockquote> <p>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 $l &lt; i &lt; j &lt; k$. Take $a &lt; b$ 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. </p>