# How to prove compactness of matrix convex sets?

I am reading a paper - "the Krein Milman theorem in Operator Convexity"; and the third section there deals with compact matrix convex sets. The first example there states that the matrix interval $[a\mathbb{I},b\mathbb{I}]$ is a compact matrix convex set in $\mathbb{C}$, the complex space. But proving this specifically is troubling me. For example how do I prove that $[aI_2, bI_2]$ is compact in $\mathbb{M}_2(\mathbb{C})$? It will suffice only to provide a hint or a reference to a theorem.

The link to the paper is http://www.jstor.org/stable/10.2307/117899

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Welcome to Math.SE! Excellent reference link. Unfortunately I cannot help you here. –  magma Mar 4 '12 at 10:32

For compactness: topologically, you can see $M_2(\mathbb{C})$ as $\mathbb{C}^4$. This means that convergence in the operator norm in $M_2(\mathbb{C})$ is simply convergence in coordinates, and it implies that Heine-Borel holds; so all you need to show is closedness. If you have a Cauchy sequence in $[a\mathbb{I},b\mathbb{I}]$, a limit matrix will exist because in each entry you will have a Cauchy sequence of complex numbers. Moreover, since the characteristic polynomial's entries depend continuously on the entries of the matrix, the characteristic polynomial of the limit will be limit of the characteristic polynomials of the sequence, which shows that the limit will again be in $[a\mathbb{I},b\mathbb{I}]$. So the interval is compact.

Convexity: the interval $[a\mathbb{I},b\mathbb{I}]$ can be characterized as those hermitian matrices $A$ such that $$a\,x^Tx\leq x^TAx\leq b\,x^Tx,\ \ x\in\mathbb{C}^2.$$ So, if $A,B\in[a\mathbb{I},b\mathbb{I}]$, $\gamma\in[0,1]$, $$a\,x^Tx=\gamma\,a\,x^Tx+(1-\gamma)a\,x^Tx\leq\gamma\,x^TAx+(1-\gamma)x^TBx=x^T(\gamma A+(1-\gamma)B)x ;$$ similarly, we obtain $$x^T(\gamma A+(1-\gamma)B)x \leq b\,x^Tx.$$

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Thank you. Basically the idea we have to use is that on finite dimensional spaces, any two norms are equivalent. On a similar question, the matrix range $W_n(T)=\{\phi(T):\phi:\mathfrak{B}(H)\rightarrow \mathbb{M}_n \; completely \; positive \; and \; \phi(I)=I_n\}$ is also compact in $\mathbb{M}_n$. So in this case I need to worry about sequence of completely positive maps. Right? –  Nirakar Neo Mar 4 '12 at 12:25
Yes, in the case of the ucp maps, you take a Cauchy sequence and you write the maps using Stinespring-Choi-Kraus as operator convex combinations, and then you use compactness to obtain a convergent subsequence on the coefficients. Or you obtain a convergent subsequence via Arveson's BW topology and then you use that in finite dimension this agrees with the pointwise norm topology. –  Martin Argerami Mar 4 '12 at 12:53
It is the image of the compact set $[a,b]$ by the continuous application $t \mapsto t \mathbb{I}$ (exercise: verify it is continuous) in the Hausdorff space $M_2(\mathbb{C})$, it is therefore compact.
Another way of seeing it is the Bolzano-Weierstrass theorem: $M_2(\mathbb{C}) \simeq \mathbb{C}^4 \simeq \mathbb{R}^8$ is a finite dimensional normed vector space (it's easy to verify that the norm induced by any norm on $\mathbb{R}^8$ induces the usual topology on $M_2(\mathbb{C})$), therefore its sequentially compact sets are its closed bounded sets, and in a metric space, sequential compactness is the same as compactness.
I think you are confused with the notation: $[a\mathbb{I},b\mathbb{I}]$ is the set of all positive semidefinite matrices with eigenvalues in $[a,b]$. –  Martin Argerami Mar 4 '12 at 10:57
(sorry, not necessarily "positive", as negative eigenvalues are allowed if $a<0$; so the right word would be "hermitian" instead) –  Martin Argerami Mar 4 '12 at 11:08