### Matrix Convex Sets

In the previous post, we introduced some objects in operator algebra. Those will occur in the examples given in this post. The purpose of this post is to define matrix convex sets.

We begin by defining what we mean by a matrix convex set. There are two definitions, and we shall show that both are equivalent. Then we shall present some examples of matrix convex sets.

Definition 1. A matrix convex set in a vector space ${V}$ is a collection ${K=(K_n)}$, where for each ${n\in\mathbb{N}}$ ${K_n\subset M_n(V)}$ is a non-empty set such that

$\displaystyle \sum_{i=1}^{k}\gamma_i^*v_i\gamma_i\in K_n,$

whenever ${v_1\in K_{n_1}, ..., v_k\in K_{n_k}}$, and ${\gamma_1\in \mathbb{M}_{n_1,n},...,\gamma_k\in \mathbb{M}_{n_k,n}}$ satisfying ${\sum_{i=1}^k\gamma_i^*\gamma_i=\mathbb{I}_n}$.

The above definition seems like a natural generalisation of convexity. But the next one is usually easier to use in specific cases.

Definition 2. A matrix convex set in a vector space ${V}$ is a collection ${K=(K_n)}$, where for each ${n\in\mathbb{N}}$ ${K_n\subset M_n(V)}$ is a non-empty set such that

For any ${\gamma\in \mathbb{M}_{r,n}}$, with ${\gamma^*\gamma=\mathbb{I}_n}$, we have ${\gamma ^* K_r \gamma\subset K_n}$.
For any ${m,n\in\mathbb{N}}$, ${K_m\oplus K_n \subset K_{m+n}}$.

We now show the equivalence of both definitions.

Definition 1 ${\Rightarrow }$ Definition 2. The first part of definition 2 follows trivially (using ${k=1}$). To show the second part, let ${v\in K_m}$ and ${w\in K_n}$; then we have to show that

$\displaystyle \begin{pmatrix} v & 0 \\ 0 & w \end{pmatrix}\in K_{m+n}$

But

$\displaystyle \begin{pmatrix} v & 0 \\ 0 & w \end{pmatrix} = \gamma_1^*v\gamma_1+\gamma_2^*w\gamma_2,$

where

$\displaystyle \gamma_1=[\mathbb{I}_m \quad 0_{m,n}] \quad\quad \text{and}\quad \quad \gamma_2=[0_{n,m} \quad \mathbb{I}_n],$

and clearly, ${\gamma_1^*+\gamma_1+\gamma_2^*\gamma_2=\mathbb{I}_{m,n}}$. Thus (2) is also satisfied.

Definition 2 ${\Rightarrow }$ Definition 1. Let ${v_i\in K_{n_i}}$, ${\gamma_i\in \mathbb{M}_{n_i,n}}$ for ${i=1,..,k}$ satisfying ${\sum_{i=1}^k\gamma^*_i\gamma_i\in K_n}$. Then by extension of (2) in definition 2, we have

$\displaystyle v = \begin{pmatrix} v_1 & & & \\ & v_2 & & \\ & & \ddots & \\ & & & v_k \end{pmatrix} \in K_{n_1+n_2+...+n_k}$

Now let ${\gamma=\begin{pmatrix} \gamma_1 & \hdots & \gamma_k \end{pmatrix}^T\in \mathbb{C}_{n_1+n_2+...+n_k,n}}$. Then ${\gamma^*\gamma=\sum\gamma_i^*\gamma_i=\mathbb{I}_n}$ and by definition 1 we have ${\gamma^*v\gamma\in K_n}$ which implies ${\sum_1^k\gamma_i^*v_i\gamma_i\in K_n}$.

An important consequence is that each ${K_n}$ is a convex set of ${M_n(V)}$. Let ${K=(K_n)}$ be a matrix convex set in ${V}$. Let ${v_1,v_2\in K_n}$. We need to show that ${\lambda v_1+(1-\lambda)v_2\in K_n}$ for ${\lambda\in [0,1]}$. Set ${\gamma_1=\sqrt{\lambda}\;\mathbb{I}_n\in\mathbb{M}_n}$, and ${\gamma_2=\sqrt{1-\lambda}\;\mathbb{I}_n\in\mathbb{M}_n}$. Then

$\displaystyle \gamma_1^*\gamma_1+\gamma_2^*\gamma_2=(\sqrt{\lambda}\;\mathbb{I}_n)(\sqrt{\lambda}\;\mathbb{I}_n)+(\sqrt{1-\lambda}\;\mathbb{I}_n)(\sqrt{1-\lambda}\;\mathbb{I}_n) = (\lambda+1-\lambda)\;\mathbb{I}_n=\mathbb{I}_n.$

Then ${\gamma_1^*v_1\gamma_1+\gamma_2^*v_2\gamma_2\in K_n}$, which is nothing but ${\lambda v_1+(1-\lambda)v_2\in K_n}$.

We present some examples of matrix convex sets. We start with the simplest vector space.

Example 2. Let ${a,b\in [-\infty, \infty ]}$. On the vector space ${\mathbb{C}}$, consider the collection

$\displaystyle [a\mathbb{I},b\mathbb{I}]=([a\mathbb{I}_n,b\mathbb{I}_n]),$

where ${[a\mathbb{I}_n,b\mathbb{I}_n]=\{\alpha\in\mathbb{M}_n: a\mathbb{I}_n\leq \alpha\leq b\mathbb{I}_n\}}$. This collection defines a matrix convex set in ${\mathbb{C}}$. We shall now verify that this indeed satisfies the conditions mentioned above (using definition 2).

Let ${\alpha\in \mathbb{M}_{r,n}}$, with ${\alpha^*\alpha=\mathbb{I}_n}$. Let ${v\in K_r}$, then ${a\mathbb{I}_r\leq v\leq b\mathbb{I}_r}$. This implies

$\displaystyle a(\alpha^*\mathbb{I}_r\alpha) \leq \alpha^*v\alpha \leq b( \alpha^*\mathbb{I}_r\alpha)$

which means ${a\mathbb{I}_n\leq v \leq b\mathbb{I}_n}$. Thus ${\alpha^*v\alpha\in K_n}$. Finally let ${v\in K_m}$ and ${w\in K_n}$. Then we show that

$\displaystyle \begin{pmatrix} v & 0 \\ 0 & w \end{pmatrix} \in K_{m+n}$

But, for ${z_1\in \mathbb{C}^m,z_2\in \mathbb{C}^n}$, we have

$\displaystyle \left\langle \begin{pmatrix} v- a\mathbb{I}_n & 0 \\ 0 & w-a\mathbb{I}_n \end{pmatrix} \begin{pmatrix} z_1 \\ z_2 \end{pmatrix},\begin{pmatrix} z_1 \\ z_2 \end{pmatrix}\right\rangle= \langle (v-a\mathbb{I}_m)z_1,z_1 \rangle+\langle (w-a\mathbb{I}_n)z_2,z_2 \rangle\geq 0$

And similarly we can show the other parts. This completes our verification. Conversely, we can show that any matrix convex set ${K=(K_n)}$ in ${\mathbb{C}}$, where ${K_1}$ is a closed convex subset of ${\mathbb{R}}$ is a closed matrix interval.

Theorem 1 Suppose that ${K=(K_n)}$ is a matrix convex set in ${\mathbb{C}}$ where ${K_1}$ is a bounded closed subset of ${\mathbb{R}}$. Then ${K}$ must be a closed matrix interval.

Proof: Since ${K_1}$ is a bounded closed and convex it must be a closed interval in ${\mathbb{R}}$, say ${K_1=[a,b]}$. Let ${\gamma\in K_n}$. Then we have to show that ${\alpha\mathbb{I}_n\leq \gamma\leq \beta\mathbb{I}_n}$. Let us first show that ${\gamma-\alpha\mathbb{I}_n\geq 0}$. Let ${\xi\in\mathbb{C}^n}$, then

$\displaystyle \begin{array}{rcl} && \langle (\gamma-\alpha\mathbb{I}_n)\xi,xi \rangle \geq 0 \\ &\Leftrightarrow& \langle \gamma\xi,\xi\rangle-\langle\alpha\mathbb{I}_n\xi,\xi\rangle \geq 0 \\ &\Leftrightarrow& \xi^*\gamma\xi \geq \alpha \xi^*\xi=\alpha \end{array}$

But from property 1, we see that ${\xi^*\gamma\xi\in K_1}$, and thus the last statement is true; hence ${\gamma\geq \alpha\mathbb{I}_n}$. Similarly, we can show ${\gamma\leq \beta\mathbb{I}_n}$. Thus ${\alpha\mathbb{I}_n\leq \gamma\leq \beta\mathbb{I}_n}$.

Conversely, let ${\gamma\in [a\mathbb{I}_n,\beta\mathbb{I}_n]}$, then ${\gamma}$ is self-adjoint and hence we can write

$\displaystyle \gamma=u^*D u=\begin{pmatrix} u_1^* & \hdots & u_n^* \end{pmatrix}\begin{pmatrix} \lambda_1 & & \\ & \ddots & \\ & & \lambda_n \end{pmatrix}\begin{pmatrix} u_1 \\ \vdots \\ u_n \end{pmatrix},$

where ${\gamma_i\in [\alpha,\beta]}$. Then by property 2, we see that ${D\in K_n}$, and by property 1, ${u^*Du\in K_n}$, that is ${\gamma\in K_n}$. $\Box$

Example 3. Let ${\mathcal{M}}$ be an operator space. Then the collection of unit balls, ${B=(B_n)}$ where

$\displaystyle B_n=\{x\in M_n(\mathcal{M}): \|x\|\leq 1\}$

is a matrix convex set. It is almost trivial to verify the conditions of matrix convexity using the abstract definition of operator space (given in this post).

Example 4. Let ${\mathcal{R}}$ be an operator system. The collection of positive cones, ${P=(P_n)}$ where

$\displaystyle P_n=\{x\in M_n(\mathcal{R}):x\geq 0\}$

is a matrix convex set. For let ${\gamma\in \mathbb{M}_{r,n}}$ with ${\gamma^*\gamma=\mathbb{I}_n}$. Let ${v\in P_r}$. Then for ${\xi\in\mathbb{C}^n}$ we have

$\displaystyle \langle(\gamma^*v\gamma)\xi,\xi\rangle=\xi^*\gamma^*v\gamma\xi=(\gamma\xi)^*v(\gamma\xi)\geq 0$

since ${v\geq 0}$. Similarly, if ${v\in P_m}$ and ${w\in P_n}$, then it is fairly obvious that ${v\oplus w\in P_{m+n}}$.

Example 5. Again let ${\mathcal{R}}$ be an operator system. Consider the collection of matrix states ${CS(\mathcal{R})=(CS_n(\mathcal{R}))}$ where

$\displaystyle CS_n(\mathcal{R})=\{\varphi:\mathcal{R}\rightarrow \mathbb{M}_n \; | \; \varphi\; \text{is completely positive and} \; \varphi(1)=\mathbb{I}_n\}.$

We say ${\varphi:\mathcal{R}\rightarrow \mathbb{M}_n}$ is completely positive if canonically amplified maps ${\varphi_r:M_r(\mathcal{R})\rightarrow M_r(\mathbb{M}_n)}$ are positive for all ${r\in\mathbb{N}}$. Then ${CS(\mathcal{R})}$ is a matrix convex set in ${\mathcal{R}^*}$. Here we have identified an ${\varphi\in CS_n(\mathcal{R})}$ as an element in ${M_n(\mathcal{R^*})}$. For verification, let ${\varphi\in CS_r(\mathcal{R})}$ and ${\gamma\in\mathbb{M}_{r,n}}$. Then we have to show that ${\gamma^*\varphi\gamma}$ is completely positive and ${\gamma^*\varphi\gamma(1)=\mathbb{I}_n}$. By definition, ${\gamma^*\varphi\gamma(x)= \gamma^*\varphi(x)\gamma}$. Then

$\displaystyle (\gamma^*\varphi\gamma)(1)=\gamma^*\varphi(1)\gamma=\gamma^*\mathbb{I}_r\gamma=\mathbb{I}_n.$

To show complete positivity, let ${[x_{ij}]\geq 0}$. Then

$\displaystyle (\gamma^*\varphi\gamma)[x_{ij}]=[\gamma^*\varphi(x_{ij})\gamma]=\Gamma^*[\varphi(x_{ij})]\Gamma\geq 0,$

where

$\displaystyle \Gamma =\begin{pmatrix} \gamma & \hdots & \gamma \end{pmatrix}^T.$

Thus ${\gamma^*\varphi\gamma}$ is completely positive since ${\varphi}$ is completely positive. Likewise, the statement for ${\varphi\oplus\psi}$ is analogously proved. ${CS(\mathcal{R})}$ is considered as the matricial version of the state space.

Example 2 and example 5 are the ones which we shall focus upon the next sections.