Let $V$ be a finite dimensional complex inner product space. If $A\_1,\dots,A\_r\in L(V)$, then define a mapping $\Phi(A\_1,\dots,A\_r):L(V)\rightarrow L(V)$ by letting $\Phi(A\_1,\dots,A\_r)(X)=A\_1XA\_1^\*+\dots+A\_rXA\_r^\*$ for all operators $X\in L(V)$.


We have $$\rho(A\_1\otimes B\_1+\dots+A\_r\otimes B\_r)\leq\rho(\Phi(A\_1,\dots,A\_r))^{1/2}\rho(\Phi(B\_1,\dots,B\_r))^{1/2}$$
whenever $A\_1,\dots,A\_r,B\_1,\dots,B\_r$ are matrices.


In particular,


$$\frac{\rho(A\_1\otimes X\_1+\dots+A\_r\otimes X\_r)}{\rho(\Phi(X\_1,\dots,X\_r))^{1/2}}\leq \rho(\Phi(A\_1,\dots,A\_r))^{1/2}$$
whenever $\rho(\Phi(X\_1,\dots,X\_r))\neq 0.$


Define a quantity $\rho\_{2,d}$ by letting
$$\rho\_{2,d}(A\_1,\dots,A\_r)$$
$$=\sup\{\frac{\rho(A\_1\otimes X\_1+\dots+A\_r\otimes X\_r)}{\rho(\Phi(X\_1,\dots,X\_r))^{1/2}}\mid \rho(\Phi(X\_1,\dots,X\_r))\neq 0,X\_1,\dots,X\_r\in M\_{d}(\mathbb{C})\}.$$


Given matrices $A\_1,\dots,A\_r$, do there necessarily exist $d\times d$-complex matrices $X\_1,\dots,X\_r$ where
$$\rho\_{2,d}(A\_1,\dots,A\_r)
=\frac{\rho(A\_1\otimes X\_1+\dots+A\_r\otimes X\_r)}{\rho(\Phi(X\_1,\dots,X\_r))^{1/2}}?$$


If $d=1$, then the supremum can be reached since
$$\rho\_{2,1}(A\_1,\dots,A\_r)=\sup\{\frac{\rho(\alpha\_1X\_1+\dots+\alpha\_rX\_r)}{(|\alpha|\_1^2+\dots+|\alpha\_r|\_1^2)^{1/2}}\mid(\alpha\_1,\dots,\alpha\_r)\neq\mathbf{0}\}$$


$$=\sup\{\rho(\alpha\_1X\_1+\dots+\alpha\_rX\_r):|\alpha\_1|^2+\dots+|\alpha\_r|^2=1\}.$$


The sumpremum may also be reached when $d\geq\text{Dim}(V)$.


The value $$\frac{\rho\_{2,d}(A\_{1},\dots,A\_r)}{\rho(\Phi(A\_1,\dots,A\_r))^{1/2}}$$ seems to be a number in the interval $[0,1]$ that is a sort of maximum value of a kind of correlation coefficient between $(A\_1,\dots,A\_n)$ and a collection of $d\times d$-matrices $(X\_1,\dots,X\_n)$. Said differently, for $d<\text{Dim}(V)$ this value seems to be a coefficient that tells one how random $(A\_1,\dots,A\_n)$ is.



Can the supremum of this quotient of spectral radii be reached?
**Yes.** The equality can in fact be reached. Our strategy will be to produce a compact set $K\_{d,r}\subseteq M\_{d}(\mathbb{C})^{r}$ such that if $(X\_1,\dots,X\_r)\in K\_{d,r}$, then $\rho(\Phi(X\_1,\dots,X\_r))=1$, and where
$$\rho\_{2,d}(A\_1,\dots,A\_r)=\max\{\rho(A\_1\otimes X\_1+\dots+A\_r\otimes X\_r)\mid(X\_1,\dots,X\_r)\in K\_{d,r}\}.$$


Let us go over a few definitions and facts to give some context to our construction. These facts can easily be found in John Watrous's 2018 book called [The Theory of Quantum Information](https://cs.uwaterloo.ca/%7Ewatrous/TQI/TQI.pdf).


A mapping $\mathcal{E}:L(V)\rightarrow L(W)$ is said to be positive if whenever $X$ is positive semidefinite, $\mathcal{E}(X)$ is also positive semidefinite. A mapping $\mathcal{E}:L(V)\rightarrow L(W)$ is said to be completely positive if the mapping $\mathcal{E}\otimes 1\_{U}:L(V\otimes U)\rightarrow L(W\otimes U)$ is positive for each finite dimensional complex Hilbert space $U$.


A mapping $\mathcal{E}:L(V)\rightarrow L(W)$ is said to be trace preserving if $\text{Tr}(\mathcal{E}(X))=\text{Tr}(X)$ for each $X\in L(V)$.


Proposition: Let $\mathcal{E}:L(V)\rightarrow L(W)$ be a linear operator.


1. $\mathcal{E}$ is completely positive if and only if there are $A\_1,\dots,A\_r$ where $\mathcal{E}=\Phi(X\_1,\dots,X\_r)$.
2. If $\mathcal{E}$ is defined by letting $\mathcal{E}(X)=A\_1XB\_1^\*+\dots+A\_rXB\_r^\*$, then $\mathcal{E}$ is trace preserving if and only if $A\_1^\*B\_1+\dots+A\_r^\*B\_r=1\_V$.


A quantum channel is a completely positive trace preserving operator $\mathcal{E}:L(V)\rightarrow L(W)$. In quantum information theory, the quantum channels are the main morphisms between quantum states.


Proposition: If $\mathcal{E}$ is a quantum channel, then $\rho(\mathcal{E})=1$.


Let $K\_{d,r}$ be the collection of all tuples $(X\_1,\dots,X\_r)\in M\_{d}(\mathbb{C})^r$ such that $X\_1^\*X\_1+\dots+X\_r^\*X\_r=1\_d$. Said differently, $K\_{d,r}$ is the collection of all tuples $(X\_1,\dots,X\_r)\in M\_{d}(\mathbb{C})^r$ such that
$\Phi(X\_1,\dots,X\_r)$ is a quantum channel. The set $K\_{d,r}$ is clearly a closed set, and $K\_{d,r}$ is bounded since $d=\text{Tr}(X\_1^\*X\_1+\dots+X\_r^\*X\_r)=\|X\_1\|\_2^2+\dots+\|X\_r\|\_2^2$ where $\|\cdot\|\_2$ denotes the Frobenius norm, so $K\_{d,r}$ is compact.


Lemma: Let $A\_1,\dots,A\_r\in L(V)$. Suppose that there is no $x\in V\setminus\{0\}$ with $A\_1x=\dots=A\_rx=0$. Furthermore, suppose that there is no subspace $W\subseteq V$ with $W\neq\{0\},W\neq V$, and $W=A\_1[W]+\dots+A\_r[W]$. Then there is a $\lambda>0$ along with a positive definite $P$ with $\Phi(A\_1,\dots,A\_r)(P)=\lambda P$.


Proof: Now, let $\mathcal{Q}$ be the collection of all positive semidefinite matrices in $L(V)$ with trace $1$, and let $F:\mathcal{Q}\rightarrow\mathcal{Q}$ be the mapping defined by letting $$F(P)=\frac{\Phi(A\_1,\dots,A\_r)(P)}{\text{Tr}\big(\Phi(A\_1,\dots,A\_r)(P)\big)}.$$ Then $\mathcal{Q}$ is convex, and $F$ is a continuous bijection, so by the Brouwer fixed point theorem, there is some $P\in\mathcal{Q}$ with
$F(P)=P$. Therefore, we have $\Phi(A\_1,\dots,A\_r)(P)=\lambda P$ for some positive $\lambda$.


Now,
$$\text{Im}(P)=\text{Im}(\lambda P)=\text{Im}(A\_1PA\_1^\*+\dots+A\_rPA\_r^\*)$$
$$=\text{Im}(A\_1PA\_1^\*)+\dots+\text{Im}(A\_rPA\_r^\*)=\text{Im}(A\_1P)+\dots+\text{Im}(A\_rP)$$
$$=A\_1[\text{Im}(P)]+\dots+A\_r[\text{Im}(P)].$$


This is only possible if $\text{Im}(P)=V$. $\square$


Let $O\_{d,r}$ be the collection of all $(X\_1,\dots,X\_r)\in M\_{d}(\mathbb{C})^{r}$ where $\Phi(X\_1,\dots,X\_r)$ is not nilpotent. Let $E\_{d,r}$ be the collection of all $(X\_1,\dots,X\_r)\in M\_d(\mathbb{C})^r$ where there is no subspace $W\subseteq\mathbb{C}^d$ with
$W=X\_1^\*[W]+\dots+X\_r^\*[W]$ and $W\neq\{0\},W\neq\mathbb{C}^d$ and where there is no $x$ with $X\_j^\*x=0$ for $1\leq j\leq r$.


The sets $O\_{d,r},E\_{d,r}$ are dense subsets of $M\_{d}(\mathbb{C})^{r}$ with $E\_{d,r}\subseteq O\_{d,r}$.


Proposition: Suppose that $(X\_1,\dots,X\_r)\in E\_{d,r}$. Then there is an invertible matrix $B$ and some positive number $\lambda$ with $(\lambda BX\_1B^{-1},\dots,\lambda BX\_rB^{-1})\in K\_{d,r}$.


Proof: Suppose that $B$ is invertible and $\lambda$ is positive. Then $(\lambda BX\_1B^{-1},\dots,\lambda BX\_rB^{-1})\in K\_{d,r}$ if and only if
$$\sum\_{k=1}^{r}\lambda^2 (B^{-1})^\*X\_k^\*B^\*BX\_k^\*B^{-1}=I$$ if and only if


$$\sum\_{k=1}^{r}\lambda^2 X\_k^\*B^\*BX\_k=B^\*B$$.


Therefore, there are $\lambda,B$ with $(\lambda BX\_1B^{-1},\dots,\lambda BX\_rB^{-1})\in K\_{d,r}$ if and only if there is a positive $\mu$ and a positive definite $P$ with $$\mu\Phi(X\_1^\*,\dots,X\_k^\*)P=P.$$
On the other hand, the existence of such a positive definite $P$ and positive $\mu$ is guaranteed by the above lemma. $\square$


Now, define a mapping $G\_{A\_1,\dots,A\_r}:O\_{d,r}\rightarrow\mathbb{R}$ by letting
$$G\_{A\_1,\dots,A\_r}(X\_1,\dots,X\_r)=\frac{\rho(A\_1\otimes X\_1+\dots+A\_r\otimes X\_r)}{\rho(\Phi(X\_1,\dots,X\_r))^{1/2}}.$$ Since the mapping $G$ is continuous, we have
$$\rho\_{2,d}(A\_1,\dots,A\_r)=\sup\{G\_{A\_1,\dots,A\_r}(X\_1,\dots,X\_r)\mid (X\_1,\dots,X\_r)\in E\_{d,r}\}.$$


Since $$G\_{A\_1,\dots,A\_r}(X\_1,\dots,X\_r)=G\_{A\_1,\dots,A\_r}(\lambda BX\_1B^{-1},\dots,\lambda BX\_rB^{-1})$$ whenever $B$ is invertible and $\lambda$ is a non-zero complex number, by the above proposition, we know that
$$\{G\_{A\_1,\dots,A\_r}(X\_1,\dots,X\_r)\mid (X\_1,\dots,X\_r)\in E\_{d,r}\}$$
$$=\{G\_{A\_1,\dots,A\_r}(X\_1,\dots,X\_r)\mid (X\_1,\dots,X\_r)\in K\_{d,r}\}.$$


Therefore, since $K\_{d,r}$ is compact, there is some $(Z\_1,\dots,Z\_r)\in K\_{d,r}$ with
$$G\_{A\_1,\dots,A\_r}(Z\_1,\dots,Z\_r)=\max\{G\_{A\_1,\dots,A\_r}(X\_1,\dots,X\_r)\mid (X\_1,\dots,X\_r)\in K\_{d,r}\}$$
$$=\rho\_{2,d}(A\_1,\dots,A\_r).$$


I have ran computations that maximize $G\_{A\_1,\dots,A\_r}$, and in these computations the maximum seems to actually be reached.


