# Can the supremum of this quotient of spectral radii be reached?

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.

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.

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.

• As a consequence, the function $\rho_{2,d}$ is actually continuous. Jun 10, 2022 at 16:16