Let $V$ be a complex finite dimensional inner product space. If $A\_{1},\dots,A\_{n}:V\rightarrow V$ are linear operators, then let $\Phi(A\_{1},\dots,A\_{n}):L(V)\rightarrow L(V)$ be the superoperator defined by letting $\Phi(A\_{1},\dots,A\_{n})(X)=A\_1XA\_1^\*+\dots+A\_nXA\_n^\*$. The completely positive superoperators from $L(V)$ to $L(V)$ are precisely the mappings of the form $\Phi(A\_{1},\dots,A\_{n})$.


If $A$ is an operator, then let $\rho(A)$ denote the spectral radius of $A$. Then


$$\rho(A\_{1}\otimes B\_{1}+\dots+A\_{n}\otimes B\_{n})\leq\rho(\Phi(A\_{1},\dots,A\_{n}))^{1/2}\rho(\Phi(B\_{1},\dots,B\_{n}))^{1/2}.$$


Is there a characterization of the systems of linear operators $((A\_{1},\dots,A\_{n}),(B\_{1},\dots,B\_{n}))$ for which
$$\rho(A\_{1}\otimes B\_{1}+\dots+A\_{n}\otimes B\_{n})=\rho(\Phi(A\_{1},\dots,A\_{n}))^{1/2}\rho(\Phi(B\_{1},\dots,B\_{n}))^{1/2}?$$



When does the Cauchy-Schwarz inequality for spectral radii of tensor products become equality?
**Yes.** We have a necessary and sufficient characterization for when the Cauchy-Schwarz inequality becomes equality.


For this post, $U,V,W$ shall denote finite dimensional complex Hilbert spaces. Suppose that $A\_1,\dots,A\_n:U\rightarrow U,B\_1,\dots,B\_n:V\rightarrow V$ are linear operators. Then define an operator $\Gamma(A\_1,\dots,A\_n;B\_1,\dots,B\_n):L(V,U)\rightarrow L(V,U)$ by letting $$\Gamma(A\_1,\dots,A\_n;B\_1,\dots,B\_n)(X)=A\_1XB\_1^\*+\dots+A\_nXB\_n^\*.$$


Observe that $\Gamma(A\_1,\dots,A\_n;B\_1,\dots,B\_n)$ is similar to $A\_1\otimes \overline{B\_1}+\dots+A\_n\otimes\overline{B\_n}$ (Here, $\overline{C}=(C^{T})^{\*}=(C^\*)^T$, so $\overline{C}$ is the matrix obtained by replacing every entry in $C$ with its complex conjugate), so the spectral radius Cauchy-Schwarz inequality
$$\rho(\Gamma(A\_1,\dots,A\_n;B\_1,\dots,B\_n))
\leq\rho(\Phi(A\_1,\dots,A\_n))^{1/2}\cdot\rho(\Phi(B\_1,\dots,B\_n))^{1/2}$$
hold in all cases. It is not too hard to prove the spectral radius Cauchy-Schwarz inequality by using the conventional Cauchy-Schwarz inequality and the characterization of $\rho(\Phi(A\_1,\dots,A\_n))^{1/2}$ given in the 1998 paper The $p$-norm joint spectral radius for even integers by Ding-Xuan Zhou.


Observe that if there is a $\lambda$ and an invertible $B$ with $B\_j=\lambda BA\_jB^{-1}$ for $1\leq j\leq n$, then
$$\rho(\Gamma(A\_1,\dots,A\_n;B\_1,\dots,B\_n))
=\rho(\Phi(A\_1,\dots,A\_n))^{1/2}\cdot\rho(\Phi(B\_1,\dots,B\_n))^{1/2}.$$


I claim that when $(A\_1,\dots,A\_n)$ and $(B\_1,\dots,B\_n)$ have no invariant subspaces, this is the only way in which the spectral radius Cauchy-Schwarz inequality becomes equality. By decomposing $(A\_1,\dots,A\_n)$ and $(B\_1,\dots,B\_n)$ according to their invariant subspaces, we obtain necessary and sufficient conditions for when the Cauchy-Schwarz inequality is actually an equality.


Recall that a linear operator $\mathcal{E}:L(U)\rightarrow L(V)$ is called positive if $\mathcal{E}(P)$ is positive semidefinite whenever $P$ is positive semidefinite. A linear operator $\mathcal{E}:L(U)\rightarrow L(V)$ is called completely positive if
$\mathcal{E}\otimes 1\_W:L(U\otimes W)\rightarrow L(V\otimes W)$ is positive whenever $W$ is a finite dimensional Hilbert space. We say that an operator $\mathcal{E}:L(U)\rightarrow L(V)$ is trace preserving if $\text{Tr}(\mathcal{E}(X))=\text{Tr}(X)$ for all $X\in L(U)$. An operator $\mathcal{E}:L(U)\rightarrow L(V)$ is said to be a quantum channel if $\mathcal{E}$ is both trace preserving and completely positive.


The completely positive mappings from $L(V)$ to $L(V)$ are precisely the mappings of the form $\Phi(A\_1,\dots,A\_n)$. Observe that any linear mapping from $L(V,U)$ to $L(V,U)$ is of the form $\Gamma(A\_1,\dots,A\_n;B\_1,\dots,B\_n)$. It is not too hard to show using the Frobenius inner product that the mapping $\Gamma(A\_1,\dots,A\_n;B\_1,\dots,B\_n)$ is trace preserving if and only if $A\_1^\*B\_1+\dots+A\_n^\*B\_n=1\_V$. In particular, the quantum channels are precisely the mappings of the form $\Phi(A\_1,\dots,A\_n)$ where $A\_1^\*A\_1+\dots+A\_n^\*A\_n=1\_V$.


Observe that if $\mathcal{E}$ is a quantum channel, then $\rho(\mathcal{E})=1$.


Let $E\_{U,n}$ be the collection of tuples $(A\_1,\dots,A\_n)\in L(U)^n$ such that there is some complex number $\lambda$ and invertible $B$ such that if $B\_j=\lambda BA\_jB^{-1}$ for $1\leq j\leq n$, then $\Phi(B\_1,\dots,B\_n)$ is a quantum channel. [By this answer](https://mathoverflow.net/a/423803/22277), $L(U)^n\setminus E\_{U,n}$ is a quite small set whenever $n>1$. In particular, if $(A\_1,\dots,A\_n)$ has no-invariant subspace, then $(A\_1,\dots,A\_n)\in E\_{U,n}$.


Theorem: Suppose that $(A\_1,\dots,A\_n)\in L(U)^n,(B\_1,\dots,B\_n)\in L(V)^n$ have no invariant subspace. Then $$\rho(\Gamma(A\_1,\dots,A\_n;B\_1,\dots,B\_n))=\rho(\Phi(A\_1,\dots,A\_n))^{1/2}\cdot\rho(\Phi(B\_1,\dots,B\_n))^{1/2}$$ if and only if there is some invertible matrix $C$ along with some complex number $\eta\neq 0$ where
$A\_j=\eta CB\_jC^{-1}$ for $1\leq j\leq n$.


Proof: Since $(A\_1,\dots,A\_n),(B\_1,\dots,B\_n)$ have no invariant subspace, there are non-zero complex numbers $\mu,\nu$ along with invertible matrices $A,B$ where if we set $R\_j=\mu AA\_jA^{-1},S\_j=\nu BB\_jB^{-1}$ for $1\leq j\leq n$, then $\mathcal{E}=\Phi(R\_1,\dots,R\_n),\mathcal{F}=\Phi(S\_1,\dots,S\_n)$ are quantum channels. Let $W$ be a complex inner product space with orthonormal basis $(e\_1,\dots,e\_n).$ Now, let $R,S\in L(V,V\otimes W)$ be the operators defined by letting
$R=\sum\_{j=1}^{n}R\_j\otimes e\_j,S=\sum\_{j=1}^{n}S\_j\otimes e\_j$. Then
$$\mathcal{E}(X)=\text{Tr}\_{W}(RXR^\*),\mathcal{F}(X)=\text{Tr}\_{W}(SXS^\*).$$


Define a mapping $\mathcal{G}$ by setting $$\mathcal{G}(X)=\text{Tr}\_{W}(RXS^\*)=\sum\_{k=1}^nR\_kXS\_k^\*.$$


Since $R\_1^\*R\_1+\dots+R\_n^\*R\_n=S\_1^\*S\_1+\dots+S\_n^\*S\_n=1\_V$, the mappings $R,S$ are isometries.


Now, let $\lambda$ be an eigenvalue of $\mathcal{G}$. Then there is some
eigenvector $X$ with $\mathcal{G}(X)=\lambda X$. Now perform a polar decomposition of $X$ to write $X=HP$ where $H$ is an isometry and $P$ is a positive semidefinite matrix. Therefore, we have
$$\lambda HP=\lambda X=\mathcal{G}(X)=\text{Tr}\_{W}(RXS^\*)=\text{Tr}\_W(RHPS^\*).$$ Now, set $T=(H^\*\otimes 1\_W)RH$. Then
$$\lambda P=\lambda H^\*HP=H^\*\text{Tr}\_W(RHPS^\*)=\text{Tr}\_W((H^\*\otimes 1\_W)RHPS^\*)=\text{Tr}\_W(TPS^\*).$$ Now let $P=\sum\_{k=1}^{n}\sigma\_ke\_ke\_k^\*$. Then $\text{Tr}(\lambda P)=\lambda\cdot\sum\_{k=1}^n\sigma\_k$. Therefore, we have
$$\text{Tr}(\lambda P)=\text{Tr}(\text{Tr}\_W(TPS^\*))=\text{Tr}(TPS^\*)=\sum\_{k=1}^{n}\sigma\_k\text{Tr}(Te\_ke\_k^\*S^\*)=\sum\_{k=1}^n\sigma\_k\text{Tr}(Te\_k(Se\_k)^\*)=\sum\_{k=1}^n\sigma\_k\langle Te\_k,Se\_k\rangle.$$


Therefore, since $$\lambda\cdot\sum\_{k=1}^n\sigma\_k=\sum\_{k=1}^n\sigma\_k\langle Te\_k,Se\_k\rangle,$$ we know that $|\lambda|\leq 1$. Furthermore, if $|\lambda|=1$, then we know that $Te\_k=\lambda Se\_k$ whenever $\sigma\_k>0$.
Therefore, we have $T|\_{\text{Im}(P)}=\lambda\cdot S|\_{\text{Im}(P)}$. In this case, we have $\lambda P=\text{Tr}\_W(TPS^\*)=\text{Tr}\_W(\lambda SPS^\*)$, so $P=\text{Tr}\_W(SPS^\*)$. Since
$P=\text{Tr}\_W(SPS^\*)$ and since $(S\_1,\dots,S\_n)$ has no invariant subspace, we know that $P$ is positive semidefinite. Therefore, $(H^\*\otimes 1\_W)RH=T=\lambda S$.


Now, if $v\in V$, then $\|RH v|=\|v\|=\|\lambda Sv\|=\|(H^\*\otimes 1\_W)RHv\|$. Therefore, if $v\in V$, then $RHv\in\text{Im}(H)\otimes 1\_W$. Therefore, since $\text{Im}(H)$ is a non-trivial invariance subspace of $U$, we know that $\text{Im}(H)=U$. Therefore, $H$ is a unitary operator.


Thus, $H^\*R\_jH=\lambda S\_j$ for $1\leq j\leq n$. Thus, $H^\*\mu AA\_jA^{-1}H=\lambda \nu BB\_jB^{-1}$. We conclude that
$$A\_j=\mu^{-1}\lambda\nu A^{-1}HBB\_jB^{-1}H^{-1}A=\mu^{-1}\lambda\nu A^{-1}HBB\_j(A^{-1}HB)^{-1}.$$


Q.E.D.


Theorem: Suppose that $A\_1,\dots,A\_n:U\rightarrow U,B\_1,\dots,B\_n:V\rightarrow V$ be linear operators. Then assign $U,V$ bases so that
$$A\_j=\begin{bmatrix}
A\_{j,1,1}&\dots&A\_{j,1,u}\\
\vdots&\ddots&\vdots\\
A\_{j,u,1}&\dots&A\_{j,u,u}
\end{bmatrix}$$
and
$$B\_j=\begin{bmatrix}
B\_{j,1,1}&\dots&B\_{j,1,v}\\
\vdots&\ddots&\vdots\\
B\_{j,v,1}&\dots&B\_{j,v,v}
\end{bmatrix}$$
and where for each $\alpha,\beta$ each of the $n$ matrices $A\_{1,\alpha,\beta},\dots,A\_{n,\alpha,\beta}$ have the same dimensions,
for each $\alpha,\beta$, each of the $n$ matrices $B\_{1,\alpha,\beta},\dots,B\_{n,\alpha,\beta}$ have the same dimensions, and where $A\_{j,\alpha,\beta}=0$ whenever $\alpha>\beta$, and where $B\_{j,\alpha,\beta}=0$ whenever $\alpha>\beta$, and where for $1\leq\alpha\leq u$, the matrices
$(A\_{1,\alpha,\alpha},\dots,A\_{n,\alpha,\alpha})$ have no non-trivial invariant subspace, and where if $1\leq\beta\leq v$, the matrices
$(B\_{1,\beta,\beta},\dots,B\_{n,\beta,\beta})$ have non-trivial no invariant subspace either.
Then $\rho(\Gamma(A\_1,\dots,A\_n;B\_1,\dots,B\_n))=\rho(\Phi(A\_1,\dots,A\_n))^{1/2}\cdot\rho(\Phi(B\_1,\dots,B\_n))^{1/2}$ if and only if there are $\alpha,\beta$ with $1\leq\alpha\leq u,1\leq\beta\leq v$ and where


1. $\rho(\Phi(A\_1,\dots,A\_n))=\rho(\Phi(A\_{1,\alpha,\alpha},\dots,A\_{n,\alpha,\alpha}))$
2. $\rho(\Phi(B\_1,\dots,B\_n))=\rho(\Phi(A\_{1,\beta,\beta},\dots,A\_{n,\beta,\beta}))$, and
3. There is an invertible matrix $C$ and some complex number $\lambda\neq 0$ such that $A\_{j,\alpha,\beta}=\lambda CB\_{j,\alpha,\beta}C^{-1}$ for $1\leq j\leq n$.


**A few observations:**


In this answer, we actually have a couple of different proofs of the spectral radius Cauchy-Schwarz inequality. To prove the Cauchy-Schwarz inequality, we need to prove that $\rho(\Gamma(R\_1,\dots,R\_n;S\_1,\dots,S\_n))\leq 1$ whenever $\Phi(R\_1,\dots,R\_n),\Phi(S\_1,\dots,S\_n)$ are quantum channels, and the general case will follow from the fact that $E\_{V,n}$ is dense in $L(V)^n$ whenever $n>1$. We have already shown that $\rho(\Gamma(R\_1,\dots,R\_n;S\_1,\dots,S\_n))\leq 1$, but there is another way of showing this using the induced trace norm.


If $\mathcal{H}:L(V)\rightarrow L(V)$, then define the induced trace norm of $\mathcal{H}$ to be $\|\mathcal{H}\|\_1=\max\{\|\mathcal{H}(X)\|\_1:\|X\|\_1\leq 1\}.$ Recall that if $R,S$ are isometries and $A$ is a complex matrix, then $\|RAS^\*\|\_1=\|A\|\_1$ whenever the matrix multiplication exists. Also, recall that $\|\text{Tr}\_{V}(X)\|\_1\leq\|X\|\_1$ whenever this inequality makes sense.


We have
$$\|\Gamma(R\_1,\dots,R\_n;S\_1,\dots,S\_n)(X)\|\_1=\|\text{Tr}\_{W}(RXS^\*)\|\_1\leq\|RXS^\*\|\_1=\|X\|\_1.$$
Therefore, $\|\Gamma(R\_1,\dots,R\_n;S\_1,\dots,S\_n)\|\_1\leq 1$, so
$\rho(\Gamma(R\_1,\dots,R\_n;S\_1,\dots,S\_n))\leq 1$ since the induced trace norm is submultiplicative.


**Empirical verification**


I have empirically verified using computer calculations that the conclusions that we have made are reasonable. Define a fitness function $F:M\_d(\mathbb{C})^n\times M\_d(\mathbb{C})^n\rightarrow\mathbb{R}$ by letting
$$F(A\_1,\dots,A\_n;B\_1,\dots,B\_n)=\frac{\rho(\Gamma(A\_1,\dots,A\_n;B\_1,\dots,B\_n))}{\rho(\Phi(A\_1,\dots,A\_n))^{1/2}\rho(\Phi(B\_1,\dots,B\_n))^{1/2}}.$$


One can maximize the value of $F(A\_1,\dots,A\_n;B\_1,\dots,B\_n)$ using gradient ascent to obtain examples of tuples $(A\_1,\dots,A\_n;B\_1,\dots,B\_n)$ with
$F(A\_1,\dots,A\_n;B\_1,\dots,B\_n)\approx 1$, but in each of these examples, we always have a complex number $\lambda$ along with some invertible $B$ where
$B\_j\approx \lambda BA\_jB^{-1}$ for $1\leq j\leq n$.


