Let $\rho(A)$ denote the spectral radius of a square matrix $A$. Let $r,d$ be positive integers. Let $X\_1,\dots,X\_r$ be $d\times d$-real matrices. Then do we necessarily have $$\rho(X\_1\dots X\_r)^{2/r}\leq \frac{d}{r}\cdot\rho(X\_1\otimes X\_1+\dots+X\_r\otimes X\_r)?$$


If $X\_1,\dots,X\_r$ are $d\times d$-real matrices with $d\leq r$ and
$$\rho(X\_1\dots X\_r)^{2/r}=\frac{d}{r}\cdot\rho(X\_1\otimes X\_1+\dots+X\_r\otimes X\_r),$$
then is $\text{rank}(X\_j)=1$ for $1\leq j\leq r$?


If $1\leq d\leq r$, then do there always exist $d\times d$-real matrices $X\_1,\dots,X\_r$ with
$$\rho(X\_1\dots X\_r)^{2/r}=\frac{d}{r}\cdot\rho(X\_1\otimes X\_1+\dots+X\_r\otimes X\_r)?$$


I have used a gradient descent algorithm to try to find counterexamples of this inequality, and the gradient descent algorithm was good at finding examples where
$$\rho(X\_1\dots X\_r)^{2/r}\approx\frac{d}{r}\cdot\rho(X\_1\otimes X\_1+\dots+X\_r\otimes X\_r),$$ but in all these examples, we have $\text{rank}(X\_j)=1$ for all $j$, and the gradient descent algorithm was unable to find any cases where $$\rho(X\_1\dots X\_r)^{2/r}>\frac{d}{r}\cdot\rho(X\_1\otimes X\_1+\dots+X\_r\otimes X\_r).$$



Is $\rho(X_1\dots X_r)^{2/r}\leq \frac{d}{r}\cdot\rho(X_1\otimes X_1+\dots+X_r\otimes X_r)$ for $d\times d$-real matries $X_1,\dots,X_r$?
The answer to all these questions is **Yes**. We shall answer this question in the more general case when $X\_1,\dots,X\_r$ are complex matrices (the inequality for this question looks a little bit different in the complex case since we need to apply conjugations and transposes). This answer shall be in the context of quantum channels. I recommend [The Theory of Quantum Information](https://cs.uwaterloo.ca/%7Ewatrous/TQI/) by John Watrous (2018) for more information on quantum information theory. Much of content of this answer is similar to the answers to [this related question](https://mathoverflow.net/q/423510/22277); fedja remarked that the idea behind the answer to that previous question also applies to this question, so fedja could have answered at least most of this question before I did.


A function $\mathcal{E}:M\_d(\mathbb{C})\rightarrow M\_d(\mathbb{C})$ is said to be positive if $\mathcal{E}(P)$ is positive semidefinite whenever $P$ is positive semidefinite. If $W$ is a finite dimensional vector space, then let $L(W)$ be the set of all homomorphisms from $W$ to $W$. We say that $\mathcal{E}:M\_d(\mathbb{C})\rightarrow M\_d(\mathbb{C})$ is completely positive if whenever $V$ is a finite dimensional complex Hilbert space, the mapping $\mathcal{E}\otimes 1\_V:L(\mathbb{C}^d\otimes V)\rightarrow L(\mathbb{C}^d\otimes V)$ is positive. We say that a linear mapping $\mathcal{E}:M\_d(\mathbb{C})\rightarrow M\_d(\mathbb{C})$ is trace preserving if $\text{Tr}(\mathcal{E}(A))=\text{Tr}(A)$ whenever $A\in M\_d(\mathbb{C})$. We say that a linear mapping $\mathcal{E}:M\_d(\mathbb{C})\rightarrow M\_d(\mathbb{C})$ is a quantum channel if it is both completely positive and trace preserving.


If $X\_1,\dots,X\_r$ are complex matrices, then define a mapping $\Phi(X\_1,\dots,X\_r):M\_{d}(\mathbb{C})\rightarrow M\_{d}(\mathbb{C})$ by letting
$$\Phi(X\_1,\dots,X\_r)(X)=X\_1XX\_1^\*+\dots+X\_rXX\_r^\*.$$ The completely positive mappings from $M\_d(\mathbb{C})$ to $M\_d(\mathbb{C})$ are precisely the mappings of the form
$\Phi(X\_1,\dots,X\_r)$, and the quantum channels $\mathcal{E}:M\_d(\mathbb{C})\rightarrow M\_d(\mathbb{C})$ are precisely the mappings of the form $\Phi(X\_1,\dots,X\_r)$ where $X\_1^\*X\_1+\dots+X\_r^\*X\_r=1\_d$.


If $\mathcal{E}$ is a quantum channel, then $\rho(\mathcal{E})=1$; quantum channels are a lot like stochastic matrices.


I claim that $\rho(X\_1\dots X\_r)^{2/r}\leq\frac{d}{r}\rho(\Phi(X\_1,\dots,X\_r))$ whenever $X\_1,\dots,X\_r$ are $d\times d$-complex matrices.


Let $O\_{d,r}$ be the set of all $(X\_1,\dots,X\_r)\in M\_d(\mathbb{C})^r$ such that
$\Phi(X\_1,\dots,X\_r)$ is not nilpotent. Let $E\_{d,r}$ be the set of all $(X\_1,\dots,X\_r)\in M\_d(\mathbb{C})^r$ such that
$\Phi(\lambda BX\_1B^{-1},\dots,\lambda BX\_rB^{-1})$ is a quantum channel for some complex $\lambda\neq 0$ and invertible matrix $B$. Then by [my answer to my other question](https://mathoverflow.net/a/423803/22277), $E\_{d,r}$ is a dense subset of $O\_{d,r}$, and clearly $O\_{d,r}$ is dense in $M\_d(\mathbb{C})^r$.


Observe that $$\frac{\rho(Y\_1\dots Y\_r)^{2/r}}{\rho(\Phi(Y\_1,\dots,Y\_r))}=\frac{\rho(\lambda BY\_1B^{-1}\dots \lambda BY\_rB^{-1})^{2/r}}{\rho(\Phi(\lambda BY\_1B^{-1},\dots,\lambda BY\_rB^{-1}))}$$ whenever $\lambda\neq 0$, $B$ is invertible, and $\Phi(Y\_1,\dots,Y\_r)$ is not nilpotent. Therefore, if $\rho(Z\_1\dots Z\_r)^{2/r}\leq\frac{d}{r}\rho(\Phi(Z\_1,\dots,Z\_r))$ whenever $\Phi(Z\_1,\dots,Z\_r)$ is a quantum channel, then $\rho(Y\_1\dots Y\_r)^{2/r}\leq\frac{d}{r}\rho(\Phi(Y\_1,\dots,Y\_r))$ whenever $(Y\_1,\dots,Y\_r)\in E\_{d,r},$
so because $E\_{d,r}$ is dense in $M\_{d}(\mathbb{C})^{r}$, we will be able to conclude that $\rho(X\_1\dots X\_r)^{2/r}\leq\frac{d}{r}\rho(\Phi(X\_1,\dots,X\_r))$ whenever $X\_1,\dots,X\_r\in M\_d(\mathbb{C})^r$.


If $\Phi(Z\_1,\dots,Z\_r)$ is a quantum channel, then $$d=\text{Tr}(1\_d)=\text{Tr}(Z\_1^\*Z\_1+\dots+Z\_r^\*Z\_r)=\|Z\_1\|^2\_2+\dots+\|Z\_r\|\_2^2$$ where $\|\cdot\|\_2$ denotes the Frobenius norm.


Therefore, by the arithmetic-geometric mean inequality, we have our main inequality, namely


$$\rho(Z\_1\dots Z\_r)^{2/r}\leq\|Z\_1\dots Z\_r\|\_\infty^{2/r}\leq(\|Z\_1\|\_\infty^2\dots\|Z\_r\|\_\infty^2)^{1/r}\leq(\|Z\_1\|\_2^2\dots\|Z\_r\|\_2^2)^{1/r}\leq\frac{\|Z\_{1}\|\_2^2+\dots+\|Z\_r\|\_2^2}{r}
=\frac{d}{r}=\frac{d}{r}\rho(\Phi(Z\_1,\dots,Z\_r)).$$


Furthermore, if $\rho(Z\_1\dots Z\_r)^{2/r}=\frac{d}{r}\rho(\Phi(Z\_1,\dots,Z\_r))$,
then $\|Z\_j\|\_\infty^2=\|Z\_j\|^2\_2=d/r$ for all $j$. However, The only way to get $\|Z\_{j}\|\_\infty=\|Z\_j\|\_2$ is if $\text{Rank}(Z\_j)=1$ since $\|W\|\_{\infty}=\|W\|\_2$ precisely when $\text{Rank}(W)=1$.


Therefore, for all $j$, there are column vectors $u\_j,v\_j$ such that $Z\_j=u\_jv\_j^\*$.


In this case, $$\|Z\_j\|\_2^2=\text{Tr}((u\_jv\_j^\*)(u\_jv\_j^\*)^\*)=\text{Tr}(u\_jv\_j^\*v\_ju\_j^\*)=
\text{Tr}(u\_j^\*u\_jv\_j^\*v\_j)=\|u\_j\|^2\cdot\|v\_j\|^2.$$ Therefore, $\|Z\_j\|\_2=\|u\_j\|\cdot\|v\_j\|=\sqrt{d/r}$.


$$\rho(Z\_1\dots Z\_r)=|\text{Tr}(Z\_1\dots Z\_r)|=|\text{Tr}(u\_1v\_1^\*\dots u\_rv\_r^\*)|
=|\text{Tr}(v\_1^\*u\_2\dots v\_r^\*u\_1)|$$
$$=|\langle v\_1,u\_2\rangle|\dots|\langle v\_r,u\_1\rangle|.$$


Since $\rho(Z\_1\dots Z\_r)=\|Z\_1\|\_\infty\dots\|Z\_r\|\_\infty$, we have
$$|\langle v\_1,u\_2\rangle|\dots|\langle v\_r,u\_1\rangle|=\|u\_1\|\cdot\|v\_1\|\dots \|u\_r\|\cdot\|v\_r\|.$$
This means that there are constants $\gamma\_1,\dots,\gamma\_r$ where
$v\_j=\overline{\gamma\_j}\cdot u\_{j+1}$. Therefore,
$Z\_j=u\_jv\_j^\*=u\_j\gamma\_j u\_{j+1}^\*=\gamma\_j u\_ju\_{j+1}^\*$.


Now, set $w\_j=\|u\_j\|\cdot v\_{j}=\|u\_j\|\cdot\overline{\gamma\_j}u\_{j+1}.$


Then $$1\_d=\sum\_{j=1}^{r}Z\_j^\*Z\_j=\sum\_{j=1}^{r}(u\_jv\_j^\*)^\*u\_jv\_j^\*=\sum\_{j=1}^{r}v\_ju\_j^\*u\_jv\_j^\*=\sum\_{j=1}^{r}\|u\_{j}\|^2\cdot v\_jv\_j^\*=\sum\_{j=1}^{r}w\_jw\_j^\*.$$


Observe that $1\_d=\sum\_{j=1}^{r}w\_jw\_j^\*$ precisely when $x=\sum\_{j=1}^{r}w\_{j}\cdot\langle x,w\_j\rangle$. We say that a tuple of vectors $(h\_1,\dots,h\_r)$ is a normalized tight frame if $1\_d=\sum\_{j=1}^{r}h\_jh\_j^\*$. We say that a normalized tight frame is equal-norm if $\|h\_1\|=\dots=\|h\_r\|$. Observe that $\|w\_j\|^2=d/r$ for all $j$, so $(w\_1,\dots,w\_r)$ is an equal-norm tight frame.


One can also obtain tuples $(Z\_1,\dots,Z\_r)$ where $\Phi(Z\_1,\dots,Z\_r)$ is a quantum channel and where $\rho(Z\_1,\dots,Z\_r)^{2/r}=d/r$ from equal-norm tight frames. Suppose that $(w\_1,\dots,w\_r)\in(\mathbb{C}^d)^r$ is an equal-norm tight frame. Then we necessarily have $\|w\_j\|^2\_2=d/r$ for all $j$. Now, suppose that $|\alpha\_j\beta\_j|^2=r/d$ for $1\leq j\leq r$. Let $v\_j=\alpha\_jw\_j,u\_{j+1}=\beta\_{j+1}w\_j$, and as always, set $Z\_j=u\_jv\_j^\*$ for $1\leq j\leq r$. Then the reader can verify that $\Phi(Z\_1,\dots,Z\_r)$ is a quantum channel with $\rho(Z\_1,\dots,Z\_r)^{2/r}=\frac{d}{r}$ and every quantum channel $\Phi(Z\_1,\dots,Z\_r)$ with $\rho(Z\_1,\dots,Z\_r)^{2/r}=\frac{d}{r}$ is of this form.


The existence of equal-norm tight frames $(w\_1,\dots,w\_r)\in(\mathbb{R}^d)^r$ is already a known result. For example, the existence of such frames is Corollary 7.1 in [An Introduction to Finite Tight Frames](https://www.math.auckland.ac.nz/%7Ewaldron/Preprints/Frame-book/draft26-7.17.pdf) by Shayne F. D. Waldron.


By applying continuity, we can actually show that $\text{Rank}(X\_j)=1$ even if we drop the assumption that $\Phi(X\_1,\dots,X\_r)$ is a quantum channel.


Let $f:O\_{d,r}\rightarrow\mathbb{R}$ be the function where
$f(X\_1,\dots,X\_r)=\frac{\rho(X\_1\dots X\_r)^{2/r}}{\rho(\Phi(X\_1,\dots,X\_r))}.$


Let $g^-:M\_d(\mathbb{C})\rightarrow\mathbb{R}$ be the function where
$g^-(X)=|\lambda\_1\lambda\_2|$ where $\lambda\_1,\dots,\lambda\_d$ are the eigenvalues of $X$ arranged in an order so that $|\lambda\_1|\geq|\lambda\_2|\geq\dots\geq|\lambda\_d|$.


Define $g:O\_{d,r}\rightarrow\mathbb{R}$ by letting $$g(X\_1,\dots,X\_r)=\frac{g^-(X\_1)+\dots+g^-(X\_r)}{r\cdot\rho(\Phi(X\_1,\dots,X\_r))}.$$


Suppose that $(X\_1,\dots,X\_r)\in O\_{d,r}$ and $f(X\_1,\dots,X\_r)=\frac{d}{r}$. Then for each $\delta>0$, there is a neighborhood $U$ of $(X\_1,\dots,X\_r)$ where if
$(Y\_1,\dots,Y\_r)\in U$, then $f(Y\_1,\dots,Y\_r)>\frac{d}{r}-\delta$. Now, let $(Y\_1,\dots,Y\_r)\in U\cap E\_{d,r}$. Then there is some $\lambda$ and invertible $B$ where if we set $Z\_j=\lambda BY\_jB^{-1}$ for all $j$, then $\Phi(Z\_1,\dots,Z\_r)$ is a quantum channel. Now, let $\lambda\_{j,1},\dots,\lambda\_{j,d}$ be the eigenvalues of $Z\_j$ ordered in a way so that $|\lambda\_{j,1}|\geq|\lambda\_{j,2}|\geq\dots\geq|\lambda\_{j,d}|$. Let $\sigma\_{j,1}\geq\dots\geq\sigma\_{j,d}$ denote the singular values of $Z\_j$.


Then
$$\frac{d}{r}-\delta<f(Y\_1,\dots,Y\_r)=f(Z\_1,\dots,Z\_r)=\rho(Z\_1\dots Z\_r)^{2/r}
\leq(\sigma\_{1,1}^2\dots\sigma\_{r,1}^2)^{1/r}$$
$$\leq\frac{\sigma\_{1,1}^2+\dots+\sigma\_{r,1}^2}{r}
\leq\frac{(\sigma\_{1,1}^2+\sigma\_{1,2}^2)+\dots+(\sigma\_{r,1}^2+\sigma\_{r,2}^2)}{r}$$
$$\leq\frac{\|Z\_1\|\_2^2+\dots+\|Z\_r\|\_2^2}{r}=\frac{d}{r}.$$


We conclude that $\sigma\_{j,1}^2\leq\frac{d}{r}$ and $\sigma\_{j,2}^2\leq\delta$. Therefore, by Weyl's inequality, we have
$$|\lambda\_{j,1}\lambda\_{j,2}|\leq\sigma\_{i,1}\sigma\_{j,2}\leq\sqrt{\frac{d\delta}{r}}.$$


Therefore, we have
$$g(Y\_1,\dots,Y\_r)=g(Z\_1,\dots,Z\_r)\leq\sqrt{\frac{d\delta}{r}}.$$


We therefore conclude that $g(X\_1,\dots,X\_r)=0$ which means that
$\text{Rank}(X\_j)\leq 1$ for $1\leq j\leq r$.


