Let $d,m$ be positive integers. Suppose that $A\_{i,j}$ is a $d\times d$-matrix with real entries whenever $i,j\in\{1,\dots m\}$.


Let $A$ be the $dm\times dm$ matrix that can be written as a block matrix as
$$A=\begin{bmatrix}
A\_{1,1} & \cdots & A\_{1,m} \\
\vdots & \ddots & \vdots \\
A\_{m,1} & \dots & A\_{m,m} 
\end{bmatrix}.$$


1. Is $$\rho(A)^2\leq\min(d,m)\cdot \rho\left(\sum\_{i,j}A\_{i,j}\otimes A\_{i,j}\right)?$$
2. For all $d,m$, can we select real matrices $A\_{i,j}$ so that
$$\rho(A)^2=\min(d,m)\cdot \rho\left(\sum\_{i,j}A\_{i,j}\otimes A\_{i,j}\right)>0?$$
3. If $$\rho(A)^2=\min(d,m)\cdot \rho\left(\sum\_{i,j}A\_{i,j}\otimes A\_{i,j}\right)>0,$$
and $d\leq m$ then is $\text{Rank}(A\_{i,j})\leq 1$ for each $i,j$?


My computer calculations suggest that the answer to these questions is 'yes'. This is a follow up of [this question](https://mathoverflow.net/questions/422925/is-rhox-1-dots-x-r2-r-leq-fracdr-cdot-rhox-1-otimes-x-1-dotsx-r-o). This question can be generalized to the case when each $A\_{i,j}$ is a complex matrix. One can rephrase the complex version of this question in terms of the completely depolarizing channel in quantum information theory.



An inequality for the spectral radius of block matrices
I'll answer #1 only, leaving the rest to you or someone else to figure out. The answer is affirmative.


Dropping the trivial case $m=1$, we may assume by density that we are in the generic position, i.e., that $A\_{ij}^T\ne 0$ and together have no non-trivial invariant subspace (this is just to avoid considering degenerate cases, which can also be done using dimension reduction but I prefer not to bother with this extra technicality).


Consider the mapping $X\mapsto F(X)=\sum\_{i,j}A\_{ij}XA\_{ij}^T$ on the cone $K$ of symmetric non-negative definite $d\times d$ matrices $X$. Notice that if $X\ne 0$ and $F(X)=0$, then all $A\_{ij}^T$ map the entire space to the kernel of $X$, which contradicts our generic position assumption. Thus the mapping $X\mapsto \frac 1{\operatorname{Tr}F(X)}F(X)$ is well-defined on the convex compact $Q=K\cap\{X:\operatorname{Tr}X=1\}$ and maps $Q$ to itself. So, by the Brouwer fixed point theorem, the linear mapping $F$ has an eigenvector in the cone with positive eigenvalue, which we can normalize to $1$ by scaling $A$, i.e., we have
$$
\sum\_{i,j}A\_{ij}XA\_{ij}^T=X\,,
$$
which, clearly, implies $\rho(\sum\_{i,j}A\_{ij}\otimes A\_{ij})\ge 1$.


I now claim that $X$ is non-degenerate. Indeed, otherwise the kernel of $X$ would be non-trivial and invariant under all $A\_{ij}^T$.


Now for a vector $y=(y\_1,\dots,y\_m)\in\mathbb R^{md}$ define its norm by
$$
\|y\|^2=\sum\_j|X^{-1/2}y\_j|^2
$$
where $|\cdot|$ stands for the usual Euclidean norm in $\mathbb R^d$. Let $y$ be a unit vector (in this norm) on which the induced operator norm of $A$ is attained.


Then $\|Ay\|^2=\sum\_{i} \langle X^{-1/2}(Ay)\_i,\xi\_i\rangle^2$ for some unit vectors $\xi\_i\in\mathbb R^d$. Now use Cauchy-Schwarz:
$$
\langle X^{-1/2}(Ay)\_i,\xi\_i\rangle^2= \left[\sum\_j \langle X^{-1/2}A\_{ij}y\_j,\xi\_i\rangle\right]^2
\\
=
\left[\sum\_j \langle X^{-1/2}y\_j,X^{1/2}A\_{ij}^TX^{-1/2}\xi\_i\rangle\right]^2
\\
\le \left[\sum\_j |X^{-1/2}y\_j|^2\right]\left[\sum\_j |X^{1/2}A\_{ij}^TX^{-1/2}\xi\_i|^2\right]
$$
This allows one to estimate the induced operator norm of $A$ and, thereby $\rho(A)$ by
$$
\rho(A)^2\le\|A\|^2\le
\sum\_{i,j}|X^{1/2}A\_{ij}^TX^{-1/2}\xi\_i|^2
\\
=\sum\_{i,j}\langle X^{-1/2}A\_{ij}XA\_{ij}^TX^{-1/2}\xi\_i,\xi\_i\rangle\,.
$$
Now let $L$ be the linear subspace of $\mathbb R^d$ spanned by all vectors $\xi\_i$. Obviously, $\operatorname{dim}L\le\min(d,m)$. Let $P$ be the orthogonal projection to $L$. Then for every non-negative definite operator $S$ in $\mathbb R^d$, we have $\langle S\xi\_i,\xi\_i\rangle\le\operatorname{Tr} (PSP)$, which yields the bound
$$
\sum\_{i,j} \operatorname{Tr}(PX^{-1/2}A\_{ij}XA\_{ij}^TX^{-1/2}P)
= \operatorname{Tr}\left(\sum\_{i,j} PX^{-1/2}A\_{ij}XA\_{ij}^TX^{-1/2}P\right)
\\
=
\operatorname{Tr}\left(PX^{-1/2}\left[\sum\_{i,j} A\_{ij}XA\_{ij}^T\right]X^{-1/2}P\right)=
\operatorname{Tr}(PX^{-1/2}XX^{-1/2}P)
\\
=\operatorname{Tr} P=\operatorname{dim}L\le\min(d,m)
$$
as required.


