If $A$ is a matrix, then let $\rho(A)$ denote the spectral radius of $A$. If $A=(a_{i,j})_{i,j}$, then let $\overline{A}=(\overline{a_{i,j}})_{i,j}$.

Suppose that $A_1,\dots,A_r\in M_{n}(\mathbb{C})$ are complex matrices where $A_1,\dots,A_r$ has no non-trivial invariant subspace. Let $(X_1,\dots,X_r)\in M_{d}(\mathbb{C})^r$ be complex matrices such that the quantity $$\frac{\rho(A_1\otimes X_1+\dots+A_r\otimes X_r)}{\rho(X_1\otimes\overline{X_1}+\dots+X_r\otimes\overline{X_r})^{1/2}}$$ is maximized. If $d\leq n$, then is $\text{Rank}(X_j)\leq\text{Rank}(A_j)$ for $1\leq j\leq r$?

My computer calculations suggest that the answer to this question is yes. This answer and this answer give special cases where the answer to this question is yes.

If we drop the assumption that $(A_1,\dots,A_r)$ has no invariant subspace, then the answer to this question is no (as a trivial example, let $A_1=\dots=A_r=0_d$ and let $X_1,\dots,X_r$ be arbitrary matrices in $M_d(\mathbb{C})^r$; one can come up with less trivial examples as well).

  • $\begingroup$ My computer experiments indicate that the optimal value of $(X_1,\dots,X_r)$ is where $X_i=\overline{UA_iV}$ for some matrices $U,V$ where $UV$ is the identity map. In this case, we always have $\text{Rank}(X_i)\leq\text{Rank}(A_i)$. $\endgroup$ Jan 13 at 21:21


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