Consider a set of $n$ by $n$ matrices $A_i$ and a set of $m$ by $m$ matrices $B_i$, with $i=1,2,...,N$.
I'm interested in the following operators:
$$O_A = \sum_{i=1}^N A_i \otimes A_i^*$$ $$O_B = \sum_{i=1}^N B_i \otimes B_i^*$$ $$O_{A \oplus B} = \sum_{i=1}^N (A_i \oplus B_i) \otimes (A_i^* \oplus B_i^*)$$
Consider the sets of eigenvalues of these operators, $\{\lambda_A\}$,$\{\lambda_B\}$, and $\{\lambda_{A \oplus B}\}$ respectively.
Is $\max (\{ |\lambda_{A \oplus B}| \}) \leq\max (\{ |\lambda_{A}| \} \cup \{ |\lambda_{A}| \})$?
This is something of a counterpoint to a recent question that I asked involving only $\otimes$ and no $\oplus$ and which additionally constrained one of the sets of matrices to be unitary.
I am hoping to show that $\max (\{ |\lambda_{A \oplus B}| \}) = \max (\{ |\lambda_{A}| \} \cup \{ |\lambda_{A}| \})$. Note that for any eigenvector of $O_A$ or $O_B$, we can construct an eigenvector of $O_{A \oplus B}$ with the same eigenvalue, so we have that $\{\lambda_A\}\cup\{\lambda_B\} \subset \{\lambda_{A \oplus B}\}$. That is, we already have $\max (\{ |\lambda_{A \oplus B}| \}) \geq\max (\{ |\lambda_{A}| \} \cup \{ |\lambda_{A}| \})$.
To make clear my conventions, by $M^*$ I mean the complex conjugate of a matrix $M$. $\oplus$ is the direct sum.
Very roughly, I'm imagining the following picture. I introduce a continuous parameter $\alpha$ so that $A_i \to \alpha A_i$. For small $\alpha$, I believe the leading eigenvalue of $O_{A \oplus B}$ will be equal to the leading eigenvalue of $O_B$. For large $\alpha$, I believe the leading eigenvalue of $O_{A \oplus B}$ will be equal to the leading eigenvalue of $O_A$. At intermediate $\alpha$, can I ever have the leading eigenvalue of $O_{A \oplus B}$ be neither of the leading eigenvalues of $O_A$ or $O_B$? I want to say no, that the leading eigenvalue of $O_{A \oplus B}$ is constrained to match that of $O_A$ or $O_B$. However, this seems a strong constraint, so I'm open to counterexamples.
I sketched a proof to my related question, and I realized a variant of that proof could work for this question.
First, I want to note that for any $j$ by $j$ set of square matrices $\{K_i\}$, $O_K = \sum_{i=1}^N K_i \otimes K^*_i$ has the same eigenvalues as $\tilde{O}_K = \sum_{i=1}^N K^*_i \otimes K_i$. This is because we can write a similarity transformation between $O_K$ and $\tilde{O}_K$ using the swap operator.
Second, we can view the operator $\tilde{O}_K$ as instead acting on $j$ by $j$ matrices $\rho$ as:
$$\bar{O}_K(\rho) = \sum_{i=1}^N K_i \rho K_i^\dagger$$
Here I use the notation $^\dagger$ for Hermitian conjugation, and I use a $\bar O$ to denote this reshaping of $\tilde{O}$. This reshaping can be viewed in terms of the isomorphism given by vectorization. (In particular, the eigensystem of of $\bar{O}$ will match that of $\tilde{O}$ on reshaping the eigenvectors of $\bar{O}$ via vectorization.)
Third, we can relate $\bar{O}_{A \oplus B}$ to $\bar{O}_{A}$ and $\bar{O}_{B}$ via projection matrices. I will choose $P_L$ to project onto the subspace that $A$ acts on, and $P_R$ to project onto the subspace that $B$ acts on. These projection matrices will actually be rectangular -- I'll take $P_L$ to be $n$ by $m+n$, and I'll take $P_R$ to be $m$ by $m+n$. By inspection, $$P_L \bar{O}_{A \oplus B}(\rho) P_L^T = \bar{O}_A (P_L \rho P_L^T) $$ $$P_R \bar{O}_{A \oplus B}(\rho) P_R^T = \bar{O}_B (P_R \rho P_R^T) $$
Here $L$ stands for the left space, namely the one that $U$ and $U^\dagger$ act on. By using cyclicity of the partial trace on this space, we can cancel $U$ and $U^\dagger$.
Fourth, consider an eigenvector $\rho$ of $\bar{O}_{A \oplus B}$: $$\bar{O}_{A \oplus B}(\rho) = \lambda \rho $$ Sandwiching by projections on both side, the third statement above implies $$\bar{O}_{A}(P_L \rho P_L^T) = \lambda P_L \rho P_L^T $$ $$\bar{O}_{B}(P_R \rho P_R^T) = \lambda P_R \rho P_R^T $$
That is, we get that an eigenvector $\rho$ of $\bar{O}_{A \oplus B}$ gives us an an eigenvector $P_L \rho P_L^T$ of $\bar{O}_{A}$ with the exact same eigenvalue if $P_L \rho P_L^T$ is nonvanishing. Similarly, an eigenvector $\rho$ of $\bar{O}_{A \oplus B}$ gives us an an eigenvector $P_R \rho P_R^T$ of $\bar{O}_{B}$ with the exact same eigenvalue if $P_R \rho P_R^T$ is nonvanishing.
The above steps are related to the question at hand. Consider the eigenspace of $\bar{O}_{A \oplus B}$ corresponding to the eigenvalue with the maximum absolute value, $\lambda_m$. If this eigenspace contains a $\rho_m$ of $\bar{O}_{A \oplus B}$ with at least one of $P_L \rho P_L^T$ or $P_R \rho P_R^T$ nonvanishing, then we can respectively construct an eigenvector of $\bar{O}_A$ or $\bar{O}_B$ with the same eigenvalue, and hence $\max (\{ |\lambda_{A \oplus B}| \}) \leq\max (\{ |\lambda_{A}| \} \cup \{ |\lambda_{A}| \})$.
Note that the eigenspace $\bar{O}_{A \oplus B}$ corresponding to maximum eigenvalue indeed contains a matrix $\rho_m$ that has nonvanishing trace, which follows from this linked answer. A nonvanishing trace of $\rho_m$ implies that at least one of $P_L \rho_m P_L^T$ or $P_R \rho_m P_R^T$ is nonzero (as if they were both zero the trace would vanish), and so we are finished.