Consider a set of $n$ by $n$ unitary matrices $U_i$ and a set of arbitrary $m$ by $m$ matrices $J_i$, with $i=1,...,N$.
Consider the eigenvalues $\{ \lambda_{U \otimes J} \}$ of the operator $O_{U \otimes J} = \sum_{i=1}^N U_i \otimes J_i \otimes U^*_i \otimes J^*_i$ and the eigenvalues $\{ \lambda_J \}$ of the operator $O_J = \sum_{i=1}^N J_i \otimes J^*_i$.
Is $\max(\{ |\lambda_{U \otimes J}| \} ) \leq \max(\{ |\lambda_{J}| \} )$ true? How could I show this?
As a little background, I've been trying to argue that $\max(\{ |\lambda_{U \otimes J}| \} ) = \max(\{ |\lambda_{J}| \} )$. I've been able to show that $\{ \lambda_J \} \subset \{\lambda_{U \otimes J}\}$ by using eigenvectors of $O_J$ to construct eigenvectors of $O_{U \otimes J}$ (and hence that $\max(\{ |\lambda_{U \otimes J}| \} ) \geq \max(\{ |\lambda_{J}| \} )$).
I'm doing this to try to understand quantum channels better. Very roughly, I'm viewing $O_{U \otimes J}$ as a quantum channel acting on a bigger space than $O_{J}$, but whose Kraus operators reduce to those of $O_{J}$ on tracing out the extra degrees of freedom that $O_{U \otimes J}$ acts on. This is just a rough picture that I have in mind, however, and I'm also trying to understand eigenvalues inequalities more generally.
To make clear my conventions, by $M^*$ I mean the complex conjugate of a matrix $M$.
I also have a companion question that explores a similar idea but with some direct sums $\oplus$ in place of some of these tensor products $\otimes$ and deals with a slightly more general set of matrices.
Here I sketch a proof.
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 $\bar{O}_{U \otimes J}$ and $\bar{O}_{J}$ via a partial trace:
$$\text{Tr}_L[\bar{O}_{U \otimes J}(\rho)] = \bar{O}_J (\text{Tr}_L[\rho]) $$ 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 $\bar{O}_{U \otimes J}$: $$\bar{O}_{U \otimes J}(\rho) = \lambda \rho $$ Taking the partial trace of both side, the third statement above implies $$\bar{O}_{J}(\text{Tr}_L[\rho]) = \lambda \text{Tr}_L[\rho] $$ That is, we get that an eigenvector $\rho$ of $\bar{O}_{U \otimes J}$ becomes an eigenvector $\text{Tr}_L[\rho]$ of $\bar{O}_{J}$ with the exact same eigenvalue if $\text{Tr}_L[\rho]$ is nonvanishing.
The above steps are related to the question at hand. Consider the eigenspace of $\bar{O}_{U \otimes J}$ corresponding to the eigenvalue with the maximum absolute value, $\lambda_m$. If this eigenspace contains a $\rho_m$ of $\bar{O}_{U \otimes J}$ with $\text{Tr}_L[\rho_m]$ nonvanishing, then we can construct an eigenvector of $\bar{O}_J$ with the same eigenvalue, and hence $\max(\{ |\lambda_{U \otimes J}| \} ) \leq \max(\{ |\lambda_{J}| \} )$.
I claim that the eigenspace $\bar{O}_{U \otimes J}$ corresponding to maximum eigenvalue contains a matrix $\rho_m$ that is positive semi-definite (this is argued in this linked answer), and so has nonvanishing trace, and so has nonvanishing partial trace. Thus we are finished.