I want to diagonalize a linear operator acting on a rectangular $n\times m$ matrix $X$. The linear equation I'm interested is of the form $ PX-XQ= \Lambda X$, where $P$ is an $n\times n$ diagonal matrix and $Q$ is an $m\times m$ diagonal matrices and $\Lambda$ is an unknown eigenvalue. $P$ and $Q$ also satisfy a condition that they are generated by a single rectangular matrix $n\times m$ matrix $G$: $P= \alpha G G^\dagger$ and $Q=\beta G^\dagger G$ where the dagger $G^\dagger$ denotes the conjugate transpose of $G$ and $\alpha$ and $\beta$ are positive real numbers.
I know this problem can be rewritten as a rather large linear equation by expressing $x$ and an $n^2$ dimensional vector, but because $P$ and $Q$ are related to each other I was wondering if there is any relationship between the eigenvalues of $P,Q$ and $\Lambda$.
The most expedient way to diagonalize the operator $f(X) = PX - XQ$ is to consider the singular value decomposition of $G$. Suppose that $G = U \Sigma V^\dagger$, with $U,V$ unitary and $\Sigma$ non-negative and diagonal of size $n \times m$. We have $P = \alpha U\Sigma\Sigma^\dagger U^\dagger$ and $Q = \beta V \Sigma^\dagger \Sigma V^\dagger$. Let $d,\phi:\Bbb C^{n \times m} \to \Bbb C^{n \times m}$ denote the maps $$ d(X) = \alpha (\Sigma \Sigma^\dagger) X - \beta X(\Sigma^\dagger \Sigma), \quad \phi(X) = U XV^\dagger. $$ We can "diagonlize" $f$ by writing it in the form $f = \phi \circ d \circ \phi^{-1}$.
We can think of $d$ as a diagonal operator in the following sense: via vectorization, we find that the matrix of $d$ (relative to the column-major "canonical basis" $\mathcal B_1$ of $\Bbb C^{n \times m}$) is given by $$ [d]_{\mathcal B_1} = \alpha I_m \otimes (\Sigma \Sigma^\dagger) - \beta (\Sigma^\dagger \Sigma) \otimes I_n, $$ where $I_n$ denotes a size $n$ identity matrix and $\otimes$ denotes a Kronecker product.
If you prefer, we find that the matrix of $f$ is diagonal relative to the eigenbasis $$ \mathcal B_2 = \{u_iv_j^\dagger : 1 \leq i \leq n, 1 \leq j \leq m\} $$ where $u_i$ denotes the $i$th column of $U$ and $v_j$ denotes the $j$th column of $V$. Traditionally, one considers the tuples $(j,i)$ in lexicographical order. This eigenbasis is orthonormal relative to the Frobenius inner product, which is defined by $\langle A,B\rangle = \operatorname{tr}(A^\dagger B)$.