I’m running into difficulty computing operator exponentials for “block matrices” of Hilbert space operators. It would be extremely useful to be able to diagonalize these block matrices. While the following example isn’t exactly the case I’m interested in, understanding it would be of great help to me.
Suppose $U,V$ are real Hilbert spaces and we have bounded linear maps $T,S:U \rightarrow V$ with adjoints $T^*, S^* $. Let $F:U \oplus U \rightarrow U \oplus U$ be the map $$ F \begin{pmatrix} X \\ Y \end{pmatrix} = \begin{pmatrix} T^*(TX - SY) \\ S^*(SY - TX) \end{pmatrix}.$$ I am interested in “diagonalizing” the operator $F$. It is pretty easy to see that $$F = \begin{pmatrix} T^* & 0 \\ 0 & S^* \end{pmatrix} \begin{pmatrix} id & -id \\ -id & id \end{pmatrix} \begin{pmatrix} T & 0 \\ 0 & S \end{pmatrix}$$ Where we think of “multiplication” as composition, and $id$ is the identity map on $U$. It would be nice to be able to use the diagonalization of the inner matrix to get a diagonalization for $F$.
For example, if $T = S$, then we can easily make the identification $U \oplus U \cong \mathbb{R}^2 \otimes U$, and write this as a product of pure tensors. Let $A = \begin{pmatrix} 1 & -1 \\ -1 & 1 \end{pmatrix}$, and let $A = QDQ^{-1}$ denote its diagonalization. Writing id for the identity map on $U$ and $I$ for $2 \times 2$ real identity matrix, we could write: $$\begin{align*} F & = (I \otimes T^*) (A \otimes id) (I \otimes T) \\ & = A \otimes T^*T \\ & = QDQ^{-1} \otimes T^*T \\ & = (Q \otimes id) (D \otimes T^*T) (Q^{-1} \otimes id) \\ & = (Q \otimes id) (D \otimes T^*T) (Q \otimes id)^{-1} \end{align*}$$ This is a “diagonalization” of $F$ in some suitable sense. But I don’t want to make the assumption that $T=S$ (and hence can write everything as a pure tensor). Is there a way to handle this more general case where $T \neq S$?