Operator Matrices

If $A$, $B$, $C$, and $D$ are pairwise commutative operators on a Hilbert space, then a necessary and sufficient condition that the operator matrix

be invertible is that the formal determinant $AD - BC$ be invertible.

If $C$ and $D$ commute, and if $D$ is invertible, then a necessary and sufficient condition that

be invertible is that $AD - BC$ be invertible. Construct examples to show that if the assumption that $D$ is invertible is dropped, then the condition becomes unnecessary and insufficient.

If $A$, $B$, and $D$ are operators on a Hilbert space $\mathbf{H}$, then the operator matrix

induces (is) an operator on $\mathbf{H} \oplus \mathbf{H}$, and if both $A$ and $D$ are invertible, then $M$ is invertible. The converse (if $M$ is invertible, then $A$ and $D$ are) is not true. If $\mathbf{H}$ and $\mathbf{K}$ are Hilbert spaces, with $\dim \mathbf{H} < \infty$, and if

is an invertible operator on $\mathbf{H} \oplus \mathbf{K}$, then both $A$ and $D$ are invertible. Consequence: the spectrum of $M$ is the union of the spectra of $A$ and $D$.