Invertibility of Identity Transformation Plus Product of Two Continuous Linear Transformations

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Theorem

Let $\struct {X, \norm{, \cdot ,} }$ be the normed vector space.

Let $I : X \to X$ be the identity mapping.

Let $\map {CL} X := \map {CL} {X, X}$ be the continuous linear transformation space on $X$.

Suppose $I + A \circ B$ is invertible, where $\circ$ denotes the composition of mappings.


Then $I + B \circ A$ is invertible, with the inverse given by:

$\paren {I + B \circ A}^{-1} = I - B \circ \paren {I + A \circ B}^{-1} \circ A$


Proof

\(\ds \paren {I + B \circ A} \circ \paren {I - B \circ \paren {I + A \circ B}^{-1} \circ A}\) \(=\) \(\ds I - B \circ \paren {I + A \circ B}^{-1} \circ A + B \circ A - B \circ A \circ B \circ \paren {I + A \circ B}^{-1} \circ A\)
\(\ds \) \(=\) \(\ds I - B \circ \paren {I + A \circ B}^{-1} \circ A + B \circ A - B \circ \paren {I + A \circ B - I} \circ \paren {I + A \circ B}^{-1} \circ A\)
\(\ds \) \(=\) \(\ds I - B \circ \paren {I + A \circ B}^{-1} \circ A + B \circ A - B \circ \paren {I + A \circ B} \circ \paren {I + A \circ B}^{-1} \circ A + B \circ \paren {I + A \circ B}^{-1} \circ A\)
\(\ds \) \(=\) \(\ds I + B \circ A - B \circ A\)
\(\ds \) \(=\) \(\ds I\)
\(\ds \paren {I - B \circ \paren {I + A \circ B}^{-1} \circ A} \circ \paren {I + B \circ A}\) \(=\) \(\ds I + B \circ A - B \circ \paren {I + A \circ B}^{-1} \circ A - B \circ \paren {I + A \circ B}^{-1} \circ A \circ B \circ A\)
\(\ds \) \(=\) \(\ds I + B \circ A - B \circ \paren {I + A \circ B}^{-1} \circ A - B \circ \paren {I + A \circ B}^{-1} \circ \paren{I + A \circ B - I} \circ A\)
\(\ds \) \(=\) \(\ds I + B \circ A - B \circ \paren {I + A \circ B}^{-1} \circ A - B \circ \paren {I + A \circ B}^{-1} \circ \paren{I + A \circ B} \circ A + B \circ \paren {I + A \circ B}^{-1} \circ A\)
\(\ds \) \(=\) \(\ds I + B \circ A - B \circ A\)
\(\ds \) \(=\) \(\ds I\)

$\blacksquare$


Sources

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