# Invertibility of Identity Minus Operator

This article needs to be tidied.Please fix formatting and $\LaTeX$ errors and inconsistencies. It may also need to be brought up to our standard house style.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Tidy}}` from the code. |

This article needs to be linked to other articles.You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by adding these links.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{MissingLinks}}` from the code. |

It has been suggested that this page be renamed.To discuss this page in more detail, feel free to use the talk page. |

## Theorem

Suppose $B$ is a Banach space and $T \in \mathfrak{L}(B, B)$, the space of bounded linear operators on $B$. If $\| T \|_* < 1$ in the operator norm $\|\cdot\|_*$:, then $I - T$ is invertible and has inverse

$$(I - T)^{-1} = I + T + T^2 + T^3 + \ldots.$$

## Proof

Define $S_n = I + T + T^2 + \ldots + T^n$. We first argue that $S_n$ converges to a bounded linear operator $S\in \mathfrak{L}(B,B)$. For any $n > m$,

\(\ds \norm{S_n - S_m}_*\) | \(=\) | \(\ds \norm{T^{m+1} + \ldots + T^n}_*\) | ||||||||||||

\(\ds \) | \(\leq\) | \(\ds \norm{T^{m+1} }_* + \ldots + \norm{T^n}_*\) | by Triangle Inequality and Operator Norm is Norm | |||||||||||

\(\ds \) | \(\leq\) | \(\ds \norm{T}_*^{m+1} + \ldots + \norm{T}_*^n\) | by repeated use of Operator Norm on Banach Space is Submultiplicative on each term | |||||||||||

\(\ds \) | \(\leq\) | \(\ds \sum_{j=m+1}^\infty \norm{T}_*^{j}\) | since each term is positive. |

Since $\|T\|_* < 1$ by assumption, this is a tail of a converging infinite series; therefore, by Tail of Convergent Series tends to Zero, this can be made arbitrarily small. Hence the sequence $S_n$ is in fact Cauchy; by Set of Bounded Linear Operators on Banach Space is Banach Space, it converges in the operator norm to some bounded linear $S = \lim_{n\to\infty} S_n$.

Next, expand $(I-T)S_k = (I-T)(I + T + \ldots + T^k)$ to yield $I - T^{k+1}$. We argue that $I - T^{k+1} \to I$ in norm:

\(\ds \norm{ (I - T^{k+1}) - I }_*\) | \(=\) | \(\ds \norm{ - T^{k+1} }_*\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \norm { T^{k+1} }_*\) | by Operator Norm is Norm | |||||||||||

\(\ds \) | \(\leq\) | \(\ds \norm { T }^{k+1}\) | by repeated application of Operator Norm on Banach Space is Submultiplicative | |||||||||||

\(\ds \) | \(\to\) | \(\ds 0\) | since $\norm{T}_* < 1$ by assumption. |

Taking limits on the left-hand side, we get $\lim_{k\to\infty} (I-T)S_k = (I-T)S = I$.

A similar computation shows us that $\lim_{k\to\infty} S_k(I-T) = S(I-T) = I$. So $S=I+T+T^2+\ldots$ is the bounded two-sided inverse of $I-T$.

$\blacksquare$