Neumann series

Suppose that T is a bounded linear operator on the normed vector space X. If the Neumann series converges in the operator norm, then Id – T is invertible and its inverse is the series:

${\displaystyle (\mathrm {Id} -T)^{-1}=\sum _{k=0}^{\infty }T^{k}}$,

where ${\displaystyle \mathrm {Id} }$ is the identity operator in X. To see why, consider the partial sums

${\displaystyle S_{n}:=\sum _{k=0}^{n}T^{k}}$.

Then we have

${\displaystyle \lim _{n\rightarrow \infty }(\mathrm {Id} -T)S_{n}=\lim _{n\rightarrow \infty }\left(\sum _{k=0}^{n}T^{k}-\sum _{k=0}^{n}T^{k+1}\right)=\lim _{n\rightarrow \infty }\left(\mathrm {Id} -T^{n+1}\right)=\mathrm {Id} .}$

This result on operators is analogous to geometric series in ${\displaystyle \mathbb {R} }$, in which we find that:

${\displaystyle (1-x)\cdot (1+x+x^{2}+\cdots +x^{n-1}+x^{n})=1-x^{n+1},}$

${\displaystyle 1+x+x^{2}+\cdots ={\frac {1}{1-x}}.}$

One case in which convergence is guaranteed is when X is a Banach space and |T| < 1 in the operator norm or ${\displaystyle \sum |T^{n}|}$ is convergent. However, there are also results which give weaker conditions under which the series converges.

Let ${\displaystyle C\in \mathbb {R} ^{3\times 3}}$ be given by:

${\displaystyle {\begin{pmatrix}0&{\frac {1}{2}}&{\frac {1}{4}}\\{\frac {5}{7}}&0&{\frac {1}{7}}\\{\frac {3}{10}}&{\frac {3}{5}}&0\end{pmatrix}}.}$

We need to show that C is smaller than unity in some norm. Therefore, we calculate:

${\displaystyle {\begin{aligned}||C||_{\infty }&=\max _{i}\sum _{j}|c_{ij}|=\max \left\lbrace {\frac {3}{4}},{\frac {6}{7}},{\frac {9}{10}}\right\rbrace ={\frac {9}{10}}<1.\end{aligned}}}$

Thus, we know from the statement above that ${\displaystyle (I-C)^{-1}}$ exists.

A corollary is that the set of invertible operators between two Banach spaces B and B' is open in the topology induced by the operator norm. Indeed, let S : B → B' be an invertible operator and let T: B → B' be another operator. If

|S – T | < |S⁻¹|⁻¹,

then T is also invertible.

Since |Id – S⁻¹T| < 1, the Neumann series Σ(Id – (S⁻¹T))^k is convergent. Therefore, we have

T⁻¹S = (Id – (Id – S⁻¹T))⁻¹ = Σ(Id – (S⁻¹T))^k.

Taking the norms, we get |T⁻¹S| ≤ 1/(1 – |Id – (S⁻¹T)|).

The norm of T⁻¹ can be bounded by

${\displaystyle |T^{-1}|\leq {\tfrac {1}{1-q}}|S^{-1}|\quad {\text{where}}\quad q=|S-T|\,|S^{-1}|.}$

The Neumann series has been used for linear data detection in massive multiuser multiple-input multiple-output (MIMO) wireless systems. Using a truncated Neumann series avoids computation of an explicit matrix inverse, which reduces the complexity of linear data detection from cubic to square.[1]

• Werner, Dirk (2005). Funktionalanalysis (in German). Springer Verlag. ISBN 3-540-43586-7.