Power series expansion of perturbed matrix inverse

Question

Is there a power series expansion for a matrix inverse of the form $\left(A+D\right)^{-1}$, where $D$ is a small perturbation of A? $A$ is invertible.

Answer 1

Yes.

You can wtite

$A + D = (I + DA^{-1})A; \tag 1$

then if $\Vert D \Vert$ is sufficiently small that

$\Vert DA^{-1} \Vert \le \Vert D \Vert \Vert A^{-1} \Vert < 1, \tag 2$

which occurs if

$\Vert D \Vert < \Vert A^{-1} \Vert^{-1}, \tag 3$

we have the series representation

$(I + DA^{-1})^{-1} = \displaystyle \sum_0^\infty (-DA^{-1})^k; \tag 4$

this series converges by vrtue of (2), being majorized by

$ \displaystyle \sum_0^\infty \Vert DA^{-1} \Vert^k = \dfrac{1}{1 - \Vert DA^{-1} \Vert} < \infty, \tag 5$

the equality here being the standard formula for the sum of an infinite geometric series.

From (1),

$(A + D)^{-1} = ((I + DA^{-1})A)^{-1} = A^{-1}(I + DA^{-1})^{-1}; \tag 6$

substituting (4) on the right-hand side yields

$(A + D)^{-1} = ((I + DA^{-1})A)^{-1} = A^{-1}\displaystyle \sum_0^\infty (-DA^{-1})^k =\displaystyle \sum_0^\infty A^{-1} (-DA^{-1})^k, \tag 7$

the desired series for $(A + D)^{-1}$.

Note Added in Edit, Friday 13 December 2019 8:11 AM PST: It is probably worth mentioning how the series (4) may be validated. Let $X$ be any operator such that

$\Vert X \Vert < 1; \tag 8$

then basic algebra shows us that

$(I - X) \displaystyle \sum_0^n X^i = (I - X)(I + X + X^2 + \ldots + X^n) = I - X^{n + 1}; \tag 9$

by virtue of (8),

$\Vert X^{n + 1} \Vert \le \Vert X \Vert^{n + 1} \to 0 \; \text{as} \; n \to \infty; \tag{10}$

therefore (9) yields, again with $n \to \infty$,

$(I - X) \displaystyle \sum_0^\infty X^i = I, \tag{11}$

whence

$(I - X)^{-1} = \displaystyle \sum_0^\infty X^i; \tag{12}$

taking

$X = -DA^{-1} \tag{13}$

in (12) yields (4). End of Note.

Answer 2

Let $A$ be nonsingular and let $\tilde{A}=A+E$ be a perturbation of $A$. It can be shown that \begin{align} \tilde{A}^{-1}={A}^{-1}-{A}^{-1}E\tilde{A}^{-1} \tag 1 \end{align} The main drawback of this formula is the presence of $\tilde{A}^{-1}$ on both sides of the equality sign. However, by expressing $\tilde{A}^{-1}$ into infinite series, it can be written \begin{align} \label{taylorExpansion_of_A} \tilde{A}^{-1}= {A}^{-1}-({A}^{-1}E){A}^{-1}+({A}^{-1}E)^2{A}^{-1}-\cdots \tag 2 \end{align} Since, ${A}^{-1}$ is a differential function of its elements, first order approximation of (\ref{taylorExpansion_of_A}) is accurate up to terms of order $||E||^2$. With this approximation (\ref{taylorExpansion_of_A}) can be written as, \begin{align}\label {tag 3} \tilde{A}^{-1}= {A}^{-1}-({A}^{-1}E){A}^{-1}. \tag 3 \end{align} It appears that the answer give by @Robert Lewis is similar, where (\ref{tag 3}) is truncaded.