How to prove $ \frac{1}{A+B} = \frac{1}{A} - \frac{1}{A} B \frac{1}{A} + \frac{1}{A}B \frac{1}{A}B \frac{1}{A} + \cdots $?

Question

For matrix $A$ and $B$ \begin{align} \frac{1}{A+B} = \frac{1}{A} - \frac{1}{A} B \frac{1}{A} + \frac{1}{A}B \frac{1}{A}B \frac{1}{A} + \cdots \end{align} How can prove above identity holds?

Answer 1

A fundamental identity for the inverse of a matrix is $$(I-A)^{-1} = \sum_{n=0}^\infty A^{n}$$ when the matrix norm of $A$ is less than 1.

Using this identity we have $$(A+B)^{-1} = (A(I-(-A^{-1}B)))^{-1} = (I-(-A^{-1}B))^{-1} A^{-1} = \left( \sum_{n=0}^\infty (-1)^n(A^{-1}B)^n \right) A^{-1}$$ Hence $$(A+B)^{-1} = \sum_{n=0}^\infty (-1)^n(A^{-1}B)^n A^{-1}.$$ Note that this holds only when the matrix norm of $A^{-1}B$ is less than 1, just like with the classical geometric series.

Answer 2

Recall Neumann's series. If $\rho(C) < 1$, we have $$ \def\I{\mathrm{Id}}(\I - C)^{-1} = \sum_{k=0}^\infty C^k $$ apply this to $C = -BA^{-1}$, giving $$ (\I + BA^{-1})^{-1} = \sum_{k=0}^\infty (-1)^k (BA^{-1})^k $$ and multiply by $A^{-1}$: $$ (A + B)^{-1} = \bigl((\I + BA^{-1})A \bigr)^{-1} = A^{-1}(\I + BA^{-1}) = \sum_{k=0}^\infty (-1)^k A^{-1}(BA^{-1})^k $$ note that this only holds if $\rho(BA^{-1}) < 1$.