I'm learning about the Shermann-Morrison formula and the way to find the inverse matrix uses a formal power series. My question is how is this formal power series done? I read in some websites and I kind of get the idea of what a formal power series is but the way to do it is still confusing me. Here's the expansion I'm trying to understand:
$$(\textbf{A} + \textbf{uv}^T)^{-1} = (1+\textbf{A} ^{-1} \textbf{uv}^T)^{-1} \textbf{A}^{-1}$$
then we make a formal power series on $ (1+\textbf{A} ^{-1} \textbf{uv}^T)^{-1} $ and get
$$(\textbf{A} + \textbf{uv}^T)^{-1} = (1- \textbf{A} ^{-1} \textbf{uv}^T + \textbf{A} ^{-1} \textbf{uv}^T \textbf{A} ^{-1} \textbf{uv}^T + \dots ) \textbf{A}^{-1}$$
after this step it is easy to get to the result. Why is the first sign a substraction? In what I've read every sign is addition unless the series is defined with changing signs but in this expansion the first sign is the only different. I would appreciate a lot to know how to do the expasion. Thanks for the help.
This is the identity $$ \frac1{1+x} = 1 - x + x^2 - x^3 + x^4 - \dots $$ with alternating signs, valid for $|x|<1$.
We prove this identity by multiplying out $(1+x)(1 - x + x^2 - x^3 + x^4 - \dots)$. If we take a product with the finite sum $(1+x)(1 - x + x^2 - x^3 + x^4 - \dots \pm x^n)$ we get $1 \pm x^{n+1}$. Provided $x^n \to 0$ as $n \to \infty$, this converges to $1$, which is what we want for the identity to hold.
You might be more familiar with the geometric series $$ \frac1{1-x} = 1 + x + x^2 + x^3 + x^4 + \dots $$ and the identity above is just the same thing with $-x$ plugged in for $x$, giving us $$ \frac1{1 - (-x)} = 1 + (-x) + (-x)^2 + (-x)^3 + (-x)^4 + \dots $$ Finally, all of this can be done with matrices in place of $x$ as well, with inverses in place of division, because nothing we've done requires $x$ to be a real number.