Prove that if $\|A\|<1$, then $(I+A)^{-1}=I-A+A^2-A^3+\cdots$.

Question

Prove that if $\|A\|<1$, then $(I+A)^{-1}=I-A+A^2-A^3+\cdots$.

I'm not sure how to do this. I know the result for $(I-A)^{-1}$, but that won't help me.

Answer 1

If you know the result for $(I-A)^{-1}$ you can employ it: if $\|A\|<1$, then $\|-A\|=\|A\|<1$; hence $I+A = I -(-A)$ is invertible and $$(I+A)^{-1} = (I-(-A))^{-1} =\sum_{n=0}^\infty (-A)^n = \sum_{n=0}^\infty (-1)^nA^n.$$

Answer 2

Suppose $X=I-A+A^2-\cdots$. Then $$ \begin{eqnarray} (I+A)X &=& (I+A)I - (I+A)A + (I+A)A^2 -\cdots \\ &=& (I + A) - (A+A^2) + (A^2+A^3) - \cdots \\ &=& I + (A - A) - (A^2 - A^2) + (A^3 - A^3) - \cdots \\ &=& I \end{eqnarray} $$

Since $(I+A)X=I$, we must have $X=(I+A)^{-1}I=(I+A)^{-1}$