Let $A$ be any $n \times n$ matrix and $\| \cdot \|$ be the matrix norm induced by vector norm on $\mathbb{R}^n$
(Euclidean
n-dimensional space).
If $\|I – A\| < 1$, then show that $A$ is invertible
and
derive the estimate
$\|A^{-1}\| < \frac{1}{ 1 – \| I – A \|}$.
Similarly
when can we expand $\frac{1}{ 1 – \| I – A \|}$ as a power series only is it if and only if the norm is less than $1$?
Thanks a lot!
Best Answer
If $A$ is singular, then 1 is an eigenvalue of $I-A$. So if the matrix norm is induced the 2-norm (i.e. the largest singular value), $\| I-A\|$ is at least 1 since the largest singular value of a matrix is not less than its eigenvalue in absolute value. Proved by contradiction.
Since $\| I-A\|<1$, we have $$\|A^{-1}\|=\|I-(I-A)\|=\|\sum_{i=0}^{\infty}(I-A)^i\|\le\sum_{i=0}^{\infty}\|(I-A)\|^i=\frac{1}{ 1 - \| I - A \|}$$
A sufficient condition for the Neumann series $\frac{1}{ 1 - x}=\sum_{i=0}^{\infty}x^i$ is $|x|<1$.