1
$\begingroup$

If $A$ is an $n\times n$ matrix such that $\|A\|<1$, show that $(I-A)$ is invertible and $$ \frac{1}{1+\|A\|} < \|(I-A)^{-1}\| <\frac{1}{1-\|A\|} $$

My trial:

First to show $I-A$ is invertible: if not, then there exists nonzero vector $x$ such that $(I-A)x =0 $ Hence $Ax=x$ So $\|x\|=\|Ax\|\le\|A\| \|x\|$ and since $x$ is a non zero vector we get $\|A\|\ge\|x\|$ which contradicts the assumption.

To prove the second part, by the triangle inequality we have $$ \|I\|-\|A\| \le \|I-A\|\le\|I\|+\|A\| $$ But the norm of the identity is equal to 1.

I stopped here since I know $\|(I-A)^{-1}\|$ doesn't equal $\|I-A\|^{-1}$

Any help?

The norm here is the spectral norm or the operator norm.

  • 0
    The first part may be the same, but I believe the second part is different enough to merit its own post. – 2017-01-06

2 Answers 2

2

Since $(I-A)^{-1}=I+(I-A)^{-1}A$, $$\|(I-A)^{-1}\|\leq 1+\|(I-A)^{-1}\|\|A\|,$$ which gives the upper bound. From $(I-A)^{-1}(I-A)=I$, we have $$1\leq\|(I-A)^{-1}\|\|I-A\|\leq\|(I-A)^{-1}\|(1+\|A\|),$$ which gives the lower bound.

1

Consider matrix $B=I+A+A^2+A^3+\ldots$ (prove that the series converges) and calculate the product $B(I-A)$. Then estimate the norm of $B$.

  • 0
    So here B to converges to the inverse, and then we can use the submultiplicative property and estimate the result by geometric series ,,thanks for your help :) – 2017-01-05