0
$\begingroup$

I am having trouble proving the following statement:

Let $\mathbf{A}\in\mathbb{C}^{n\times n}$ be a square matrix such that $\|\mathbf{A}\|<1$, for some induced norm $\|.\|$. Then, $\|(\mathbf{I}-\mathbf{A})^{-1}-\mathbf{I}\|\leq\dfrac{\|\mathbf{A}\|}{1-\|\mathbf{A}\|}$.

I have been able to prove that $\|(\mathbf{I}-\mathbf{A})^{-1}\|\leq\dfrac{1}{1-\|\mathbf{A}\|}$ by using the absolutely converging geometric series $(1-z)^{-1} = \sum_{n=0}^{\infty}{z^n}$, having $|z|<1$, together with Neumann's theorem $(\mathbf{I}-\mathbf{A})^{-1} = \sum_{n=0}^{\infty}{\mathbf{A}^n}$, but I can't prove the above "extension" statement from the inequalities that emerge after using the triangle inequality: $\|(\mathbf{I}-\mathbf{A})^{-1}-\mathbf{I}\|\leq\|(\mathbf{I}-\mathbf{A})^{-1}\|+\|\mathbf{I}\|$.

Can anyone please provide me with a hint/help on how to proceed? Thanks in advance!

Note The formulation to be proven appears in Generalizations of the Condition Number by Predrag S. Stanimirovic.

$\endgroup$
2
$\begingroup$

$(I-A)^{-1}=I+A+A^{2}+\cdots$ (the series converging in the norm since $\|A\|<1$) and so $(I-A)^{-1}-I=A+A^{2}+\cdots$. So $\|(I-A)^{-1}-I\| \leq \|A\|+\|A\|^{2}+\cdots=\frac {\|A\|} {1-\|A\|}$.

$\endgroup$
0
$\begingroup$

It is :

$$\mathbf{\|(1-A)^{-1}\| \leq \frac{1}{1-\|A\|}}$$

But, also, as you proved by using the triangle inequality :

\begin{align*} \|(\mathbf{1}-\mathbf{A})^{-1}-\mathbf{1}\| &\leq\|(\mathbf{1}-\mathbf{A})^{-1}\|+\|\mathbf{1}\|\\ & \leq \mathbf{\frac{1}{1-\|A\|}} + \|\mathbf{1}\| \leq \mathbf{\frac{\mathbf{\|A\|}}{1-\|A\|}} \end{align*}

Important note : Since we are talking about matrices, try to be careful not to make things tangled with the number $1 \in \mathbb R$ and the $\mathbf{1} \equiv \mathbf{I}$ matrix which is essentialy the element $1$ but in $C^{n \times n}$.

$\endgroup$
  • $\begingroup$ Thanks, but I think the $\mathbf{1}$ in the denominator of the right hand side should be a scalar $1$ since $\|\mathbf{A}\|$ is a nonnegative scalar. $\endgroup$ – Im YoungMin Dec 2 '18 at 1:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.