# Upper bound of the norm of a matrix difference using an absolutely converging geometric series and Neumann's theorem

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.

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

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 $$C^{n \times n}$$.

• 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. – Im YoungMin Dec 2 '18 at 1:00