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

Question

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.

Answer 1

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

Answer 2

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 $C^{n \times n}$.