Theorem Let $A$ be a square matrix, then $$\lim_{k \to \infty } A_{k} =0 \rightarrow \rho(A)<1$$ Moreover, the geometric series case $\sum_{k=0}^{\infty} A_{k}$ is convergent iff $\rho(A) < 1$. In such a
$$\sum_{k=0}^{\infty} A_{k} = (I - A)^{-1}.$$ As a result, if $rho(A) < 1$ the matrix $I - A$ is invertible and the following inequalities hold $$\frac{1}{1+||A||} \leq ||(I-A)^{-1}|| \leq \frac{1}{1 - ||A||}$$
where $||\cdot||$ is an induced matrix norm such that $||A|| < $1.
From $$(A + \delta A)(x+ \delta x) = b + \delta b$$ it follows that $$(I+A^{-1}\delta A)(x+ \delta x)= x+A^{-1}\delta b.$$ Moreover since $\gamma \mu(A) < 1$ and $||\delta A|| \leq \gamma ||A||$ it turns out that $I-A^{-1}\delta A$ is non singular. Taking the inverse of such a matrix and passing to the norms we get $$||x+ \delta x || \leq ||(I-A^{-1}\delta A)^{-1}|| (||x|| + \gamma||A^{-1}|| \cdot ||b||)$$.
From theorem then it follows that: $$||x+ \delta x || \leq \frac{1}{||(I-A^{-1}\delta A)||} (||x|| + \gamma||A^{-1}|| \cdot ||b||) $$
Substituting $||b|| \leq ||A|| \cdot ||x||$ then i obtain:
\begin{align*} ||x+ \delta x || \leq & \frac{1}{||(I-A^{-1}\delta A)||} (||x|| + \gamma||A^{-1}|| \cdot ||b||) \\ \leq & \frac{1}{||(I-A^{-1}\delta A)||} (||x|| + \gamma||A^{-1}|| \cdot ||b||) \\ \leq & \frac{1}{||(I-A^{-1}\gamma ||A||)||} (||x|| + \gamma||A^{-1}|| \cdot ||A|| \cdot ||x||) \\ \leq & \frac{1}{||(I-\gamma \mu(A)||} (||x|| + \gamma \mu(A) \cdot ||x||) \\ \leq & \frac{1}{||(I-\gamma \mu(A)||} (||x|| + \gamma \mu(A) \cdot ||x||) \\ \leq & \frac{1}{1-\gamma \mu(A)} (||x|| + \gamma \mu(A) \cdot ||x||) \\ \leq & \frac{||x|| + \gamma \mu(A) \cdot ||x||}{1-\gamma \mu(A)} \end{align*}
But the true inequality is :
$$\frac{||x+ \delta x ||}{||x||} \leq \frac{1 + \gamma \mu(A) }{1-\gamma \mu(A)}$$
and this $||x||$ cannot avoid it.What I am doing wrong?Any help?
\begin{align*} ||x+ \delta x || \leq & \frac{1}{||(I-A^{-1}\delta A)||} (||x|| + \gamma||A^{-1}|| \cdot ||b||) \\ \leq & \frac{1}{||(I-A^{-1}\delta A)||} (||x|| + \gamma||A^{-1}|| \cdot ||b||) \\ \leq & \frac{1}{||(I-A^{-1}\gamma ||A||)||} (||x|| + \gamma||A^{-1}|| \cdot ||A|| \cdot ||x||) \\ \leq & \frac{1}{||(I-\gamma \mu(A)||} (||x|| + \gamma \mu(A) \cdot ||x||) \\ \leq & \frac{1}{||(I-\gamma \mu(A)||} (||x|| + \gamma \mu(A) \cdot ||x||) \\ \leq & \frac{1}{1-\gamma \mu(A)} (||x|| + \gamma \mu(A) \cdot ||x||) \\ \leq & \frac{||x|| + \gamma \mu(A) \cdot ||x||}{1-\gamma \mu(A)} \\ \leq &\frac{||x||(1 + \gamma \mu(A)) }{1-\gamma \mu(A)} \\ \frac{||x+ \delta x ||}{||x||}\leq &\frac{1 + \gamma \mu(A) }{1-\gamma \mu(A)} \end{align*}