I have an orthogonal matrix $A \in \mathbb{R}^{10^{5}\times 10^{5}}$ and a perturbation matrix $\delta A = \epsilon A, \epsilon = 10^{-16}.$How can I calculate the range of values of parameter $\gamma \in \mathbb{R}^{+}$ such that the relative perturbation error to be less than $10^{-7}$?.Given that $||b||_{2}=1$ and $\delta b = [\epsilon,0,\dots,0,\epsilon]^{T}$ and $$\frac{||x + \delta x ||}{ ||x||} \leq \frac{1+\gamma K(A)}{1- \gamma K(A)} $$
where $K(A) = ||A|| * ||A^{-1}||$.
I feel like I'm missing something here but supposing $\kappa_{2}(A) = \|A \|_{2}\|A^{-1}\|_{2}$ and $A$ is orthogonal then $\kappa_{2}(A) = 1$ and your formula becomes
$$ \frac{1+\gamma}{1-\gamma} $$
Now you should realize the left-hand side is non-negative since $\frac{\| x + \delta x \|}{\|x\|}$ is strictly positive since $\| x\|$ can't be $0$. In that case we know that it's $0$ when $\gamma = -1$ and undefined when $\gamma = 1$. If you solve from when the expression is equal we have
$$ \frac{1+\gamma}{1-\gamma} = 10^{-7} $$
we get
$$ (1+\gamma)(10000000) = 1-\gamma $$
$$ 10000001\gamma = -9999999 \implies \gamma = \frac{-9999999}{10000001}$$
So the values that work are where $\gamma \in (-1, \frac{-9999999}{10000001})$.