I am trying to show the following logic:
if $\lambda \in \mathbb{R}$ and $||(\lambda I-A)^{-1}||_2$ is $||(\lambda I-A)^{-1}||_2 > \frac{1}{\eta}$ where $\eta$ is machine precision, then $\lambda$ is an eigenvalue of $A+\delta A$ for some $||\delta A ||_2 \le \eta$.
So I tried to show it like this:
For some real number $\lambda$, I know from the problem statmeent that the quantity $||(\lambda I-A)^{-1}||_2$ is $||(\lambda I-A)^{-1}||_2 > \frac{1}{\eta}$ where $\eta$ is machine precision. The condition number of $(\lambda I-A)^{-1}$ is $k((\lambda I-A)^{-1}) = ||((\lambda I-A)^{-1})||*||(\lambda I-A)||$. Then, the condition number must be extremely high $k((\lambda I-A)^{-1}) = \infty $. Therefore $(\lambda I-A)^{-1})$ must be singular. And if $(\lambda I-A)^{-1})$ is singular, then $\lambda$ is an eigenvalue of $A$.
Now, I am getting stuck working the $||\delta A ||$ in there. How do I work that in there? and, is my logic correct thus far?