The following is from problem 26.1 in Trefethen and Bau:
Given $A\in\mathbb{C}^{m\times m}$ with spectrum $\Lambda(A)\subseteq\mathbb{C}$ and $\epsilon>0$, define the 2-norm $\epsilon$-pseudospectrum of $A$, $A_\epsilon(A)$, to be the set of numbers $z\in\mathbb{C}$ satisfying any of the following conditions:
- (i) $z$ is an eigenvalue of $A+\delta A$ for some $\delta A$ with $||\delta A||_2\leq\epsilon$;
(ii) there exists a vector $u\in\mathbb{C}^m$ with $||(A-zI)u||_2\leq\epsilon$ and $||u||_2=1$;
(iii) $\sigma_m(zI-A)\leq\epsilon$;
(iv) $||(zI-A)^{-1}||_2\geq\epsilon^{-1}$.
Prove that conditions (i)–(iv) are equivalent.
Edit: I now have the professor's solution, but I still can't make much sense of it, although it looks like I was somewhat on the right track before. Here are the troubling bits:
Since $||(zI-A)^{-1}||_2\geq\epsilon^{-1}$, there is a vector $v$ with $||v||_2=1$ such that if $w=(zI-A)^{-1}v$, then $||w||_2\geq\epsilon^{-1}$.
So far so good.
Multiplying each side by $zI-A$, we have $(zI-A)w=v$. It follows that \begin{equation} \bigg(zI-A-\frac{vw^*}{w^*w}\bigg)w=0\,, \end{equation}
I can't make the "it follows" leap here. How is the fraction constructed?
so $z$ is an eigenvalue of the matrix $A+\delta$, where $\delta A=(vw^*)/(w^*w)$.
If $z$ is such an eigenvalue, then by the above equation $w$ must be an eigenvector. How do we know that about $w$?
All that is done is multiplying by 1 in disguise, followed by some rebracketing of the matrices: $$ v = v \frac{w^*w}{w^*w} = \frac{v(w^*w)}{w^*w}=\frac{(vw^*)w}{w^*w}$$ From here we note that $vw^*$ is indeed a square matrix, and thus so is $\frac{vw^*}{w^*w}$. Make the definition $\delta A:= \frac{vw^*}{w^*w}$. Then note that we can use this to rewrite $(zI - A) w = v$ as $$(zI - (A+\delta A) ) w = 0$$ this is the equation that 'followed'.