Condition numbers: Find vector pair for which the equality holds I have that $Ax = b$. I would like to find a vector pair $(b, \delta b)$ for which the following equality holds:
$$\frac{||\delta x||_2}{||x||_2} = \kappa_2(A) \frac{||\delta b||_2}{||b||_2},$$
where $\kappa_2(A)$ = $\frac{\sigma_1}{\sigma_m}$
Generally we have LHS $\leq $ RHS, but there should exist some $b$ and $\delta b$ for which equality is achieved. I am supposed to express the vectors $(b, \delta b)$ in terms of left singular vectors.
So far I have (or at least think I have) that $b$ is a multiple of the first column of $U$, where $A = U \Sigma V^T$. But that's as far as I've come.
I have been trying to somehow show that $||b||_2 =  \sigma_1 ||x||_2$ and that $||\delta b ||_2 =  \frac{||\delta x||_2}{\sigma_m},$ because then the equality would follow, but I haven't succeeded. 

If $u$ and $v$ are left and right singular vectors of $A$, with $\|u\|=\|v\|=1$ and singular value $\sigma$, then
$$Av = \sigma u$$
$$\|Av\| = \sigma \|u\| = \sigma = \sigma \|v\|.$$
Does that help?
