The condition number of a matrix gives a sharp estimate of the sensitivity of $x$ with respect to perturbations of $b$ when solving $Ax = b$, this means there exists a right hand side $b$ and a perturbation $δb$ such that
$\dfrac{\|δx\|_2}{\|x\|_2}=k_2(A)\dfrac{\|δb\|_2}{\|b\|_2}$
Where $k_2(A)$ is the condition number of $A.$
Give a vector pair $(b, δb)$ for which this equality holds.
This is how far i get:
We know that $k_2(A)=\dfrac{\sigma_1}{\sigma_m},$ because we are in the 2-norm. ($\sigma_1$ and $\sigma_m$ are the maximal resp. minimal singular value of $A.$
By theorem, we know that we achieve equality when $b$ is a multiple of a left singular value of $A$ corresponding to the maximal singular value $\sigma_1$. Then let $b=ku_1$ for some scalar $k$. When solving for $Ax=b$ using the SVD we get $x=\frac{k}{\sigma_1}v_1$, where $v_1$ is a right singular vector corresponding to $\sigma_1$.
So now this is what i have:
$x=\frac{k}{\sigma_1}v_1$
$b=ku_1$ for some scalar $k$.
$\|db\|_2=\sigma_m \|dx\|_2$
But now im completely lost into finding the vector $db$. There is another thread with this question but i do not get anything from the answers there. Any help is appreciated, thanks!