The problem of computing $b$, given $x$, has a condition number $k(A) = ||A||\frac{||x||}{||b||} \le ||A||||A^{-1}||$.
It is said that if the second norm is used then the inequality becomes equality when $x$ is a multiple of a right singular vector of $A$ corresponding to the minimal singular value $\sigma_m$.
$||A||_2 = \sigma_1$, the largest singular value of $A$ and $||A^{-1}||_2 = 1/\sigma_m$. $||x||_2 = ||av_m^T||_2, ||b||_2 = ?$
Can you help me to understand how to get equality?
Best Answer
$Ax=b$ so $x = A^{-1}b$. To get the equality, let $b = u_m$, the $m$-th left singular vector of $A$. Then $$ x = A^{-1}b = V\Sigma^{-1} U^*u_m = \frac{1}{\sigma_m}v_m, $$ so $\|x\|_2= 1/\sigma_m = \|A^{-1}\|_2$.