Prove that $cond(A)\ge \frac{||A||}{||A-B||}$ for any induced matrix norm

Question

Prove that for any induced matrix norm: $cond(A)\ge \frac{\left\lVert A \right\rVert}{\left\lVert A-B \right\rVert}$

Where $A$ is an invertible matrix, and $B$ is a singular matrix.

The condition number is: $cond(A) := \left\lVert A \right\rVert \left\lVert A^{-1} \right\rVert$

I have tried to prove that $\left\lVert A \right\rVert \left\lVert A-B \right\rVert \ge 1$ ,but I'm not sure how to use the fact that $B$ is singular.

Answer 1

Let $z\in \ker B$ be a unit vector. It is easy to see that $$ \|Az\|=\|Az-Bz\|\le \|A-B\|. $$ On the other hand, we have $$ 1 = \|z\|=\|A^{-1}Az\|\le \|A^{-1}\|\|Az\|. $$ Therefore, we have $$ \|A^{-1}\|^{-1}\le \|Az\|\le \|A-B\|, $$ and $$ \frac{\|A\|}{\|A-B\|}\le\|A\|\|A^{-1}\|=\text{cond}(A) $$ follows.