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

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.

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.