$\|\cdot\|$ is the induced norm for $n\times n$ matrices in $\mathbb{C}$, with respect to some vector norm ($\mathbb{C}^n\to\mathbb{R}$). $A$ and $B$ are $n\times n$ matrices where $A$ is non-singular (invertible) and $B$ is singular. $\kappa(A)$ is the condition number of a non-singular matrix defined by $\kappa(A)=||A||\cdot||A^{-1}||$
Prove: $\dfrac{1}{\kappa(A)}\leq\dfrac{||A-B||}{||A||}$
Work so far:
I know that $1=\|I\|=\|AA^{-1}\|\leq\|A\|\cdot\|A^{-1}\|$ from the sub-multiplicative property of norms so that $\frac{1}{\kappa(A)}\leq1$
We know that $\|A\|>0, \|B\|\geq0$
$\|B\|=0$ iff $B$ is the zero matrix in which case the property holds. Assuming B is not the zero matrix, then $\|B\|>0$. So I just need to show $\|A\|\leq\|A-B\|$ for a singular B.
$\|A\|-\|B\|<\|A\|=\|A-B+B\|\leq\|A-B\|+\|B\|$
This is where I get stuck. I think I'm missing some property of induced norms or I've made an incorrect assumption somewhere.
Best Answer
Your bound $\kappa(A)\geq 1$ is too crude. Instead, appeal directly to the definition of $\kappa(A)$ to get $$\dfrac{1}{\kappa(A)}\leq\dfrac{\|A-B\|}{\|A\|} \iff \|A-B\|\cdot\|A^{-1}\|\geq 1. $$ Now look at the effect of $(A-B)A^{-1}$ (or $A^{-1}(A-B)$) on a suitable vector to conclude $\|A-B\|\cdot\|A^{-1}\|\geq 1$.