Condition number inequality Let A be an invertible n x n matrix, 
How to show that:
$K(A) \ge \dfrac{\|A\|}{\| B - A \|}$
where $K(A)$ is the condition number of the matrix $A$ and
for any $B$ being an $n\times n$ singular matrix.
Thank you!
 A: $K(A) = \|A\| \|A^{-1}\|$, so the inequality in question is $\|A - B\| \ge \frac 1{\|A^{-1}\|}$.
Pick a non-zero vector $x \in \ker(B)$. Then $(A - B)x = Ax$, so $\|A - B\| \ge \frac{\|Ax\|}{\|x\|}$.
Next, let $y = Ax$. Then $\frac{\|Ax\|}{\|x\|} = \frac{\|y\|}{\|A^{-1}y\|} \ge \frac{1}{\|A^{-1}\|}.$
A: Let $x$ be a nonzero vector in $\ker B$ and let $X$ be the square matrix in which every column is equal to $x$. By a suitable scaling of $X$, we may assume that $\|X\|=1$. Therefore
\begin{align*}
\kappa(A) &= \frac{\kappa(A)\|B-A\|}{\|B-A\|}
= \frac{\|A\|\|A^{-1}\|\|B-A\|\|X\|}{\|B-A\|}\\
&\ge \frac{\|A\|\|A^{-1}(B-A)X\|}{\|B-A\|}
= \frac{\|A\|\|0-X\|}{\|B-A\|}
= \frac{\|A\|}{\|B-A\|}.
\end{align*}
Note: The above proof only makes use of the submultiplicativity of a matrix norm. We deliberately avoid using the inequality $\|M\|\ge\|Mx\|/\|x\|$. While it is true that for any submultiplicative matrix norm $\|M\|$ for square matrices, there exists a compatible vector norm $\|x\|$ such that $\|M\|\ge\|Mx\|/\|x\|$ for any $M$ and any $x\not=0$, this fact is usually not taught in classes.
