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.
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$K(A) = \|A\| \|A^{-1}\|$, 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 $\frac{\|Ax\|}{\|x\|} = \frac{\|y\|}{\|A^{-1}y\|} \ge \frac{1}{\|A^{-1}\|}.$ |
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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. |
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