# For any non-singular $A$, $\frac{1}{\kappa(A)}\leq\frac{\|E\|}{\|A\|}$ if $E+A$ is singular [closed]

Let $$A$$ be a non singular, and $$A+E$$ be singular matrix. Prove that $$\frac{1}{\kappa(A)}\leq\frac{\|E\|}{\|A\|}$$.

I thought of assuming $$\|A+E\|=0$$, but it doesn't seem to be right.

• "cond A" stands for the condition number of A?
– Set
Commented Dec 24, 2020 at 20:01
• @Thoth yes that what I mean
– peru
Commented Dec 24, 2020 at 20:05

I suppose that $$\|\cdot\|$$ is an operator norm, i.e. $$\|M\|=\sup_{\|x\|=1}\|Mx\|$$. By the given assumptions, $$I+A^{-1}E$$ is singular. Therefore $$A^{-1}Ex=-x$$ for some unit vector $$x$$. Hence $$\|A^{-1}\|\|E\|\ge\|A^{-1}E\|\ge\|A^{-1}Ex\|=\|-x\|=1$$ and $$\frac{\|E\|}{\|A\|}\ge\frac{1}{\|A\|\|A^{-1}\|}=\frac{1}{\kappa(A)}$$.
It suffices to show that $$\|E\|$$ is greater than or equal to the smallest singular value of $$A$$, since $$\|A\|=\sigma_{\max}(A)$$ and $$\kappa(A) = \sigma_{\max}(A) / \sigma_{\min}(A)$$.
This follows immediately from a Weyl inequality (see this question and references therein, and/or Corollary 8.6.2 in "Matrix Computations" by Golub and van Loan) that states $$|\sigma_{\min}(A+E) - \sigma_{\min}(A)| \le \|E\|.$$