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

Question

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.

Answer 1

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 $\frac{\|E\|}{\|A\|}\ge\frac{1}{\|A\|\|A^{-1}\|}=\frac{1}{\kappa(A)}$.

Answer 2

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 $$|\sigma_{\min}(A+E) - \sigma_{\min}(A)| \le \|E\|.$$