Let $A \in \mathbb{R}$ be non-singular. Assume that for some induced matrix norm $$\frac{||E||}{||A||} \leq \frac{1}{\kappa(A)}$$ Prove that $A+E$ is non-singular, where $A + E$ is the perturbation of A and $\kappa(A)$ is the condition number of A.

How can I go about proving the above statement? I'm new to numerical linear algebra and would appreciate any help on this problem.


1 Answer 1


I assume you mean $A\in \mathbb{R}^{n\times n}$ and that the inequality is strict (otherwise the statement is wrong). We want to show that $A+E$ is non-singular, i.e., that $A+E$ is invertible. Since $A$ is assumed to be invertible, we can write

$A+E=A(I+A^{-1}E)$, so it remains to show that $(I+A^{-1}E)$ is invertible. But for any $B\in \mathbb{R}^{n\times n}$, we know by the Neumann series that $(I+B)$ is invertible as long as $\|B\|<1$. But by assumption, we have

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

Thus $(I+A^{-1}E)$ is invertible, which yields the desired claim.


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