# If $A$ is non-singular (invertible) and $A+E$ is singular (not-invertible) show $K_p(A)\geq \dfrac{||A||_p}{||E||_p}$

$$A\in\mathbb{R}^{n\times n}$$ is a non-singular (invertible) matrix and we denote its $$p$$-norm with $$||A||_p$$ which is: $$||A||_p = \underset{x\neq 0}{\text{max}} \dfrac{||Ax||_p}{||x||_p}$$ which $$x\in \mathbb{R}^n$$. Also $$K_p(A)$$ is its condition number equal to $$||A||_p||A^{-1}||_p$$. Now assume we add matrix $$E$$ to $$A$$ such that $$A+E$$ is a singular (not-invertible) matrix. Show $$K_p(A)\geq \dfrac{||A||_p}{||E||_p}$$ This is what I tried:
\begin{align} K_p(A)&\geq \dfrac{||A||_p}{||E||_p}\\||A||_p||A^{-1}||_p&\geq \dfrac{||A||_p}{||E||_p} \end{align} Here, if $$||A||_p$$ is zero, we are done. Else: \begin{align} ||A^{-1}||_p&\geq \dfrac{1}{||E||_p}\\ ||A^{-1}||_p||E||_p&\geq 1 \end{align}

Here's where I'm stuck. I also know: \begin{align} 1=||I||_p=||AA^{-1}||_p\leq ||A||_p||A^{-1}||_p \end{align}

But I can't go further to prove what question is asking for. I don't know how to connect that $$A+E$$ is singular to these inequalities.

Any help or hint is immensely valuable to me!

• When you say $p$-norm, do you mean induced $p$-norm? Commented Nov 11, 2023 at 14:44
• @V.S.e.H. added the definition to the body. Commented Nov 11, 2023 at 14:50

Note that for $$x\in N(A+E)$$ nonzero, where $$N(A+E)$$ is the nullspace of $$A+E$$ (remember that $$A+E$$ is singular) we have $$0 = (A+E)x\implies Ax=-Ex\implies x=-A^{-1}Ex$$ Thus $$1=\frac{\|A^{-1}Ex\|_p}{\|x\|_p}\leq\|A^{-1}E\|_p\leq\|A^{-1}\|_p\|E\|_p.$$
• It looks nice but I didn't get the part you said $x\in N(A+E)$ nonzero, so you concluded $0=(A+E)x$. What is $N(A+E)$ notation? Null space? Commented Nov 11, 2023 at 14:58