On invertibility of $A+E$ where $||E||_2<$ smallest singular value of $A$ and $||A^{-1}E||_2<1$

Question

Let $A,E \in M_n(\mathbb C)$ . Suppose $\sigma_\min >0$ be the smallest singular value of $A$ and $||E||_2 < \sigma_\min$. Suppose $||A^{-1}E||_2 <1$. Then how to show that $A+E$ is invertible ?

My work : Going by contradiction; assume ,if possible, $\det (A+E)=0$, then $\det (I+A^{-1}E)=0$. So $-1$ is an eigenvalue of $A^{-1}E$, so $1=|-1|\le ||I+A^{-1}E||_2$, so $||A^{-1}E||_2 <1 \le ||I+A^{-1}E||_2$ ; but I am unable to proceed further.

Please help.

NOTE: Here $||M||_2:=\sup_{||x||_2=1}||Mx||_2=\sigma_\max$

Answer 1

Suppose that $(A+E)(x)=0$, $x\neq 0$, this implies that $A(x)=-E(x)$ and $x=-A^{-1}E(x)$, we deduce that $\|x\|=\|A^{-1}E(x)\|\leq \|A^{-1}E\|\|x\|<\|x\|$. Contradiction. This implies that $Ker(A+E)=\{0\}$ and $A+E$ is invertible.