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

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.

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