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

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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.

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  • $\begingroup$ ah, so simple ... I was thinking the wrong way ... thanks ! $\endgroup$ – user521337 Jan 12 at 5:03

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