# Question about eigenvalue and invertibility problem

I found a post with a question like this: enter link description here

Assume A is invertible, Let $$\lambda$$ be an eigenvalue of $$A$$. Prove that $$\lambda^{-1}$$ is an eigenvalue of $$A^{-1}$$.

given that $$A$$ is invertible, $$Ax=\lambda x$$, $$A$$ is invertible, and $$\lambda\neq 0$$, we have
$$Ax=\lambda x\implies A^{-1}Ax=A^{-1}\lambda x\implies x=\lambda A^{-1}x\implies \frac1\lambda x=A^{-1}x.$$
My question is why are we assuimg $$\lambda \neq 0$$. And for a $$\lambda \neq 0$$, can we always claim that it has an inverse?
• If $\lambda =0$ them $\lambda^{-1}$ does not exist ; otherwise it exists. – Kavi Rama Murthy Feb 20 '20 at 6:03
We can assume $$\lambda \neq 0$$ since $$A$$ is invertible if and only if $$0$$ is no eigenvalue of $$A$$. See this question for more details.
If $$\lambda \neq 0$$, then $$\lambda$$ always has a multiplicative inverse in the corresponding field, so $$1/\lambda$$ exists.