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 $A^{-1}$.
The answer goes:
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?
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.