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How to solve this $A-\lambda I_n$ is invertible implies $\lambda\in\mathbb{C}\backslash\left\{0,1\right\}$

Let $A\in\mathbb{C}^{n\times n}$ be such that $A-\lambda I_n$ is invertible. How to prove $\lambda\in\mathbb{C}\backslash\left\{0,1\right\}$?

I can prove the statement under the assumption that $A
eq I_n$ by the fact that:

$A-\lambda I_n$ is invertible implies $\text{det}\left(A-\lambda I_n\right)
eq0$
we can show $\text{det}\left(A-\lambda I_n\right)=\text{det}\left(A-I_n\right)-\lambda\text{det}\left(A-I_n\right)$
$\text{det}\left(A-I_n\right)
eq0$ by the assumption.
$\text{det}\left(A-I_n\right)
eq0$ means $A$ is invertible.

Can I assume $\text{det}\left(A-I_n\right)
eq0$ then the $A-\lambda I_n$ is invertible implies $\lambda\in\mathbb{C}\backslash\left\{0,1\right\}$. However, it seems invalid.

A:

It is true, because $\lambda\in \mathbb C\setminus\{0,1\}$ implies $\det\left(A-\lambda I_n\right)
eq0$, as the determinant is a continuous function in $\lambda$.

Q:

Are neutrinos created and annihilated pairs?

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