Proof Expla.: $\lambda$ is an eigenvalue of $A\in M_{n\times n}(F)$ iff $\det(A-\lambda I)=0$?

Question

I don't understand how Theorem 2.5 works in the following proof

And my idea of this proof is that: If $\large(A-\lambda I_n)$ is invertible then $v$ has to be $\mathbf0,$ which contradicts the assumption that ["]there exists a nonzero vector $v$[..."]. Is this correct?

Here is theorem 2.5

Answer 1

The idea is:\begin{align}\text{exists }v\neq0\text{ such that }Av=\lambda v&\iff\text{exists }v\neq0\text{ such that }(A-\lambda\operatorname{Id})v=0\\&\iff A-\lambda\operatorname{Id}\text{ is not one-to-one}\\&\iff A-\lambda\operatorname{Id}\text{ is not invertible}\end{align}and it is in this last equivalence that theorem 2.5 is used, since it says that a linear map between vector spaces of the same (finite) dimension is one-to-one if and only if it is onto and that therefore it is one-to-one if and only if it is invertible.