Proof correctness Eigenvalues and Isomorphisms Prove that $\lambda$ is an eigenvalue of $T \iff$ the map represented by $T-\lambda 1$ is not an isomorphism.
Proof:
$\rightarrow$ 
Suppose $\lambda$ is an eigenvalue of $T$, then  we have $(T-\lambda 1)v = 0$ where $v\neq 0$. It is enough to show that $T$ is not one-to-one. 
 By contradiction if $T$ were one-to-one then $(T-\lambda 1) = 0 \implies v = 0 \rightarrow \leftarrow$.
$\leftarrow$
By contraposition, Suppose $T$ is an isomorphism. We must show that $\lambda$ is not an eigenvalue of $T$ where $(T-\lambda 1)v = 0$. Since $T$ is an isomorphism then $T$ is one-to-one and we have $(T-\lambda 1)v = 0 \implies v = 0 \implies v$ is not an eigenvector and hence $\lambda$ is not an eigenvalue of $T$.


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*Is the above "proof" correct?

 A: I think that my proof would be useful for you.
Theorem: 
$T$ is endomorphism of finite-dimentional vector space $V$. $\lambda$ is eigenvalue of $T$ if and only if the endomorphism 
$$T-\lambda E:V\mapsto V$$ 
is not an isomorphism.
Proof: 
In proof I'm going to use three lemmas (I think you know them well)


*

*$f:V\mapsto W$ is injective if and only if $Ker(f)$ is zero subspace of $V$.

*$f:V\mapsto W$ is surjective if and only if $Im(f)=W$.

*$dim(Ker(f))+dim(Im(f))=n$ where n is dimension of space.


$\rightarrow$ 
Suppose $\lambda$ is eigenvalue of $T$ and $v\neq0$,$v\in V$, then
$$Tv=\lambda v  \implies (T-\lambda E)v=0$$
So, $v\in Ker(f)\implies Ker(f)\neq \{0\}$.
Therefore, from the first lemma we have that $f=T-\lambda E$ is not an isomorphism.
$\leftarrow$ $f=T-\lambda E$ is not an isomorphism then $Ker(f)\neq \{0\}$ or $Im(f) \neq V$ (from first and second lemmas) 
But we have lemma 3, so at any rate $Ker(f)\neq \{0\}$. And so we can choose $v\in Ker(f)$. Then
$$(T-\lambda E)v=0 \implies Tv=\lambda v ,$$
so $\lambda$ is eigenvalue of $T$
$\blacksquare$
