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 $T$.
- Is the above "proof" correct?