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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 $T$.

  1. Is the above "proof" correct?
  • 3
    You prove the same thing twice: $ P \Rightarrow Q$ and ~$Q \Rightarrow $~$P$ (which are equivalent statements), so you need either a "Suppose $T- \lambda 1$ is not an isomorphism" ($Q$), or a "Suppose $\lambda$ is an eigenvalue of $T$" (~$P$).2012-12-16

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