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Prove that $\lambda$ is an eigenvalue of $T \iff$ the map represented by $T-\lambda 1$ is not an isomorphism.



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 $(T-\lambda 1) = 0 \implies v = 0 \rightarrow \leftarrow$.


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?
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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 $T$" (~$P$). – andybenji Dec 16 '12 at 6:02

I think that my proof would be useful for you.


$T$ 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)

  1. $f:V\mapsto W$ is injective if and only if $Ker(f)$ is zero subspace of $V$.
  2. $f:V\mapsto W$ is surjective if and only if $Im(f)=W$.
  3. $dim(Ker(f))+dim(Im(f))=n$ where n is dimension of space.


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 $T$


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