Eigenvalues/Endomorphism/Under Isomorphism/Fact/Proof

Proof

(1). Let ${}v\in V$ be an eigenvector of ${}\varphi$ for the eigenvalue ${}a$. Set

${}w:=f(v)\,.$

Then

${}\psi (w)=(f\circ \varphi \circ f^{-1})(f(v))=(f\circ \varphi )(v)=f(\varphi (v))=f(av)=af(v)=aw\,.$

The reverse statement holds in the same way. (2) and (3) follow directly from (1).

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