Eigenvalues/Endomorphism/Under Isomorphism/Fact/Proof

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

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

Then

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

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