Eigenvalues/Endomorphism/Under Isomorphism/Fact

Let

$\varphi \colon V\longrightarrow V$

denote an endomorphism on a ${}K$-vector space ${}V$, and let

$f\colon V\longrightarrow W$

denote an isomorphism of ${}K$-vector spaces. Set

${}\psi =f\circ \varphi \circ f^{-1}\,.$

Then the following hold.

1. A vector ${}v\in V$ is an eigenvector of ${}\varphi$ for the eigenvalue ${}a\in K$ if and only if ${}f(v)$ is an eigenvector of ${}\psi$ for the eigenvalue ${}a$.
2. ${}\varphi$ and ${}\psi$ have the same eigenvalues.
3. The mapping ${}f$ induces for every ${}a\in K$ an isomorphism

   $f\colon \operatorname {Eig} _{a}{\left(\varphi \right)}\longrightarrow \operatorname {Eig} _{a}{\left(\psi \right)}.$

Proof, Write another proof