let $\displaystyle \varphi :V\rightarrow V$ be a linear transformation.

let $\displaystyle u$ be eigenvector of $\displaystyle \varphi$ with the eigenvalue $\displaystyle \lambda$

let $\displaystyle v$ be eigenvector of $\displaystyle \varphi$ with the eigenvalue $\displaystyle \mu $

I need to prove that if $\displaystyle u+v$ is an eigenvector of $\displaystyle \varphi$, then $\displaystyle \lambda=\mu$

this is what I tried:

$\displaystyle \varphi(u+v)=a(u+v)=au+av$

on the other hand:

$\displaystyle \varphi(u)+\varphi(v)=\lambda u+\mu v$

but how can I conclude that $\displaystyle a=\lambda$ and $\displaystyle a=\mu$?