Past exam of the mathematics course of the University of Oxford / 2017 / Prelims / Mathematics I / 4 / a / ii

A real number $\lambda$ is an eigenvalue of $T$ if $Tv=\lambda v$ for some nonzero vector $v\in V$. Its eigenspace isIt is a vector subspace because it contains $0$ and $T(av+bw)=aTv+bTw=\lambda (av+bw)$ whenever $v,w\in E_{\lambda }$ and $a,b\in \mathbb{R}$.

If $v\in E_{\lambda }\cap E_{\mu }$, then $\lambda v=Tv=\mu v$, so $(\lambda -\mu )v=0$. Since $\lambda \ne \mu$, this forces $v=0$. Therefore