= Solution
{c}
A real number $\lambda$ is an <eigenvalue> of $T$ if $Tv=\lambda v$ for some nonzero <vector> $v\in V$. Its <eigenspace> is
$$
E_\lambda=\{v\in V:Tv=\lambda v\}=\ker(T-\lambda I).
$$
It 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
$$
\boxed{E_\lambda\cap E_\mu=\{0\}.}
$$
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