Let $T$ be a linear operator on $\mathbb C^n$ where $n<\infty$ and let $U$ be the eigenspace associated with eigenvalue $\lambda$. For simplicity, assume $\lambda$ is the only eigenvalue of $T$.
Let $(v, (T-\lambda I)v, \dots,(T-\lambda I)^{k-1}v)$ be a Jordan chain of length at least two. Let $w=(T-\lambda I)^{k-1}v$ and $W=\operatorname{span}(w)$. Then does every Jordan basis for $T$ contain a vector in $W$? This is the sense in which I mean that the Jordan chain "singles out" the subspace $W\subseteq U$ as mentioned in the question title.