Can there be just one eigenvalue on the invariant subspace (generalized eigenspace) associated with an eigenvalue?

Question

In the proof of Jordan decomposition here, once I know that an indecomposable subspace $V$ is of the form $V=Ker((f-\lambda Id)^n)$, can there be an other eigenvalue $\mu$ for $f\vert_V$?

Answer 1

No. If $(f-\lambda I)^n = 0$ on $V$ and $f v = \mu v$ for some $v \in V$, then $(f -\lambda I) v = (\mu - \lambda) v$ so $0 = (f - \lambda I)^n v = (\mu - \lambda)^n v$, which implies either $\mu = \lambda$ or $v = 0$.