TITLE: Invariant subspace of cyclic space is cyclic
QUESTION [5 upvotes]: Let $V$ be a finite dimensional vector space and let $T:V\rightarrow V$ be a cyclic linear operator, that is, there exists $v \in V$ such that $\{v, Tv, T^2v, \dots\}$ generates $V$.
Let $W\subset V$ be a $T$-invariant subspace, that is, $T[W]\subset W$. I'm trying to see that $T|W$ is also $T|W$-cyclic, that is, there exists a $w \in W$ such that $W=\langle w, Tw, T^2w, \dots\rangle$.

REPLY [5 votes]: Let's write $W=<w_{1},\dots,w_{r}>$.
Since $v$ is cyclic vector of $V$, there exist $p_{1},\dots,p_{r}\in K[X]$ such that $w_{i}=p_{i}(T)v$.
We have: $W=<w_{1},\dots,w_{r}>=\{(q_{1}p_{1}+\dots +q_{r}p_{r})(T)v:q_{1},\dots,q_{r}\in K[X]\}$ (one contention is obvious; the other ocurres because $W$ is $T$-invariant).
Now, $\{q_{1}p_{1}+\dots+q_{r}p_{r}:q_{1},\dots,q_{r}\in K[X]\}=<d>_{K[X]}$, where $d=gcd(p_{1},\dots,p_{r})$
Then, $W=\{q(T)(d(T)v):q\in K[X]\}$, and $d(T)v$ is a cyclic generator of $W$