Invariant subspace of cyclic space is cyclic 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 $W=\langle w, Tw, T^2w, \dots\rangle$.
 A: 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 $W$
A: I would tend to think about this in terms of the following fact.

Fact: The minimum polynomial $p$ of a linear transformation $T: V \to V$ always satisfies $\deg(p) \leq \dim(V)$, and equality holds precisely when $T$ is cyclic.

If $T:V \to V$ is a linear transformation and $W \subset V$ is a $T$-invariant subspace, then $T$ descends to a map on the quotient $V/W \to V/W$. Let $p_V,p_W$ and $p_{V/W}$ denote the minimum polynomials of $T$ itself, the restriction of $T$ to $W$, and the map $T$ induces on $V/W$. Note that 
$$\begin{align}\deg(p_W) + \deg(p_{V/W}) \geq \deg(p_V) && && (1)\end{align}$$
because $p_W \cdot p_{V/W}$ is a monic polynomial of degree $\deg(p_W) + \deg(p_{V/W})$ annihilating $T$ whereas $p_V$ is, by definition, the unique monic polynomial of smallest possible degree annihilating $T$. To see $p_W \cdot p_{V/W}$ annihilates $T$, recall $p_{V/W}(T)$ is zero, as a map of $V/W$,  and so $p_{V/W}(T) v \in W$ for any $v \in V$. Thus, $p_W(T) \cdot p_{V/W}(T) v = 0$ for any $v \in V$, and we are done.
If $T$ is cyclic, then the map it induces on $V/W$ is clearly cyclic too; if $v \in V$ is cyclic for $T$, then $v+W$ is cyclic for the map on the quotient. So, from the fact and the inequality (1), we get
$$\deg(p_W)  \geq \deg(p_V) - \deg(p_{V/W}) =  \dim(V) - \dim(V/W)  = \dim(W)$$
which, again by the fact, tells us that $\deg(p_W) = \dim(W)$, and so the restriction of $T$ to $W$ is also cyclic.
A: First of all, apologies for a previous seriously incorrect answer.
Let $f(x)$ be the minimal polynomial of $T$ over the underlying field $F$.
As noted in another answer (to which this one is related), $T$ is cyclic iff the degree of $f(x)$ equals the dimension of $V$.
Then as a vector space $V \cong F[x]/(f(x))$, where the action of $T$ on $V$ corresponds with multiplication by $x$ on the right. Thus a $T$-invariant subspace of $V$ corresponds to an $F[x]$-submodule of $F[x]/(f(x))$.
Since $F[x]$ is a PID, the correspondence theorem implies that $W$ corresponds to a submodule $(g(x))/(f(x))$, where $g \mid f$. Clearly this module is also cyclic, generated by $g(x)$.

In $V$ this means the following. As noted in the other answer, $T$ is also cyclic on $V/W$. Let $g$ be the minimal polynomial of (the map induced on $V / W$ by) $T$. Then $W$ is cyclic, generated by $g(T) v$.
