T is cyclic iff there are finitely many T-invariant subspace

[Eclair_de_XII]

Homework Statement

"Let $T$ be a linear operator on a finite-dimensional vector space $V$ over an infinite field $F$. Prove that $T$ is cyclic iff there are finitely many $T$-invariant subspaces.

Homework Equations

T is a cyclic operator on V if: there exists a $v\in V$ such that $\langle T,v \rangle = V$
A subspace W of V is T-invariant if: for all $w\in W$, $T(w)\in W$

The Attempt at a Solution

We prove the trivial case first. Suppose $T$ is a cyclic operator on $\{0\}$ the zero subspace. Then $T(0)=0\in \{0\}$, and so, $\langle T, 0 \rangle$. Thus, there is only one T-invariant subspace of the zero subspace.

I can't prove it the other way, though, and I'm not sure how to proceed.

 

[fresh_42]

What is $\langle T,v \rangle$ and how can it determine the entire vector space?

 

[Eclair_de_XII]

$\langle T,v \rangle=\{f(T)(v)|f(x)\text{ is a polynomial}\}$

And I am guessing that it can determine an entire vector space $V$ if every element of $V$ can be written as a linear combination of the elements of $\langle T,v \rangle$ and vice versa. In other words, if $v\in V$, then $T(v),T^2(v),T^3(v),...\in V$ and moreover, $V$ consists of these elements only as well as the zero vector.

 

[mathwonk]

v is cyclic for T if every vector in V is a linear combination of vectors of form v, Tv T^2v, T^3v,...i.e. if every vector in V is of form f(T)(v) where f is a polynomial.

notice that if v is cyclic for T and a subspace W contains v and W is T - invariant, then that subspace W is the whole space V.

Intuition suggests to me that the property that the minimal polynomial f of a cyclic operator has degree equal to the dimension of the full space is relevant here, and the factors of f should be useful in describing the invariant subspaces. I.e. if W is an invariant subspace then the restriction of T to W has minimal polynomial some factor of the minimal polynomial f of T on all of V.

Indeed I would conjecture that if T is cyclic, with minimal polynomial f, then the only invariant subspaces are those of form kernel.g(T), where g is one of the finitely many monic factors of f.

In the other direction, to produce lots of invariant subspaces (in the non cyclic case), note that given any vector v, the vectors of form h(T)(v) for all polynomials h, is a T invariant subspace.

So the point seems to be to show that in the non - cyclic case lots of subspaces can have the same (restricted) minimal polynomial, but not in the cyclic case.

of course over a finite field there are always only a finite number of invariant, and non invariant for that matter, subspaces for any operator.

 

[Eclair_de_XII]

First of all, I'm sorry because I couldn't quite understand what's being said, here, @mathwonk. But I do have another possible idea for a solution. Basically, I let $dim(V)=n$, then construct an $n$-dimensional basis for $V$, $(v_1,...,v_n)$. Then I let $v=\sum_{i=1}^n a_iv_i$. From there, I use the hypothesis that $\langle T,v \rangle=V$, and then use the linearity of $T$ to argue that $V$ is the direct sum of $\langle T,v_i \rangle$. Is this wrong? The main problem, as I see it, is that there's no way to guarantee that each basis vector generates a T-invariant subspace.