# Necessary and sufficient conditions for a module to have finitely many invariant subspaces

I have had trouble answering the following question which is from a study guide to a qualifying exam I will be taking later this summer. I am thinking this question has something to do with cyclic vectors but I have not been able to put the two definitions together.

Definition: If $\alpha$ is any vector in $V$, the $T$-cyclic subspace generated by $\alpha$ is the subspace $Z(\alpha;T)$ of all vectors of the form $g(T) \alpha$, $g \in F[x]$. If $Z(\alpha; T) = V$, then $\alpha$ is called a cyclic vector for $T$.

Let $V$ be a finite-dimensional vector space over an infinite field $F$ and let $T:V\rightarrow V$ be a linear operator. Give to each $V$ the structure of a module over the polynomial ring $F[x]$ by defining $x \alpha = T(\alpha)$ for each $\alpha \in V$

1. In terms of the expression for $V$ as a direct sum of cyclic $F[x]$-modules, what are necessary and sufficient conditions in order that $V$ have only finitely many $T$-invariant $F$-subspaces?

2. Every linear operator I have encountered has finitely many $T$-invariant subspaces. Is there a good example of one that has infinitely $T$-invariant $F$-subspaces?

I was thinking that one direction might require $T$ not to have any cyclic vectors but I dont think this is the only hypothesis we need in order to answer even one direction for part 1.

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If $T$ has an eigenvalue $\lambda$ with geometric dimension at least $2$, and $\mathbf{v}$ and $\mathbf{w}$ are two linearly independent eigenvectors corresponding to $\lambda$, then for every scalar $\alpha$ you have that $\mathrm{span}(\mathbf{v}+\alpha\mathbf{w})$ is $T$-invariant. They are all pairwise distinct, so that gives you infinitely many distinct $T$-invariant subspaces. –  Arturo Magidin Aug 2 '11 at 9:19
The identity has infinitely many invariant subspaces as long as the space has dimension at least 2. –  Tobias Kildetoft Aug 2 '11 at 9:20
What is a $T$-cyclic vector? A $T$-cyclic subspace is a subspace of the form $\mathrm{span}(\mathbf{x},T(\mathbf{x}),T^2(\mathbf{x}),\ldots)$, and they always exist for any $T$. –  Arturo Magidin Aug 2 '11 at 9:21
I find your claim in 2 that all operators you have encountered have only finitely many distinct $T$-invariant subspaces rather hard to believe, in view of the above examples. I think you've just never noticed. –  Arturo Magidin Aug 2 '11 at 10:12
I think that the easiest way to get examples of infinitely many $T$-invariant subspace is to let $T$ be the all zero map. In that case I invite you to prove that any subspace of $V$ is $T$-invariant. How many 1-dimensional subspaces does the $xy$-plane have again? –  Jyrki Lahtonen Aug 3 '11 at 20:39
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We claim that necessary and sufficient condition that $V$ has only finitely many $T$-invariant $F$-subspaces is that $V$ has a cyclic vector for $T$.

Let $A = F[X]$. $V$ is regarded as an $A$-module as explained by the OP. A $T$-invariant $F$-subspace of $V$ is none other than an $A$-submodule of $V$. Suppose $V$ has a cyclic vector $v$ for $T$. This is equivalent to saying that $V = Av$.

Define an $A$-homomorphism $\psi:A \rightarrow V$ by $\psi(g) = gv$. Let $I$ = Ker($\psi$) = {$g \in A$; $gv = 0$}. $I$ is an ideal of $A$. Hence $I$ is generated by a polynomial $f(X)$. $V = Av$ is isomorphic to $A/(f(X))$ as an $A$-module. Let $n$ be the dimension of $V$ over $F$. Since $1, Tv, T^2v, \dots T^nv$ are linearly dependent, there exists a polynomial $g(X)$ of degree $n$ such that $g(X)v = 0$. Hence $f(X)$ is not zero. Every $A$-submodule of $A/(f(X))$ is of a form $(g(X))/(f(X))$, where $g(X)$ is a factor of $f(X)$. Since the number of monic factor polynomials of $f(X)$ is finite, the number of $A$-submodule of $A/(f(X))$ is finite. Hence $V$ has only finitely many $T$-invariant $F$-subspaces.

Conversely suppose that $V$ has only finitely many $T$-invariant $F$-subspaces. Let $v$ and $w$ be vectors of $V$. We consider the set $\Gamma$ = {$A(v + tw)$; $t \in F$}. Since $V$ has only finitely many $A$-submodules, $\Gamma$ is finite. Let $\sigma:F \rightarrow \Gamma$ be the map $\sigma(t) = A(v + tw)$. Since $F$ is infinite, $\sigma$ cannot be injective. Hence there exist distinct elements $s$, $t$ of $F$ such that $A(v + sw) = A(v + tw)$. Since $(s - t)w = v + sw - (v + tw) \in A(v + sw)$, $w \in A(v + sw)$. Hence $v \in A(v + sw)$. Hence $Av + Aw = A(v + sw)$.

Inductively, if $v_1, ..., v_n \in V$, there exist $t_2, ..., t_n \in F$ such that $Av_1 + ... + Av_n = A(v_1 + t_2v_2 + ... + t_nv_n)$.

Since $V$ is finitely generated over $A$, $V = Av$ for some $v \in V$. Hence we are done.

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