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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.2011-08-02
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    The identity has infinitely many invariant subspaces as long as the space has dimension at least 2.2011-08-02
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    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 $T$.2011-08-02
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    Sorry I should have included the definition. I have edited the post.2011-08-02
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    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*.2011-08-02
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    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?2011-08-03

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