Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
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

1 Answer 1

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.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.