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T is cyclic iff there are finitely many T-invariant subspace
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[QUOTE="mathwonk, post: 6074812, member: 13785"] 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. [/QUOTE]
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T is cyclic iff there are finitely many T-invariant subspace
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