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Irreducible linear operator is cyclic

  1. Jan 1, 2015 #1
    I´m having a hard time proving the next result:
    Let [itex] T:V→V [/itex] be a linear operator on a finite dimensional vector space [itex] V [/itex] . If [itex] T [/itex] is irreducible then [itex] T [/itex] cyclic.
    My definitions are: [itex] T [/itex] is an irreducible linear operator iff [itex] V [/itex] and {[itex] {\vec 0} [/itex]} are the only complementary invariant subspaces.
    T is cyclic linear operator iff V is a cyclic vector space (i.e. there is a vector [itex] \vec v ∈ V [/itex] such that [itex] V [/itex] is generated by the set of vectors {[itex] { \vec v, T(\vec v), T^{2}(\vec v),...} [/itex]}

    I was trying to do it by contradiction: suppose [itex] T [/itex] is not a cyclic linar operator then [itex] ∀ \vec v ∈ V [/itex] the generated space by the set {[itex] { \vec v, T(\vec v), T^{2}(\vec v),...} [/itex]} is not equal to [itex] V [/itex] also if [itex] \vec v \neq \vec 0 [/itex] then [itex]span [/itex] { [itex]\vec v, T(\vec v), T^{2}(\vec v),... [/itex] } is not equal to [itex] {\vec 0} [/itex].

    Moreover [itex]span [/itex] {[itex] \vec v, T(\vec v), T^{2}(\vec v),... [/itex] } is invariant (I´ve already proven it); also I know that every subspace has a complement that is : [itex] ∃ W [/itex] subspace of V such that [itex] W ⊕ span[/itex]{[itex] \vec v, T(\vec v), T^{2}(\vec v),... [/itex]} [itex] = V [/itex]

    Then I think a just need to prove that the complementary subspace [itex] W [/itex] is invariant that is : [itex] T(W)⊆ W [/itex] but this is the part that I´m having trouble.

    Any comment, suggestion, hint would be highly appreciated
     
  2. jcsd
  3. Jan 2, 2015 #2

    lavinia

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    If the span of a non-zero vector,v, is not the whole space then there is a vector ,w ,which is not in the span of v. The span of w is a second cyclic subspace. Both of these subspaces are irreducible and invariant. So what is their intersection?
     
  4. Jan 2, 2015 #3
    But how can we know that T(w) is not in span{v,T(v),...} ?
     
  5. Jan 3, 2015 #4

    lavinia

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    By assumption, w is not in the span of v.
     
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