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Cyclic subspaces question.

  1. May 31, 2006 #1
    prove that Z(v,T)=Z(u,T) iff g(T)(u)=v, where g(t) is prime compared to a nullify -T of u. (which means f(t) is the minimal polynomial of u, i.e f(T)(u)=0). (i think that when they mean 'is prime compared to' that f(t)=ag(t) for some 'a' scalar).

    i tried proving this way:
    suppose, g(T)(u)=v and suppose v belongs to V v doesn't equal 0, so f(T)(u)=0=av
    a=0, i need to show that some polynomial function of v belong to V too, but i don't really know how?
    and i need help on other way proof (suppose, Z(v,T)=Z(u,T)).
     
  2. jcsd
  3. May 31, 2006 #2

    matt grime

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    What is Z(v,T), what is T (we can guess from context but shouldn't have to), what is a 'nullify -T of u' (that makes no sense as a sentence, nullify is a verb)?

    For what its worth two polynomials f and g are prime if the ideal they span is all of the polynomial ring, or equivalently there exits polys h and k such that fh+gk=1 (or they have no common factors) and does not mean one is a scalar multiple of the other.
     
  4. May 31, 2006 #3
    Z(v,T) is a T cyclic subspace of V which is spanned by v. v is a vector which belongs to V, and T is a linear operator T:V->V.
    nullify is a unique polynomial f(t) (f(t)=t^k+...+a1t+t) of lowest rank which its highest degree coefficient is 1 such that: f(T)(v)=0.
     
  5. May 31, 2006 #4

    matt grime

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    Z(v,T) is the subspace *generated* by T and v ie the smallest subspace invariant under T and containing v. It is *not* 'spanned' by v, it is spanned by v, Tv, T^2v,.. T^kv (and these are a basis, where k is the k in your expression for f). Span and generation have subtly different meanings, and I think you should keep them separate.

    Nullify is still a verb. f(t) is the minimal poly which nullifies v (apparently) but that does not mean you can call it 'the nullify'.
     
    Last edited: May 31, 2006
  6. May 31, 2006 #5
    ok, now that matters are cleared any other hints on this question?
     
  7. May 31, 2006 #6

    matt grime

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    Yes, as I put in my first post, start with the definition of coprime polynomials too see what is going on.

    (note that your assertion that prime meant f(t)=ag(t) can't be what was meant, since f(T)u=0 by definition, hence ag(T)u=0, and Z(0,T)=0 which is not in general Z(u,T)).
     
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