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Vector space, basis, linear operator

  1. Dec 5, 2008 #1
    1. The problem statement, all variables and given/known data
    Let V be a vector space of dimension n. And the linear operators E=A^0, A^1, A^2, ... A^(n-1) are linearly independent. Prove that there exists a v in V such that V=<v, Av, A^2v, ..., A^(n-1)v>


    2. Relevant equations



    3. The attempt at a solution
    Here are something that I tried.
    the degree of the minimal polynomials p(t) such that p(A)=0 is larger than n-1. I wanted to start the proof from here but have no idea how to proceed.
    assume V is over complex field C so that there is an eigenvector. However, it seems that this is also not the disired v to make them independent.
    Any hint? Thanks a lot
     
  2. jcsd
  3. Dec 5, 2008 #2

    morphism

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    Are you using <,> to denote span? If so, then it suffices to find a v such that {v, Av, ..., An-1v} is linearly independent. Stated differently, we want to make sure that whenever k0, k1, ..., kn-1 are scalars that are not all zero, then

    [tex] k_0 + k_1 Av + \cdots + k_{n-1} A^{n-1}v \neq 0[/tex].

    Or, equivalently,

    [tex] (k_0 + k_1 A + \cdots + k_{n-1} A^{n-1})v \neq 0 \iff v \not\in \ker(k_0 + k_1 A + \cdots + k_{n-1} A^{n-1})[/tex].

    But what can you say about ker(k0+ k1 A + ... + kn-1 An-1)?
     
  4. Dec 5, 2008 #3
    thanks for the hints! Here's what I think,

    so k0E+k1A+...+kn-1An-1 (called [tex]B_{k_{1},k_{2},..,k_{n-1}}[/tex])cannot be the 0 operator (since E, A, ..,An-1 are linearly independent), which implies that [tex]dim(Ker\! B_{k_{1},k_{2},..,k_{n-1}}) \leq n-1[/tex]

    (**) suppose we have V1, V2 as two proper sub space of V, then we can find a vector in V which is neither in V1 nor V2
    (assume we cannot, find two vectors such that [tex]x\in V_{1} \: x\notin V_{2} \: y\in V_{2} \: y\notin V_{1}[/tex] x+y will give a contration)

    If we have only finitely many [tex]B_{k_{1},k_{2},..,k_{n-1}}[/tex] I can find a v such that [tex]B_{k_{1},k_{2},..,k_{n-1}}v\neq 0 \: for\,all\: B_{k_{1},k_{2},..,k_{n-1}}[/tex] according to (**). Does this remain true if there are countably many or uncountably many subspace? I've no idea...

    Am I thinking right?.....

    Thanks
     
  5. Dec 5, 2008 #4

    HallsofIvy

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    What is your definition of "linearly independent" for linear operators?
     
  6. Dec 5, 2008 #5

    morphism

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    Presumably linear independence refers to independence in the vector space L(V) of linear operators on V.

    No, but fortunately we only need to consider finitely many [itex]B_{k_{1},k_{2},..,k_{n-1}}[/itex]. Consider the irreducible factors of the minimal polynomial of T.
     
  7. Dec 5, 2008 #6
    it's the same as other things I think...All the linear operators on V over field K forms a vector space, called L(V), and A_1,...,A_s in L(V) are linearly independent iff k_1*A_1+...k_s*A_s=0 implies k_1=...=k_s=0...

    Edit:
    I'll have a look at the irreducible factors...Thanks
     
    Last edited: Dec 5, 2008
  8. Dec 5, 2008 #7
    can you explain it a little more? I can not figure it out...
    Thanks
     
  9. Dec 5, 2008 #8

    morphism

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    You're going to want to use the rational canonical decomposition from here.
     
  10. Dec 5, 2008 #9
    Ah...it seems that there are lots of materials under the topic "rational canonical" that I cannot understand quite fast right now. I'll back to this question later. Anyway, Thanks a lot for your help!
     
  11. Dec 6, 2008 #10

    HallsofIvy

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    Which is the same, then, as saying that if v is any non-zero vector, {A_1v, A_2v, ..., A_s v} are independent in V.
     
  12. Dec 6, 2008 #11
    No, this is not true. if some Ai v=0, then they are not independent.
     
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