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Linear algebra - need to show deg of minimal poly = dimension of V

  1. Feb 13, 2014 #1
    1. The problem statement, all variables and given/known data

    if V = C_x for some x belongs to V then show

    deg(u_L) = dim(V)

    here,

    L: V -> V linear operator on finite dimensional vector space

    C_x = span {x, L(x), L^2(x),.............}

    u_L = minimal polynomial



    2. Relevant equations



    3. The attempt at a solution

    since C_x = V, it spans V. if it spans v, then degree of minimal poly that we somehow obtain from L should be dim(V). i dunno.

    i can't figure out how to connect C_x and L with minimal poly.
     
  2. jcsd
  3. Feb 13, 2014 #2
    If you were to prove that dim(Cx) = deg(uL) then you would be done. Does this seem like an easier problem?
     
  4. Feb 13, 2014 #3

    Dick

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    Homework Helper

    I'll add one more clue to kduna's suggestion. If ##n=deg(U_L)## then ##L^n(x)=0## for all x in V.
     
  5. Feb 14, 2014 #4
    If ##n=deg(U_L)## and ##L^n(x)=0## for all ##x \in V##, this would imply that ##U_L(x) = x^n##. Which is certainly not always true.

    However it is true that ##L^n(x) \in \text{span}\{x, L(x), ... , L^{n-1}(x) \}##. This is a big hint. Can you see why this is true?
     
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