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Linear operator proof

  1. Oct 12, 2009 #1
    1. The problem statement, all variables and given/known data

    Let V be a finite-dimensional vector space over the field F and let S and T
    be linear operators on V. We ask: When do there exist ordered bases a and b
    for V such that a = [T]b? Prove that such bases exist if and only if there is
    an invertible linear operator U on V such that T = USU-1. (Outline of proof:
    If a = [T]b, let U be the operator which carries a onto b and show that
    S = UTU-1. Conversely, if T = USU-1 for some invertible U, let a be any
    ordered basis for V and let b be its image under U. Then show that a = [T]b.
    where [T]b means T with relative to b

    3. The attempt at a solution

    I'm not sure where to start. Is there a particular theorem to use here?
     
  2. jcsd
  3. Oct 12, 2009 #2

    jbunniii

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    [Note: I think the [itex]U[/itex] and [itex]U^{-1}[/itex] should be reversed in the problem statement.]

    I don't think you need any particular theorem. The proof outline looks good. So let's suppose that there do exist two bases, [itex]a[/itex] and [itex]b[/itex], such that [itex]a = [T]b[/itex].

    Let's give the basis elements some names:

    [tex]a = \{a_1,\ldots,a_n\}[/tex]
    [tex]b = \{b_1,\ldots,b_n\}[/tex]

    As the outline suggests, let us define a new operator U as follows:

    [tex]U(a_j) = b_j, \textrm{ for }j=1,\ldots,n[/tex]

    Since we have specified [itex]U[/itex] on a basis for [itex]V[/itex], that uniquely defines [itex]U[/itex] on all of [itex]V[/itex]. By the way, note that [itex]U[/itex] is invertible.

    Now consider the operator [itex]U^{-1}TU[/itex]. In particular, can you find a useful way to express this operator in terms of the basis [itex]a[/itex]?

    Hint: you want to express [itex][U^{-1}TU]_a[/itex] as the product of three matrices, where the middle one is [itex][T]_b[/itex].
     
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