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Linear Algebra: Invariant Subspaces

  1. Oct 11, 2007 #1
    The problem statement, all variables and given/known data

    Prove or give a counterexample: If U is a subspace of V that is invariant under every operator on V, then U = {0} or U = V. Assume that V is finite dimensional.

    The attempt at a solution

    I really think that I should be able to produce a counterexample, however even if I find an appropriate subspace, I have NO idea how to show that U is invariant under every operator on V. There are an arbitrary number of them!

    One thing I was thinking was use a one dimensional subspace for my U, because then every transformation of a vector u in U would be a constant times u. In other words, T(u) = cu.

    Does anyone know if I'm on the right track?
     
  2. jcsd
  3. Oct 11, 2007 #2

    morphism

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    Hint #1: You're not going to be able to produce a counterexample.

    Hint #2: Suppose U is a nontrivial subspace of V that's invariant under every operator on V, and let {u_1, ..., u_n} be a basis for U. A linear operator is completely determined by its action on the basis.
     
  4. Oct 12, 2007 #3
    My textbook states that for operators on complex vector spaces with dimension greater than one, and real vector spaces with dimension greater than two, that there will be invariant subspaces other than {0} and V.

    Maybe the book means for a particular operator?
     
  5. Oct 13, 2007 #4

    morphism

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    It doesn't matter if a particular operator has other invariant subspaces. This problem is asking you to consider subspaces that are invariant under EACH operator.
     
  6. Oct 13, 2007 #5

    HallsofIvy

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    morphism's point: Given [any subspace U of V, of dimension m, 0< m< n= dim V, there exist a basis for V, [itex]v_1, v_2, ..., v_n}[/itex] such that the first m vectors, [itex]{v_1, v_2, ..., v_m} form a subspace for U. Now CONSTRUCT a linear transformation on V such that U is NOT invariant under that linear transformation.
     
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