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Unitary, invariant subspaces

  1. May 7, 2007 #1
    1. The problem statement, all variables and given/known data

    Let U be a unitary operator on an inner product space V, and let W be a finite-dimensional U-invariant subspace of V. Prove that

    (a) U(W) = W
    (b) the orthogonal complement of W is U-invariant
    (for ease of writing let the orthogonal complement of W be represented by W^.

    2. Relevant equations

    Unitary: U*U = UU* = I

    3. The attempt at a solution

    (a) first show that U(W) is contained in W, and then show that W in contained in U(W).
    - U(W) is contained in W because W is U-invariant
    - show W is contained in U(W)
    choose x in W and show it is contained in U(W)
    U(x) is in W
    Is this circular thinking?
    Should I instead show that the range of U (restricted to W) is W itself? Or that the nullity of U (restricted to W) is 0?

    (b)
    Note: U restricted to W (let’s call it U_w) is also unitary.

    W^ = {x in V : <x, y>=0 for all y in W}

    Now show that U(W^) is contained in W^

    I’m not sure what to do now.

    Thanks for your help!
     
  2. jcsd
  3. May 7, 2007 #2

    AKG

    User Avatar
    Science Advisor
    Homework Helper

    Do the last one.
    Use the definitions. You want to say that if x is in W^, then so is U(x). Well x is in W^ iff for all y in W, <x,y> = 0, and U(x) is in W^ iff for all y' in W, <U(x),y'> = 0.
     
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