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Proof regarding the image and kernel of a normal operator

  1. Jun 28, 2017 #1
    1. The problem statement, all variables and given/known data
    Show that if T is normal, then T and T* have the same kernel and the same image.

    2. Relevant equations

    3. The attempt at a solution
    At first I tried proving that Ker T ⊆ Ker T* and Ker T* ⊆ Ker, thus proving Ker T = Ker T* and doing the same thing with Im T, but could not find a way of doing so. I am relatively certain this has to do with discomposing T into direct sum subspace or doing something with the orthonormal basis of V comprised of eigenvalues of T but I cannot seem to figure it out.
    I would love some assistance on the matter.
  2. jcsd
  3. Jun 28, 2017 #2


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    Gold Member

    You understand that ##T## is unitarily diagonalizable here, right? You really can't ask for much more as the mutually orthonormal eigenvectors form a partition (and indeed are a proper, length preserving, coordinate system).

    ##T = U D U^*##

    ##T^* = \big(U D U^*\big)^* = U D^* U^*##

    with respect to the the right nullspace, how would you identify it in ##T##, above? What would it look like in ##T^*##? If you are so interested, what would the left nullspace look like for both of them?
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