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Homework Help: Linear Algebra - Invariant Subspaces/Adjoint

  1. Apr 10, 2008 #1
    1. The problem statement, all variables and given/known data
    Suppose T is in L(V) and U is a subspace of V. Prove that U is invariant under T if and only if Uperp is invariant under T*.

    2. Relevant equations
    V = U [tex]\oplus[/tex] Uperp
    if v [tex]\in[/tex] V, u [tex]\in[/tex] U, w [tex]\in[/tex] Uperp, then v = u + w.
    <Tv, w> = <v, T*w>

    3. The attempt at a solution
    If U is invariant under T, this means that if u [tex]\in[/tex] U, Tu [tex]\in[/tex] U. Basically the same thing for Uperp. Not really sure where to go from here. Any ideas? Thanks!
  2. jcsd
  3. Apr 10, 2008 #2


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    Rember that <u, w> = 0 for any u in U, v in Uperp. If U is invariant under T, then Tu is in U so <Tu, w>= 0= <u, T*w>, for any u in U. What does that tell you about T*w?
  4. Apr 10, 2008 #3
    Does it say that T*w must be in Uperp, since <u, T*w> = 0 for any T*w?
  5. Apr 10, 2008 #4
    Just a bump to see if I am understanding this correctly:

    If T is invariant under U, then <Tu, w> = 0 since Tu is in U, w is in Uperp. But <Tu, w> = <u, T*w> = 0, which means that T*w is in Uperp. This proves that T* is invariant under Uperp.

    If T* is invariant under Uperp, then <u, T*w> = 0 since u is in U, T*w is in Uperp. But <u, T*w> = <Tu, w> = 0, which means that Tw is in U. This proves that T is invariant under U.

    Is that correct, or am I missing something?
  6. Apr 11, 2008 #5


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    Actually it proves that Uperp is invariant under T*!

    No, your second part looks so much like the first part because T and T* are "dual".
  7. Apr 11, 2008 #6
    Err... that's what I meant. :redface: It was kind of late. Thanks for your help! :smile:
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