1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Staff Emeritus
    Science Advisor

    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


    User Avatar
    Staff Emeritus
    Science Advisor

    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:
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Linear Algebra - Invariant Subspaces/Adjoint