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Linear Algebra - Invariant Subspaces/Adjoint

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1. Homework Statement
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. Homework 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!
 

Answers and Replies

HallsofIvy
Science Advisor
Homework Helper
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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?
 
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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?
Does it say that T*w must be in Uperp, since <u, T*w> = 0 for any T*w?
 
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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?
 
HallsofIvy
Science Advisor
Homework Helper
41,736
894
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.
Actually it proves that Uperp is invariant under T*!

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

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