Linear Algebra - Invariant Subspaces/Adjoint

[steelphantom]

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^*.

Homework Equations

V = U \oplus Uperp
if v \in V, u \in U, w \in Uperp, then v = u + w.
<Tv, w> = <v, T^*w>

The Attempt at a Solution

If U is invariant under T, this means that if u \in U, Tu \in U. Basically the same thing for Uperp. Not really sure where to go from here. Any ideas? Thanks!

 

[HallsofIvy]

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?

 

[steelphantom]

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?

 

[steelphantom]

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]

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".

 

[steelphantom]

Err... that's what I meant. It was kind of late. Thanks for your help!