Self-Adjoint Operators problem

[genjuro911]

Homework Statement

T a linear operator on inner product space V and W a T-invariant subspace of V. Then if T is self-adjoint then Tw is self-adjoint.

Homework Equations

Thm: T is self-adjoint iff $$\exists$$ an orthonormal basis for V consisting of e-vectors of T.

The Attempt at a Solution

Let $$\beta$$1 be a basis for Tw and by thm can extend to a basis $$\beta$$ for V, s.t. $$\beta$$1$$\subseteq$$$$\beta$$. But by above thm, $$\beta$$ is ON and consists of e-vectors of T, so then $$\beta$$1 is also ON and consists of e-vectors of T, and Tw is self-adjoint.

Does my proof make any sense?? Thanks everyone!

 

[Dick]

Tw is the restriction of T to W, right? Which makes sense as an operator on W since W is invariant? If so, then the proof is immediate using the definition of 'adjoint'. As for your proof... If you start with a basis for W and extend to V, how can you know the basis you started with can be selected to have orthonormal eigenvectors if you haven't proved Tw is self-adjoint yet?

 

[genjuro911]

Tw is the restriction of T to W, right? Which makes sense as an operator on W since W is invariant? If so, then the proof is immediate using the definition of 'adjoint'.

Would it be possible to elaborate a bit more? <T(v),w> = <v,T*(w)> for v,w in W is the def from what I remember but how is that immediate?

As for your proof... If you start with a basis for W and extend to V, how can you know the basis you started with can be selected to have orthonormal eigenvectors if you haven't proved Tw is self-adjoint yet?

I was kind of thinking the same thing but don't the basis vectors for Tw have to come from the basis for T?

Thanks

 

[Dick]

Would it be possible to elaborate a bit more? <T(v),w> = <v,T*(w)> for v,w in W is the def from what I remember but how is that immediate?


I was kind of thinking the same thing but don't the basis vectors for Tw have to come from the basis for T?

Thanks

Nooo. Take V=R^2. Take v=(1/sqrt(2),1/sqrt(2)). Take W to be the subspace t*v for real t. One orthonormal basis for V is e1=(0,1) and e2=(1,0). v isn't in that basis. A basis of a subspace isn't automatically a part of the basis of the containing space. You have to arrange it to be so. You are having a hard time seeing the obvious solution because it's, uh, obvious. That does make things hard to see.

T is self adjoint in V. So T=T* in V. Let Tw be the restriction of T to W. Which makes sense because T(W) is contained in W. So Tw:W->W. But W is contained in V. So if <v,T(w)>=<T(v),w> (T is self adjoint) for v,w in V, then <v,T(w)>=<T(v),w> for v,w in W. What does this tell you about the relation between Tw and (Tw)*?