## 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
Homework Helper
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?

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
Homework Helper
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)*?