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!

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?

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