Question about adjoint transformations- is this a valid proof

[Zoe-b]

Homework Statement

Q is an invertible self-adjoint linear transformation on an inner product space V. Suppose Q is positive definite. I have already shown that inv(Q) is self-adjoint, that all eigenvalues of Q are positive, so there exists S s.t. S^2 = Q.
Now suppose P is any self-adjoint linear transformation on V. I have shown inv(S)*P*inv(S) is also self adjoint. The bit I'm trying to do is: deduce or prove otherwise that there are linearly independent vectors e₁...e_n and scalars a₁...a_n such that for i between 1 and n:
Pe_i = a_iQe_i

Homework Equations

Spectral theorem.

The Attempt at a Solution

I have been staring at this for ages- I've done this but I'm not sure at all that its valid so if its complete rubbish please let me know!

inv(S)*P*inv(S) is self-adjoint so by the spectral theorem there exists an orthonormal (and therefore linearly independent) basis of eigenvectors e₁...e_n with associated eigenvalues a₁...a_n s.t

inv(S)*P*inv(S)e_i = a_ie_i for i = 1..n

(possibly dodgy step coming up!)
Now consider the transformation inv(S)*P*inv(S) restricted to span(e_i)
On this subspace of V, inv(S)*P*inv(S) = a_i*I
pre-multiply by S, post multiply by S
to get P = a_i*Q on this span
so Pe_i = a_iQe_i

Is that correct?
Thanks.

 

[Zoe-b]

Can anyone help? So sorry for bumping but even a 'yes' or 'no' answer as to whether the proof is valid would be extremely useful.
Thanks again.

 

[Dick]

Can anyone help? So sorry for bumping but even a 'yes' or 'no' answer as to whether the proof is valid would be extremely useful.
Thanks again.

Well, no. It can't be a valid proof. You can only restrict the action of S to span(ei) if ei is also an eigenvector of S. That much should be pretty clear. I'm still not clear whether your final conclusion is true or not though. I'll tell you one thing I think is true though. Pick Q=[[1,0],[0,4]] and P=[[0,1],[0,1]]. I think you can find two independent vectors such that Pei = aiQei. But they aren't orthogonal. Clearly I'm still fishing around with this problem.

 

[Zoe-b]

Thanks.. dodgy step was indeed dodgy then. Yeah the vectors being linearly independent should obviously be easier to prove than being orthogonal (which may not be true at all) so I will look at that. Just struggling at the moment to relate inv(S)Pinv(S) and Q since there is no guarantee (infact its unlikely surely?) that Q is diagonal over the same basis as inv(S)Pinv(S) is. I have in my notes that two matrices are simultaneously diagonalisable iff they commute but here that seems to be what I'm trying to prove anyway.

 

[Dick]

Thanks.. dodgy step was indeed dodgy then. Yeah the vectors being linearly independent should obviously be easier to prove than being orthogonal (which may not be true at all) so I will look at that. Just struggling at the moment to relate inv(S)Pinv(S) and Q since there is no guarantee (infact its unlikely surely?) that Q is diagonal over the same basis as inv(S)Pinv(S) is. I have in my notes that two matrices are simultaneously diagonalisable iff they commute but here that seems to be what I'm trying to prove anyway.

Ok, think I see it finally. As you said, you have inv(S)*P*inv(S)ei = aiei. S is invertible. So pick vi=inv(S)ei. Since the ei are a basis, so are the vi (but not necessarily orthogonal). The vi are the vectors you want. Do you see it?

 

[Zoe-b]

yep I do thank you! Was getting rather frustrated because the rest of the question I could prove if I had proved this bit first. Thanks again.