1. Jan 3, 2012

Zoe-b

1. The problem statement, all variables and given/known data
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 independant vectors e1...en and scalars a1...an such that for i between 1 and n:
Pei = aiQei

2. Relevant equations
Spectral theorem.

3. 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...en with associated eigenvalues a1...an s.t

inv(S)*P*inv(S)ei = aiei for i = 1..n

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

Is that correct?
Thanks.

2. Jan 4, 2012

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.

3. Jan 4, 2012

Dick

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.

Last edited: Jan 4, 2012
4. Jan 5, 2012

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 guarentee (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.

5. Jan 5, 2012

Dick

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?

6. Jan 5, 2012

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.