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kai_sikorski
kai_sikorski is offline
#5
Feb24-12, 08:09 PM
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Quote Quote by TranscendArcu View Post
1. The problem statement, all variables and given/known data



3. The attempt at a solution
So I observed:

T(B) = λB
T-1(B) = λ'B

Also,

T-1(T(B)) = λ'λB = B

This implies,

λ'λ = 1

And so, there should be a relation

[itex]λ = \frac{1}{λ'}[/itex].

Is that right?
Your conclusion is correct but I'm not sure I agree with your proof. In particular this
T-1(B) = λ'B
is only true if B is an eigenvector of T, and thats not really clear a priori. Just say that B is an eigenvector of T so

T(B) = λB

Apply T-1 to both sides
T-1(T(B)) = T-1(λB)

T is a linear operator (I assume) ... I'll let you finish the argument. It's pretty trivial.

Also as to the diagonalization comment, while I agree that's a good way to think about it intuitively it's not a good technique for a proof. Obviously there are issues with some finite dimensional space maps being non-diagonalizible; but even more fundamentally for a Hilbert space I think you would need the operator to be compact to have any hope of such an approach working.