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PF Gold
P: 162
 Quote by TranscendArcu 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 $λ = \frac{1}{λ'}$. 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.