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
[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 nondiagonalizible; 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.