Let V be the vector space of all sequences (n) of complex numbers such that the 'sum of squares' n is finite. V has a well-defined inner product given by ((In), (Un)) := nyn n=1 (you do not have to prove that this is a well-defined inner product). Define S: VV, (1, 2, 3, ...) (0, 1, 2, 3, ...). This S is a linear map, and an isometry (you do not have to prove this either). Compute S*, and show that S*S=I, but that SS* I (here I is the identity operator).