1. The problem statement, all variables and given/known data Prove or give a counterexample: if S ∈ L(V) and there exists an orthonormal basis (e1, . . . , en) of V such that llSejll = 1 for each ej , then S is an isometry. 2. Relevant equations 3. The attempt at a solution Can't think of a counterexample. I am assuming that since ej is a member of an orthonormal basis, llejll HAS to be 1 by definition. And if llSejll=1, then llSejll=llejll. I mean, considering that llSejll=1 for ALL ej, all ej have 1 as an eigenvalue. Should I go ahead and do the proof, because I am having trouble seeing how S is not an isometry given the information I posted above.