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evilpostingmong
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Homework Statement
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.
Homework Equations
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.