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Let H be a separable Hilbert space with the standard orthonormal basis (en).
(a) The unilateral shift operator T(a1, a2, ¼) = (0, a1, a2, ¼) on H is
injective and the closure of its range is the closed linear span {e2, e3, ¼} which doesn't contain e1.
(b) If S = T*, then S(a1, a2, ¼) = (a2, a3, ¼). So S is surjective but Ker(S) ¹ {0}, since it is the linear span of e1.