A bounded operator on a Hilbert space which has no square root.

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Suppose that T is the operator T (x_{1}, x_{2}, ⋯) = (x_{2}, x_{3}, ⋯) on l^{2} ( in fact T is the adjoint of the unilateral shift operator). If T has a square root S, then S^{2} = T and KerS ⊆ KerT = Cξ_{1} in which ξ_{1} = (1, 0, 0, ⋯). Since T is not one to one we conclude that S is not one to one. So that KerS = Cξ_{1}. T is surjective, hence S is onto. So there exists an element η such that Sη = ξ_{1}. Since T η = S^{2}η = Sξ_{1} = 0, we have η = λξ_{1} for some λ ∈ C and hence ξ_{1} = Sη = λSξ_{1} = 0, a contradiction.

Comment. There is an open subset of L(H) consisting of invertible operators with no square roots. Ref. J.B. Conway and B.B. Morrel, Roots and logarithms of bounded operators on Hilbert spaces, J. Funct Anal, 70(1987), 171-193.