I t is well known that every non-empty closed subset of the complex plane is the spectrum of a normal operator in a HUBERT space. This result cannot be extended to the case of BANACH spaces. Indeed, it is shown in [17] that the spectrum of a spectral operator in a GR~THENDIECX space with the DUNFORD-PETTIS property is a finite set; the argument is based on the fact that such a BANACH space has no SCHAUDER decomposition. The purpose of this note is to show that, in a locally convex spice, the existence of an unconditional SCHAUDER decomposition ensures the property that every non-empty closed subset of the complex plane is the spectrum of a scalar-type spectral operator (see Theorem 6). As a consequence, a BANACH space has the above property if and only if it has an unconditional SCHAUDER decomposition (see Corollary 7).Throughout this note, let X be a sequentially complete, locally convex &us-DORFP space and X' its dual space. Let L ( X ) denote the space of all continuous linear operators from X into itself. If the space L ( X ) is equipped with the topology of pointwise convergence on X, then it will be denoted by L , ( X ) . The identity operator on the space X will be denoted by I . Let T be a linear operator defined on a subspace of X with values in X. Suppose that 1 is a complex number such that the operator 1I -T is an injection with dense range whose inverse (U -T)-1 defined on the range of 1I -T is extendable to a continuous operator from X into itself. Then, the extension of (11-T)-1 to X will be denoted by R (1.; T). The resolvent equation
R(1; T)-R(C; T ) = -( I -[ ) R ( I ; T ) R([; T )holds whenever I and i are complex numbers for which the operators R(A; T ) and R(C; T ) exist.The resolvent set e(T) of a linear operator T from a subspace of X into X consists of those complex numbers 1 such that there is a neighbourhood U of 1 in the