Abstract. The author showed in [10] that a certain self-dual subnormal operator is represented as the sum of mutually commutative normal operator and a quasinormal operator and also showed that the converse of this theorem is not true, i.e., a pure operator $S$ which is the sum of mutually commutative normal operator and a quasinormal operator is not necessarily a self-dual subnormal operator, although $S$ is subnormal. In this paper we examine a pure operator $S$ of this type. We observe that this decomposition is unique. Moreover we show a necessary and sufficient condition for a subnormal operator to be of this type. This is an improvement of Brown's theorem [2] concerning quasinormal operators. We also give a necessary and sufficient condition for an operator of this type to be a self-dual subnormal operator.

AMS Subject Classification (1991): 47B20

Received August 11, 1993. (Registered under 5577/2009.)