Abstract. It is shown that a closed algebra A of compact operators is triangularizable if and only if the algebra A/rnd(A) is commutative.An algebra A of operators on a Banach space X is triangularizable if there exists a maximal chain of subspaces of X invariant for A [7]. If X is finite dimensional, this reduces to the usual concept, since in this case A is triangularizable if and only if there exists a basis for X with respect to which all the operators in A have upper triangular matrices. Recently the concept of triangularizabüity has attracted some attention; see [2,6,7]. Also, for some earlier finite dimensional results see McCoy [9, 10]. In this paper we show that if A is a closed algebra of compact operators, then A is triangularizable if and only if A/rad(A) is commutative (where rad(A) denotes the Jacobson radical of A). It follows that if A is radical or commutative then A is necessarily triangularizable.Notation. X will always denote a complex Banach space, 9> (X) the algebra of bounded linear operators on X, and %(X) the ideal of compact operators. The spectrum and spectral radius are denoted by a and r respectively. Subspaces of X are always assumed to be closed. A always denotes a complex Banach algebra. If
A G 9) (X), a subspace M of X is invariant for A if T(M) G M (T G A). Finally if S, T G 9> (X), [S, T] denotes their commutator, [S, T] = ST -TS.Definition. A closed subalgebra A of 9> (X) is triangularizable if there exists a maximal chain of subspaces of X each of which is invariant for A.Before stating the next lemma we need to establish some notation. Let £ G 9> (X), and 'S a maximal chain of subspaces, each invariant for T. Proof. Ringrose [11] proves that ct(£)\ {0} = {aM(T): M E<5}\ {0}, from which the above result is immediate.