Let A = B(X), the Banach algebra of bounded linear mappings from a Banach space X into X, and T Î Sing(A). For y Î X and g Î X, the rank one operator y[`(Ä)] g Î B(X) is defined by (y[`(Ä)] g)(z) = g(z)y  (z Î X).
If T isn't 1-1, then there exists an element x ¹ 0 such that Tx = 0. So if f Î X^# and f(x) = 1, then T(x[`(Ä)] f) = Tx[`(Ä)] f = 0. So T is a left divisor of zero.
If T(X) ¹ X and T(X)^- ¹ X, then by the Hahn-Banach theorem there exists a non-zero functional f such that f(T(X)) = 0. Therefore (x[`(Ä)] f)T = 0, for all x Î X. Thus T is a right divisor of zero.

Finally if T(X) ¹ X and T(X)^- = X, then there exists a sequence (y_n) in X satisfying ||y_n|| = 1 and Ty_n® 0. If f Î X^# with ||f|| = 1 and U_n = y_n[`(Ä)] f, then ||U<sub>n</sub>|| = ||y<sub>n</sub>|| ||f|| = 1 and ||TU<sub>n</sub>|| = ||Ty<sub>n</sub>[`(Ä)] f|| £ ||Ty_n|| and hence TU_n® 0. Thus T is a right topological divisor of zero.