### An extreme positive operator on a polyhedral cone

A. Guyan Robertson
1988 Glasgow Mathematical Journal
In , R. Loewy and H. Schneider studied positive linear operators on circular cones. They characterised the extremal positive operators on these cones and noticed that such operators preserve the set of extreme rays of the cone in this case. They then conjectured that this property of extremal positive operators is true in general. Shortly afterwards R. C. O'Brien  gave a counterexample to the conjecture, using a highly non-trivial result in  . Some time later the present author
more » ... sent author remarked in [4, Lemma 3.1] that a certain biquadratic form of M. D. Choi and T. Y. Lam  gives rise to an extremal positive operator on the cone of positive semidefinite matrices of order 3 under which no extreme ray of the cone is mapped to an extreme ray. Again, the proof of the relevant result is far from easy. The purpose of the present note is to give a very simple example which shows that the conjecture of Loewy and Schneider fails even for polyhedral cones. Let e 1; e 2 , e 3 , e 4 denote the usual basis vectors for R 4 and let e 5 = e, + e 2 -e 3 -e 4 = (1,1, -1 , -1). Let K be the cone generated by e l5 e 2 , e 3 , e 4 and e 5 . It is easily verified that e t ,. . . , e s are extremal in K. As usual, we shall write x 3= y if x-yeK and A 2= B if A and B are linear operators satisfying Ax 2= Bx for all xe K. We will need to use the fact that the cone K 45 generated by e 4 and e 5 is a face of K. That is: if x, y e K and x + ye K 4S , then x and y are in K 45 . In order to see this, suppose that K 45 contains the vector 5 X Ufa = {oc x + a 5 , a 2 + a 5 , a 3a 5 , a 4a 5 ), where 0)2=0 (l=£/=£5). Since the vector is a linear combination of e 4 and e 5 it follows that or, + a 5 = <x 2 + &5 = -(«"3 -<*s)-Hence ar t = a 2 = a 3 = 0, because a lt a 2 , ar 3 are non-negative. If follows easily from this that /Ct 5 is a face of K, as we asserted. We now construct an extremal positive linear operator on K which does not map extreme vectors to extreme vectors. The operator A is denned to have diagonal matrix diag(l, 1,1,0) relative to the standard basis e x ,e 2 , e 3) e 4 . Then Ae 5 = (1,1, -1,0) = e 4 + e 5 , so that A is a positive operator on K and maps the extreme vector e 5 to e 4 + e 5 , which is not extreme. Finally we claim that A is an extreme positive operator on K. For suppose that A 5= B, where B is a positive operator on K. If 1 ^; « 3 then e, = Aej 3= fie ; , so that Be, = j3 ; e y for some /?, ^ 0. Also 5e 4 = 0 since i4e 4 = 0. Now e 4 + e 5 = Ae 5 3= Be 5 . Therefore Be 5 = j3 4 e 4 + j8 5 e 5 , where ]3 4 , /3 5 2=0, since K45 is a face of K. Thus /3 4 e 4 + j3 5 e 5 = B(ej + e 2 -e 3 -e 4 ), Glasgow Math. J. 30 (1988) 347-348. https://www.cambridge.org/core/terms. https://doi.