Mr Bromwich, Theorems on Matrices and Bilinear Forms. 85 



Then by Cauchy's theorem, as there are no poles of the 

 function other than r = a, b, c, ... in the finite part of the r-plane, 

 we have 



A(s) + B(s) + G(s)+:.. = 1. 



But, considering the mode of formation of A (s), it is clear that 

 A (s) is divisible by all the factors of yjr (s) except (s— a) a ; or 

 A (s) is divisible by (s — by (s — c)?. ... In the same way B (s) will 

 be divisible by (s — a) a (s — c)v ... and so on. 



It follows that [A (s) . B (s)] and every similar product must 

 be divisible by ^(s); hence by the relation between A (s), B (s), 

 G (s), ... we see that [A (s)] 2 — A (s) is also divisible by yjr (s). 



Since A (s), B (s), C (s), . . . are rational algebraic functions of s, 

 we may put A for s in each of them ; let the resulting forms be 

 denoted by A a , A b , A c , ...\ 



Then since -v/r (A) = we have 



A£ = A a , A a A b = 0, 



and A a + A b + A c + ... = E. . 



Further, we have that (s — a) a A (s) is divisible by yjr (s) and so 

 on replacing s by A 



(A-aE) a A a = 0. 

 But we have 



(r — a) a — {s — a)° 



s — a 

 + 7 c„ + 



+ 



(s - a) c 



(r — a) a (r — s) r — a (r — a) 2 ' (r — a) a ' 



and combining this with our last result we see that 



E 



aE (A - aEy- 1 ' 



+ ... + v 



(rE-A)-iA a = +. 



r —a (r — ay (r — a) a 



Remembering that %A a = E, we now see that 



(rE-A)-\=2A a 



E 



i — a 



A-aE 



(r - a) 2 



+ 



(A - aE)*-* 

 (r — a) a 



for A a and A are commutative, as A a is a function of A only. 



Now taking the sum of the residues of (rE — A)~ l f{r) we see 

 that 



f{A) = XA a 



Ef(a) + (A-aE)f'(a) + ... 



(a — 1)1 J x ' 



(«-l)! 



1 It is easy to see that A a = A x , A b = B l ,... where A lt B lf C 1 ,... are clefineil as 

 before (p. 80). In fact Frobenius's proof in Crelle starts from the functions defined 

 as Aj , B 1 , C x , . . . have been. 



