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,h, c, ... in the finite part of the r-plane, 
we have 
A(s) + B(s)+C(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)> ... and so on. 
It follows that [ A ( s ) . B (s)] and every similar product must 
be divisible by 'fr(s); hence by the relation between A ( s ), B ( s ), 
C ( s ), ... we see that [ A (s)] 2 — A ( s ) is also divisible by (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 , ... 1 . 
Then since yjr (J.) = 0 we have 
A a 2 =A a , A a A b = 0, 
and A a + A b + A c + . . . — E. 
F urther, we have that ( s — a) a A (s) is divisible by (s) and so 
on replacing s by A 
(A-aE)*A a = 0. 
But we have 
(r — a) a —(s — a) a 
1 s — a (s — a) a ~ x 
+ 7 ^ 2 +...+ V } 
(r — a) a (r — s) r — a (r — a) 2 ’ “ (r — a) 
and combining this with our last result we see that 
(rE-A)~ l A a = 
E A — aE (A — aEY _1 " 
T 7 ^ + ... + v 
r — a (r — a) 2 
(r — a) a 
A a . 
Remembering that 2A a = E, we now see that 
(rE-A)~i=2A a 
E A - aE (A — aEy- 1 
r — a (r - a) 2 (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)~ x f(r) we see 
that 
f{A) = ^A 
Ef{a) + (A-aE)f\a) + ... 
+ 
(.4 — aEy -1 
(“-!)! 
/-'(a) • 
1 It is easy to see that A a = A l , A b = B 1 ,... where A lf B lt are defined as 
before (p. 80). In fact Frobenius’s proof in Crelle starts from the functions defined 
as A lf B 1 , Cj , . . . have been. 
