Mr Bromwich, Theorems on Matrices and Bilinear Forms. 87 
It is known that a substitution P can be found 1 such that 
A = P(A X + A 2 ) P~\ 
where, corresponding to the invariant-factor (r — a) a , we have 
A 1 = a (x x y x + ... + x a y a ) + x 2 y x + x 3 y 2 + . . . + x a y a _ x 
= aE x + G x say, 
and A 2 does not contain any of the variables x ly . .., x a , y x , . .., y a . 
Now we have 
f(A) =f [. P (A, + A,) P-] = Pf(A x + A 2 ) P~\ 
so that f(A) has the same invariant-factors as f(A 1 + A 2 ). Again, 
since A ly A 2 have no common variables, we have 
f(A x + A 2 ) =f(A x ) +f(A 2 ) =f(A 2 ) +f (aE x + C x ) 
= f(A 2 ) + E x f (a) + G x f' (a) + ~f" (a) + . . . + f a ~ l («)• 
To see that the last line is correct we have only to calculate 
(rE x — A x )~ y and expand in powers of (r — a). We find that 
(rE x - A,)- 1 = 
G x 
* + 
E x 
r — a 1 (r — of ' (r — a) 
+ 
(r - a) a 
and so our expression for f(A x ) follows by what has been proved 
before. 
But in the present case we have 
f'(a) = 0,f"(a) = 0, = 0, 
and thus here 
f(A l + A 2 ) =/( Ad + EJ(a) + ^/* («)+••• + (“!)! 
So we have to calculate the invariant-factors of the deter- 
minant 
| rE -/( A x + A) l = | [rE, - f{A , )J + [vE, -f{A 2 )} |; 
since [rE x — /(Aj)} and {rE 2 — f(A 2 )} have no common variables 
we may calculate their invariant-factors separately. So consider 
1 For the actual determination of P we may consult Jordan, Cours d' Analyse, 
t. 3, p. 175 et seq. ; also a series of papers by Burnside, Baker, Dixon and the author 
in volumes 30 — 32 of the Proceedings of the London Math. Society (1899 — 1900). 
