Mr Bromiuich, Theorems on Matrices and Bilinear Forms. 87 



It is known that a substitution P can be found 1 such that 

 A = P(A 1 + A 2 )P~\ 



where, corresponding to the invariant-factor (r — a) a , we have 



A x = a (x 1 y 1 + ... + x a y a ) + x 2 y x + x z y 2 + ... + x a y a _ x 



= aE x -f- Cj say, 



and A 2 does not contain any of the variables w 1} ..., x a , y 1} ..., y a . 

 Now we have 



f(A) =f[P(A 1 + A 2 )P-*] = Pf{A, + A 2 ) P~\ 



so thaty(-A) has the same invariant-factors asf(Ax + A 2 ). Again, 

 since A 1} A 2 have no common variables, we have 



f(A 1 + A 2 ) =f(A 1 ) +f(A 2 ) =f(A 2 ) +f(aE 1 + C\) 



=f{A 2 ) + E x f(a) + C\f (a) + %f" (a) + ...+ (^/^ («)■ 



To see that the last line is correct we have only to calculate 

 (rEi — AJ' 1 and expand in powers of (?^ — a). We find that 



(rE, - A,)- 1 = — i- + -. i-^ + -. !— +...+■ 



r — a (r — a) 2 (r — of (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 



/'(a) = 0,/"(a) = 0, ...,/*- (a) = 0, 



and thus here 



/(A, + A 2 ) =f(A 2 ) + EJia) + |*/* (a) + . . . + (^y/^ 1 <«)• 



So we have to calculate the invariant-factors of the deter- 

 minant 



| rE -/(A, + A 2 ) | = | {rE, -/(A,)} + [rE 2 -f{A 2 )} |; 



since [rE 1 —f(A 1 )} 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). 



