70 Br Baker, On the Invariant Factors of a Determinant. 



§ 4. To prove that the number A of linearly independent sets 

 of solutions of the equations D l x = is at least I, under the 

 hypothesis that (p — d) 1 is the highest power of p — in the deter- 

 minant \a — p\, we may proceed as follows. 



Denote D l = (a— 6) 1 by c ; the matrix of n rows and A columns 

 of which any column is formed by the constituents of one of the A 

 sets of solutions of esc = 0, by £ ; take another matrix r\ of n rows 

 and n — A columns arbitrarily so that the matrix a = (£, rj), of 

 n rows and columns, is of non- vanishing determinant. Every 

 solution cc of ex = can be written x = gt, where t is a row of A 

 elements ; and every matrix £', such as £, formed of A columns 

 constituted by linearly independent sets of solutions, can be 

 written £' = £t, where t is a matrix of A rows and columns. In 

 particular, since from c£ = follows aci; = ca% = 0, there is a 

 matrix /3 of A rows and columns such that a£ = £/3, which, if the 

 polynomial of order A in p, given by |/S — p\, be denoted by H (p), 

 satisfies the equation H (/3) = 0. Also the matrix cr -1 aa- is of the 

 form 



where 7, k are matrices of (n — X.) columns and respectively A and 

 n — A rows ; for the condition for this is 



»(fc *) = (£*)(£ *), 



namely a£ = £/3 together with 



which determines 7 and « ; and gives also 



car] = (c£ C17) r J = (0, en) (Jj = ct;« ; 



and we have 



\a — p\ = \/3 — p\ \k — p\. 



Let (p— 0) r be the highest power of p — 6 dividing \fi — p\ or 

 H (p), and hence dividing \a— p\, so that r 5 Z, r 5 A; we can find 

 two integral polynomials A (p), B (p) such that 



(p-ey=A(p)( P -ey + B(p)H(p); 



if then x= ^t be a set of solutions of c# = 0, or (a — 6) 1 x = 0, since 

 from a% = |/3 we infer a m ^= %fi m and hence H(a)% = %H (/3) = 0, 

 we have 



(a-6) r x = A (a) (a - 6) l x + B(a) H (a) %t = 



