418 HITCHCOCK. 



that not more than q' + N — 2 of the E's shall be polynomials of the 

 same standard set. For when several E's belong to the same standard 

 set they are functions of a set of m + 1 of the a's not including a,. 

 Since any element of the determinant contains every one of a set of 

 m + N — 2 of the a's (exclusive of a/), and also contains x, it follows 

 that the corresponding b's will possess N — 3 of the a's in common. 

 That is, they have in common A 7 — 2 arbitrary vectors not belonging 

 to the reference system of the E's which occur in the same elements. 

 Now by theorem IV, if q' + 1 of the E's belong to a standard set 

 they satisfy an identity 2L(ay)£(a,-) = 0, but the L's are functions 

 of those a's only which occur in the E's. By writing for the E's an 

 identity of the form (32) it is evident that even when the L's contain 

 N — 2 arbitrary vectors we shall have q' + 1 + (N — 2) of the E's 

 (which now correspond to the C's of (32)), connected by a relation 

 2LE = 0. Hence only q' + N - 2 of the E's can belong to the 

 same standard set, as was to be shown. 



We are how in a position to prove that the b's may in general be 

 selected by the following rule : 



Rule for choice of b's. Select any m + N — 2 from the given 

 set of n — \ vectors ar • -a„_i. Choose q' + N — 2 polynomials 

 Ei, Ei, • • • out of the standard set of degree K — 1 based on m + 1 

 of these vectors, which we may number from 1 to m + 1. 



Make a second selection of m + 1 a's by omitting ai from the first 

 selection and adjoining a m+2 , choose q' -\- N — 3 polynomials of the 

 new standard set based on this second selection of a's. 



If we have not yet q polynomials, make a third selection of m + 1 

 a's by omitting a m+ 2 from the second selection and adjoining a^+z', 

 choose q' + A T — 3 polynomials from a third standard set based on 

 this third selection of a's. 



If we have not yet q polynomials, proceed in a similar manner to 

 form new groups until q polynomials have been obtained: every selec- 

 tion of a's consists of ao, a3, • • -a^i together with one other. Each 

 of the b's contains every a not in its own ' E ' but included in the 

 original choice of m + N — 2 of the a's. 



We have to prove that the determinant whose elements are b*a ; 

 E(&j) does not vanish identically, and that the rule is always possible. 



We note that, by the rule, ai occurs only in the E's numbered from 

 2 to q' + N — 2 inclusive. Hence ai occurs in all the b's except 

 those so numbered. If the determinant vanishes identically a set of 

 numbers cr • -c q exists such that Scb»a,-jE(a,-) = 0, the c's being inde- 

 pendent of a ; and not all zero. Now let a, = x + g&i where g is 



