QUADRATIC VECTORS. 453 



Case 2. Lot p = 0. As before, p'\ cannot be zero. Hence r = 0. 

 And as before, '\i q = 1 p'i= a, r'i= y, r"\= z, and p"\= 0. 



We cannot have r"i= 0. For if so we cannot also have q" \= 0, 

 for dp would be a monomial. Then //'= 0, (by 4), and /"= 0, 

 (by 11). This makes u = 0, (by 6), which is contrary to hypothesis. 



We must then have //'= 0,(by o) and ajjain 3 = 0, (by 4), or /•"] = 0, 

 just proved impossible. Hence tliis case is impossible. 



Case 3. Let r'i= 0. We cannot have //i= 0, by the same reason- 

 ing as before. We must then ha^•e /• = 0, (by 8), ;/ — 0, (by 7), and 

 r'= 0, (by 2), as before. Then r'\= 0, (by 14), and z = 0, (by 13). 

 But not both // and z are zero for this leaves (251) unchanoed. Hence 

 this case is impossible. Hence no term p^bp can be found. 



It remains to consider the possibility that (195) shall not possess 

 an axis other than /3i. If we write ^ = bi^i + 62182 + l^s^s, 'din\ 

 II — ("liSi + c-ilS-i + f3/33, the vector (195) becomes 



(3x(l) i.r2.r3+^i.»'2''+ ".rs"^) +i82(/>2-J"2-i'3+''2-'"2") +|83(63.r2.r3+ f';i.r2-+ a.v 1.C2) , 



(252) 



^^'hich we may denote as usual by Fp. If p = iSi.v\ + /32-'"2 + /S.-i-i's, the 

 vector equation I pFp = defines the axes of Fp, and, by multiplying- 

 out, is equivalent to the three scalar equations 



•T2(63.i'2.J'3 + f'3-1'2" + 'i->'\-r-i) — .r:i{h2.V2.V3 + c.r2"-^) = 0, 



.^3(61X2.1-3 + CiX2^ + uxs^) — .ri(/>3.r2.r3 + C3.r2- + a.r 1.1-2) = 0, 



.r 1(62.1*2.^-3 + CoXi^) — .V2(hi.v-\V:i + ci.vn' -\- ii.Vs^) = 0. 



If there is no axis except /3i, these ec[uations have no solution when .v-i 

 is not zero. P'urthermore, if .To is not zero, any solution of the first 

 and third equations simultaneously must be a solution of the second. 

 By elimination of .ri from the first and third equations we haxe the 

 cubic 



62^3:3^+ (262C2 — 6063 — au).i-2.C3-+ (co- — 63C2 — 6-)C3 — a6i)a:2^.r3 



— (aci+C2C3).r2^=0. (253) 



If this equation can be sohed for .rs, we can find .ii from the first of the 

 above cubics, and so have an axis other than /3i. That no such axis 

 exist it is necessary that 69 = 0, and hence also that au = contrary 

 to hypothesis. Hence such an axis must exist. This completes the 

 study of (195), and hence of irreducible quadratic vectors. 



