QUADRATIC VECTORS. 453 



Case 2. Let p = 0. As before, p'l cannot be zero. Hence r = 0. 

 And as before, if g = 1 j)'i= a, r\= y, r"\= z, and ?j"i= 0. 



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

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

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



We must then have p"= 0,(by 5) and again 2=0, (by 4), or r"] = 0, 

 just proved impossible. Hence this case is impossible. 



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

 ing as before. We must then have r = 0, (by 8), y = 0, (by 7), and 

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

 But not both y and z are zero for this leaves (251) unchanged. Hence 

 this case is impossible. Hence no term pS8p can be found. 



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

 an axis other than /3i. If we write f — hi^\ + 1)2^2 + ^sjS.?, and 

 11 = ri/3i + Co^o + CsiSs, the vector (195) becomes 



i8i(6i.r2.r3+ri.r2-+"-r3-)4-/32(62.r2-f3+f'2.f2^)+/33(/^3-»'2J"3+C3.r2"+a.ViX2), 



(252) 



which we may denote as usual by Fp. If p = /Si.i'i -(- (32X2 + PsXs, the 

 vector equation VpFp = defines the axes of Fp, and, by multiplying 

 out, is equivalent to the three scalar equations 



^2(63^^2-^3 + f3.r2" + r/.ri.ro) — .rs{b2X2J-3 -\- cx-^) = 0, 



Xi{h\X2Xz + ClXo- + nx{~) — X\{}hX2Xz + CiX-? + axiX2) = 0, 



.Ti(62.r2.r3 + C2-f2") — X2{bxX2Xi + ri.r2^ + ux:^) = 0. 



If there is no axis except jSi, these equations have no solution when .r2 

 is not zero. P'urthermore, if X2 is not zero, any solution of the first 

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

 By elimination of Xi from the first and third equations we have the 

 cubic 



'^2"-i"3''+ (26202 — 6263 — aM).r2.r3"4- (c2^ — ^sCa — 6203 — 061)0- 2^X3 



— (aci+ C2C3)x2^ = 0. (253) 



If this equation can be solved for X3, we can find Xi from the first of the 

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

 exist it is necessary that 62 = 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. 



