QUADRATIC VECTORS. 449 



and since r is not zero it follows that p"q — pq" = 0. But from 4, 

 writing for //'i and q'\ their values, cp and cq, 



- c{p"q - pq") = 9 - z. 



Therefore g = z = 0, contrary to fact, since if g vanishes the quad- 

 ratic vector is reducible. Thus if r'l = 1, p = 0, which is equivalent 

 to saying that either p or r'l must vanish. 



Case 2. Let p = 0. We need not now assume r'l = 1. By 8, 

 either p'l or r must vanish. Suppose p'i = 0. By 9, we have a = 0, 

 But if a = the vector (241) has, at jSi, a quadruple axis of the first 

 kind, contrary to hypothesis. 



Suppose p = r = 0. Then, by 9, neither p'l nor q can vanish. If 

 we take 9 = 1, we have p'l = a, (by 9), and r\ = y, (by 7). Also 

 r"i = z, (by 13), and p"i = 0, (by 15). Now r"i cannot vanish, for if 

 so we have, by 6, either q"i = 0, or r" = 0; but if q"i = 0, g = 0, 

 by 4; and if r" = 0, p" = by 11, (we cannot have r'l = 0, since 

 not both 2/ and z vanish), and again g = (by 4). Hence r"i is not 

 zero. 



We therefore have p" = 0, (by 5), and z = g, (by 4). Equation 6 

 now becomes 



qr"r" - gq" = 0, (A) 



which will serve to determine q" . There remain the six equations 

 1, 2, 3, 10, 11, 12. Substituting values, these become, 



1. q\r' - q'y = h, 10. q\r" - q'g + q",r' - q"ii = 



2. p'y — ar' = — y 11. p'g — ar" = — g 



3. aq' - p'q'i = 63 12. aq" - p'q'\ = g^ - y. 



By elimination of q" and p' from 11, 12, and (A), we have 



q"i = gi - >/. (B) 



By elimination of q' and /' from 1, 2, and 3, 



y{q\ - h) = abi. (C) 



Since neither a nor bi can be zero, y cannot. Hence this equation 

 gives a value for q'l. By 11, we have r" in terms of p'. It is then an 

 easy calculation to substitute, in 10, all other unknowns in terms of p' 

 and y. We find all coefficients of powers of y cancel out, giving 

 big = 0, which is impossible since the vector is irreducible. Hence 

 we cannot have p = 0. 



