QUADRATIC VECTORS. 403 



which is therefore an axis; coinciding with /Si or 182 if either Au or A23 

 vanish. Therefore conditions (93) are both necessary and sufficient. 

 To introduce double axes we now form the polar vector of Vp Fp by 

 operating with Sp'V or its equivalent, yielding 



Vp' { ^i{AnX2X3 + AnXsXi + AnXiXi + BiXs^) 



+ ^2(A2lX2X3 + A22XZX1 + A2iXiX2 + ^a^s^) 

 + ^3{A3iX2X3 + ^322-3X1 + + BzXz^) ] 



+ Vp{^i[An{x'2X3 + X2x'i) + Anix'zXi + Xzx'^ + Aii{x'iX2 + Xix'2) 



+ 2B,x'zXz] 

 + /32[A2i(x'2a:3 + X2x'^ + ^22(a;'3a;i + a:3a;'i) + A2z{x'iX2 + xix'a) 



+ 2B2x'3X,] 

 + /33U3i(a:'2ar3 + iC2.r'3) + ^132(0:^1 + Xzx\) + 



+ 2Bzx'zXzl (96) 



That a direction p shall be a double axis is the same as saying that the 

 polar vector equated to zero shall yield at most one scalar equation. 

 At /3i we have X2 = x^ = 0, leading to the vanishing of the determinant 

 (63), independently of the 5's. But this leads to Az2 = 0, because ^23 

 is not zero, while Azz vanishes. In a similar manner we have Az\ = 0. 

 That is, all three ^'s of the third line of (92) are zero. 



To make the direction /3i^i3 + i32^23 a double axis, we write, in the 

 polar vector, p = ^lAu + ^2A2z, Xi = An, X2 = ^23, ^-3 = 0. We 

 have also p = i8ia;'i + /32a;'2 + iSs^^'s. Substituting these values and 

 dropping accents we must have 



V{^iAn + 182^23} {i3iUu^23a-3 + ^i2-4 130:3 + ^13(^232^1+^13.1:2)] 

 + i82U2i^23a:3 + A22A1ZXZ 4- ^23(^233:1+ v4 130:2)] 



+ V{^,Xi + i32.T2 + ^3X3} 1/31^13^23 + ^2AizA2z'} = (97) 



equivalent to a single scalar equation in p. The coefficient of VlSz^i 

 is X3^i3M23, so that if neither ^13 nor ^23 vanish the required scalar 

 equation must be 0:3 = 0. This is the same as saying that the tan- 

 gent plane to the cones (3) at the required axis must be the plane of j8i 

 and ^2, (in so far as these cones do not possess double lines there). 

 The coefficient of F/32i33 yields nothing new. In the coefficient of 

 V^i^2, on the other hand, we have terms in 0:3 together with 



^13^23(^230:1 — ^130:2). (98) 



That the coefficient of Fj8i/32 shall be a multiple of 0:3 is thus impossible. 

 This completes the proof. 



