QUADRATIC VECTORS. 435 



will lie in sets of three on five planes, no three planes having a com- 

 mon axis, except those through ^3. 



Case 2. (147) = 0. If /3i, jS^, and /Se are diplanar, choose them to 

 be zeros. Of the four others, ^-2, 184, 185, /Sy, (247) and (257) cannot be 

 zero, since (237) = 0. (457) cannot be zero since (147) = 0. (245) 

 cannot be zero since (345) = 0. The problem is then solved. But if 

 (136) = choose /So, 84 and /Se to be zeros, if diplanar. Then of the 

 four others /3i, ^3, /Ss, ^-, we cannot have (135) nor (137) zero because 

 (136) = 0. We cannot have (157) nor (357) zero since (567) = 0. 

 The problem is then solved. But if (246) = 0, choose |8i, /So, and /Ss 

 to be zeros. Of the four others, ^3, 184, /3g, 07, we cannot have (346) 

 nor (467) zero, since (246) = 0. AVe cannot have (347) = nor 

 (367) = since (237) = 0. The problem is then solved if (125) 

 does not vanish. But the three vectors 81, 02, 05 are the intersections 

 of planes through the other four axes taken in pairs. It has just been 

 shown that no three of these four can be coplanar. Therefore (125) 

 cannot vanish, and the proof is complete. The axes now lie in sets 

 of three on six planes, the four not chosen as zeros being at the inter- 

 section of three planes each. 



The form (197) is then always possible for a quadratic vector of the 

 first or general class, having just seven distinct axes. 



34. If an irreducible quadratic vector possesses one or more double 

 axes, but no triple axis, it may still be thrown into the form (197), 

 but the manner of obtaining a proper (^ and is somewhat different. 

 Let, at first, each double axis be replaced by two single axes, one coin- 

 ciding with the original double axis, the other lying in the original 

 tangent plane to the cones (3) at the double axis. Then let the selec- 

 tion of three axes proceed as in Art. 33, these three to be diplanar, 

 and of the four others no three to be coplanar. Then let the axes in 

 the respective tangent planes approach their original positions. 

 We now distinguish two cases. 



Case 1. If the three axes selected remain single axes, let them be 

 reduced to zeros, the vector then taking the form of the right member 

 of (201). The vectors ai, ao, a^, cannot be coplanar, and the factoriza- 

 tion into the form (197) proceeds as before. 



Proof. Suppose ai, 02, and as to be coplanar. Put 



as = viai + na2. 

 The vector Fp takes the form 



aiixoXs + 7)1X1X2) -\- ao(.i-3a;i + nx 1.1-2). 



