664 MOORE. 



when Mi are mutually perpendicular unit simple planes and ?/;, num- 

 bers. We can then derive ^J — 1 complex 2-vectors as follows 



A = (MxM) ■ M = 'IZm on^HU i ^ j 



1 



B = (MxMxM)-(MxM) = QZmimj^mK^Mi i 7^ j 7^ k. 

 (20) 



P = (MxMxM. . .p factors) • (JfxJfxM . . . {p-l) factors) 

 = p!m\m2. . .inpZmiino. . . ??? ,:_i??^+i . . .vipMi. 



We have then all together p equations to solve for the j) plane il/i, 

 3l2,...Mp- and the solution will be unique unless these equations 

 prove not to be linearly independent, that is imless the determinant A 

 of the system vanishes. If we observe how the columns of this 

 determinant are made up we see that it contains each of the m's as a 

 factor. x\lso if mi = ^^nij, A = 0. Therefore we can write A in the 

 factored form 



A = •p'.{v-iim2. ■ .7/?p)x(?»r — mf). 



Now if ???i = ±/?i2 say, and the other ins, are all different then a value 

 of X can be found so that 



A + \M 



will lie in a space of 2p — 4 dimensions and by the above argument 

 this new complex can be resolved into the sum of p — 2 mutually per- 

 pendicular plane vectors. The remaining 4-space will be completely 

 perpendicular to the space in which A + \M lies and that part of M 

 lying in it can be resolved into the sum of two perpendicular planes 

 in CX3 2 ways. Hence the whole complex can be resolved into the sum 

 of p mutually perpendicular planes in 00- ways. The same argument 

 of course applies to any pair of equal roots. If 7?u = ^mi = ^niz 

 the other m's being all different and different from mi, a value of X 

 can be found so that A -\- \M will lie in a space of 2^ — 6 dimensions 

 and the part of M lying in this space can be resolved into p — 3 mutu- 

 ally perpendicular planes. -The whole complex can then be resolved 

 into the sum of p mutually perpendicular planes provided the complex 

 2-vector in space of six dimensions such that (MxM) • M = \M can be 

 resolved into the sum of three mutually perpendicular planes. Con- 

 tinuing the argument we see that the resolution is always possible 

 provided that in a space of 2m dimensions in which 



{MxM)-M = \M 



