THE DYADICS IN THREE DIMENSIONS. 421 



the p's being complexes or lines and the X's numbers. This reduction 

 can be accomplished by the same transformation that reduces the 

 quadratic form 



i:>aikXiXh (A) 



to a sum of squares. Suppose, in fact, the transformation 



iji = ^fXikXk (B) 



reduces (A) to the form 



•When we replace the y's in (C) by the values from (B) and expand, 

 the coefficient of XiXk will be 



O-ik + 0-ki 



as in (A). Now let 



Pi = ^ixikqk (D) 



Then 



Xipipi + ^iViPi + + Xepe^e (E) 



will be equal to rs. For when we replace the p's in (E) by their values 

 from (D) and expand, the sum of the coefficients of pipk and pkPi in 

 the result (as in the case of the quadratic form) will be 



(^ki + O-ik = 2aifc. 



Also from the symmetry of (E) it is clear that the coefficients of 

 PiPh and pkPi will be equal. Hence each is equal to a a and so (E) 

 is equal to rs. In the manipulation of the dyadic and the quadratic 

 form the principal difference is that XiXk = XkXi while pipk and Pkpi 

 need not be equal. The above discussion shows that the reduction 

 to the sum of squares does not require that the individual products 

 be commutative but merely that the whole dyadic be self conjugate. 

 Any self-conjugate two-two dyadic can be expressed in the form 



rs = \{pipi + piPi) + At(p2P5 + piPi) + vipspe + PePs) 



where pi, p2- ■ -Pq are complexes {or lines) and X, tx, v numbers. To show 

 this, first reduce rs to the form 



rs = \iqiqi + \2q2q2 + ....+ Xe^e? 6- 



