THE DYADICS IN THREE DIMENSIONS. 413 



Let 



Bn = 'EfJLnmAm- (E) 



Using these values, let 



XBnBn = l^VikAiA,. (F) 



Since (F) is obtained from {E) in the same way that (D) is from (C) 

 except that AiAk is not equal to AkAi we must have 



vik + Vki = Xtfc + Xti =^ 2Xifc. (G) 



Also, since the left side of (F) is symmetric 



Vik = Vki- (H) 



From (6r) and {H) we get 



Vik = Xifc. 

 Therefore (F) is equivalent to 



BB = liBnBn = ^'KikAiAk. 



A seJj-conjugaie dyadic represents a polarity. For the transform of a 

 plane ^ is the point 



X = B(B^) = B,{B,^) + B^iBo^) + 53(53^) + B,{B,^). (36) 



This point is the pole of ^ with respect to the quadric surface 



{B,^y + {B.2^y + (B^^y + (54^)^ = 0. 



The points Bi, B2, B3, Bi form a self-polar tetrahedron with respect 

 to the quadric. For, the pole of the plane [Bi B2 B^, from (36), is 



BiiBiB.BoBs) + BoiB^B.B.Bs) + BziBzB.B.Bz) + B.iBiBiB.B^) = 



—Bi{BiB'2B3Bi). 



Thus the pole of [BiB-^Bs] is the point Bi and similarly with the other 

 faces of the tetrahedron. 



An anti-self -con jugate dyadic represents a null-system. For if, 



BC = B\C\ -\- B2C2 ~i~ B3C3 -\- BiCi 



■ is anti-self-conjugate, 



i:BiCi = -i:CiBi = iS(J5iCi - dBi). 

 Hence 



(|£)(Q) = ^2{(^B,)(C.-^) -i^Ci){B,^)} = 0. 



