THE DYADICS IN THREE DIMENSIONS. 429 



by a symbolic multiplication of B and CDE analogous to the outer 

 product. 



Again, let pq be the two-two idemfactor. Then 



[Bp] [aq] 



is a three-three dyadic which transforms any plane ^ into a plane 



r= [Bp]{^q-^)^ B-pq-^^= [B-m 



To interpret this let A', Y, Z be any three non-collinear points on ^ 

 and X', Y', Z' their transforms by B(3. Join X' to YZ, Y' to XZ, 

 Z' to A'}'. By the general theorem of the preceding section these 

 planes intersect in a point on ^ . This is true, whatever points X, Y , Z 

 are taken on ^. This involves the following geometrical theorem. 

 Take four points A"^, Y, Z, W in a plane, no three of which are col- 

 linear, and four points X' , Y' , Z' , W in a second plane. A dyadic 5/3 

 can be found which will transform A", Y, Z, W, into A'', Y' , Z' , W. 

 Joining the points A', Y, Z and A"', Y' , Z' , W as above we get a point. 

 Similarly, }', Z, W and Y', Z' , W determine a second point etc. 

 In this way we get four points which lie in a plane, namely, in the 

 plane into which [Bp] [/3g] transforms [ATZ]. 

 If we write B^ in the form 



5/3 = B[CDE], 



[Bp] [•/3g] = [Bp][{qCD)E - (qCE)D + iqDE)C} 

 = B-pq-{[CD]E - [CE]D + [DE]C] 

 = [BCD]E - [BCE]D + {BDE)C. 



This can be regarded as a symbolic product of B and [CDE] analogous 

 to the outer product. 



Finally let aA be the three-one idemfactor. The double product 

 of this and the dyadic 5/3 is 



(bo) (J3A) = - b-aA-13 = - (6/3). 



When this is zero the scalar, or linear invariant, of the dyadic vanishes. 

 When the double product of two dyadics vanishes, we shall call them 

 apolar. To interpret this let A\, A2, A3, Ai be the vertices of a tetra- 

 hedron and ai, ai, as, 04 the opposite planes. Let B(3 transform 

 Ai, Ao, As, Ai into the points Ai'. A-/, A3, A4. The general theorem 

 states that {Ai'ai), (A2'a2), (As'as), (^14^4) will satisfy a linear relation. 



