ROTATIONS TX HYPERSPACE. 657 



that A = ± M or the complement of M is ± M. (From the defini- 

 tion of complement, the complement of M is m-yMi + miM2). If we 

 indicate complement by * we have the relations for 4-space. 



A.' 12 = "347 "■ 13 = A'42, K 14 = JCoS, k 23 = /i'l4, A" 24 = />"3i, /i'*34 = A'lo. 



Then if a complex is such that M* = ± M it can be written in the form 



M = (aioA'io + ank\3 + (lukii) ^ {(hsii'i-i + «i4A-23 + «i2A-34). 



The expressions in brackets are completely perpendicular planes and 

 in this case we have a resolution into completely perpendicular planes. 

 Hence: A complex 2-vector can always be resolved into the sum of two 

 completely perpendicular planes. 



By a proper choice of axes M can be written in the form 



M = niikn + moht 



and if M* = ± M the above resolution is not unique. We shall now 

 proceed to investigate this case. We saw that a complex could always 

 be resolved into the sum of tw'o planes one of which was arbitrary. 

 Let the arbitrary plane be X and write 



M = X + K. 



If r is any vector in A', consider the transformation 



/ = r-M = r-X -{- r-K. 



r-X is SL vector in X and if r' also lies in X, r-K must vanish since X 

 and K have no vector in common. But since r was any vector in X 

 the vanishing of r-K for all values of r means that K must be com- 

 pletely perpendicular to A'. Hence the resolution of M into the sum 

 of two perpendicular planes resolves itself into finding the planes left 

 invariant by the transformation / = r-M. We shall therefore find 

 the planes left invariant by this transformation. 



The invariants of this transformation are more easiiy found by 

 expressing it in the form of a dyadic. From formula (1) we have 



r-ikii) = (r-ki)k,- (r-ki)ki 



which can be ^\Titten as the dot product of /• into the dyadic kjki — 

 kikj. The transformation 



/•'= r-M 

 then takes the form 



