'Column operations' and 'Gram-Schmidt' on quaternion sums A classic problem with quaternion equations is how to 'simplify' *linear* sums such as A1 Z B1 + A2 Z B2 + A3 Z B3 + ..., where Ai, Bi, Z are all general quaternions. The game here is to avoid converting quaternion equations into sets of linear equations by appealing to an existing 4D coordinate system -- e.g., 1,I,J,K. If these letters Ai, Bi, Z all denoted complex numbers, then the solution is trivial: A1 Z B1 + A2 Z B2 + A3 Z B3 + ... = (A1 B1 + A2 B2 + A3 B3 + ...) Z Unfortunately, the quaternions aren't commutative, so this plan fails. Nevertheless, we can still perform certain types of simplification -- e.g., I want to 'eliminate' the 'scalar part' of A in the first term of sum AZB+CZD. If s(C)=0, then simply exchange the terms: CZD+AZB and we're done. On the other hand, if s(C)/=0, then let R=A-C*s(A)/s(C); clearly s(R)=0. Then A=R+C*s(A)/s(C), so AZB+CZD = (R+C*s(A)/s(C))ZB+CZD = RZB + CZB*s(A)/s(C) + CZD = RZB + CZ(D+B*S(A)/s(C)) We now have a 2-term sum with s(R)=0. We can perform an analogous operation to eliminate the scalar part of B in the first term of the sum AZB+CZD. If s(D)=0, then we exchange terms as before. If s(D)/=0, then let S=B-D*s(B)/s(D); clearly s(S)=0. AZB+CZD = AZ(S+D*s(B)/s(D))+CZD = AZS + (s(B)/s(D))AZD + CZD = AZS + (C+A*s(B)/s(D))ZD We now have a 2-term sum with s(S)=0 (and since A didn't change, it will also preserve s(A)=0 if we previously assured that property). But we're just warming up. What if R=A-(A.C)/(C.C)*C ? (where (A.C) denotes the 4D dot product: (A.C)=s(AC').) Then A=R+(A.C)/(C.C)*C, and AZB+CZD = (R+(A.C)/(C.C)*C)ZB+CZD = RZB + CZB*(A.C)/(C.C) + CZD = RZB + CZ(D+B*(A.C)/(C.C)) So now R.C=0, and we have R orthogonal to C. We can do the same thing with B,D: Let S=B-(B.D)/(D.D)*D. Then B=S+(B.D)/(D.D)*D, and AZB+CZD = AZ(S+(B.D)/(D.D)*D)+CZD = AZS + (B.D)/(D.D)*AZD + CZD = AZS + (C+(B.D)/(D.D)*A)ZD So S.D=0, and we have S orthogonal to D. So we have a type of 'Gram-Schmidt' orthogonalization process. What can we do with these types of operations? 1. Given a sum of n terms of the form Ai Z Bi, we can assure that at most one of the Ai's and at most one of the Bi's (not necessarily for the same i) has a non-zero scalar part. 2. Given a sum of n terms of the form Ai Z Bi, we can assure that all of the Ai's are orthogonal, OR all of the Bi's are orthogonal (making Bi's orthgonal messes up the orthogonality of the Ai's, and vice versa). Since the dimension of the quaternion space is 4, these operations produce sums having at most 4 non-zero terms. This is messy for symbolic algebra, but works just fine for numerical quaternions. 3. These operations extend to quaternions over other fields -- e.g., Zp -- even when the characteristic is 2.