836] 
ON THE QUATERNION EQUATION qQ — Qq' = 0. 
301 
Instead of directly working out the condition in the manner indicated above, I 
present the investigation in a synthetic form as follows: 
Taking v, v for the vector parts of the two given quaternions, so that q = w + v, 
q' = w' + v', I write for shortness 
0 = w — w', 
a= v 2 + v' 2 , = — x 2 — y 2 — z 2 — x' 2 — y' 2 — z 2 , 
/3 = v 2 — v 2 , = — x 2 — y 2 — z 2 + x' 2 + y' 2 + z 2 , 
D — — 6 ( a — d 2 ), 
A = /3 - 0 2 , 
B = /3 + 0 2 ; 
so that 0, a, (3, D, A, B are all of them scalars. With these I form a quaternion 
Q = (D + Av)(D + Bv'); I say that we have identically 
qQ — Qq' = [D — v. v 2 + v.v 2 + vv'. 6) (6* — 2a6 2 + /3 2 ). 
It of course follows that, if 6 i — 2a6 2 + /3 2 = 0, then qQ — Qq' = 0, viz. — 2a0 2 + y8 2 = 0 
is the condition 0 = 0, for the existence of the required quaternion; and this condition 
being satisfied, then (omitting the arbitrary scalar factor) the value of the quaternion 
is Q = (D + Av) {D + Bv'), a value giving T(Q) = 0, that is, W 2 + X 2 + Y 2 + Z 2 = 0. If 
0=0 (that is, w = w'), then the condition becomes /3 = 0, that is, x 2 + y 2 + z 2 — x 2 — y' 2 — z' 2 = 0; 
and these two conditions being satisfied, Q ceases to have the determinate value given 
by the foregoing formula: it has a value involving an arbitrary parameter, and is 
no longer such that W 2 + X 2 + Y 2 + Z 2 = 0. 
The identical equation is at once verified; we have 
qQ - Qq' = (w + v) Q - Q {w - v) 
= 0 (D 2 + DAv + DBv' + ABvv') 
+ v (D 2 + DAv + DBv' + ABvv) 
— (D 2 + DAv + DBv + ABvv') v 
= 0D 2 + DAv 2 — DBv 2 
+ v (DA0 + D 2 — ABv' 2 ) 
+ v (DB0 + ABv 2 — D 2 ) 
+ vv'(0AB+DB -DA ). 
The first line is here = D {D0 + Av 2 + Bv' 2 }, viz. the term in { } is 
- (a - 0 2 ) 0 2 + (/3 - 0 2 ) v 2 - (/3 + 0 2 ) v 2 , 
= - OL0 2 + 0 i + /3. /3 - 0 2 . a, 
= 0 4 - 2a^ 2 + /3 2 ; 
and similarly each of the other lines contains the factor 0 4 — 2a0 2 + /3 2 , and the equation 
is thus seen to hold good.