Full text: The collected mathematical papers of Arthur Cayley, Sc.D., F.R.S., late sadlerian professor of pure mathematics in the University of Cambridge (Vol. 12)

302 
ON THE QUATERNION EQUATION qQ — Qq'= 0. 
[836 
The tensor of Q is = (D 2 — A 2 v 2 ) (D 2 — B 2 v' 2 ); and we have 
J)2 _ ¿2 V 2 = (a _ 02)2 02 _ y2 _ 02), 
which, observing that 
= 6 6 — (2a + v 2 ) 6 4 + (a 2 + 2/3v 2 ) d 2 — v 2 (3 2 , 
a 2 + 2[3v 2 = /3 2 + 2ar 2 is =(6 2 — v 2 ) (6 4 — 2a6 2 + /3 2 ); 
and similarly 
I) 2 - B 2 v' 2 is = (6 2 - v' 2 ) (6 4 - 2<x6 2 + /3 2 ); 
hence the tensor is 
TQ = (6 2 - v 2 ) (6 2 - v' 2 ) (6 4 - 2a6 2 + /3 2 ) 2 , 
which, in virtue of 6 4 — 2a6 2 + y8 2 = 0, is = 0. 
The particular case is when Tq — Tq' — 0, that is, w 2 — v 2 — w' 2 + v 2 = 0, or say 
w 2 — w' 2 = v 2 — v' 2 , that is, 6 (w + w') = /3. Combining with this the general condition 
0 4 — 2a6 2 + /3 2 = 0, we find 6 2 {6 2 — 2a + (w + v/) 2 ) = 0, or in the second factor, for 6 and a 
substituting their values, we have 20 2 (w 2 - v 2 + w' 2 — v' 2 ) = 0, that is, 26 2 (Tq + Tq) = 0. 
Attending to the assumed relation Tq — Tq — 0, the second factor can only vanish if 
Tq = 0, Tq' = 0; hence, disregarding this more special case, the factor which vanishes must 
be the first factor, that is, 6=0; or the equations Tq — Tq' = 0 and 6 4 — 2a# 2 + ¡3 2 = 0 
give /3 — 0 and 6 = 0, that is, we have as already mentioned w — w = 0, and 
a? + y* + z 2 = x' 2 + y' 2 + z 2 , viz. the given quaternions have their scalars equal, and the 
squares of their vectors also equal. The equation here is vQ — Qv — 0; and writing 
v 2 = v' 2 = — p 2 , we see at once that a solution is 
Q = (— p 2 + vv) + A (v + v'), 
where A is an arbitrary scalar; in fact, with this value of Q, we have at once vQ 
and Qv' each 
= — p 2 (v + v) + A (—p 2 + vv'); 
and the equation vQ — Qv' = 0 is thus satisfied. The value of the tensor is easily 
found to be 
TQ = 2 (A 2 + p 2 ) (p 2 + scoc' + yy' + zz'), 
which is not = 0. 
In accordance with a remark in the introductory paragraphs, the solution 
Q = — p 2 + vv' + A (v + v') 
is not comprised in the general solution. As to this, observe that, in the case in 
question 6 = 0, ¡3 = 0, we have from the general theorem the form 
Q 
oM 2 
, ol6 , 
+ VV - -g (v + V); 
that is,
	        
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