93 
of Edinburgh, Session 1866-67. 
Hence 9i9^PiPh = S <P;\?p. 2 
= S P, P> 2 > or = Sp^Vi, 
because (p is its own conjugate. 
But fp 2 = y 2 2 p 2 , 
fpi = 9 1 Pi j 
and therefore 
9i9*&PiP* = ^Spip 2 = pfSp 1 p 2 ; 
which, as ^ and p 2 are by hypothesis different, requires 
Spi p 2 = 0. 
Similarly Sp 2 p 3 — 0, Sp 3 p x = 0. 
If two roots be equal, as y 2 ,p 3 , we still have, by the above proof, 
Spip 2 = 0, and Sp 1 p 3 = 0. But there is nothing farther to determine 
p 2 and p 3 , which are therefore any vectors perpendicular to p x . 
If all three roots be equal, every real vector satisfies the equation 
{<p-g)p = 0. 
Next, as to the reality of the three roots when the function is 
self- conjugate. 
Suppose g 2 + - 1 to be a root, and let p. A + a-.J — 1 be the cor- 
responding value of p, where g 2 and h 2 are real numbers, p 2 and cr 2 
real vectors, and V — 1 the old imaginary of algebra. 
Then <p(p 2 + a.y - 1) = (g 2 + T) (p 2 + oV - 1), 
and this divides itself, as in algebra, into the two equations 
( Pp i = 9-2p-2- 
f>cr 2 = h 2 p 2 + g 2 (r 2 . 
Operating on these by Scr 2 , Sp 2 respectively, and subtracting the re- 
sults, remembering our condition as to the nature of <f> 
= Sp a pcr 2 , 
we h ave h(cr 2 + p 2 2 ) = 0. 
But, as cr 2 and p 2 are both real vectors, the sum of their squares 
cannot vanish. Hence h 2 vanishes, and with it the impossible part 
of the root. 
