82 Proceedings of the Royal Irish Academy. 



Now, x^'^p = ViqViqp = ^p, or generally $ = i//^. Therefore, the 

 equation satisfied by ij/ is simply the square of that satisfied by ij/, 

 and the roots of $ are equal in pairs. This agrees with the conclu- 

 sions of the last article. 



12. A discussion of the roots and axes of the linear vector ij/ 

 presents some points of interest. 



If ^ is an axis, and i the corresponding root, 



il/j3 = hl3, and ^/3- = SI34f(3 = S^qfS = 0. 



Hence, either h or (3 = 0. But all the roots are not zero ; so if 5 does 

 not vanish, y8 must be of the form a + ha', in which a and a' are two 

 real vectors at right angles to one another and of equal lengths, and h 

 is the imaginary y^- 1 of algebra.^ ^ is thus the vector to a circular 

 point at infinity in the plane of a and a'. /? being imaginary, h must 

 be imaginary also, and - b must be its conjugate, as is evident on 

 inspection of the symbolic equation in the last article ; so 5 is of the 

 form hff, where g is real. 



Replacing b by hy (where h = y^- 1), it is evident, as 



if/{a + ha) = hg (a + ha'),1 



that '/'(«■"" ^^^') = - ^iy {^ ~ ha), 



also. Hence, adding and subtracting, 



ij/a = - ya', and ij/a' = ya; 



also, i/^^a = - yij/a' = - y'^a, and ij/'^a' = - g'^a . 



Again, if for some other root, b^ (= hy-^, 



xpjBi = hy.jBi, and ^/5'i = - hy^p\ 

 we have 



SjB^P, = hy^SlSIBi = + ^^Sif ^ = - >S/?i./^/5 = - hy 8(3,(3, 



and if y is not - y,, S(3^i = 0. In like manner, 



S(3'fi, = S(3(3\ = S(3'(3\, and if (3, = a, + ha\, 



Saai = Saa'i — Sa'a, = Sa'a'i = 0, 



SO the planes of aa' and of aia'i are hyper-perpendicular, or every line 

 in one plane is perpendicular to every line in the other. 



1 Veiificatiou is easy ; for 6- = (a + ha')'^ = o'^ + h'^a''^ = a^ - a'^. 



