378 PROCEEDINGS OF THE AMERICAN ACADEMY. 



Then ^ _ u 



© ft © = ft, ® O O = ft, ©o 2 = 1 ; 



and, if 6' is properly chosen, —1 is not a latent root of ©, which is 

 then given by Cayley's expression. 



Since © is skew symmetric, 6' ft, B' ft, ft 0', ft 6', are commutative. 

 Therefore, o 



© ©a = © © 

 Whence follows 



(® © ) 2 = © 2 © 2 =1. 

 Let now 



ft® = r ; 



then f = — 1\ and r ft T = ft. We also have 

 and therefore 



r 4 = l. 



A symmetric matrix rn of non-zero determinant can be found to 

 satisfy the equation r w T = m . And if &? denotes any symmetric 

 square root of nr, the last equation may be written 



(^) _1 r^. bt* T (st*)" 1 = 1; 

 which, if ij/ = iu*T (nr*) -1 , becomes 



$ijs = 1. 

 Therefore, 



T= (tjl)- 1 ifr tat, 



in which if/ is orthogonal. But, since T ft V = ft, if/ is commutative 

 with (n7=) -I ft (zu?)~ l . Therefore, by (9) ip is equal to the product of 

 two expressions of the form 



[(^)- ! ft (nr*)" 1 - Y']- 1 [O^)- 1 ft (cr*)- 1 + Y'] 

 = 57* (ft - bt*Y' ari)- 1 w* • (bt*) -1 (" + srfY'sri) (bj*)- 1 

 = st* (ft - Y)- 1 (ft + Y) (w*)- 1 , 



in which Y' and therefore Y==E7£Ysr*, is symmetric. Whence it 

 follows that r = (zzi)~ l \p vfi is the product of two of Cayley's 

 expressions. Therefore, 



<£ = O <£ = n- 1 © = ft" 1 ® <£ 



= — r®<£ 



is equal to the negative of the product of four of Cayley's expressions. 

 Since — <f> is also a solution of the equation </>ft <£ = ft, it is also ex- 

 pressible thus. Whence it follows that $ is given by the product of 

 four of Cayley's expressions. 



