376 PROCEEDINGS OF THE AMERICAN ACADEMY. 



7. If <f) is symmetric, 6 ft + ft — 0, and therefore <j> = <f> . Let 



© = O <t> , 



then © =: — ®, ®fi® = fi; moreover, © 2 = —1. Consequently, © is 

 given by Cayley's expression ; and therefore 



is the product of three of Cayley's expressions. 



8. If cf> is orthogonal, it is commutative with ft ; therefore 6 and 

 are commutative with ft. Consequently, 



<£ = e -hn(9 + d) — e - .'(9 + 0)0 _ <£-l 



From the last equation follows 



ft (f) Q = cfiQ ft. 



Let now 



® = ft<£ ; 



then © = —0, ©ft© == ft, and ® 2 = — 1. Consequently, © is given 

 by Cayley's expression. 



We have </> = ft _1 ®<£; therefore any orthogonal matrix satisfying 

 the equation <£ft</) — ft is the product of ft -1 into two of Cayley's 

 expressions. But ft -1 </> is orthogonal, and is also a solution of this 

 equation, and consequently can be thus expressed. Whence it follows 

 that <£ is given by the product of two of Cayley's expressions.* 



9. This theorem holds if ft 2 dp — 1. In this case, if <£ ft cf> = ft 

 and <f> is orthogonal, let w be any fourth root of — ft 2 expressible in 

 powers of ft 2 . Then 



v, ft 



to «i to — to <£ to =ft. 

 to 



* The two matrices given by Cayley's expression whose product is equal to 

 (p caji both be taken orthogonal. For in the equation <f> — il _1 &<p the matrices 

 and <p are orthogonal. Similarly, if n _1 <p = ty — Xl -1 &'$, the two matrices 

 0' and \p given by Cayley's expression are orthogonal. Therefore, <p — n ^ is 

 the product of two orthogonal Cayleyan solutions of the equation $ n <p = H. 



If the expression (fl — T) -1 (ft + T) is orthogonal nT = Tfl. Therefore, 

 the orthogonal subgroup of solutions of the equation </>n4>=n in which 

 fl- = —1 is generated by the totality of such expressions for all symmetric 

 matrices T commutative with fl and such that | fl — Y | dp 0. 



