TABER. — AUTOMORPHIC LINEAR TRANSFORMATION. 179 



If now 



4> = (1 - 2 $) J 



then — 1 is not a latent root of <f>, and is a solution of the equation 

 <f>Q (f> = Q. The matrix <£ — (1 — 2 4>) is also a solution of this 

 equation; and we have <j>o = 1. Therefore 



</)q li = 12 <£o . 



But if 



tl<f> = if/ 

 we then have 



^ = — if/. 



Since <£ = O -1 if/, and since <£ 2 = 1, we have 



fi-^fi- 1 ^ = 1. 

 Therefore, 



if/ Q, \f/ = — if/ O if/ 



= \f/U~ l if/ 



= fi. 



Consequently if/ is also a solution of the equation. 



The matrix <f> has no latent root equal to — 1. The matrix <i>, and 

 therefore <f> , are real ; if/ is then a real skew symmetric matrix, 

 and consequently has no latent root equal to — 1. Therefore hoth 

 (f> and if/ may be represented by Cayley's expression. Therefore the 

 most general real solution of the equation <f>Cl cf> = 12 , in which 12 is 

 a real skew symmetric matrix such that £2 2 = —1, is 



= fi- 1 (Q - Y)- 1 (Q + Y) (O — Y')" 1 (« + Y'), 



in which Y and Y' are arbitrary real symmetric matrices. The ex- 

 pression 



(fi — Y)- 1 (O + Y) (O — Y')" 1 (fl + Y>) 



is equally general. 



Clark University, Worcester. 



# = [{* + !)» -(1 + 1)"]™ F t (0) F 2 (0) F,(tf>) 



(- 1)» (2)— ^i(-l) F 2 (-l) ' • Fd-1)* 



Since is real, its imaginary latent roots occur in pairs which are conjugate 

 imaginary, and have the same multiplicity. From which it follows that * is 

 real. 



