TABER. — LINEAR TRANSFORMATIONS. 375 



Since & &" = #" #', we have 



= Or 1 (e*')- 1 e*" n 



— n- 1 e-*'+*"n 



_ e n-i(-t>' + d")n 



But —d / + &" is a polynomial in odd powers of Yfi -1 ; and 

 since Q -1 (Y£2 -1 ) 2r + 1 for any positive integer r is symmetric, 

 Cl~ x {— &' + &") CI is the product of a symmetric matrix into Q. 



Whence it follows that every solution of the equation <£ CI </> = CI 

 given by Cayley's expression is the product into itself an infinite 

 number of times of a solution which differs only infinitesimally from 

 the matrix unity.* 



6. Assuming that Cl is a polynomial in <£ satisfying the equation 

 $n<l> = Q, let = ei(« + *)n and </> = gi(»-«)n. Then, by (3), 



<f> = <£<£<> = <£o0; 

 and, by (3) and (4), 



— — "-" 



(f)Ci(f> = ci, 4> ci4> = ci, <£ 2 = l. 



By (4), $ is the product of matrices given by Cayley's expression. 

 But the matrix may be so chosen that no integer multiple of 

 7r V— 1 other than zero is a latent root of £ (6 + 0) 12, in which case 

 <^> is given by Cayley's expression.! 



* In the three preceding sections (3), (4), (5), the assumption ft 2 = —1 has 

 not been employed in the demonstration. The theorems therein given, there- 

 fore, hold for any skew symmetric matrix whose determinant does not vanish. 



t It is readily shown that <p is given by the square of Cayley's expression. 



For if some integer multiple of 2iy-l other than zero is a latent root of 

 6 ft, a matrix 0' ft also a polynomial in <p can be found, of which no integer 

 multiple of 2 tt -y/ — 1 other than zero is a latent root, such that <p = e#n — e fl'fi. 

 But then no integer multiple of 2t y -1 other than zero is a latent root of 



ft 0' and therefore of 6' ft. Consequently, no integer multiple of — i other 



8 f ft 0' ft 



than zero is a latent root either of — or of Whence it follows that —1 



4 4 



is not a latent root either of e$ 0' Dore^'Q; and since these matrices are com- 



mutative, —1 is not a latent root of their product. Therefore, e 4 (0' + e ') n is 



given by Cayley's expression, and <£ = e* (0' 4- 0') Q by the square of Cayley's 



expression. 



