DIFFERENTIAL EQUATIONS OF ASTRONOMICAL INTEREST. 
IG7 
and, therefore, \fs 0 being the conjugate complex of \\r 
Hence 
axx 0 = — \Js 0 {3xx 0 . 
■7 + 7 -) = 0 , 
W w 
showing that + = 0 > which proves that \f is a pure imaginary. 
Writing X for \(r~ 1 , the equations above are the same as 
(a _ 1 /3—X) x = 0 ; 
we prove that the invariant factors are linear by showing* that it is not possible 
to find a row of 2 n quantities y x , y 2 , ..., such that 
For this would involve 
(a 1 /3—X) y — x. 
(/3—a\)yx 0 = aXX 0 , 
of which the right side is real, so that, X being a pure imaginary, either of these 
would be equal to 
(f3+a\)y 0 x; = (p + Z\)xy 0 = (~/3 + a\)xy 0 , 
of which the last is zero in virtue of 
x = 0 . 
4 
As axx 0 is not zero; the assumed equation for y is impossible, and the invariant 
factors are linear. 
From this fact it follows that it is possible to find a matrix h such that 
h ] /3 Wi = 
iir x 0 0 0 
0 — i(T X 0 0 
0 0 Vjo 0 
0 0 0 — i<j 2 
where 7 , <r 2 are real. Then the given differential equations, which are of the form 
if transformed by the linear substitution 
(ii, >h, 1 2 5 >h) = h (X l5 Y 15 X 2 , Y s ). 
* See, for example, ‘ Proc. Camb. Phil. Soc.,’ XII. (1903), p. 65. 
