164 Mr Bromwich, “ Ignoration of coordinates ” 
Solve for the y s in terms of the a?’s and y's, then we get 
y = — S~ l Rx + S~ l rj ; 
of course the determinant |'S | is supposed not to be zero, for if 
it were zero the quantities x, r\ would not form an independent 
set of variables, so that the method of transformation would fail 
entirely. Substituting in the f’s, we have now 
f = (P - QS-'R) x + QS-% 
y = — S~ l Rx + S~h ], 
which solves our problem. But, for the dynamical investigations, 
it is important to reduce the final substitution to a symmetrical 
one, so as to connect it with a quadratic form ; now since V was 
a quadratic form, the original substitution was symmetrical, and so 
P' = P, Q' = R, R' = Q, S' = 8, 
where accented letters denote the matrices conjugate {or transposed) 
to the unaccented matrices. Using these facts we get 
(QS~f = S~ J Q' = S~'R, 
(P _ QS-iRy = P' - R'S-'Q’ = P - QS-'R, 
(S-f = s-\ 
and we see that the substitution giving f, y in terms of x , r) is 
not symmetrical as it stands, but can be made so by writing it 
in the form 
e = (P~QS-'R)w + QS-% 
— y = S~ x Rx — S~ l r). 
The matrix of this substitution ( A , say) is 
( P-QS-'R\QS~' \ _ ( 1 1 - QS~ l \ (P\ 0 \ ,_L \~0_\ 
t S-'R I - S-'J VO | S- 1 j V 0 | - Sj {- S-'M | S~v ’ 
where the zeros denote matrices all of whose elements are zero, 
and the unities are square matrices with units along the principal 
diagonal and zeros in all other places. This equation is trans- 
formable to the shape 
A = C'BC, 
for clearly the first matrix is the conjugate of the third; and it 
is to be observed that C is the matrix of the substitution ex- 
pressing x , y in terms of x, y. 
Now consider the quadratic form derived from V, 
