164 Mr Bromwich, Pignoration of coordinates" 



Solve for the y's in terms of the xs and t/s, then we get 



y = - S~ l Rx + S-hi ; 



of course the determinant \S\ is supposed not to be zero, for if 

 it were zero the quantities x, 17 would not form an independent 

 set of variables, so that the method of transformation would fail 

 entirely. Substituting in the £'s, we have now 



£ = (P - QS-tR) x + QS-% 



y= - S- l Rx + 8'% 



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' = S, 



where accented letters denote the matrices conjugate (or transposed) 

 to the unaccented matrices. Using these facts we get 



(QS- 1 )' = S-'Q' = S-'R, 

 (P _ QS-iR)' = F - R'S-'Q' = P - QS-W, 



(s-y = s-\ 



and we see that the substitution giving £, y in terms of x, v is 

 not symmetrical as it stands, but can be made so by writing it 

 in the form 



%=(P-QS^R)x + QS-> v , 

 -y= S~*Rx - £- V 

 The matrix of this substitution (A, say) is 

 (P - QS-iR I QS-i \ /l J - QS~ l \ (P J \ / 1 I 



V s-*r \-s-y \o\ s- 1 J \0\-SJ K-S-'RIS 



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 = G'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, v. 



Now consider the quadratic form derived from V, 



