588 THE BELL SYSTEM TECHNICAL JOURNAL, MAY 1952 



Let h be such that Cb > 0. Any 6-rowed minor of \f/Z* is a sum of 

 products of 6 elements of \f/Z*. That b-rowed minor which has in it a 

 term with a lowest possible exponent of (p — po) is the upper left 6-rowed 

 minor. Even this minor has a term with exponent 



(r - el) + • • • + (r - ei) (7) 



for (p — Po), this term being the product of the main diagonal elements. 

 Hence the highest common factor of all 6-rowed minors of \I/Z* has an 

 exponent for (p — po) not less than (7). Hence 



(r - f i) + • • • + (r - f,) (8) 



is not less than (7), since this is the exponent of (p — po) in the product 

 of the first h diagonal elements of the normal form of \f/Z*. The in- 

 equality between (8) and (7) is just the conclusion claimed in the lemma. 

 5.281 Proof of 5.27: Let 



W(p) = A(p)Z{p)B(p) 



be the normal form of Z(p). Then 



W = AZiB + AZoB. (9) 



If we expand all three terms here in Laurent series about po , the term 

 AZ'iB contributes no negative powers. It follows then from the diagonal 

 form of IF that the matrix 



Z* = AZxB 



satisfies the conditions of 5.28. The Ck of that lemma are, from (9), 

 just the exponents of {p — po) in the successive diagonal elements of 

 W, the normal form of Z, and the Ck of 5.28 are the similar fiuantities 

 for the normal form of Z* = AZiB. But the normal form of AZiB is 

 the same as that of Zi (5.16). Therefore in the inequality of 5.28 we 

 may interpret all of the e's as exponents in the respective normal forms 

 of Z and Z\ . 

 Now 



Zi(p) = Z{p) + (-Z,(p)) 



and —Z-iij)) is again finite at po • Hence we may conclude by the argu- 

 ment just used that if fj, > also 



f 1 + • • • + f6 > £l + • • • + f 6 . 



Hence if either of Ch or £& is non-negative 



€\-\- ■ ■ ■ -\-Jb=j:i + • • • -\-Jh . 



