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A SIMPLE PROOF OF THE TUCKER THEOREM
*Takao F u j i m o t o
1 . The Tucker theorem i n l i n e a r s y s t e m s c a n p r o d u c e a l o t o f u s e f u l c o r o l l a r i e s and i s i n d i s p e n s a b l e i n t h e a n a l y s i s o f l i n e a r economic m o d e l s . The r e a d e r may r e f e r t o Koopmans③ ,Howe② ,Nikaido勺 , 9 〕e t c . Even i n n o n l i n e a r programming, t h e Kuhn‑Tucker theorem① d e p e n d s on t h e Minkowski‑Farkas lemma, which i n t u r n i s e a s i l y d e r i v e d from t h e Tucker t h e o r e m . The o r i g i n a l p r o o f due t o Tucker (1 ゜ 〕 i sp u r e l y a l g e b r a i c and e l e m e n t a r y ( h i s THEOREM 1 i n (1
叩p .8 ) . N i k a i d o⑦ p r e s e n t e d an i n t e r e s t i n g p r o o f which u t i l i z e d t h e Stiemke t h e o r e m . The p r o o f by N i k a i d o i n C , 〕u s e st h e t h e o r y o f c l o s e d convex c o n e , which i s e l e g a n t b u t a l i t t l e s o p h i s t i c a t e d .
2 . I n t h e s e p r o o f s , o p t i m i z a t i o n d o e s n o t a p p e a r e x p l i c i t l y . Now i t i s w e l l known t h a t e x i s t e n c e p r o b l e m s , s e p a r a t i o n theorems and o p t i m i z a t i o n p r o c e d u r e s a r e c l o s e l y r e l a t e d i n some o f economic t o p i c s . We may r e f e r t o a s e r i e s o f p r o o f s o f t h e e x i s t e n c e o f von Neumann growth e q u i l i b r i u m . Another good example i s t h a t o f N i k a i d o ③ ,whose method o f p r o o f seems t o be q u i t e f r u i t f u l . I n Morishima‑Fujimoto ⑤ ,the Kuhn‑ Tucker theorem i s a p p l i e d t o a m i n i m i z a t i o n problem t o p r o v e a n o n l i n e a r e x t e n s i o n o f t h e F r o b e n i u s t h e o r e m , a v o i d i n g t h e f i x e d ‑ p o i n t t h e o r e m s .
* T h i s n o t e i s a r e v i s i o n o f my m i m e o . ( 1 . 〕
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The p r o o f i n t h i s n o t e i s performed through a s i m p l e m i n i m i z a t i o n problem and i s s i m i l a r t o t h a t u s e d by Morishima ( ( 的 p . 1 6 4 ) t o p r o v e t h e Minkowski‑Farkas lemma.
3 . The Tucker theorem s t a t e s :
THEOREM Given any m
Xn r e a l matrix A, t h e f o l l o w i n g two p r o b l e m s ( i ) and ( i i ) have a t l e a s t o n e p a i r of s o l u t i o n s s u c h t h a t
ぷ 十
p'A>Oand x;(p'A)i=O for a l l i . ( i ) f i n d
Xs u c h t h a t Ax=O and x~o.
( i i ) find p s u c h t h a t p'A~O.
I n t h e t h e o r e m , x and p a r e a r e a l n‑column v e c t o r and a r e a l m‑column v e c t o r r e s p e c t i v e l y , t h e prime i n d i c a t i n g t r a n s p o s i t i o n . S u b s c r i p t s i ' s a r e u s e d t o s t a n d f o r t h e i ‑ t h e l e m e n t o f r e s p e c t i v e v e c t o r s . The i n e q u a l i t y s i g n s a r e u s e d i n a u s u a l s e n s e . The above p r e c i s e r e s t a t e m e n t o f t h e Tucker theorem i s found i n N i k a i d o⑦ p . 1 2 6 , C , 〕p .3 7 .
4 . Now h e r e i s a s i m p l e p r o o f :
PROOF I f t h e problem ( i ) h a s a s o l u t i o n which i s s t r i c t l y p o s i t i v e , x>O, then p u t P=O and we have t h e d e s i r e d r e s u l t .
N e x t , s u p p o s e t h a t t h e problem ( i ) h a s no s o l u t i o n s u c h t h a t x>O.
e minimum number o f z e r o Then, t h e r e i s a x =(ふ,……云 n ) which h a s t h
・e l e m e n t ( s ) among t h e s o l u t i o n s o f ( i ) . Write t h i s minimum number a s r a n d , w i t h o u t t h e l o s s o f g e n e r a l i t y , we s u p p o s e
み=
0 f o r l~j~r, and
幻>0 f o r
r+l~j三n.Of c o u r s e , r may be e q u a l t o n , i n which c a s e t h e r e e x i s t s t h e z e r o s o l u t i o n o n l y . Let
( 1 ) M=x'A'Ax.
Now c o n s i d e r t h e f o l l o w i n g m i n i m i z a t i o n p r o b l e m .
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Problem: Minimize M s u b j e c t t o Xj~O f o r e a c h j , and ( 2 )
正j=1.
j‑1
Note t h a t t h e v a r i a b l e s a r e x/s and t h a t t h e summation ( 2 ) i n t h e con‑
s t r a i n t r a n g e s from j = 1 t o r .
S i n c e t h e above problem i s q u a d r a t i c , i t h a s a s o l u t i o n v e c t o r x * . I t c a n be shown t h a t t h e minimum v a l u e o f M i s p o s i t i v e , M*>O. F o r , i f i t i s z e r o , Ax*= 0 and we c a n form a new v e c t o r x 。=云十x * , which s a t i s f i e s Ax 。 =0, X 。 : 2 : : 0 and h a s a l e s s number o f z e r o e l e m e n t ( s ) than云 , t h u s a c o n t r a d i c t i o n . Now we a p p l y t h e Lagrange m u l t i p l i e r theorem with non‑
n e g a t i v e v a r i a b l e s .
Form a L a g r a n g i a n f u n c t i o n , L=M‑1 (区巧ー 1 ) .
j=l
At t h e s o l u t i o n p o i n t x * , t h e r e c a n be found t h e m u l t i p l i e r v a l u e
入*and we h a v e ,
( 3 J 2x*'A'A‑
入* e 。 こ 0 , ( 4 )
2x*'A'Ax* —入*e。x*=O,where e 。 = ( 1 , . , 1 , 0 , . , 0 ) , i . e . t h e f i r s t r e l e m e n t ( s ) a r e u n i t y , o t h e r s b e i n g z e r o . Using ( 1 ) and ( 2 ) , t h e a b o v e e q u a t i o n ( 4 ) l e a d s t o
2M* —入*= o.
Thus, we f i n d
入*i sp o s i t i v e . Then, by p u t t i n g p'= 2 x * ' A ' , i t f o l l o w s from ( 3 ) t h a t p'A~O, where t h e f i r s t r e l e m e n t s a r e p o s i t i v e . T h e r e f o r e , we o b t a i n 云 '+P'A>O. Moreover, s i n c e Ax=O, t h e v e c t o r x* +云 i s a g a i n a s o l u t i o n whose l a s t n‑r e l e m e n t s a r e s t r i c t l y p o s i t i v e . S u p p o s i n g x* i s a l r e a d y s u c h a s o l u t i o n , then from ( 3 ) and ( 4 ) ,
(元A);=Of o r
r+l~i印n.Thus,
ふ(元