UBC Theses and Dissertations

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UBC Theses and Dissertations

Linear transformations on matrices. Purves, Roger Alexander 1959

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LINEAR TRANSFORMATIONS ON MATRICES by ROGER ALEXANDER PURVES B.A., U n i v e r s i t y o f B r i t i s h C o l u m b i a , 1957 A THESIS SUBMITTED IN PARTIAL FULFILMENT OF THE REQUIREMENTS FOR THE DEGREE OF MASTER OF ARTS i n t h e Department of MATHEMATICS We a c c e p t t h i s t h e s i s as c o n f o r m i n g t o t h e r e q u i r e d s t a n d a r d . THE UNIVERSITY OF'BRITISH A p r i l , 1959 COLUMBIA ABSTRACT In t h i s t h e s i s two p r o b l e m s c o n c e r n i n g l i n e a r t r a n s -f o r m a t i o n s on M^, t h e a l g e b r a of n - s q u a r e m a t r i c e s o v e r t h e complex numbers, a r e c o n s i d e r e d . The f i r s t i s t h e d e t e r m i n a t i o n o f t h e s t r u c t u r e o f t h o s e t r a n s f o r m a t i o n s w h i c h map n o n - s i n g u l a r m a t r i c e s t o n o n - s i n g u l a r m a t r i c e s ; t h e s e c o n d i s t h e d e t e r m i n a t i o n o f t h e s t r u c t u r e o f t h o s e t r a n s f o r m a t i o n s w h i c h , f o r some p o s i t i v e i n t e g e r r , p r e s e r v e t h e sum o f t h e r x r p r i n c i p a l s u b d e t e r m i n a n t s o f e a c h m a t r i x . In what f o l l o w s , we use E t o d e n o t e t h i s sum, and t h e p h r a s e " d i r e c t p r o d u c t " t o r e f e r t o t r a n s f o r m a t i o n s o f t h e f o r m T(A) = cUAV f o r a l l A i n M n o r T ( A ) = cUA'V f o r a l l A i n M n where U, V a r e f i x e d members o f M and c i s a complex number. . n The m a i n r e s u l t o f t h e t h e s i s i s t h a t b o t h non-s i n g u l a r i t y p r e s e r v e r s and E ^ - p r e s e r v e r s , i f r > k, a r e d i r e c t p r o d u c t s . The c a s e s r = l f 2 , 3 a r e d i s c u s s e d s e p a r a t e l y . I f r = l , i t i s shown t h a t E^ p r e s e r v e r s have no s i g n i f i c a n t s t r u c t u r e . I f r=2, i t i s shown t h a t t h e r e a r e two t y p e s o f l i n e a r t r a n s f o r m a t i o n s w h i c h p r e s e r v e Eg, and w h i c h a r e n o t d i r e c t p r o d u c t s . F i n a l l y , i t i s shown t h a t t h e s e c o u n t e r examples do not g e n e r a l i z e t o t h e c a s e r=3• These r e s u l t s and t h e i r p r o o f s w i l l a l s o be f o u n d i n a f o r t h c o m i n g p a p e r by M. Marcus and JR. P u r v e s i n t h e C a n a d i a n J o u r n a l of M a t h e m a t i c s , e n t i t l e d L i n e a r T r a n s f o r m a t i o n s of  A l g e b r a s of M a t r i c e s : I n v a r i a n c e o f t h e E l e m e n t a r y Symmetric  F u n c t i o n s . In presenting t h i s t h e s i s i n p a r t i a l f u l f i l m e n t of the requirements f o r an advanced degree at the U n i v e r s i t y of B r i t i s h Columbia, I agree that the L i b r a r y s h a l l make i t f r e e l y a v a i l a b l e f o r reference and study. I f u r t h e r • agree that permission f o r extensive copying of t h i s t h e s i s f o r s c h o l a r l y purposes may be granted by the Head of my Department or by h i s r e p r e s e n t a t i v e s . I t i s understood that copying or p u b l i c a t i o n of t h i s t h e s i s f o r f i n a n c i a l gain s h a l l not be allowed without my w r i t t e n permission. The U n i v e r s i t y of B r i t i s h Columbia, Vancouver 8, Canada. Date 2-6 > /?(Pf ACKNOWLEDGEMENTS W i t h o u t t h e c o n t r i b u t i o n o f M a r v i n Marcus i t i s d o u b t f u l w hether t h i s t h e s i s w o u l d have a p p e a r e d . Over a p e r i o d o f f i v e y e a r s he has shown a c o n s t a n t w i l l i n g n e s s t o e x p l a i n p a t i e n t l y t o me t h e e l e m e n t a r y m a t h e m a t i c a l i d e a s I have f o u n d p u z z l i n g , t o r e v e a l f r e e l y h i s t e c h n i q u e s f o r p o s i n g and s o l v i n g new p r o b l e m s , and t o keep a t a n o t e n t i r e l y e f f i c i e n t s t u d e n t t o c o m p l e t e t h e more mundane r e q u i s i t e s o f o b t a i n i n g an e d u c a t i o n . F o r t h i s , a l o n g w i t h t h e f a c t t h a t t h e m a j o r i d e a s c o n t a i n e d i n t h e t h e s i s a r e h i s , I w o u l d l i k e t o e x p r e s s my deep g r a t i t u d e t o Dr. M a r c u s . I w o u l d l i k e t o t a k e t h i s o p p o r t u n i t y t o t h a n k t h e N a t i o n a l R e s e a r c h C o u n c i l o f Canada f o r f i n a n c i a l a s s i s t a n c e g i v e n d u r i n g t h e w r i t i n g o f t h i s t h e s i s . TABLE OF CONTENTS 1. INTRODUCTION 1 2. NON-SINGULARITY PRESERVERS 2 3. E -PRESERVERS 6 ^ r 4- Eg, 17 5. BIBLIOGRAPHY 37 1• I n t r o d u c t i o n Let M n denote the algebra- of n-sguare m a t r i c e s over the complex numbers, and GL^ the set of n o n - s i n g u l a r members of M . In the f i r s t p a r t of t h i s t h e s i s we determine the n s t r u c t u r e of those l i n e a r transformations- T of M i n t o M n n having the p r o p e r t y that T(GL n) _• GL^. In the second p a r t , we determine the s t r u c t u r e of those l i n e a r t r a n s f o r m a t i o n s of M i n t o M which, f o r some n n ' r _ 4f preserve the sum of the r x r p r i n c i p a l subdeterminants. of the members of M . The remaining cases of r = 1, 2 or 3 are c o n s i d e r e d i n the f i n a l s e c t i o n . The f o l l o w i n g n o t a t i o n w i l l be used throughout the t h e s i s : A», J>(A), t r ( A ) , (A) ± , det (A), denote r e s p e c t i v e l y the transpose of A, the rank of A, the t r a c e of A, the entry at p o s i t i o n ( i , j ) , the determinant of A, where A i s a member of M . In a d d i t i o n , 0 , I , E.,, denote r e s p e c t i v e l y the n , * n' n l j ' n x n zero m a t r i x , i d e n t i t y m a t r i x , the n x n m a t r i x with a 1 at p o s i t i o n ( i , j ) and zeros elsewhere. The c o l l e c t i o n of p - t u p l e s of p o s i t i v e i n t e g e r s Q i s d e f i n e d as the set of a l l p - t u p l e s w = where 1 < <••-< i ^ < n . We w i l l o f t e n be l e d to l i n e a r t r a n s f o r m a t i o n s of M i n t o M of the form n n T(A) = cUAV or T(A) = cUA»V f o r U, V f i x e d members of M , and c a s c a l a r . Such t r a n s -* n' formations we w i l l speak of l o o s e l y as " d i r e c t p r o d u c t s " . The m o t i v a t i o n i s t h a t , f o r the f i r s t a l t e r n a t i v e , r e g a r d i n g - 2 -2 t h e members of M as n d i m e n s i o n a l column v e c t o r s as i n n £l] t h e m a t r i x r e p r e s e n t a t i o n o f T i s t h e d i r e c t p r o d u c t o f V T and U. 2> N o n - s i n g u l a r i t y p r e s e r v e r s We f i r s t e s t a b l i s h i n t h r e e lemmas t h a t a l i n e a r t r a n s f o r m a t i o n m a p p i n g n o n - s i n g u l a r m a t r i c e s t o non-s i n g u l a r m a t r i c e s i s i t s e l f n o n - s i n g u l a r . Lemma 2.1. I f A i s a member o f M d i f f e r e n t f r o m z e r o  n . t h e n A i s s i m i l a r t o a m a t r i x w i t h a l l d i a g o n a l e n t r i e s  d i f f e r e n t f r o m z e r o . P r o o f . L e t J ( A ) d e n o t e t h e J o r d a n c a n o n i c a l f o r m o f A. I f J ( A ) i s d i a g o n a l and has n o n - z e r o t r a c e , by a r e s u l t o f M. Mar c u s and J . L . McGregor [2] we have J ( A ) i s s i m i l a r t o a m a t r i x whose d i a g o n a l e n t r i e s a r e n t r ( A ) . I f J ( A ) i s t h e m a t r i x E-j^z we p r o c e e d t o c o n s t r u c t t h e s i m i l a r i t y t r a n s f o r m as f o l l o w s . S e t D<*> - J n U ( 2 ) - -(n-1) • J n _ x where J K i s t h e k-component column v e c t o r w i t h a l l e n t r i e s 1. and a r e o r t h o g o n a l . N o r m a l i z e and and d e t e r m i n e v e c t o r s U^ 3^,...,U^ n^ so t h a t t h e s e t o f v e c t o r s l / 1 ) . . . l / n ) i s o r t h o n o r m a l . Then t h e ( i , i ) e n t r y o f UE-^U* i s u.,u._ w h i c h i s n o t z e r o . 11 i2 We c o m p l e t e t h e p r o o f by i n d u c t i o n . The c a s e n =l f o l l o w s by d e f i n i t i o n . I f A i s i n M ^ and J ( A ) i s d i a g o n a l - 3 -w i t h z e r o t r a c e we can assume t h e ( l , l ) e n t r y i s d i f f e r e n t f r o m z e r o . The s u b m a t r i x B o b t a i n e d by d e l e t i n g t h e f i r s t row and column o f J(A) i s . n o t z e r o s i n c e j ( A ) has a t l e a s t two n o n - z e r o e i g e n v a l u e s i f i t s t r a c e i s z e r o . By t h e i n d u c t i o n h y p o t h e s e s t h e r e i s a V'"£ s u c h t h a t VBV 1 has a l l n o n - z e r o d i a g o n a l e n t r i e s . The p r o d u c t d i a g ( l , V ) j ( A ) d i a g ( l , V - 1 ) = d i a g (A,, V B V - 1 ) where X i s t h e ( l , l ) e n t r y o f J ( A ) , i s a s i m i l a r i t y t r a n s -f o r m o f A and has a l l n o n - z e r o d i a g o n a l e n t r i e s . I f J(A) i s n o t d i a g o n a l , we can assume t h e (1,2) e n t r y o f J(A) i s 1 and t h e s u b m a t r i x B, d e f i n e d as above i s not t h e z e r o m a t r i x . Then, as b e f o r e , we have a V i n M n s u c h t h a t t h e p r o d u c t , P = d i a g ( l , V ) j ( A ) d i a g ( ^ V - 1 ) has a l l n o n - z e r o e n t r i e s on t h e d i a g o n a l , i f t h e ( l , l ) e n t r y o f P i s not z e r o . I f t h e (1,1) e n t r y i s z e r o , and b ^ i s t a k e n as t h e (1,1) e n t r y o f VBV 1 t h e r e i s a U s u c h t h a t U /0 * \ i f 1 1 ° b l J has n o n - z e r o d i a g o n a l e n t r i e s . Then t h e p r o d u c t d i a g (U, I n _ 1 ) P d i a g ( i f 1 , I ^ ) r e p r e s e n t s a s i m i l a r i t y t r a n s f o r m o f A and has a l l n o n - z e r o d i a g o n a l e n t r i e s . Lemma 2.2; F o r any A £ 0 i n t h e M t h e r e i s a Z i n M — — — — — — n n - 4 -s u c h t h a t A + Z has no e i g e n v a l u e s i n common w i t h t h e e i g e n v a l u e s of Z. P r o o f . By t h e above lemma t h e r e i s a P i n M s u c h t h a t — — - rt P ^AP has a l l d i a g o n a l e n t r i e s d i f f e r e n t f r o m z e r o . D e f i n e X = ( x , . ) as f o l l o w s . x . . = 1 i = l , . . . . n x.. = - ( P _ 1 A P ) . . i > j i j ' i j x. . = 0 i < j Then X has n e i g e n v a l u e s e q u a l t o 1 w h i l e P ^AP + X has e i g e n v a l u e s 1 + (P ^ A P ) ^ i = l , . . . , n , none o f w h i c h a r e e q u a l t o 1. I t i s t h e n e a s i l y v e r i f i e d t h a t Z = PXP""1 has t h e r e q u i r e d p r o p e r t y . Lemma 2.3. Any T w h i c h p r e s e r v e s n o n - s i n g u l a r i t y i s i t s e l f n o n - s i n g u l a r . P r o o f . We have t h a t i f d e t [ x l - [ T ( I ) ] " 1 T ( A ) ] = 0 f o r some x, t h e n d e t ( x l - A) = 0 f o r t h a t x. In o t h e r words, t h e d i s t i n c t e i g e n v a l u e s of [ T ( l ) ] ~ 1 T ( A ) a r e a s u b s e t o f t h e d i s t i n c t e i g e n v a l u e s o f A. Now s u p p o s e t h e r e i s an A /= 0 s u c h t h a t T (A) = 0. Choose a Z s u c h t h a t A + Z, Z have no e i g e n v a l u e s i n common. We have [ T ( I ) ] - 1 T ( A + Z) = [ T ( I ) ] " 1 T ( Z ) . - 5 -From t h i s , e q u a t i o n we can c o n c l u d e t h a t t h e d i s t i n c t e i g e n v a l u e s o f [ T ( l ) ] 1 T ( Z ) a r e a s u b s e t o f t h e d i s t i n c t e i g e n v a l u e s of A + Z. But we a l s o have t h a t t h e d i s t i n c t e i g e n v a l u e s of [ T ( l ) ] 1 T ( Z ) a r e a s u b s e t o f t h e d i s t i n c t e i g e n v a l u e s o f Z. T h i s c o n t r a d i c t s t h e c h o i c e of Z and t h e lemma i s p r o v e n . Theorem 2.1. I f T p r e s e r v e s n o n - s i n g u l a r i t y and T ( l ) = I t h e n T p r e s e r v e s a l l e i g e n v a l u e s . P r o o f . I f T ( A ) has a s e t o f n d i s t i n c t e i g e n v a l u e s t h e n A has t h e same s e t of e i g e n v a l u e s . I f T ( A ) does n o t have ( r ) -. n d i s t i n c t e i g e n v a l u e s , l e t [B } be a se q u e n c e o f members ( r ) o f M n s u c h t h a t B x has n d i s t i n c t e i g e n v a l u e s f o r a l l r and l i m B ^ = T ( A ) . r—> <*» . By t h e c o n t i n u i t y o f T - 1 , as B ^ — T ( A ) , T - 1 ( B ^ r ^ )-> A. Then we have as r -—> oo e.v. B r 5> e.v. T ( A ) = e.v. T ~ 1 ( B r ) • -> e.v. A and t h e e i g e n v a l u e s o f T ( A ) a r e t h e e i g e n v a l u e s o f A. C o r o l l a r y 2.1. I f T p r e s e r v e s n o n - s i n g u l a r i t y , t h e n t h e r e e x i s t non-s i n g u l a r U, V s u c h t h a t e i t h e r 1 T ( A ) = UAV o r T ( A ) = UA»V - 6 -P r o o f . I f T p r e s e r v e s n o n - s i n g u l a r i t y , t h e t r a n s f o r m a t i o n [ T ( l ) ] 1 T p r e s e r v e s e i g e n v a l u e s . By Theorem 3 °f [3] t h e r e e x i s t s a U s u c h t h a t e i t h e r [ T ( I ) ] _ 1 T ( A ) = U A t T 1 f o r a l l A i n M n o r [ T ( I ) ] " * 1 T ( A ) = UA'U" 1 f o r a l l A i n M n and t h e c o r o l l a r y f o l l o w s by m u l t i p l y i n g on t h e l e f t by T ( l ) . 3 • E P r e s e r v e r s — i . In t h i s s e c t i o n we f i r s t show t h a t i f r > 2 t h e l i n e a r t r a n s f o r m a t i o n s T on w h i c h p r e s e r v e t h e sum o f t h e r x r p r i n c i p a l s u b d e t e r m i n a n t s o f e v e r y member o f form a g r o u p . I f r = l , t h e r e a r e l i n e a r t r a n s f o r m a t i o n s w h i c h p r e s e r v e t h e t r a c e b u t w h i c h have no i n v e r s e . An example i s a p r o j e c t i o n o n t o t h e s u b s p a c e o f c o n s i s t i n g o f a l l m a t r i c e s w i t h z e r o s a t some f i x e d s e t o f o f f - d i a g o n a l p o s i t i o n s . I f A i s i n M we s h a l l w r i t e E (A) t o d e n o t e t h e sum n r of t h e r x r p r i n c i p a l s u b d e t e r m i n a n t s . Lemma 3.1. I f r > 2 and E (A) = E T ( A ) f o r a l l A i n M " '' * ~ r r v n th e n T i s n o n - s i n g u l a r and E r ( A ) = E r ( T _ 1 ( A ) ) . P r o o f . S u p p o s e T ( A ) = 0 , A 0. Then (1) E r ( A + X) = E r ( T ( A + X)) = E r ( T ( X ) ) = E r ( X ) . By Lemma 2.1 A i s s i m i l a r t o a m a t r i x w i t h a l l n o n - z e r o d i a g o n a l e n t r i e s . - 7 -L e t X = ( x . . ) be d e f i n e d as x,, = x i = l , . . . , r - l x. . = 0 i = r , • . . - n i i ' ' x. . = 0 i < j i j x. . = - ( P _ 1 A P ) , . i > j I J y i j J where P "*"AP has a l l d i a g o n a l e n t r i e s d i f f e r e n t f r o m z e r o . By ( l ) E (P""1AP + X) = 0. T h e r e f o r e t h e c o e f f i c i e n t o f r — 1 x must be z e r o . T h i s i s t h e sum of t h e l a s t ( n - r ) + 1 e n t r i e s on t h e d i a g o n a l o f P "**AP. Thus, by c h o o s i n g X s u i t a b l y , we can show t h a t t h e sum o f any ( n - r ) + 1 d i a g o n a l e n t r i e s o f P "*"AP i s z e r o . Now, a s s u m i n g n-r+1 < n o r 2 < r , l e t a,B be any two d i a g o n a l e n t r i e s o f P -^AP. Then l e t be t h e sum o f n-r+1 d i a g o n a l e n t r i e s w h i c h i n c l u d e s a and e x c l u d e s B, and Sp t h e same sum e x c e p t t h a t B r e p l a c e s (X. Then, f r o m s a " S B = 0 we have a = 8, and a l l d i a g o n a l e n t r i e s of P "''AP a r e e q u a l , and e q u a l t o z e r o . Thus A must be z e r o , and T has an i n v e r s e . To show T 1 p r e s e r v e s E f : E_( Y ) = E r T ( T - 1 ( Y ) ) = E r T _ 1 ( Y ) . I t f o l l o w s e a s i l y f r o m t h i s lemma t h a t t h e s e t o f l i n e a r t r a n s f o r m a t i o n s w h i c h p r e s e r v e E , f o r some r > 2, f o r m a g r o u p . We now p r o c e e d t o d e t e r m i n e t h e s t r u c t u r e of s u c h a g r o u p f o r r > 4» T h i s i s t h e c o n t e n t o f - 3 -Theorem 3.1. I f f o r some r s u c h t h a t 4 < r < n-1, E (A) = E T ( A ) f o r a l l A i n M , t h e n t h e r e e x i s t U, V X J . It i n M s u c h t h a t e i t h e r n T ( A ) = UAV f o r a l l A i n M n o r T ( A ) = UA»V f o r a l l A i n M n and UV = e I where r6 =a. 0 (mod 277") . By Lemma 3«1 we know t h a t a T w h i c h p r e s e r v e s E^, f o r some r > 2, i s n o n - s i n g u l a r . Our f i r s t f o u r lemmas show t h a t a l i n e a r t r a n s f o r m a t i o n w h i c h p r e s e r v e s E^, f o r r > 4/ maps m a t r i c e s o f r a n k 1 t o m a t r i c e s o f r a n k 1. Lemma 3.2. L e t A £ 0 be a f i x e d member o f M . Then n d e g ( d e t ( x A + B) ) < 1 f o r any B i n Mn i f and o n l y i f A i s of r a n k 1. P r o o f . We can assume A i s i n J o r d a n f o r m . I f A i s o f r a n k 1 t h e n A has a t most one n o n - z e r o e n t r y . I f A has a l l z e r o e i g e n v a l u e s t h i s f o l l o w s i m m e d i a t e l y . I f A has non- 1 z e r o e i g e n v a l u e s i t c a n n o t have more t h a n one. Assume t h i s t o be i n t h e ( l , l ) p o s i t i o n . T h e r e c a n n o t be a 1 on t h e s u p e r d i a g o n a l . A 1 a t p o s i t i o n (1,2) i s i m p o s s i b l e w i t h o u t t h e e i g e n v a l u e a t ( l , l ) b e i n g a l s o a t ( 2 , 2 ) ; and a 1 e l s e -where on t h e s u p e r d i a g o n a l w o u l d mean t h a t t h e r a n k o f A i s g r e a t e r t h a n 1. As A has o n l y one n o n - z e r o e n t r y i t f o l l o w s e a s i l y t h a t deg d e t ( x A + B) < 1 f o r a l l B. - 9 -In the other d i r e c t i o n , we f i r s t show that A has at most one non-zero e i g e n v a l u e . Suppose A has the set of non-zero eigenvalues A,, , ...,A,. and that A i s i n Jordan form with eigenvalue A.. i n p o s i t i o n 1 j ( i . . f i j ) . Choose B d i a g o n a l with 0*s at p o s i t i o n s ( i j / i j ) j=l#"««#k and l ' s elsewhere. Then det(xA,+ B) + B) = TT (\. x) - ( // X. ) x k j = l j j = l j TT I f k > 1, det(xA + B) would have degree g r e a t e r than 1 i n x, and so A has at most one non-zero eigenvalue X* We now assume X t 0 and show that a l l super-d i a g o n a l e n t r i e s are zero. Let i Q be the l a r g e s t i n t e g e r such t h a t there i s a 1 at p o s i t i o n ( i Q , i + l ) . If n=2, t h e r e i s a zero at p o s i t i o n (1,2-) of the Jordan form of a m a t r i x with eigenvalues X,0, and A i s c l e a r l y of rank 1. Let n > 2 and B = (b,.) be d e f i n e d as f o l l o w s b . . = 0 i i b. , = 1 1 1 1 = 1 i = i +1 o i£ i i£ i +1 o o l +1 , 1 o ' o b,, = 0 elsewhere. Then det(xA + B) = -A.x - x. If X t 0, i n order t h a t deg det(xA + B) < 1, there must be a zero at ( i , i + l ) . Repeating t h i s procedure, - 10 -we can show t h a t t h e r e a r e no l * s above t h e d i a g o n a l , i f 7V. * 0. Now assume A, = 0 and t h a t t h e (1,2) e n t r y o f A i s 1. Once a g a i n , l e t i be t h e l a r g e s t i n t e g e r s u c h t h a t t h e r e i s a 1 a t p o s i t i o n ( i ,'i +l) . I f i > 2 v o' o o d e f i n e B = ( b . . ) as b.. = 0 i = l , 2 , i +1,i . i i ' * o ' o b ^ = 1 e l s e w h e r e on t h e d i a g o n a l . b. . . = 1 1 + 1 , 1 o ' o b21 - 1 = 0 e l s e w h e r e o f f t h e d i a g o n a l . Then d e t ( x A + B) = x 2 . In t h i s way, a l l l * s i n p o s i t i o n s ( i , i + l ) , 2 < i < n-1, can be e l i m i n a t e d . To show t h a t t h e r e c a n n o t be a 1 at (2,3) s e t B = d i a g ( E 3 ] / I n A 3 ) . We have shown t h a t d e g ( d e t ( x A + B)) < 1 f o r a l l B i m p l i e s t h a t t h e J o r d a n form o f A has a t most one non-z e r o e n t r y , and t h u s A i s o f r a n k 1. Lemma 3.3 L e t r be an i n t e g e r s u c h t h a t 3 < r < n, and A £ 0 a f i x e d member of M^. I f deg E r ( x A + B) < 1 f o r a l l B i n M , t h e n A has at most one n o n - z e r o e i g e n v a l u e , - 11 -P r o o f . Assume A i s i n J o r d a n f o r m w i t h e i g e n v a l u e s A,, , . . . .A, . L e t B = d i a g ( z . . . . . . z ) . 1 ' n \ 2 f ' n Then E r ( x A + B) = S TT (xA.. + z. ) / . . \,_ „ k=l k k w= ( 1 . . . . . 1 ;eQ We have r r TT (XA.. + z. ) - *y c v T IV • TTox* k=l k xk — - —> a e s " Bew-S? t = 0 S t ^ w where S i s t h e sum o v e r a l l s u b s e t s o f w = ( i l t . . . , i r ) w i t h t members and S q i s t h e empty s e t . We can now w r i t e r . PEW- -t=0 weQ S. <ZW r n t - ^ and f o r t > 2, we have a t - y y TT v TT z = o T Z x ^ * aeS a BE W-S P w £ Q S,<Z- w z z r n t — f o r any c h o i c e o f z. , . . . . z . I n T h i s sum can be c o n s i d e r e d as a p o l y n o m i a l i n t h e z. as f o l l o w s : T T " N - ^ TT ;^  s t = "TT ZB ( y N TT x ; X ^ ^ pev P ^ * aeS. a ve<2„ .„ s <S(Q -v) 1 r^-tn t v nn 7 From t h i s we can c o n c l u d e t h a t t h e t - t h e l e m e n t a r y s y mmetric f u n c t i o n o f any n - ( r - t ) of t h e A.^,...,A.n i s z e r o . - 12 -In p a r t i c u l a r , t h e s e c o n d e l e m e n t a r y s y m m e t r i c f u n c t i o n of any n - ( r - 2 ) o f t h e A»'s i s z e r o . Now i f a l l t h e A,*s a r e e q u a l t h e y a r e e q u a l t o z e r o . Assume t h a t A,c £ A.^  and t h a t n'-(r-2) < n. Then, s i n c e r > 3 and s e t t i n g k = n - ( r - 2 ) , we can c h o o s e a s e t o f e i g e n v a l u e s A.. ,...,A,. where i . £ c, i . d f o r j = l , . . . , k - l . xl . 1 k - l 3 3 We have 0 = E 9(A. ,A.. ,...,A, ) = A. E 1(A,. ,...,A. ) ^ c 1l x k _ i c i i 1 i k - 1 + E (A.. , . .. ,A. ) * 1 k - l 0 = E9(A,,,A,. ...A. ) = A-.E^A.. , ...,A., ) xl 1 k - l a ± i 1 i k - 1 + E ?(A. , ... ,A. ) A 1 k-1 We t h e n have (A. - A, ,) E.. (A.. , . . . ,A.. ) = 0 - c d 1 i , ' ' l , n ' 1 k-1 I f r > 3 t h e n k-1 < n-2 and t h e e i g e n v a l u e s A,^, i£ c , d a r e e q u a l t o z e r o by t h e argument o f Lemma 3.1. Now we can w r i t e A, A, , = 0 e d and i t f o l l o w s t h a t A can have a t most one n o n - z e r o e i g e n v a l u e . I f r = 3 we have t h a t t h e s e c o n d s y m m e t r i c f u n c t i o n A. o f any ( n - l ) o f t h e A., i s z e r o . L e t E.(A.,) d e n o t e j i j E . (A,, , • • ., A,. , A.. . , . . . ,A, ) i = l , 2 . We f i r s t show i i * j-1 j +!' ' n ' - 1 3 -A, j E ^  (A, j ) j = l , . . . , n i s z e r o . A. A E 2(A. 1,... /A. n) = Tv-.E^A,..) + E 2(A..) = A.. E. (A.. ) J 1 J Summing t h e s e n e q u a t i o n s n E 2 ^ 1 ' * * " = 2 E 2 ^ 1 ' * ' * A S i n c e n > 2 E (A. ,...,A, ) = 0 and A,.E.(A..) = 0, *~ l n j X j j = l , . . . , n . S e t t i n g A.. j = l we have A.2 = A.. s J J A,. (Tv.. - s) = 0 J J Thus t h e n o n - z e r o e i g e n v a l u e s o f A a r e e q u a l . I f t h e r e were more t h a n one n o n - z e r o e i g e n v a l u e t h e s e c o n d s y m m e t r i c f u n c t i o n o f any n - 1 of t h e A, Ts w o u l d not be z e r o . T h i s c o m p l e t e s t h e c a s e t h a t r = 3 . Lemma 3.4. L e t A £ 0 be a f i x e d member o f M „ and r an n+3 i n t e g e r s u c h t h a t r < r < n + 3 . Then d e g ( E r ( ' x A + B) < 1 f o r a l l B i n M i f and o n l y i f A i s o f r a n k 1 . n+3 P r o o f . We can assume A i s i n J o r d a n f o r m . I f A i s o f r a n k 1 , A has a t most one n o n - z e r o e n t r y . I t f o l l o w s t h a t any p r i n c i p a l s u b d e t e r m i n a n t o f (xA + B) i s a t most a l i n e a r f u n c t i o n . Thus deg E ^ ( x A + B) < 1 . - 14 -We now proceed by i n d u c t i o n on the "only i f * part of the lemma, S ( n ) . S ( l ) i s t r u e by lemma 3 - 2 . Let A be i n M ,. We have an r such that 4 < r < n + 4 and n+4 — — E (xA + B) i s l i n e a r i n x f o r each B i n M .,. If r v ' • n+4 r = n + 4# A i s of rank 1 by lemma 3 « 2 . If r < n + 4# by lemma 3 - 3 A has at most one non-zero eigenvalue A.. In A take A. to be i n p o s i t i o n ( l , l ) and the ( 2 . 3 ) e n t r y to be e(=l or 0 ). Let B have a 1 at p o s i t i o n s ( 3 / 2 ) and r - 3 l * s p l a c e d i n any of the d i a g o n a l p o s i t i o n s ( i , i ) i > 3, with zeros i n a l l remaining p o s i t i o n s . Then E r ( x A + B) = -A.- £• x 2 If A, ± 0 and £ = 0 , or A. = £ = 0 , we have that the second row and column of A have only zero e n t r i e s . If we r e s t r i c t B to those m a t r i c e s with zero e n t r i e s i n the second row and column, we see that the hypotheses of S(n+3) are s a t i s f i e d . T h erefore the submatrix of A ob t a i n e d by d e l e t i n g row and column 2 of A i s of rank 1 and hence A i s a l s o . If A. = 0 , we show f i r s t t h a t A cannot have £^ = £,-, = 1 where £^ i s the entry at ( 1 , 2 ) and the entry at (n+ 3 ,n+ 4 ) . Let B be d e f i n e d as f o l l o w s : b 2 1 = bn+4,n+3 hi ± = 1 3 < i < r - 2 b.. = 0 elsewhere. - 15 -Then E r ( x A + B) = e^c^x and we pan assume ~ °-Then the i n d u c t i o n argument a p p l i e s as before to give the rank of A as 1. Lemma 3 .5. I f f o r some r such that 4 < r < n E T(A) = E (A) f o r a l l A i n M then matrices of rank 1 r r n are mapped to m a t r i c e s of rank 1 by T. Proof. Let A be of rank 1. Then by the p r e c e d i n g lemma, deg(E r(xA + B) < 1 f o r a l l B i n ^ . S i n c e E r T ( A ) = E r ( A ) , deg E^(xT(A) + T(B)) < 1. Since T i s no n - s i n g u l a r , t h i s holds f o r a l l members of M , and thus n' T(A) i s of rank 1. Lemma 3 .6. If f o r some r such that 4 < r < n f o r a l l A i n M n Proof. Let A be of rank r . We can w r i t e r i = l where C. i s of rank 1 1 = 1 , • • . , r . Then r r i = l i = l S i n c e T e x i s t s and preserves E we have - 16 -We a r e now i n a p o s i t i o n t o p r o v e t h e main t h e o r e m o f t h i s s e c t i o n . By c o r o l l a r y 2.1, s i n c e T p r e s e r v e s r a n k we have e i t h e r T(A) = UAV o r T ( A ) = VUA'V. S i n c e T has an i n v e r s e U,V a r e n o n - s i n g u l a r and we can w r i t e , f o r a l l B i n M n E r ( U ( V B V ~ 1 ) V ) = E ^ V B V " 1 ) = E ^ (B) S e t t i n g UV = P o u r c o n d i t i o n becomes E r ( P B ) = E r ( B ) o r t r ( C r ( P B ) ) = t r ( C r ( B ) ) o r t r ( C r ( P ) - I) C r ( B ) ) = 0 I f (C^CP) - I) t 0 we can assume t h a t t h e ( i , j ) e n t r y i s d i f f e r e n t f r o m z e r o . Then c h o o s e B s u c h t h a t t h e ( j , i ) e n t r y of C^CB) i s 1 and a l l o t h e r e n t r i e s a r e z e r o . B w i l l be t h e m a t r i x w i t h a l l z e r o e n t r i e s e x c e p t f o r t h o s e on t h e d i a g o n a l o f t h e s u b d e t e r m i n a n t a t ( j , i ) p o s i t i o n o f C r ( B ) , w h i c h we s e t e q u a l t o 1. Then t r ( ( C r ( P ) - I) C r ( B ) ) * 0. By c o n t r a d i c t i o n we c o n c l u d e C r ( P ) = I w h i c h i m p l i e s [4] P = e l 6 I , r6 ==0(27r) . F o r l a t e r r e f e r e n c e we s t a t e t h i s p a r t of t h e p r o o f o f t h e o r e m 3»1 as - 17 -Lemma 3.7. S u p p o s e E (UAV) = E (A) f o r a l l A i n M r r where r i s f i x e d and 2 < r < n. Then UV = e l 6 I where r 6 = 0 ( 2 i r ) . 4 • E^, E 3 " In t h e p r e c e d i n g s e c t i o n , we d e t e r m i n e d n e c e s s a r y and s u f f i c i e n t c o n d i t i o n s t h a t a l i n e a r t r a n s f o r m a t i o n of M n .preserve E , f o r some r > 4. F o r r = l , 2 , 3 , t h e r e a s o n -i n g u s e d f o r r > 4 f a i l s . In t h e c a s e s r = l , 2 , t h i s i s c l e a r . F o r r=3, lemma 3«4 i s n o t t r u e . The 4 x 4 m a t r i x E12 + E34 l s ° f r a n k 2 a n d E 3 ^ E 1 2 + E34^ + B^ i s l i n e a r f o r a l l B i n M. . 4 The q u e s t i o n t h a t now a r i s e s i s w h e t h e r o r not t h e r e a r e l i n e a r t r a n s f o r m a t i o n s w h i c h p r e s e r v e E r f o r some r=l,2,3# and w h i c h a r e not d i r e c t p r o d u c t s . We a r e g o i n g t o l o o k f o r s u c h t r a n s f o r m a t i o n s i n two s u b s e t s of t h e s e t o f l i n e a r t r a n s f o r m a t i o n s on M . The n f i r s t o f t h e s e i s t h e s e t o f l i n e a r t r a n s f o r m a t i o n s S w h i c h a r i s e f r o m p e r m u t a t i o n s o f e n t r i e s o f m a t r i c e s . An example h e r e i s t h e i n t e r c h a n g e o f two e n t r i e s . The s e c o n d i s t h e s e t o f l i n e a r t r a n s f o r m a t i o n s C c o n s i s t i n g of t h e Hadamard p r o d u c t o f e v e r y m a t r i x w i t h some f i x e d m a t r i x . That i s C : ( a . , ) -—> (c, . a . . ) - 18 -f o r a l l ( a . ..) i n M and some f i x e d ( c , ,) i n M . Mote t h a t 1 3 n v i j ' n C ^ ( c , .) h e r e . C i s a l i n e a r t r a n s f o r m a t i o n on M . ( c , ,) 1 J n 1 j i s a member of . We now p r o c e e d t o c h a r a c t e r i z e t h e t r a n s -f o r m a t i o n s C and S w h i c h p r e s e r v e E ^ , Eg. E ^ : Here any S w h i c h l e a v e s e n t r i e s on t h e main d i a g o n a l w i l l p r e s e r v e t h e t r a c e . The same h o l d s f o r any C s u c h t h a t t h e f i x e d ( c ^ j ) h a s a l l d i a g o n a l e n t r i e s 1. E ^ : Lemma U.l. I f S p r e s e r v e s Eg t h e n S = ^ - ^ ^ ^ where ir^ i s a p e r m u t a t i o n o f d i a g o n a l e n t r i e s o n l y , 7Tg i s a p e r m u t a t i o n on t h e s e t o f p a i r s of s y m m e t r i c a l l y l o c a t e d e n t r i e s , and 7T^, at a s u b s e t o f t h e s e t o f s y m m e t r i c a l l y l o c a t e d p o s i t i o n s , i n t e r c h a n g e s s y m m e t r i c a l l y l o c a t e d e n t r i e s . P r o o f . C o n s i d e r t h e e f f e c t of S on a d i a g o n a l m a t r i x D w i t h two l * s and t h e r e m a i n i n g e n t r i e s z e r o on t h e d i a g o n a l . In o r d e r t h a t S p r e s e r v e Eg i t i s n e c e s s a r y t h a t SD be a d i a g o n a l m a t r i x w i t h two l T s and r e m a i n i n g e n t r i e s z e r o . Thus S i n v o l v e s a p e r m u t a t i o n o f t h e d i a g o n a l e n t r i e s 7T ^ . C o n s i d e r t h e e f f e c t o f S on t h e m a t r i x E , . + E , . . i j 3 1 In o r d e r t h a t S p r e s e r v e Eg i t i s n e c e s s a r y t h a t s ( E . j j + ^j ±~) have two l ' s a t s y m m e t r i c a l l y l o c a t e d p o s i t i o n s . Thus S moves t h e e n t r i e s at ( i , j ) , ( j , i ) t o some o t h e r p a i r of s y m m e t r i c a l l y l o c a t e d p o s i t i o n s and t h e n e i t h e r i n t e r c h a n g e s o r l e a v e s f i x e d t h e s e e n t r i e s . - 19 -We can w r i t e any matrix (a,.) = A i n M as the 1 j n f o l l o w i n g sum A = D + y P, . <_- a. lj 3 i < j where D = d i a g ( a 1 1 # . . . , a ) ^ 1 1 nn P. . = (a. .E, , + a . ,E . . ) Then <-—^  ij. SA = SD i < 3 sp.;. i < j The e f f e c t of S on P., i s known. We have e i t h e r S(a.,E.. + a,.E..) = a,.E.-, + a..E f. i j i j j i j i i j kZ j I -ik or S (a . . E . . + a..E.,) = a , . E, „ + a, J . , i j i j j i ] i j i kl 13 -Zk We can c o n s i d e r the e f f e c t of S on P,. to take p l a c e i n the f o l l o w i n g order. F i r s t S takes the p a i r of e n t r i e s at to (k,-?) , (-2, k) and then interchanges a ^ and a ^ , or not, depending on S. Then we can w r i t e ^ \ S P i j a s 'wl'r2^ ^ P i j ) where T2 i s a permutation i < J i < j - 20 -on t h e s e t of p a i r s o f s y m m e t r i c a l l y l o c a t e d e n t r i e s and i n t e r c h a n g e s o r l e a v e s f i x e d th.e e n t r i e s a t ( i , j ) , ( j , i ) d e p e n d i n g on t h e e f f e c t o f S on P ^ j - Then, s i n c e t h e e f f e c t of S on t h e d i a g o n a l e n t r i e s i s i n d e p e n d e n t o f t h e e f f e c t o f S on t h e o f f - d i a g o n a l e n t r i e s we have f o r a l l A i n M n S ( A ) = V^^W^K Lemma 4.2. Suppose C : ( a ^ ) =>(c^ja^^) p r e s e r v e s E g . Then c . . = c,. 1 and e i t h e r c , . = 1, i = l , . . . , n o r i j j l , 1 1 c . , = — 1, i = l , . . . . n . i i ' P r o o f . L e t i , j be f i x e d i n t e g e r s between 1 and n. D e f i n e A as f o l l o w s : a . . = x . . a.. = x., n i l i j i j a . , = x . , a . . = x , . J J J J J i J i a. . = 0 e l s e w h e r e , k l Then E _ ( A ) = x..x,. - x..x.. and E „ ( c ( A ) ) = c . . c . , x , . x . . -2 V i i J J i j j i 2 V n J J n J J c . . c . . x . . x . . . S e t t i n g t h e s e e q u a l we have t h e f o l l o w i n g i j j i i j j i e x p r e s s i o n i d e n t i c a l l y z e r o f o r any x,. , x , . , x . . , x . , . n j j i j j i ( c . . c , , - 1 ) x , , x . . + ( l - c . . c . . ) x , . x , , = 0 i i J J i i J J i J J i I J J i T h i s g i v e s us c , . c , , = 1 f o r any i f i . i i JJ - 21 -As n > 2 we have t h e e q u a l i t i e s c . , c , , = 1 i i 33 c . . c. , = 1 n kk c, , c . , = 1 kk 33 f o r any i , j , k d i s t i n c t . T h e s e i m p l y c,, = c,. = c,, a n d t h a t e i t h e r c.. = - 1 , i = l , . . . , n , o r c., = +1, i = l , . . . , n . i i I I I 1 X I I We now show t h a t t h e t r a n s f o r m a t i o n s 7T^,ir^,ir^ a r e n o t d i r e c t p r o d u c t s . T h i s i s done by e x h i b i t i n g , n o n -s i n g u l a r m a t r i c e s N w h i c h a r e mapped t o s i n g u l a r m a t r i c e s by iri»7r2,7r3' T h i s i s s u f f i c i e n t s i n c e by lemma 3-7 a d i r e c t p r o d u c t p r e s e r v i n g p r e s e r v e s n o n - s i n g u l a r i t y . S u p p o s e TT^ maps t h e e n t r y a t ( j , j ) t o p o s i t i o n ( i ^ , i j ) . T h e r e i s a p e r m u t a t i o n 7T o f l , . . . , n s u c h t h a t 7T ( j ) = j and i r ( i ) ^ i f o r a l l i £ j . L e t N be a p e r m u t a t i o n m a t r i x w i t h l * s a t t h e p o s i t i o n s ( i , 7 r ( i ) ) f o r i = l , . . . , n . Then ir^U' i s s i n g u l a r s i n c e row j c o n s i s t s o f z e r o s . S u p p o s e ir^ maps e n t r i e s a t ( i , j ) ( j , i ) t o (k,J?) C^,k). L e t N h a v e l f s a t ( i , j ) a n d ( j , i ) a n d z e r o s i n t h e r e m a i n i n g o f f d i a g o n a l p o s i t i o n s . On t h e d i a g o n a l l e t N have z e r o s a t ( i , i ) ( j , j ) and l * s e l s e w h e r e . Then ^ s s i n g u l a r a n d N i s n o n - s i n g u l a r . S u p p o s e 7T 1 i n t e r c h a n g e s e n t r i e s a t ( i , j ) , ( j , i ) and l e a v e s f i x e d t h o s e a t ( k , ^ ) , C ^ , k ) . I t i s n o t d i f f i c u l t t o - 22 -f i n d a n o n - s i n g u l a r N i n M„ o r M s u c h t h a t 7T.. N i s s i n g u l a r . 3 4 1 F o r example, s u p p o s e 7T^ i n t e r c h a n g e s t h e e n t r i e s a t (.1,2) ( 2 , l ) and h o l d s f i x e d t h o s e at ( 4 , 3 ) ( 3 , 4 ) . Then E = E., + E „ . + 1 4 <• 1 E 0 0 + E„„ + E, „ i s n o n - s i n g u l a r and 7T. E i s s i n g u l a r . £j 4 J 1 We p r o c e e d t o c o n s t r u c t an N i n M f o r n > 4-n . S u p p o s e none o f t h e f o l l o w i n g e q u a l i t i e s o c c u r : i = k i = 1 j = k j = 1. Then, s e t = E ^ + E.. ^  + E ^ + E^, k. We can f i n d a p e r -m u t a t i o n IT o f l , . . . , n s u c h t h a t 7T i n t e r c h a n g e s 1 and j , 2 and i , and l e a v e s r e m a i n i n g numbers f i x e d . Then, s e t t i n g P e q u a l t o t h e p e r m u t a t i o n m a t r i x d e r i v e d by p e r m u t i n g t h e rows o f t h e i d e n t i t y m a t r i x a c c o r d i n g t o 7T we w i l l have ^ i V " E 1 2 + E 2 1 + V + S i m i l a r l y we can f i n d a Q s u c h t h a t Q P i r ^ P . Q . = E 1 2 + E 2 1 • E 3 4 • E ^ . We now show t h a t t h e t r a n s f o r m a t i o n 7T» : A — ? R7T, (A)R' 1 1 where R = PQ, maps a n o n - s i n g u l a r m a t r i x t o a s i n g u l a r m a t r i x . We have t h a t tr% i n t e r c h a n g e s t h e e n t r i e s at ( 1 , 2 ) ( 2 , l ) and h o l d s f i x e d t h e e n t r i e s at ( 3 / 4 ) ( 4 , 3 ) • Then we can f i n d a 4 x 4 n o n - s i n g u l a r m a t r i x C s u c h t h a t 7T* d i a g ( C , I n ^) i s non-s i n g u l a r . I t f o l l o w s t h a t 7 T ^ d i a g ( C , I n _ ^ ) i s s i n g u l a r . ' - 23 -I f any o f t h e e q u a l i t i e s l i s t e d above h o l d we can f i n d , as b e f o r e , an R s u c h t h a t t h e m o d i f i e d t r a n s f o r m a t i o n IT1 : A—>R77\ (A)R» 1 1 i n t e r c h a n g e s t h e e n t r i e s at ( 1 , 2 ) and ( 2 , l ) and h o l d s f i x e d t h e e n t r i e s at ( 3»l) ( l » 3 ) - We can p r o c e e d e x a c t l y as above t o f i n d a n o n - s i n g u l a r N s u c h t h a t 7r'N i s s i n g u l a r . 1 Thus we have l i n e a r t r a n s f o r m a t i o n s S w h i c h p r e s e r v e E^ and w h i c h a r e n o t d i r e c t p r o d u c t s . A t r a n s f o r m a t i o n C w h i c h p r e s e r v e s E^ and i s not a d i r e c t p r o d u c t i s - g i v e n as f o l l o w s C : ( a . .) 5> ( c . a . . ) (a. .) e M c^2 = *^ °21 = *^ ^ A. £ 1 c . . = 1 e l s e w h e r e . C maps t h e m a t r i x d i a g ( A , I n _ ^ ) where A i n i s d e f i n e d as A = ^ l E 1 2 + E 2 3 + E 3 1 Then d e t A = A."1 w i t h d e t C(A) = 1. C i s not a d i r e c t p r o d u c t , f o r a c c o r d i n g t o lemma 3-7/ i f a d i r e c t p r o d u c t p r e s e r v e s E^ i t p r e s e r v e s t h e a b s o l u t e v a l u e o f t h e d e t e r m i n a n t : The f o l l o w i n g t h r e e lemmas w i l l be u s e d t o show t h a t we c a n n o t f i n d an E ^ - p r e s e r v i n g l i n e a r t r a n s f o r m a t i o n amongst t h e t r a n s f o r m a t i o n s S w h i c h i s n o t a d i r e c t p r o d u c t i n t h e s e n s e o f t h e o r e m 3»1« - 24 -Lemma 4 .3 . I f A e M has n l ' s and t h e r e m a i n i n g e l e m e n t s a r e z e r o , t h e n f o r n > r > 1 E A = ( n ) i f and o n l y i f A = I . r r J P r o o f . S i n c e e v e r y r x r s u b m a t r i x of A has i n t e g e r d e t e r m i n a n t we have det (X r) < ( d e t ( X r ) ) 2 E (A) < d 2 A r — Z . — i . r we where t h e sum e x t e n d s o v e r a l l r x r s u b d e t e r m i n a n t s . We a l s o have j > J d 2 ( A ) = j | C r ( A ) | | 2 » t r C r(A) C r ( A * ) = t r C (AA*) = E ( a 2 , . . . , a 2 ) r r 1 n I f A i s of r a n k k < r we have 0 = E ^ ( A ) < ( n ) . So t a k e k > r and c o n t i n u e : E r ( a 2 a 2 ) = E r ( a 2 , a 2 ) < ( \ ) k ~ r E ^ ( a 2 , . . . , a 2 ) ( 1 ) = ( * ) k " X ( t r ( A A * ) ) r = ( \ ) k - r | | A | | 2 r The i n e q u a l i t y o f ( l ) i s j u s t i f i e d by theorem 5.2 o f [ 5 ] « S i n c e l | A | | 2 r = n r we have (2) E r ( A ) < ( * ) k " r n r We c o n s i d e r two c a s e s : ( i ) k = n. A i s a p e r m u t a t i o n m a t r i x and a l l e i g e n v a l u e s o f A a r e o f modulus one. L e t Tv., , . . . ,X, be t h e 1' ' n - 25 -e i g e n v a l u e s o f A. We have t h a t E (A,, , . . . ,A. ) = ( n ) r 1' ' n r w h i c h s t a t e s t h a t a sum o f ( ^ ) complex numbers e a c h o f modulus one i s ( ^ ) • We show t h i s i m p l i e s e a c h summand i s one. S e t m = ( ^ ) and l e t a_. + iB.., j = l , . . . , m be t h e summands. S i n c e a , < 1. j = l , . . . , m ; a . < 1 f o r J j o some j would i m p l y m —i / „ a, < n . * J j = l Thus a. = 1 and B. = 0, j = l , . . . , m . J J Now we show a l l t h e e i g e n v a l u e s a r e e q u a l . L e t A.c, A.^ be a p a i r o f e i g e n v a l u e s . In t h e e x p a n s i o n o f E (A., , ... ,A, ) we w i l l have two t e r m s r l ' n ^ c ^ i " " ' Xi 1 i * 2 ' 3 = 1 / ' . - / r - l 1 r - l J A, ,A,. • • - A,, d l I N r - l S i n c e t h e e i g e n v a l u e s a r e not z e r o and t h e two t e r m s a r e e q u a l t o one we have by c a n c e l l a t i o n A, = A. , c d A l s o , A i s n o r m a l , so we have a u n i t a r y U s u c h t h a t UAU""1 i s d i a g o n a l . L e t A, be t h e e i g e n v a l u e o f A. Then UAU" 1 = \l A = A.I S i n c e A c o n s i s t s o n l y o f l T s and O's - 26 -( i i ) k < n. If k = 1 then r = 1. If E^ (A) = n, A = I which contra-d i c t s k = j J(A) = 1. If k > 2 ( I ) k " r n r < ( » ) and so, from (2) E r ( A ) < ( * ) Lemma 4.4. If E„S(A) = E 0 ( A ) f o r a l l A e M where n > 4, J i n — S can be d e s c r i b e d as f o l l o w s . The e n t r i e s at any p a i r of symmetrically l o c a t e d p o s i t i o n ( i , j ) and ( j , i ) are e i t h e r interchanged or l e f t f i x e d . Proof. In order that E ^ S ( l ) = ( * ) we must have S ( l ) = I by lemma 4-3« T h i s i m p l i e s that e n t r i e s on the main d i a g o n a l remain on the d i a g o n a l . Thus we can modify S by p r e - and p o s t - m u l t i p l y i n g S(A) by P and P T r e s p e c t i v e l y , where P i s a permutation m a t r i x chosen so that the t r a n s f o r m a t i o n A ——?-PS(A)P» leaves the d i a g o n a l elements of A f i x e d . We denote such a m o d i f i c a t i o n of S by vr. We c o n s i d e r the e f f e c t of T on the m a t r i x : N 0 « diacf ( 0 n _ 2 , J ) 1° wh e r e J = - 27 -We wish to show 7TN = N o o We assume 7TN £ N o o E i t h e r ( i ) 7TNo has a 1 at some (k,.|) such that (k < /) and ( k , p t ( n - l , n ) or ( i i ) irN o has a 1 at some (k , , f ) such that k > and ( k j ) * ( n , n - l ) . In ( i ) l e t D £ M n_ 2 be d i a g o n a l with a 1 at ( k , k ) , n-3 O's elsewhere on the d i a g o n a l . Then E 3 d i a g ( D , j ) = -1 Since 7r leaves d i a g o n a l e n t r i e s f i x e d , the number of non-zero rows of 7 r d i a g(D,j) i s at most 2. Therefore E^ 7T d i a g ( D , j ) = 0 and we can conclude t h a t a l t e r n a t i v e ( i ) cannot o c c u r . In a s i m i l a r manner we can e l i m i n a t e a l t e r n a t i v e ( i i ) . Thus we can conclude that ir e i t h e r interchanges or leaves f i x e d the e n t r i e s at (n - l ; n ) ( n , n - l ) . Now we c o n s i d e r a m a t r i x with l T s at ( i , j ) and ( j , i ) and zeros elsewhere. We c a l l t h i s m a t r i x N^ and assume TN1 t N 1 There i s a P such t h a t PN P» = N., o 1 - 28 -Then TT ( P N P » ) £ P N P » o o or P»7r(PN P»)P £ N v o ' o This i n e q u a l i t y leads us to the a l t e r n a t i v e s ( i ) ( i i ) with »7rN o» r e p l a c e d by "P»7r(PNoPt )P«». The c o n c l u s i o n f o l l o w s : P»7r(PN P')P = w o o or irN^ - N 1 Lemma 4.5. If there e x i s t U,V i n M such that e i t h e r — ' n S(A) = UAV f o r a l l A i n M n or S(A) = UA»V f o r a l l A i n M n then U,V are permutation m a t r i c e s . 2 Proof. C o n s i d e r i n g members of M as n column v e c t o r s as n d e s c r i b e d i n the i n t r o d u c t i o n we must have the m a t r i x r e p r e s e n t a t i o n of S as a permutation m a t r i x . Thus the d i r e c t product of V f and U must be a permutation m a t r i x . Suppose V c o n t a i n e d an entry d i f f e r e n t from one or z e r o . Then every non-zero entry of U must be the i n v e r s e of t h i s element, which, i n t u r n , i m p l i e s every non-zero element of V i s the same. If U c o n t a i n e d two non-zero elements i n one row or column so would the d i r e c t product of V1 and U. Thus U = A,P and V = A. "*"Q where P i s a permutation m a t r i x and Q c o n t a i n s only 0 Ts and l T s . S i n c e the d i r e c t product - 29 -of V T and U i s n o n - s i n g u l a r , Q i s n o n - s i n g u l a r and c o n t a i n s no rows or columns of z e r o s . If Q had two l * s i n any row or column so would the d i r e c t product of V T and U. Thus Q i s a permutation m a t r i x . Theorem 4.1. If E^S(A) = E ^ A ) f o r a l l A i n M n + 2 then e i t h e r ( i ) S(A) = PAQ or ( i i ) S(A) = PA»Q where P, Q are permutation m a t r i c e s . Proof. By i n d u c t i o n on the statement of the lemma. I f n = l , E ^(A) f o r A i n i s the determinant. Thus S becomes a l i n e a r t r a n s f o r m a t i o n which preserves the determinant. By c o r o l l a r y 2.1 we have a U, V i n such that e i t h e r S(A) = UAV or S(A) = UA»V. It f o l l o w s from lemma 4«5 that U, V are permutation m a t r i c e s . As i n lemma 4«4 we use the m o d i f i c a t i o n , 7T, of S i n showing the i n d u c t i v e s t e p . Since 1T leaves d i a g o n a l elements of the members of M f i x e d and pr e s e r v e s E „ , by lemma 4«4 we can w r i t e , f o r any C i n ^ n + 2 , 7T d i a g ( 0,C) = d i a g ( 0 , T T C ) E 3 (TTC) = E 3 ( d i a g ( 0 , i r C ) ) = E^tT d i a g ( 0 , C ) ) = E 3 ( d i a g ( 0 , C ) ) = E 3 ( c ) - 30 -m By the i n d u c t i o n hypotheses, 7TC = P C Q or 7TC = P C Q . However, s i n c e v l e a v e s d i a g o n a l elements f i x e d P = Q = I and we have 7TC = C or 7TC = C t . S i m i l a r l y we can w r i t e IT diag(D,0) = diag(7r D,0) and conclude 7rD = D or 7rD = D T. NOW set m = n+2 and, f o r n o t a t i o n a l convenience only, l e t A be any member of M such that a-. = a. = 0. Define C = (c,,) and D = (d..) ml lm 1 j I j as f o l l o w s : C : C 1 2 = ° C i j = a i + l j+1 i , j = l , . . . , m (i,j)£(l,2) Ds d 3 2 = 0 d 2 3 - a 3 3 d.. = a . . j=l,...,m d., = a.. i=l,...,m d.. = 0 elsewhere, i i i i ' ' i j Now A = d i a g (0, C ) + d i a g (D, 0) 7TA = IT d i a g ( 0 , C ) + TT diag(D,0) = d i a g (0 , 7 T C ) + d i a g (TT D,0). By lemma 4«3 w e know that e i t h e r 7T interchanges a 3 2 , a 2 3 or leaves them f i x e d . In the former case, s i n c e TT interchanges °12' C21' ~ a n c * s l n c e TT interchanges d 2 3 and 7TD = D * . and - 31 -Therefore 7TA = diag(0,C») + diag(D»,0) = A» i n the same way the second a l t e r n a t i v e leads us to conclude TA = A. With the exce p t i o n of the e n t r i e s at (m,l) and (l,m) the e f f e c t of w i s determined. • We e l i m i n a t e t h i s case with the t e s t m a t r i x E 1 0 + E„ n + E ., ., + E , „. 12 2,m+l m+1,1 m+1,2 If 7T does not interchange the e n t r i e s at (1,2) and ( 2 , l ) and interchange the e n t r i e s at (l,m+l) and (m+l,l) we have a change in E^ from 1 to 0. A s i m i l a r change occurs i f ir interchanges the e n t r i e s at (1,2) and (2,1) and does not interchange the e n t r i e s at (l,m+l) and (m+l,l). Thus we can conclude that e i t h e r 1TA = A or 7TA = A' and from the d e f i n i t i o n of TT the theorem i s proven. We now s t a t e and prove Theorem 4.2. I f E_C(A) = E 0 ( A ) f o r a l l A i n M then there =s 3 3 n are d i a g o n a l U,V i n M so that C(A) = UAV f o r a l l A i n M n We w i l l use Q^n to denote the set of 3 ~ " t u P l e s of p o s i t i v e i n t e g e r s w = ( i ^ i ^ ) where 1 < i ^ < i 2 < i ^ < n - 32 -i s to be the subspace of r e s u l t i n g from s e t t i n g equal to zero the e n t r i e s of each A i n complementary to the p o s i t i o n s of the 3 x 3 p r i n c i p a l submatrix determined by = (i., i _ i 0 ) , and h the isomorphism from L to M„ • J- j> w J> w Lemma 4.6. Let C be a Hadamard t r a n s f o r m a t i o n such that E_C(A) = E 0A f o r a l l A i n M . Then f o r a l l w £ Q 0 there 3 3 n 3n e x i s t d i a g o n a l U V i n M. such t h a t ^ w w 3 C : A E L > h " 1 ( U h(A)V ) W W W Proof. S i n c e C p r e s e r v e s E^ we have det h(A) = det hC(A) f o r a l l A i n L . The t r a n s f o r m a t i o n w R : h(A) > (h)C(A) induced on i s l i n e a r and p r e s e r v e s determinant. Thus, by c o r o l l a r y 2.1 t h e r e e x i s t U , V such t h a t W W R : h(A) *r U h(A)V W W which i m p l i e s C(A) = h _ 1 ( U h(A)V ). W W We have yet to prove that U , V are d i a g o n a l . In o r d e r to do t h i s we r e s t r i c t our a t t e n t i o n to R and employ the f o l l o w i n g n o t a t i o n R : h(A) = (a..) 7>(r..a..) i , j = l , 2 , 3 . U = U. V = V. W W - 33 -We have, where U^"^ i s the i - t h column of U, Since R : E . . >• r . . E . . 1 1 1 1 1 1 we have v . . U ( i ) - r . . e ( i ) v..U ( i> =0, j * i i i n I j Now r . . £ 0 as C i s n o n - s i n g u l a r and so i i y u, . = 0 h £ i h i v . j = 0 j * i . T h i s holds f o r any i so U, V are d i a g o n a l . We are now i n a p o s i t i o n to prove theorem 4*2. It i s s u f f i c i e n t to show (c..) i s of rank 1. For, then I j ' every row of ( c ^ ) i s a m u l t i p l e of some row, say the i Q -row. I f we r e - l a b e l the e n t r i e s of the i - t h row as o d, = c, , j =1,...,n J o J then c . . = c . d . i /= i U l j o where c. i £ i i s the m u l t i p l i e r . 1 o If we choose U = d i a g ( c 1 , c 2 < . . . , c , _ l l l , c , + 1 , . . , , c n ) V = d i a g ( d 1 , d 2 , . . . , d n ) i t f o l l o w s that f o r a l l A i n M n C(A) = UAV. - 34 -In order to show (c..) i s of rank 1 we show every 2 x 2 submatrix i s s i n g u l a r . We designate the 2 x 2 sub-ma t r i x i n v o l v i n g rows a^ ,a,> and columns P^ r P ^ b y (a^Pj)« The set {a^; a 2, B ^  ., B 2 ] co n t a i n s e i t h e r f o u r d i s t i n c t i n t e g e r s or l e s s . If the l a t t e r holds we are assured we can f i n d a 3 x 3 p r i n c i p a l submatrix which i n c l u d e s (cx^p\) as a p a r t . Suppose y = (y^ y^ y^) i s the set of rows and columns of t h i s p r i n c i p a l submatrix. By the p r e c e d i n g lemma we know C : Jy * h " 1 [ U Y h ( J 3 ) V y ] where i s the n x n m a t r i x with l * s at p o s i t i o n s of the p r i n c i p a l submatrix determined by y and zeros elsewhere; and Uy = d i a g ( u 1 # u 2 , u ^ ) = d i a g ( v x / v 2 , v ^ ) . From t h i s i t f o l l o w s t h a t f o r some i ^ i 2 j ^  j 2 c ~ a = u . v , c ^ D = u . v . a i p l Z l J l a i P 2 X l J 2 C Q = U , V , C D = U . V , a 2 P 1 x2 3 l a 2 B 2 i 2 j 2 and thus (cx.^p\) i s s i n g u l a r . If fa^ #a 2,B^,P 2 #} c o n t a i n s f o u r d i s t i n c t i n t e g e r s we c o n s i d e r the two p r i n c i p a l submatrices d e r i v e d from the two sets of d i s t i n c t i n t e g e r s p, = [a^,a 2,B^}, V = [a.^,a 2 /B 2} - 35 -By the p r e c e d i n g lemma, we have C : J 3 — • _ . - ! [ h ( J 3 ) y C : J • > h~ 1[U v. h ( J 3 ) V R ] where = d i a g ( u ^ u 3) Uy = diag(u» u» u» ) V = d i a ^ ( v l V2 V = diag(v» v* v,» ) . From t h i s i t f o l l o w s that f o r some 1 1 » J 1 # i 2 c n = u , v. c = u . v. V l x l J l a i a i X l x l C n = U , V , C = U . V , a 2 * V X2 J l a 2 t t l X2 Xl and f o r some n^ n^ m 2 c D = u' v' c = u T v' a 1 B 1 n j L m 2 a_a 1 n ; L C o = u* v' c = U T V * a 2 B 2 n 2 m 2 a 2 a 1 n g From t h i s set of e q u a l i t i e s i t i s easy to e s t a b l i s h that V i and thus (o.^Bj) i s s i n g u l a r . T h i s completes the proof of Theorem 4«2. - 36 -Theorems 4»1 and 4«2, i n c o n j u n c t i o n with lemma 3 . 7 f a l l o w us to conclude that the only t r a n s f o r m a t i o n s C and S which preserve are d i r e c t products as i n Theorem 3«1. The quest i o n remains: Are a l l l i n e a r t r a n s f o r m a t i o n s which preserve E^ d i r e c t products? Concerning Eg p r e s e r v e r s , we were able to e x h i b i t two types of l i n e a r t r a n s f o r m a t i o n s which p r e s e r v e d Eg and which were not d i r e c t p r o d u c t s . It i s not known whether or not these t r a n s f o r m a t i o n s and d i r e c t product t r a n s f o r m a t i o n s exhaust the c l a s s of l i n e a r t r a n s f o r m a t i o n s which p r e s e r v e E_. BIBLIOGRAPHY 1. M. Marcus A l l l i n e a r o p e r a t o r s l e a v i n g the u n i t a r y group i n v a r i a n t , to appear i n Duke Math. J . 2. M. Marcus and J.L. McGregor Extremal p r o p e r t i e s of  Hermitian m a t r i c e s . Canadian J . Math., v o l . 8, 1956, pp. 524-531. 3« M. Marcus and L i n e a r t r a n s f o r m a t i o n s on B.N. Moyls algebras of m a t r i c e s . Canadian J . Math., v o l . 11, 1959, PP. 61-66. 4« J . Williamson M a t r i c e s whose s-th compounds are equal. B u l l e t i n of the American Mathematical S o c i e t y , v o l . 39, •1933, PP. 108-111. 5- G.H. Hardy, J.E. L i t t l e w o o d , and G. P o l y a I n e g u a l i t i e s . 2nd ed., Cambridge U n i v e r s i t y Press, 1952. 

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