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# Eigenproblem: Orthogonality Proof

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For the s y m m e t r i c g e n e r a l i z e d e i g e n p r o b l e m A y =&#61472;&#61548; B y , l e t ' s p r o v e t h e o r t h o g o n a l i t y s t a t e m e n t s y JT B y i = 0 , J&#8800; i a n d yJT A y i = 0 , J &#8800; i f o r &#61548;J =&#61548; i . F i r s t , w r i t e t h e e i g e n e q u a t i o n f o r y i a n d t h e e i g e n e q u a t i o n f o r yJ .

(1) A y J=&#61472;&#61548;J B yJ
(2) A y i =&#61472;&#61548; i B y i

S e c o n d , p r e m u l t i p l y b o t h s i d e s o f t h e f o r m e r b y y JT a n d b o t h s i d e s o f t h e l a t t e r b y y i T , w h i c h p r o d u c e s t w o s c a l a r e q u a t i o n s .
(3) y i T A y J=&#61472;&#61548;J y i T B yJ
(4) y JT A y i =&#61472;&#61548; i y JT B y i

T h i r d , t a k e t h e t r a n s p o s e o f b o t h s i d e s o f t h e s e c o n d s c a l a r e q u a t i o n a n d s u b t r a c t t h e r e s u l t f r o m t h e f i r s t s c a l a r e q u a t i o n .

(5) y J T (A T -A)y i =&#61472;&#61548;I yJ T B y i&#61472;&#61485;&#61548;J yJ T B T y i

F o u r t h , e m p l o y a k e y c h a r a c t e r i s t i c o f A a n d B i n o r d e r t o g e t a n e q u a t i o n t h a t s e t s t h e s t a g e f o r p r o v i n g t h e o r t h o g o n a l i t y s t a t e m e n t f o r B ;

A and B are symmetric (this is what we need to suppose), i.e. A T = A, BT = B so we get:
(6) 0 =&#61472;&#61480;&#61548;I -&#61548;J ) yJ T B T y i

t h e n j u s t d o i t !

Since &#61548;I /= &#61548;J , we have 0 = yJ T B T y i .

A r m e d w i t h t h i s p r o o f , r e t u r n t o t h e f i r s t s c a l a r e q u a t i o n t o p r o v e t h e o r t h o g o n a l i t y s t a t e m e n t f o r A .

Follows directly from equations (3) & (4).

W h a t h a p p e n s t o y o u r p r o o f s i f &#61548;J =&#61548; i ?

They don't work!