UBC Theses and Dissertations

UBC Theses Logo

UBC Theses and Dissertations

Characterization of subspaces of rank two grassmann vectors of order two Lim, Marion Josephine Sui Sim 1967

Your browser doesn't seem to have a PDF viewer, please download the PDF to view this item.

Item Metadata

Download

Media
831-UBC_1968 A1 L54.pdf [ 4.21MB ]
Metadata
JSON: 831-1.0080623.json
JSON-LD: 831-1.0080623-ld.json
RDF/XML (Pretty): 831-1.0080623-rdf.xml
RDF/JSON: 831-1.0080623-rdf.json
Turtle: 831-1.0080623-turtle.txt
N-Triples: 831-1.0080623-rdf-ntriples.txt
Original Record: 831-1.0080623-source.json
Full Text
831-1.0080623-fulltext.txt
Citation
831-1.0080623.ris

Full Text

CHARACTERIZATION  OF  SUBSPACES  OF RANK TWO GRASSMANN  VECTORS  OF ORDER TWO  MARION-JOSEPHINE SUI SIM L I M B.Sc, of  A THESIS  M. S c . ,  Wellington,  Victoria New  University  Zealand/  1963  SUBMITTED I N : P A R T I A L F U L F I L M E N T OF  THE REQUIREMENTS FOR THE DEGREE OF  DOCTOR OF PHILOSOPHY  in  the  Department of  MATHEMATICS  We a c c e p t t h i s to  The  thesis  as  conforming  r e q u i r e d ^standarid  University  of  British  December  1967  Columbia  In p r e s e n t i n g t h i s  thesis  i n p a r t i a l f u l f i l m e n t o f the  advanced d e g r e e -at t h e U n i v e r s i t y  Library  of B r i t i s h  Columbia,  f o r an  I agree that  the  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 r e f e r e n c e and  agree that  study.  I  further  permission f o r extensive copying of t h i s thesis f o r s c h o l a r l y  p u r p o s e s may  be g r a n t e d  by t h e Head o f my  that  Department o r by h i s  tatives.  It i s understood  financial  g a i n s h a l l not be a l l o w e d w i t h o u t my  copying or p u b l i c a t i o n  Depa rtment The U n i v e r s i t y o f B r i t i s h V a n c o u v e r 8, Canada  Date  requirements  . ^  -  Columbia  written  represen-  of t h i s t h e s i s  permission.  for  ii.  Supervisor:  D r . R.  Westwick.  ABSTRACT  Let algebraically space  be  U  an n - d i m e n s i o n a l v e c t o r 2  closed  field.  Let  denote  A U  spanned by a l l Grassmann p r o d u c t s Subsets of  Cg'Cu) > a r e  C  2 1  as  2 A U  of  the'  an  ( )-dimensional 2  e U .  X^AX^ ,  2 C^(U)  denoted by  and  follows  2  (U)'  p C  defined  vectors  space over  2 ^  = {z  € A U / o == j z = x-^Ax^; x ^ , X g 2  = {z e A U / o 4=  =  z  z  z  U ]  C  l^ ^  p  i + 2 > i> 2  z  e  z  e  U  a n d  z 4 C*(U) } A vector pure  or  which i s  decomposable.  rank one.  Similarly A subspace  space I f  H -  [o}  is  In this two  subspaces. If  p  C^(U)  Each vector each v e c t o r of  H  of  contained  thesis  in  dim U > 6 ,  exists  that  nonzero  in  2  i s zero 2 C ^ U ) is  C (U) is  then  f  € H  has  called  is  called  s a i d to rank  have  two.  a r a n k two  sub-  p  C (U). 2  we a r e - c o n c e r n e d are  as  with investigating  subspace has  the  rank  follows:  d i m H < n-3  a 2-dimensional vector  or  in  2 A U  The m a i n r e s u l t s  then there every  in  .  If  <x,y> form  d i m H _> 4 , of  U  such  iii.  f  = xAx in  The are  also  + yAy  f  rank  U  f  {x,y,x ,y }  where  f  f  is  independent  .  two  characterized.  subspaces  of  dimension  less  than  four  iv.  TABLE  OP  CONTENTS  PAGE  1  INTRODUCTION CHAPTER I .  Preliminaries  CHAPTER I I .  Representation  CHAPTER I I I ,  The r a n k dimension  CHAPTER I V ,  The r a n k dimension  CHAPTER V.  The r a n k dimension  CHAPTER I V . BIBLIOGRAPHY  8 19  Theorems  two s u b s p a c e s when  U  has  5 two s u b s p a c e s when  32 U  has k]J  6 two s u b s p a c e s when 7  The main r e s u l t s  U  has 79 92 10J  V.  ACKNOWLEDGEMENTS  I am indebted to my supervisor. Dr. E. Westwick, f o r h i s generous and valuable assistance i n the research and w r i t i n g of t h i s paper.  Much g r a t i t u d e i s due to Dr  B, N. Moyls, who  not only read t h i s t h e s i s and o f f e r e d h e l p f u l suggestions, but guided me i n t o the f i e l d of M u l t i l i n e a r Algebra.  A l s o , thanks  i s due t o Dr- J . L. Brenner who provided some e x c e l l e n t suggestions. I am g r a t e f u l to the U n i v e r s i t y of B r i t i s h and the National Research Council f o r t h e i r f i n a n c i a l  Columbia support.  Last, but not l e a s t , I wish to thank Misses S a l l y Bate and Doreen Mah and Mrs. Carol Gerlach f o r typing the  thesis.  INTRODUCTION  The o b j e c t  of  t h i s paper  subspaces o f .Grassmann v e c t o r s algebraically In  closed this  Product Space, products,  use  is  that  In F  a short  the  shall  summary o f the  of  this  M ,  (b)  There  paper,  Let  vector  [E : i  f  some v e c t o r  over  an  the  the  Grassmann  of  Grassmann  paper.  G r a s s m a n n P r o d u c t S p a c e we  shall  Some p r e l i m i n a r y d e f i n i t i o n s  and a  U  denotes an n - d i m e n s i o n a l  { E ^ : 1 <_ i  <_ r ]  spaces and l e t  is  1 <_ i  <_ r }  following  exists  such t h a t  If  2 ,  all  space  over  .  space w i t h the (a)  define  known p r o p e r t i e s  be r TT E .  1=1  product  and o r d e r  of  needed.  P r o p o s i t i o n 1. dimensional  someof  N. Bourbaki.  are  characterization  field.  d e f i n i t i o n of  of  proposition  a field  state  the  rank 2  i n t r o d u c t i o n , we  and g i v e The  of  is  M  ,  Let  M  a f a m i l y of be  be  a finite-dimensional  properties:  a multilinear transformation is  spanned by  N ,  Cartesian  1  <j>( TT E . ) i=l  then there  exists  cj>  of  r ir E . i=l  into  .  any m u l t i l i n e a r t r a n s f o r m a t i o n  space  the  finite-  of  r T E. i=l  into  a unique l i n e a r  trans-  2.  formation  g  of  M  into  N  such t h a t  f = g-<j> .  Then every v e c t o r space having the above two p r o p e r t i e s M . (p. 1 8 , [ l ] ) .  i s isomorphic to Definition '2.  Let  P r o p o s i t i o n 1.  M  be the v e c t o r space described i n  i s the tensor'product of the f a m i l y r {E,: 1 < i < r} , and. i s denoted by ® E . In the case where  E, = ... = E 1  Then  M  = U , then r  x-^®. . . ® x  r ® U ;  in  r  D e f i n i t i o n 3. . .  r ® E.  1=1  r is ® U . i - l  1  x^ € U ,  Every vector  i s c a l l e d a pure tensor«  A r - l i n e a r transformation  f  r ® U  of  into a  i = 1  i s called alternating i f f ( x , . . . , x ) = o for r x = (x.) € ® U having at l e a s t two members of the i=l  vector space  ¥  every vector  1  r  1  ( x : 1 < i < r}  equal.  i  [1]).  (p.58,  We note that every m u l t i l i n e a r a l t e r n a t i n g t r a n s formation  f  i s skew-symmetric;  sgn a . f ( x , . . . , x ) r  x  D e f i n i t i o n 4.  .  f (x^^y..  .,x^ ^) r  =  (Seep. 58, [ l ] ) . U  Let  i.e.,  be a v e c t o r space and  r, €> U  be the  1=1  tensor product of  U  for  r > 2 .  generated by the pure t e n s o r s  denoted by  1=1  x-j®. . ® x v  r  N  be the subspace  f o r which at l e a s t  {x^} are e q u a l . Then the Grassmann Product Space r r A U , i s the q u o t i e n t space of ® U of N . Thus  two of the  r f : ® U  Let  1=1  r  — » A U  •  such that  f (x, ®...®x ) = x , A . . . A x ^ ± r 1 r  i s alternating  3.  By convention,  P .  AU  i s t h e space  itself,  U  AU  i s the f i e l d  63).  ( f l ] , p.  r Every vector 1 <_ i  <_ r , i s c a l l e d a p u r e Let  < x , . .'. j > >  (Grassmann)  r  = o .  If  x ,...,x  x A,..Ax  r  4= o .  If  a  x  Let  4b. r AU  .  C T  Then3  Y  A  If  - A x  x-^A. . . A x  •  r  r  f o r some  r  Definition  5.  o f pure  vectors.  then  = sn a . x A . . . A x 1  r  be'nonzero  = y A...Ay 1  ( f l ] , ' p . .'64). pure  implies  r  {1,. . . ,r] , t h e n  .  s  then  vectors  <x ,...,x > 1  r  i n =  ( U ] ,P. 05).  1  1  generated  are l i n e a r l y independent,  r  y-^A. . . A y ^ r  space  i s a permutation of  a(r) ,  r  <x ,...,x >  y A...Ay  1  x^A...Ax  <y± ,• • • y > 4c.  (l) -•  x^ e U ,  •  r  x A...Ax 1  the vector  are l i n e a r l y dependent,  r  e AU ,  r  vector.  the f o l l o w i n g properties  x-^,...,x  1  X-^A.. . A x  denote  r  x - ^ , . . • >^  We n o t e If  x  1  by the vectors  4a.  of the form  Let  = <y ,...,y > x  ,  r  y e P ,  then  =  X A...Ax ±  r  ( [ l ] , p . 95) •  z = x^A...Ax  r  be a n o n z e r o p u r e v e c t o r  in  r AU  .  Then  <x^,...,x  U(z)  > . Let  x  defines  V  P r o p o s i t i o n 6.  i s defined  By property  t o be t h e r - d i m e n s i o n a l  o 4= x = x A . . . A x 1  i f  U(Z) r  4.D,  r  € AU .  is  space  well-defined. T h e n we w i l l  say that  V = U(x) . Let, V  and  W be two s u b s p a c e s  of  U  of  4.  of in  dimension r and r AU d e f i n i n g V ,  W .  I n order  that  VAW  the  subspaee  Corollary vector v  =[= o  that  U(x) .  this  v  be  r AU  in  paper,  we  to  consider  algebraically  in  necessary  and  stated,  closed  sufficient r*t s  in  VAW  A U  defines  97) r  = y-^A.,.Ay^  be in  s a i d to have  a nonzero U  such r-tl  in  order  two,  r  we w i l l  field.  assume  In order  that to  of  F  in will  higher  Is  show t h i s  condition  we w i l l  show  not  closed,  then the  r a n k two  subspaces  different  from the  ones o b t a i n e d  definition this  below)  c a n be  ( E x a m p l e 11.22) t h a t  order.  an  not unreasonable, algebraically  that  .  a l t h o u g h i n C h a p t e r I we  Grassmann v e c t o r s  pure  A U .  s h a l l be m a i n l y i n t e r e s t e d  find  otherwise  p.  is  order  Unless  vector  be a p u r e v e c t o r s AU defining  V A X ^ A . . , A x ^ = vAy-^A. . . A y ^ =f o  of  necessary  is  v  a nonzero vector  Grassmann v e c t o r s it  it  x = x-^A...Ax  Any v e c t o r In  a pure  ([1],  .  Let  Then  w  let  nonzero pure' v e c t o r  Let  r AU .  in  and  (V + W)  7.  respectively;  VfiW = o ,  the  5  s  if  F  is  is  (see in  paper. r We d e f i n e  C^(U)  ,  subsets  i n d u c t i v e l y as  of  vectors  of  AU ,  denoted  by  follows;  C^(U)  = {z  e AU I o = ' }= x-j^A. . . A x  c£(U)  = {z  e AU | z = z + z 1  and  2  where  k-1  z | U  r  where  x^ e U , i  z _ e c£(U) ]  ,  z  2  = 1, . . . , r ]  e C^_ (U) 1  cT(U)}  1=1 C, (U)  is  the  set  of  Grassmass  vectors  of  rank  k  and  5.  order  r  . We d e f i n e 2 Rg(U)  the  set;  = {H / H  is  and  Then  a subspace  i s contained i n  H - {o}  H e R^u)  contained i n  i s c a l l e d a r a n k two  2  AU Cg(U)  }  subspace.  2 We show i n t h i s p a p e r n > 6  and  F  is  ( T h e o r e m VT.100). basis  algebraically Also i f  bases f o r  if  the  if  H s R^U)  closed,  then  d i m H _> 4 ,  ( D e f i n i t i o n 11.33).  we show t h a t  that  d i m U <_ 5 ,  H  has  a  (1,1)  these main r e s u l t s ,  then  dim H _ 3 .  of  dimension three  rank 2 subspaces  dim U = n  d i m H <_ n-3  then  In a d d i t i o n to  ,  We a l s o  obtain  and l e s s  than  three. In Chapter I ,  we show t h a t  a necessary  and  sufficient  s condition for  the  vector  ( E x.Ay.) 1=1 1  the v e c t o r s  x^,y^,...,x ,y s  be  g  to have  rank  s  is  that  1  independent  F r o m t h i s , we o b t a i n the r e s u l t t h a t  i f  ( T h e o r e m 1.13).  C (U) 2  4 °  > then  d i m u _> 4 . I n C h a p t e r 1I  3  rank 2 vector  by v e c t o r s  any i n d e p e n d e n t p a i r o f show t o o  that  dimension one, if  F  if  we o b t a i n some r e p r e s e n t a t i o n s in  U ,  vectors  dim U = 4 ,  provided  F  is  of  and a l s o r e p r e s e n t a t i o n s i n a'rank  then every  two s u b s p a c e .  e.g.,  for We  r a n k two s u b s p a c e  algebraically closed.  i s not a l g e b r a i c a l l y closed,  any  F a Reals  has  However, ,  then  }  6.  E x a m p l e 1 1 . 2 2 shows space  when  there  exists  a 2 - d i m e n s i o n a l r a n k two s u b -  d i m II = 4 . I n C h a p t e r I I I , we c o n s i d e r  the case  dim U = 5 .  T h e o r e m 111.40, we show t h a t  t h e rank 2 subspaces  < 3 •  bases f o r the 3-dimensional  T h e o r e m 111,45 g i v e s  two s u b s p a c e s . basis  ( D e f i n i t i o n 11,33)  show t h a t {f  ,f ,f  x.  has  We show a l s o  d  y  that  for  dimension rank  dim U = n , every  H e R | (U)  for  have  (1,1)  has dimension <n-3.  I n C h a p t e r I V , we c o n s i d e r t h e c a s e  dim U = 6 .  every  H  }  3 - d i m e n s i o n a l r a n k 2 subspace  such t h a t  x  dim[ 1 U ( f , ) ] 1 = 1  a basis  of pairwise-Pg vectors  Theorem I V . 6 l ) .  i n Remark I V . 5 3 *  kinds  Theorems  = 6  with  We  basis  ( D e f i n i t i o n 1.12)  ( D e f i n i t i o n 11.19 a n d  Using this result,  w h i c h i s one o f t h r e e  we show  H  (Theorem I V . 6 2 ) .  IV,57> I V . 5 9 .  has a b a s i s  These  Combining  are given these  r e s u l t s w i t h those  o f C h a p t e r s I I a n d I I I , we show t h a t t h e  r a n k two subspaces  have  d i m e n s i o n <: 3  I n C h a p t e r V , we c o n s i d e r show t h a t  dim H = 4 ,  H  this result,  we f i r s t  and  has a  (1,1)  basis  show t h a t  d i m T, U ( f . ) = 7 , i=l  i f H  then  (Theorem I V . 72).  the case  [H]  the rank 2 subspaces  have  dim U = 7 .  We f i n d  (Theorem V . 7 3 ) ;  this basis  ( T h e o r e m V.85). dim H = 3 , has a basis  To o b t a i n  H = <f ,f ,f^> 1  those  2  of pairwise-P,D  i s one o f 2 p o s s i b l e  representations  g i v e n i n Theorems V . 7 4 , 75.  We  dimension < 4 , and i f  1  vectors.  In  f o r these  basis  U s i n g t h e above  o f C h a p t e r s I I t o I V , we show  kinds  members a r e  r e s u l t s and  dimH < 4 , and i f  7.  dim H = 4 , i t has a  (1,1)  basis.  I n C h a p t e r V I , we o b t a i n consider  first  <f^,fg,f^> dim  the case  8 , then { f - ^ f ^ f ^ }  pairwise-Pg vectors  (1,1)  and  basis  for  We n o t e  L  is a  <f^,f^,f^>  H o <f- ,f ,f-j> 2  .  Also,  We  H  and i f basis of  Furthermore, i f then  H  has a  t h e n show t h a t  dim U _> 6 (Theorem V I . 100) i f dim H _> 4 ,  We  that i f  (1,1)  (as above),  (Theorems VI.86, 88).  dim H 1 n-j5 when results.  dim U = 8 .  i s a ^ - d i m e n s i o n a l rank. 2 subspace  3 s U(f.).=  H e R|(U)  t h e main r e s u l t s .  has a  using  (1,1)  the above basis.  8.  CHAPTER  I  PRELIMINARIES  The sufficient  that  the  aim  In  also  result  i  the  = 1, . . . , k  defined z  that  s  We  process that  of  z  the  space  +  U(z)  T h u s we  , we  x^,...,x can =  j  ;  z^  }  = U ( ) + • • - + U ( z can  associate  U(z)  of  U  k  i.e.,  the  e )  with  above  (Theorem  above r e s u l t , C^(U)  e c  we  obtain  C^(U)  each  .  U , then f o r  1  <n  as  1  r  x  l  x  any  r  1—  a(i-^,. . •, i ) c P  the  f o r us  .  to regard  image o f  each  coefficient  an  alternating function  as  follows  r ct :  -• F  Definition  cartesian  ,  E  1.1  product  =  {1, . . . , n j  Let  E  =  of  E  ,  {1,2,  1  . . . ,n}  < r  < n  and  .  let  r TT E 1=1  ,  is well-  ) x . A...Ax.,  a(i ,...,i  1  convenient )  the  i s a b a s i s of  S  where  a,(i^, . . . , i  z  that  and  write  —  be  +  (U)  i s independent  g  obtaining  o  e C  ^  show i n f a c t  z = ]_ * ''  i f  a necessary  s £ x Ay,)  (  g  K i , <• • • < i  will  .  a p a r t i c u l a r sub s p a c e  z  It  is to.find  a vector  rank,  (Theorem I.10).  e C^(U)  r € AU  chapter  {x-^y-j , . . . , x , y ]  , then  If z  has  i s that  1.13)• the  this  condition  vector  condition  of  be  the  9-  r E -> F p : ir i=l  We define  to be alternating i f i t i s  r-liriear and p(k,,...,k ) = 0 whenever at least two members r  of  {k,,...,k }(5 E) r  skew-symmetrie; where  a  are equal. ' We note that such a  i.e.,  P( (i)> ••• k  a  i s a permutation of  j k a  p is  ( r ) ^ = sgn a.p(k ,...,k ) 1  r  (See p. 5 0 , [ 5 ] . )  {l,...,r} .  The following known result t e l l s us when z e A*U i s zero or has rank one. Theorem 1.2  Let x-,, . . ., x^  be a basis of U .  Let  p ( i . . . , i ) x A...Ax e AU l<L<...<i <n 1 — 1 r— where p i s an r-linear alternating function. z =  X J  Then  z  r  i  ±  i s a pure vector i f and only i f  (1) ^ ( - l ) p ( i , . . . , i _ , J ) p ( J , . . . , J _ , J M  0  1  r  for a l l sequences taken from  1  M  (i,,...,i  o  u  ) and  (  1  ( i + 1  ,-.- J ) = 0 5  j ,...,J ) Q  r  r  of integers  {1,. . ., n} .  Furthermore, there are (n-r) independent equations (See p. 2 8 9 , p. 312 of [2].)  i n the system of equations ( l ) .  It w i l l be necessary for us to know that i f U' c U  and o ^ z e c£(U') , then  z e c£(U)  .  In order to  obtain this result, we w i l l require the following lemma and i t s corollary.  The latter w i l l also be useful for proving a  later result. Lemma I . 3  Let z =  £ p(i , . . . , i l<i <...<i <n 1  Let  s and m  be integers,  r  ) x . A...Ax € C^(U) 1 r  o <_ s <^ r , o <_ m <_ n , and l e t  10.  z  E  =  s+l<_i  Then e i t h e r  Proof; i  We  p(l,...,S,i  <-  • -<i <m  z'  e C^(U)  g + 1  set  J  ...,i )x A...Ax Ax r  1  s  or  5  and  p' (!]_, • • • , ± ) = ° • r  A  A  z' = o .  p'(i^,...,1  = 1 ...,i . = s  i  s+1  r  ) = p(i ,..., i  )  i f  < m  .  1  s+1  < i  ,<•••<i  r  Otherwise  Then p'(±  1 3  .,.,i )x r  A...Ax  i  ±  K i , < . . . < i <n — 1 r— Now  consider  £  We  show t h i s  C a s e 1. the  • • • > i - _2_ > Jj^ ) P (  (-l)^p  2.  Case i  1  .. .<. J  Q>  ~}.  > m  or  since every  r  and  .  r  Again  f o r some  t .  and i n  1  <_ m  1 ...,s  p'(i ,...,l _ ,j ) 1  r  1  u  p ' (3 Q >' • > > except But  i f  ' •" * ^ R )  p o s s i b l y when J  D  = v{\>>  In either  case  .  r  .  1,.. ., s  s i n c e each term  Also  ••^r.ijJj) P( J i>' 0  ±  t  < m  i s zero.  ( t = 1,. .. , r - l )  case,  a  n  d  ' ^ - l ' ^ + i ' * •»  i s one o f t h e i n t e g e r s  i s one o f t h e i n t e g e r s  are present  are present i n  In this  =  are p r e s e n t i n  s  1,. .., s  J ,...,J  (t = o,.. . , r )  s  term i s z e r o .  the sum i s z e r o  A l lthe integers  , . . . A_  s  > m  Not a l l of the i n t e g e r s  i .  ^|i+l ' '° ^r ^ "  a n d a p p l y T h e o r e m 1.2.  ...,1^ ^ , o r n o t a l l o f the i n t e g e r s in  3  sum i s z e r o  sum i s z e r o  Case  • • •>^ ( j - i  r  l,...,s  ,  then  * J  r  )  >  1,...,s .  ±  J  11.  p(i ,...,i _ ,J 1  r  1  fi,,...,i^  ( J  )p(j ,...,j 0  ^j^]  we h a v e  f j l  _ ,J 1  | J + 1  ,...,d )  since  =o  r  a r e p e t i t i o n of  in  .  r  -.•,^x^)?'  H e n c e p=o £ (-l) p' ( i , , M  • ••>J  (3 > Q  u  -  1  *  • ••* J ) r  r [1=0  since z'  z e C^(U)  (Theorem 1 . 2 ) .  Therefore  by Theorem  1.2,  e cJ(U) .  C o r o l l a r y IA.  Let  z =  E p(l,,. . .,1 1 <!-[_< • • • <i <_n  )x  Let  s  and  s+l<_i  s+1  m  be  z'  Theorem 1.5  i  e c£(U)  z =  a p p l y Lemma 1 . 3  E y 1=1  z  i  Let  e cJ(U') , K  +  " '  ,  +  z  k  U' c U  o < m < n .  r  1  s  then  some  » ± z  e  if  y±  Let i  A...AX, s+1  c  z^ .  i( ) u  >  e C£(U')  1  " 1?••- A  Note t h a t  }  i  Then  z' = (z,+ -•'+z^.)'  c £ ( U ' ) c c£(U)  = 1,...,k  such  that  &  be a b a s i s  be a n e x t e n s i o n o f  r  and  E y , e G£(U) . 1=1 .  x-,,...,x 1 s  .  i < k ,  be a s u b s p a c e .  1  Let x,,...,x  k  ,...,i )x A..•Ax Ax  s + 1  for  to each term  We show t h a t  1  ,  e C (U) r  r  Write  Proof:  o < s < r  E p(l,...,s,i < - • • <i <m  Then Proof:  integers  A...Ax,  ±  1  r  of  this basis  U'  and  let  to a b a s i s  of  U .  ,  12.  Suppose k  t  £ y 1=1  =  ±  £ z 1=1  Clearly  i'  Z  Since  J  z' •  =  J  1  ,  £ >_ k  : AU-  We w i l l  follows  whenever  e cF(U')  .  .  Since  be  the  •  A...Ax. r>  £ z' J=l  =  £ z, j=l  J  ¥e h a v e will  l  x  {1,..•,s}  $  r  )x.  \  that  [ i ^ . ..,i )  1  =  3  £ y, j=l  ,  1  result  convenient  to  1.6 J  For  have  the  following  such  drop  the  z € c £ ( U ) , we d e f i n e  that index  R (z)  = k  p  r  z = y -t• • -+y 1  = z +- • • +z  k  1  k  >;  to  y ,z i  U(y )+-•-+U(y )  = U(z )+-•-+U(z )  associate  z € C^(U)  1  this  k  1  with  e n d we n e e d If £ K i , < - • -<i  — 1  some  x e U ,  k  1  r  and o n l y  show  notation  definitions r z e AU , r  .  In  and  such  ,  i.e.,  z e c£(U)  ,  z e c£(U) 1 < i  t h i s way, subspace  .  of  < k , we  , then  can U .  To  lemmas. that  )x. A . - • Ax. l r x  if  € C^(U)  i  if  = k  results.  that  a particular  p(I-. , . . . , i <n r—  if  R (z)  w h e n no c o n f u s i o n  We s h a l l now p r o c e e d  z =  ii t  ,  ,  rank.  Definition r  a  < i  /-,  x  z'  .  let  J  It  R  ,  £ p( ^(i,,...,i K i , < - • • < ! <s — i r—  3  for  l > k  , . . . ,1 ) = o  1  B y Lemma 1.3, then  show  1=1,...,*,  l  P^(1T,...,1)X.  C ^ ( U ' )  £ p( ) ( i j=l Let  To  ^  e  cf(U) ,  €  x  .  =  i  z  *  £ <_ k  y  ,  e CJ(U)  1  1  ,  where  X-, , . . . , x  is  13.  a basis  of  U ,  t h e n we s h a l l w r i t e  X A Z for  the  £ p ( i - , , . . . , i )xAX. A. . .AX, l<i,<-••<i <n 1 v  xAz =  this  Lemma 1.7  r x e U , y € 0,(11)  •  Then f o r  Furthermore Proof:  A U .  c o i n c i d e s w i t h t h e u s u a l d e f i n i t i o n when z e C ^ ( U )  Let  x A y = xAz =|= °  e  x  r  Note that  vector  Since  some  r , z € AU r-l  v e  A U ,  and  suppose  z = y + xAv .  R(xAv) < R ( z ) .  x A y =(= o , we c a n f i n d a b a s i s  U such that x = x, and y = XgA...Ax , . z = £ p ( i , , . . . , 1 ) x . A...Ax. . 1 < 1 < . . . < 1 <n 1 r r +  x,,...,x  of  n  Let Since  r  1  xAy = x A z , we m u s t h a v e other  p's  w h i c h c a n be  among t h e a s s o c i a t e d Z =X A...AX 2  r + 1  p(2,...,r+l) different  {!,,...,i } +  where  v =  r  L  1  ±  r  £ p ( l , i , . . . , i 2<i <-«.<i <n ^  Let  and suppose Then  ) x . A...Ax. e 2 r  p  r  s = 1, m = n  R(z) = k  Lemma 1 . 8  1  XAV  Taking if  only  Therefore 2  2  that  the  E _ p ( l , i , . . . ,i )x- AX ^A. . . AX 2<i <•..<i <n " ^2 2  y +  Also,  f r o m zero must have .  r  = 1 .  , then y,+...+y  X A  i n C o r o l l a r y 1.4  R(XAV)  <_ k ,  € c£(U)  R  , y,  i.e.,  E yi ^ = o , x e U . 1=1  x € U(y,)  ;  i  = l,...,k  R ( x A v ) <_  e c£(U)  k  .  , we s e e  r-l A u .  easily  R(Z) .  (i = l,...,k)  14.  Proof:  Suppose  x | U(y, ) .  Then  (- 2 y,) ^ o . 1=2 r - l v € A U and  xAy. = x A  1  1  1  k - £ y, = y. 4- X A V , where  By Lemma 1.7,  1=2 R(xAv) < k-1 .  1  1  X A V ^ = -(y +-•-+y )  But  1  which i s a c o n t r a d i c t i o n .  implying  k  Hence  x € U(y ) .  R(xAv) = k  Similarly  1  x e U ( y ) , i = 2,...,k . ±  Lemma 1 . 9  ±  z  € c£(U)  ±  ( i = l,...,k) ,  x e U , x 4 U(z^)+«•'+U(z^) .  and  X A Z € c£ (U) .  Then Proof: y  e c£(U) j  E z  Let z =  +1  Suppose  R(xAz) = ^  and: that  xAz = y-^+'-'+y^ ,  6 •C^ (U) i 1 < i < ( . ' C l e a r l y I < k . We now show +1  1  Let x = x Then  1  x^,...,x  , and z =  x ...,x 2 J  i s a b a s i s of 1  r  U( z )+. • .+TJ(z ) . 1  k  x  x^ A ( ^ y ) = o  and  ±  x-j^ 6 U(y.) , j = 1,..,,!,  y_+- • «+y^ e C ^ ( U ) , +1  ]  (Lemma 1.8).  o 4 y j e C ^ ( U ) , j = 1,. . . ,-t . <t  j = 1,...,t .  2  n  Hence  = v(  .,i )x  , p^^Ci2<i <- • -<i <n E  3  !  1  r  — 1  - •  x,Az = E y. . 3  A .. .AX, l ^  p^. ' (i , • • • , i ) x A x < i <n i., • ' 1 . r— )  1  2<in<-  i  x  J  J=I  y ^ = XjA-Wj  v^ 6 c£(U) ,  U ( v ) + . . •+U(v ) c <x ,...,x > ; 1  Moreover  Therefore  +1  Now  so chosen t h a t  r  Since  where  U  £ p ( i , ,. . . ,i )x. A... Ax, 2<i <-•-<i <s l r 1  then  g  he a h a s i s of  -6 _> k .  r  1  1  A. . .AX. r  N  Hence  E p . J = I  w  1  ' ( i , ,. . . , i ^  J = o  .  unless  15.  fi,,...,!„}  c  {2,...,s]  .  c  {2,...,s)  ;  ...,l ] r  Therefore  y'  d  =  ,  k  = t  and  where  J  Let y  y +•••+y 1  x  = U(  k  z  .  Z l  ±  e C|(U)  x  u  that  and  exists  = R(xA(y +-•-+y ))  k  By  = 1,. . . , k .  Let Then  Theorem I . 1 0 ,  ,  1  U(z)  is  generality,  such  e  i s defined  then  ,  Since  c  k  (  u  )  »  t o be  z  e  ±  c  i (  u  a necessary  and  ) »  U( z ) + - • . +,U( z ) ±  well-defined.  We w i s h now t o f i n d  that  we h a v e a c o n t r a d i c t i o n .  z = z +---+z^ U(z)  < k-1  (Lemma 1 . 9 ) . k  k  of  the case,  k  2  k  1.11  then  Then  x e U(y^)  +-•-+y ))  xA(y^+.•.+y ) = X A ( Z ^ + « • - + z )  i  a vector  B u t i f t h i s were  R(xA(z +-•«+z )) = k  Definition  A...Ax, i  (i = l,...,k)  Without loss  .  1  k > £ ,  x  k  k  there  k  y i  )x, l  x  that  1  4 U()+•••+U(z ) R(xA(  ,...,x, r  )+-"+U(z ) .  U ( y ) + « • • + U ( y ) 4= ( z ) + ' * • + U ( z ) . we c a n a s s u m e  Since  e c£(U) .  k  on the contrary  k  .  1  r  i . e . . k <_  = z +-•-+z  k  1  1  T  r J  € C^(U) ,  ±  Suppose  x  = k  U(y )+...+U(y ) Proof:  p(i ,...,± )  A . . .Ax,  £ p( )(i 2<i <-•-<i <s 1  R(xAz)  such t h a t  )x. Ax,  1  y" =  R(z) < I ;  Theorem I . 1 0  . . , i  x  J  =  Y  , where  v  J  p ( ^ ( i ,  J y* 0=1  which Implies  1  E y!, = x A z  ,1=1  £  z =  J p(^(i ,...,± )  then  2 < L < . • .<i„<s — 1 r—  3  Hence  E y . =  1=1  I n the c a s e where  sufficient  f c  .  16.  condition  .  S  (  that the vector  .  £ x.Ay 1=1 1  ) e C  1  2/ \ (U)  .  We p r o v e  first  s  the f o l l o w i n g .  Lemma 1.12 dim  Let  €  U)  , i = 1,. .. ,k , a n d l e t  (U(z )+-•-+U(z )) = rk . 1  Then  k  R(2,+...+z ) = k . k  Proof:  S u p p o s e t h e lemma i s f a l s e .  integer  f o r which i t f a i l s .  z  Let of  l  +  k  f z  =  x € U ( ) k  ,  XA(Z +. 2  z  •  2 '** +  .  +  +  i  y  e  Then k  6  G  k  /  C  i ^  ) > ^1  €  c  •  .  Let  u  .  k  By the c h o i c e  H e n c e , b y Lemma 1.9  k  assumed  ,  since  = XA(Z,+'•*+z ) = x A ( y + * • - + y ^ )  B u t we  the smallest  i ( ) , 1 = 1,.. . ,<l .  2  , and  +  k  be  | U ( z ) + °•°+U(z )  ^  u  U  k  k >_ 2  Clearly  x  € C ^(TJ)  k  >_ k - 1  + Z  •+z^)  2  l "'  .  xA(z +'•«+z ) I  y  Let  , we  1  i < k  .  Therefore  must  i = k-1  have .  B y Lemma I . 1 0 , U(xAz )+« • -+U(xAz ) = U ( x A y ) - f • •+U(xAy _ ) 2  Hence x'  k  1  k  1  < x > + U ( z ) + « • ' + U ( z ) = <x>+U(y ) + • • * + U ( y _ ) 2  e U(z ) 1  ,  k  x'  independent  <x'>+U(z )+-•-+U(z ) 2  k  intersections,  we  k  By a s i m i l a r  7  ±  =  of  x  k  .  Then  = <x'>+U(y ) + « • •+U(y _ ) 1  k  = U(y )+..-+U(y _ ) 1  k  1  1  .  argument,  U( )+...+U(z _ )+U(z Z l  1  1  = U(y )+...+U(y _ ) . 1  k  1  1 + 1  1  .  again  obtain  U(z )+...+U(z ) 2  1  )+-•-+U(z ) k  .  Taking  Let  Hence  k  i  I x.Ay 1 i  =  ^  S  x  Then  y  f  If  a  If  a  ±  ±  1  s E x.Ay,  =  *j£ [ x A (  that  y  ±  ± o , then  x A(y  x  i  1  <_ s-1 .  ±  (  A  in  y  i  i  "  f  .y_}  i f  s  .  i  = ( x + ^ ±  ±  s  ±  i  s  s  X  s  1  x  ) y A  s  1  •  = a x ) ±  ±  s  •  }  1  s  5  Therefore,  is  i  i  {x ,y , • •• x , y } i f  is  g  that dependent,  R(f) = s ,  then  independent. i f  {x ,y ,...,x ,y ] 1  8  X  « <x ,y > ±  i  ,  i  f r o m Lemma 1.12  1  g  . y „ > ] = 2s  n  X  follows  a  necessary.  i n t h e a b o v e w a y , we h a v e  d i m [<x, ,y >+••'+<x  i  vectors.  f  8  Conversely,  U(x Ay )  s  ±  s  )  o  - a x ) + P x Ay  i  R ( f ) <_ s-1 < s .  S  i  - a.x ) + P x A ( y  Hence,  {x,,y,,...,x  s  + ^ ^ ^7  ±  ±  ¥ s  +  *  - a x ) + P x Ay ]  i  x Ay  i  e  s s-1 E a,x Ax, + E p,x Ay, —.j^ i s i  £ X,Ay. + i i  = o , then  1  impossible.  , i.e.  s-1  =  s  the condition i s g  ±  Adding the terms  ±  n (  1  =  = (  then  i *ependent  n  e <x ,y ,,..,x >  g  = x.A(y  1  which i s  i f and only i f  2  a  We show f i r s t  Suppose  = {o}  ±  e C (U) s  1  f l ' ^ " l ^ • ' * *xs,ys^ Proof:  1  follows.  T h e o r e m I.13  then  H Y i=l  u  1  The r e s u l t  R(f)  k  U(y )+«•'+ (y _ ) =  Independent,  Since  S  = l,...,s that  .  is  g  ;  and  x Ay ±  ±  e C^(U)  i t  R( 2 x.Ay.) = s .  1=1  i C o r o l l a r y 1.14  dim  1  Let f =  S x.Ay, 1=1 s  k < s .  <x ,y ,...,x ,y > < 2k ; 1  1  s  Then  , and  s  R ( f ) <_ k-1 .  Proof: . I f R ( f ) = k , then  f = y-[+..«+y  k  where  [U(y()+...+U(y ) c <x ,y ,...,x ,y > , y^ € cf(U) , k  i = l,...,k ,  1  1  8  By Theorem 1.13, we can see e a s i l y t h a t  dim [U(y()+.-.+U(y£)] = 2k . < x  l'^"l' • •• s  H ( f ) < k-1 .  , x  s  > y  s  >  w  n  i  c  n  h  a  But [U(y£)+.•-+U(y£)] c s  dimension  < 2k .  Hence  19.  CHAPTER  I I  R E P R E S E N T A T I O N THEOREMS  This  chapter  i s made  up o f f o u r  sections.  The  first  p  section The  i s concerned with  other  vectors  three  representations  a r e concerned* w i t h  for  f € C|(U) .  representations  ofpairs of  2 / \  i n  H e R (U) . 2  As  p f e C (U)  any  2  h a s many r e p r e s e n t a t i o n s  i n  p Cg(U)  (we s h a l l  illustrate  obtain  some p a r t i c u l a r r e p r e s e n t a t i o n s  t o u s e when c o n s d i e r i n g greater  Definition  be  which w i l l  two s u b s p a c e s  be s i m p l e r  o f dimension  main r e s u l t s a r e contained  i n Theorems  11.26 and  and C o r o l l a r i e s 11.21 and I I . 3 0 .  S E C T I O N 1.  any  rank  i ti s essential f o r us t o  t h a n one. The  11.31  later),  Representations  I I . 15  f 6 Cg(U) .  L e t f e c|(U)  .  L e t f = x.jAx  o f f e'C^U)  .  Then  representation  the 4-dimensional  Definition  for  subspace  <x^,...,x^>  I.11 a n d T h e o r e m 1.13*  fact  coincides  with  that  used  t h e same n o t a t i o n  U(f)  U(f)  .  + xy\x^  be  i s defined t o  We n o t e  that by  i s well-defined  i n D e f i n i t i o n I.11.  for i t .  2  We h a v e  and i n  hence  20. p  I t i s easy t o see t h a t ations;  f = x-^AXg  e.g.,  f e C (U) = x A(x  X^AX^  +  1  The f o l l o w i n g theorem shows t h a t i f any complementary subspace o f  y-^  in  where  Theorem II.16  Let  and l e t  basis of  U(f) .  Then  f = y-]_Au + V A W  Proof;  Since  f =  U'  1  ) A X ^  is  <u,v,w> = U' . fy^,...,y^}  f  has a r e p r e s e n t a t i o n  where  <u,v,w> = <y ,y^,y^>  be any  2  2 f e A <y^,...,y^> , t h e n  £  a, y A y 1  l<i<j<4  l  ^  , a, , e F  1 3  1  = y,A( E n7i) ^2<j<4  +  a  1  1  which i s o f the form  3  y^  +  A u  1  v  A  w  u  one ( C o r o l l a r y 1.14).  Now  J  ( £ a, .y.Ay, 2<i<j<4  J /  2 £ a. .y.Ay. e A <y ,y ,y > 2<i<j<4 ^ ^  Theorem 1.13,  X  U ( f ) , then t h e r e i s a  f = y-j_Au + V A W 2  + x^) + ( x ^ -  2  € U ( f ) , and  representation  f e C (U)  has many r e p r e s e n t -  2  >  J  1  3  since  i s n e c e s s a r i l y of rank a t most  s i n c e R ( f ) ,= 2 , the  VAW  ^ o ,  By  {u,v,w] i s i n d e p e n d e n t and hence <u,v,w>=<y ,y^,y^>. 2  The f o l l o w i n g theorem i n d i c a t e s how  any  f e C (U) 2  may be r e p r e s e n t e d i n terms o f any 2 - d i m e n s i o n a l subspace o f U(f). Theorem 11.17  Let  f e Cg(U)  d i m e n s i o n a l subspace o f (i)  there e x i s t o =}= y  e  F  5  v,w o  r  in  U(f). U(f)  and l e t Then  <x x > 1 >  2  be any 2-  either  such t h a t  f = yx^Ax^  + vAw  ,  21.  ( i i ) There e x i s t Proof:  i n U ( f ) such t h a t  v',w'  L e t x^,...,x^  I I . 16,  f  be any b a s i s o f  has a r e p r e s e n t a t i o n  <u,v,w> = <x ,x^,x^> .  so  x-^AXgAf ••- o .  ctx-^ + p x a 4= °  Now  impossible  Then  f o r some  would i m p l y and hence  x  f = i x  be  Then 2  [X A(CV 1  1.14 a n d t h e f a c t t h a t Hence If A  x  2  +  f  zero. This i s  Therefore  f  <x^,x^> 0 <v,w> .  But  has form  (ii).  i s z e r o , and  u = ax-j^ + b x + c v + dw 2  2  X^AX^  + dw) + vAw] .  By C o r o l l a r y  [ x ^ f c v + dw) + vAw] h a s  has form ( i ) . has a r e p r e s e n t a t i o n  , t h e n t h e f o l l o w i n g theorem shows how  i n terms o f any 2 - d i m e n s i o n a l  which i n t e r s e c t s  <x- ,x > L  Theorem I I . 1 8  Let f = x Ax  a 2-dimensional  subspace o f  V n <x^,x^>  i s n o n z e r o and  a, p e F , n o t both  R(f) = 2 ,  f e C (U)  represented  U(f)  2  Therefore  f = bx-jAXg +  r a n k one.  <x^,x > "H <v,w>  e <v,w> .  2  15  b 4= ° •  with  where  2  <x ,u,v,w> = < x x , v , w > 1  f = x-^Au + vAw  By Theorem  x ^ e <x ,v,w> c <u,v,w> .  x-^AXgA f 4= o .  Case 2: hence  c <v,w>  2  U(f) .  Then we have two c a s e s :  2  Case 1:  f = x^Av' + XgAw' .  1  2  2  and  V , there i s a r e p r e s e n t a t i o n  may  subspace o f  <x^,x^> .  + x^Ax^ e C ( U ) . 2  U ( f ) such t h a t  a r e both nonzero.  f  Let V  V fi < x x >  Then f o r any b a s i s  l 3  2  be and  {v-^Vg}  of  f = v-jAu + VgAw , where  < v v , u , w > = <x ,...,x > . 13  2  Proof:  1  v-^AVgAf = o .  result follows.  1+  We have Case 1 o f Theorem I I . 17 and t h e  22.  S E C T I O N 2.  Pairs  o f Rank 2 V e c t o r s .  In this pairs  of vectors  representations consider  section,  we a r e I n t e r e s t e d  i n  {f^,fg}  in  We s h a l l  for  ^1^2  the space  Definition  i  H € Rg(U) . '  ^2^ ^  n  U  [U(f ) + U(f )]  o  d  o  =  k  { i t  l  i  t  } / f  2  » f  1  °2  e 2  ( U )  a  n  dim[U(f )+U(f )] 1  {f^,. ..,?•£} f f , , f .} e P ( U ) j  i  V  2 j we s a y t h a t  Let d i m U _> 4 i  = 1,2,  fore  f  for  a  w  ^l- -  e  1  C (U) 2  d  i f each  < k' •  pair  I n t h e case where  k'  i s a P^-pair.  Since  dim U ( f ) = 4 ,  t o be n o n - t r i v i a l .  dim [ U ( f ) 1  P^-pairs  dim U ( f ) = 4 ,  Also  ±  + U(f )]  < 8 .  , where  4 <_ k <_ 8 .  g  then  We n e e d  there-  We c o n s i d e r  P^-pair. Let  Then Let f  H e Rg(U)  {f-^fg}  is  ^X^AX^+X^AX^.  T h e o r e m 11.16, where  + i > 1 < i»3 {f-^fg}  only  T h e o r e m 11.20  Proof:  >  = k }  2  are pairwise-P^  e cJ(U) .  and so  consider  first  o  11.19  P (U)  is  s  obtain  .  2  1  T  independent  fg  <Xg,x^,x^> ,  {f-^fg}  Then x ^ , . . . , x ^ i s  has a r e p r e s e n t a t i o n Since  f  H .  d i m <v,w> D <x^,x^> _> 1 .  a basis of U(f  = x-^Au + V A W  <v,w> ,  subspaces of t h e 3 - d i m e n s i o n a l then  a P^-pair i n  dependent.  < u , v , w > = <Xg,x^,x^> .  2-dimensional  and  <x^,x^>  are  subspace By property  )  and  23.  4.c  ( I n t r o d u c t i o n ) , we  e <v,w> o < ^i i).> > and  x^  2  u =  S b.x, 1=2  V  =  ^  that  z = \f±  now  has  x  y  2  + b x 2  2  Now  1 3  2  r(\,f ,f )  Since  u,w'  Thus  r(x,f  g(x)  in  2  are 1 3  x  V  i  This  H 5  .  For  +  b x^)  + b^x^)  ,  =  = x  2  x^  1  0  0  \+b  0  0  0  d  + x(d  0 5 1  b  2  +  3  a non- zero  (b d 2  value  - d b ) = g(x)  4  2  X  of  = o , a n d therefore.. r(x,f-j_,f ) '= o  X ,  R(z)  .  A  2  C o r o l l a r y 11.21 Then  Hence Let dim H  ff^fg} dim U = 4 ,  = 1  .  nonzero  4  .  4 ° •  polynomial  g(x)  .  determinant  X+d^  2  and  <_ 1  be  4  (t> d^ - dgb^)  is a non-trivial  be  0  0  f ) 2  + d^x^  2  0  b  2  independent,  \ e F ,  c o n d i t i o n t h a t the  2  that  x^, 2  + b )  4  then i m p l i e s  .  , \x^ + d x  t o the  such  4  + d^)  2  ±  c o n d i t i o n t h a t the v e c t o r s  f )  1  \ € F  5  2  independent i s equivalent  r(\,f  e x i s t s a nonzero  + t> x  2  + d x  k  + bye^  2  Then  ^  there  + b x  2  + x .A(\x  (\x  € F .  b.,d  ;  Now  1  independent i n  1  The  .  r a n k a t most one.  = x A( x  2  + x^Aw'  J  show t h a t  cannot be  + f  ^  +  + fg  {f-^fg}  1  J V  of . g e n e r a l i t y assume  f g = x-^Au  S d.X, i=2,4';  w  V  We  hence  x  x  let  f  can w i t h o u t ' l o s s  are He  in  F  For  t h i s value  dependent i n R (U) 2  .  such  H .  that of  24.  The f o l l o w i n g example algebraically H € R|(U) Example  shows t h a t  c l o s e d and dim U = 4 ,  if  F  then there  i s not exists  an  of dimension 2.  11.22 U = <x^,...,x^> x,y  e c|(U)  X = X^AXg + y  =  X  For  1  ;  F a Reals.  : XjAX^  A ( X ^ + X ^ )  +  ( X ^ - X  2  ) A X ^  .  X e F  z = \x + y = x A(xx +x^+x ) + 1  T(X,x,y)  Now  i f  1  0  0  0  0  X  1  1  0  -1  x+i  0  0  0  0  1  = (X + 1 / 2 )  g(x)  ( X X  2  + 3/4 > o  5  + X ^ - X  2  ) A X ^  = X(X+1)  + 1 =  i n Reals.  g(x)  Hence  p R(z)  =2  for a l l  X e P .  Hence  H = <x,y> e R ( U ) , a n d 2  dim H = 2 . I n Theorem 11.20 necessary and s u f f i c i e n t  and Example 1 1 . 2 2 ,  condition for  x,y  we s e e t h a t  t o be  the  independent  p in  H € R (U) ,  be n o n z e r o r(x.,x,y) that  in  F .  F  for a l l  i s that X .  that  be n o n z e r o g(X)  the determinant  Since  i s a quadratic polynomial  r(X,x,y)  condition in  dim U = 4,  2  for a l l  this  g(x) X  in  P(x,x,y)  determinant F ,  the c o n d i t i o n  i s equivalent  be an i r r e d u c i b l e , q u a d r a t i c  to  the  polynomial  25-  i t i s i n t e r e s t i n g t o note t h a t i f a f i e l d i r r e d u c i b l e quadratic polynomial then we can f i n d two independent  F  s  .  has an  h(X) over i t , dim U = 4 vectors  x,y £ II  }  H e R (U) >  3  2  r(X,x y)  such t h a t the corresponding determinant h(x)  F  i s equal t o  5  In t h i s way,, given such a n o n - a l g e b r a i c a l l y c l o s e d f i e l d  dim U = 4  we can f i n d an  }  H  dim K =. 2 .  i n R (U) w i t h 2  The method used i s the f o l l o w i n g : Let  dim U = 4 3  h( X) = X -f a^X -i- a  U = Oc-^ ..>  ..X;,  > . Let  be i r r e d u c i b l e i n F  Q  (non-algebraically  closed). 0  The companion m a t r i x o f h(X) i s B =  -a X  XI - B =  Now  a  Q  X-!-a  1  1 0 0 0 0 u  0 K  'X 0 a  u  O  0  z = Xx + y ,  where  T  JL,  x^y e C|(U) ., X e F  y = x- A(-x ) + X ^ A ( a x  rank  it  H € R (U) , 2  Example  1  -1 = h(uf X )\ \ X 1 0 0 X+a-,  >  2  -a  0  We now take t h i s determinant t o be T ( X , x y )  in  Q  -I  det ( X I - B) =  L  1  n  since  p  + a. x^) .  Then  corresponding t o x =  a  y  x-^AXg +  X^AX _  are independent  z = X - A ( Xx -x^) + X ^ A ( Xx^+a^g+a-jX^) L  1 (  2  has  f o r a l l X e F . We noxi i l l u s t r a t e by an example. U = <2p...,x > , 1|  dim U = 4  3  F = R a t i o n a l s . h(x) = \  r c  26.  Then  x = x-^Ax^ 4- xyNx^ y =  X ^ A ( - X ^ )  (-2)x^Ax  +  are independent i n H e R ( U )  2  2  since:  \ e F  For  ; z = Xx + y  = x A(X.x 1  1  (^x,y)  r  Hence In  0  0  X  0 0  0  -2  -  = X -2  1  2  1  0  0  X  assume  F  i s algebraically  {f-^f )  i sa  2  independent v e c t o r s i n H e R ( U ) , 2  k  in F  as i n a l l e x c e p t t h e d i s c u s s i o n  We now show t h a t i f  that  4= o  dim <x,y> = 2 .  i m m e d i a t e l y p r e c e d i n g , we  We have a l r e a d y  2  • 0  0  0  the r e s t o f t h i s t h e s i s ,  are  - 2x )  - x^) + x y \ ( x ^  2  seen t h a t  k  cannot be  closed.  P ^ - p a i r , and  then  k = 5 or 6 .  4 .  We must show  J or 8 .  cannot be  p Lemma 11.23 and  Let f  z = a f + p>f  Proof:  1  2  1  ?  .  2  e C (ll) 2  Then  be a  R(z) = 4  Let  {f^fg}  a,p 6 F , b o t h nonzero Then  ,  be a z = af^  .  P^-pair + f3f  f^ = x^Ax  1  2  1.13.  i n C (U) , 2  .  2  R(z) = 3 . By Theorem I I . 16 , we c a n  L e t U ( f ) fl U ( f ) = <x > .  find a basis  o =}= a , ( 3 € F ,  Pg-pair;  T h i s f o l l o w s i m m e d i a t e l y from Theorem  Lemma 11.24  Proof:  f  2  {x-^ ...,x^}  1  of  U ( f ) such  + x ^ A x ^ , and a b a s i s  1  {x^ x^,x^ Xj} }  )  that of  U ( f ) such 2  27.  that dim  f  2  = x Ax ±  + XgAx  5  [U(f-j_) 0 U ( F ) ] = 1 2  For  o + o ^ P e F ,  z  =  af  4- ax^Ax^ + pxgAx^ ;  .  7  ,  _^  Now s i n c e dim [ U ( f ) + U ( f ) ] 1  and _  +  and  {x-^Xg,. . . ,x } +  = 3  R(z)  Q  X  g  = 7  i s independent.  )  b y Theorem 1.13.  We make t h e f o l l o w i n g d e f i n i t i o n f o r c o n v e n i e n c e . D e f i n i t i o n 11.25  Let  f, f  g  s Cg(U) .  We  define  ,. \ f ( t t ) = u ( f ) n U ( f ) . l t  2  1  Theorem 1 1 . 2 6 independent (i) (ii)  Proof:  3  L e t H € R?(U) .  subset >  dim ^  H .  1  f  Since  £ U ( f , ) < dim ^ i=i ~ 1=1  Lemmas 1 1 . 2 3 ,  [ U ( f . ) + U ( f .)] = 5 , 6 ,  [L,...^, }  1  U(f ) j L  = 4 ,  U(f.) + 2 . 1  i < i < k .  24,  1 < i < j < k .  W(f,,f^) = U ( f ) n U ( f j ) , the r e s u l t ±  Finally,  be an  f o r 1 <_ i < j <_ k  e Cg(U) , dim  ±  Let  Then  U ( f ) < dim  1=1  Since  of  dim W(f. , f .) > 2  By Theorem 1 1 . 2 0 , dim  2  ( i ) follows.  ( i i ) f o l l o w s d i r e c t l y from ( i ) . Theorem 1 1 . 2 6  shows t h a t any p a i r o f i n d e p e n d e n t  p vectors i n H € R (U) 2  must be a  P,.-pair o r a P g - p a i r .  We  s h a l l now o b t a i n r e p r e s e n t a t i o n s o f such a p a i r o f v e c t o r s , and these w i l l be e x t r e m e l y  useful f o rfinding  t h e b a s i s o f an  H e R|(U) . S e c t i o n 3 d e a l s w i t h a P ^ - p a i r o f Independent v e c t o r s in fact  H ;  S e c t i o n 4 deals with the Pg-pairs.  show t h a t such p a i r s can be e x p r e s s e d  We s h a l l i n  i n a form we c a l l a  28. (1,1) form, d e f i n e d  D e f i n i t i o n 11.27  thus:  2  z e A U , we  For  define  z has a r e p r e s e n t a t i o n V(z)  ,| { U U } j  =  Definition  1 5  11.28  form i f there dim  [U  1 1  1  fU  =1  ]  2  2  and  = <x ,x >  U  {f-^fg}  exist  n U  z = x Ax x  2  and  x  2 i  {f  1 ?  f )  e C (U) 2  4  , U  2  dim  [U  2  = <x^,x^>  c a n be e x p r e s s e d  ) e V(f^) ,  Mote t h a t i f { f ^ , ^ } form, t h e n  5  e c|(U)  ,U  1 1  1  + x Ax  ~]  2 1  n U  2  i = 1,2 2  j  in (l,l)  such t h a t  ] = 1 .  c a n be e x p r e s s e d  i n (1,1)  have r e p r e s e n t a t i o n s  2  f-^ = X-j^A^ + X g A U g f  2  = x^Av + XgAv 1  SECTION J .  2  where  <x >  =  fi  and  <x >  = U  2  1  2  1  U^  2  D Ug  2  The P - P a i r s . c  I n t h i s s e c t i o n we o b t a i n r e p r e s e n t a t i o n s P^-pair  o f independent v e c t o r s i n  Theorem 11.29  Let  H e R (U) 2  f  f  l  =  y  4  = y  2  where [g(f ) + U(f )] 1  2  A  l  u  A 5  ^  +  u  + 2  2 u  A u  i  2  {f-^fg}  Then t h e r e  be a  A  u  3  <u ,u ,u > x  2  3  i s some b a s i s o f = W(f ,f ) . x  2  P^pair  i s a representation:  3  {u^u^u^y^y-p.} and  H e R (U) .  and l e t  of independent v e c t o r s i n H .  f o ra  29. Proof: x  Let  Let  U  = W(f ,f ) n  e U(f _)  1  , x  ]  .  0  U  1 ±  and x  Q  Then b y T h e o r e m 11.16, t h e r e l  f  f  =  =  2  where  l  x  x  A  w-  A  2  they  that  Then s i n c e  are dependent;  f o r some  contrary  - Xufg =  to hypothesis. Now  since  are both nonzero, both  Vg and  Wg =» c v  1  Thus  + dw^ .  is  y 4= o  <v ,w > 1  Clearly  (since  impossible).  Let  otherwise  y  5  - g - dw^ + c p _  1  Y  1  - av )  1  ?  x  l  V]  "  b  _  l  Y  C  l  V  '  U  2  =  w  obtain the representations: t  1  3  » y ^ A ^ + UgA^ =  y  5  A U  2  +  u  l  A  u  3  (Lemma  has rank one,  < v , w > D <w ,w^> 1  1  g  of generality Vg = a v ^ + b w  that 1  ,  Finally,  <w^,Wg,w^> = <v^,w^> , w h i c h  Now we s e t  - b " c" (x  q  w^ = a v ^ + pw^ + yv^  implies  4  f  .  1  b ^ o , c ^ o .  y  u  and  and  1  are i n  z e U  Z)AW-^  we may a s s u m e w i t h o u t l o s s  Wg  - XjiWgAw^) = o,  are independent.  < v , w > n <Vg,v^> 1  o ^ \ e F .  v^AVgAv-^ =  f o r some  v-^w-^  on the  ,  V ^ A ( VgAv^  - X^iXg -  w^ e <v^,Vg,v^> = <v^,w^,v^> where  hence  - XuWgAw^ = w.^Az  B u t then  =  , we h a v e  1  o 4= u e F ;  VgAv^  Suppose  i.e.  <w ,Wg,w^>  1  Implies  .  O  v^^w^ a r e i n d e p e n d e n t .  <v ,Vg,v-^> =  juw^AWgA^^  .  Q  + WgAw^ ,  L  1  contrary  | U  3  v  1>  that  1.8).  V  +  ?  are representations  < v V g , v ^ > = <w ,Wg,w^> = U  We show f i r s t  This  l  v  e U(fg) , x  2  '  l ' 5 U  =  b  v  3  ,  30.  I I . 30  Corollary of  Independent  in  (1,1)  vectors  f  = y^Au  2  =  i n  (1,1)  section  {  f 1  >  f 2  ^  a  P 5  c a n be  -P  a  (1,1)  expressed  form.  deals mainly We  relation"  Let  Let  a Pg-pair  Corollary  o f independent  c a n be  expressed  II.32 r e v e a l s a strong  i n the Pg-pair.  H € Rg(U) .  vectors  i n ( l , 1)  with  show t h a t a P g - p a i r  (Theorem II.31).  independent  i n  H .  Let Then  {f^,f^} {f^,fg}  be a  Pg-pair  c a n be  form.  <x > c W(f fg) . 1  t  B y T h e o r e m II.16,  there are  representations f ^ = x^Au 4- vAw , f  2  = X^AU'  3-dimensional  = (x ,...,Xg> 1  r  5  2  form  expressed  are  i  4- U g A ( - y )  1  H e R (U) .  T h e o r e m II.31  Proof:  Then  e  The P g - p a i r s .  "structure  of  .  b  x  are i n  This  in  H  l e t {f^fg}  + u Au^  5  2  4.  2  (-U- )AU  {f-j^jf }  vectors  i n  and  11.29, there a r e r e p r e s e n t a t i o n s  By Theorem  SECTION  H e R^U)  form.  Proof:  Thus  Let  .  4- v ' A W ' , w h e r e  subspaces  of  <u,v,w>  <x ,...,Xg>;  F o r a,B, e F ,  2  and  <u',v',w'>  [ U ( f ) 4- U ( f ) ] 1  2  z = af  + pf  1  = x A(au  + pu') + avAW + pv'Aw'  1  ,  and  <au + pu',av,w,pv',w'> 5 <x ,...,Xg> . 2  We now show contrary a',p'  that  <v,w> n < v ' ,w'> = o .  , the vector  fx ,a'u  T h e o r e m 1.13.  for a l l  contradicting  Now some  = X-^Au 4- vAw ,  11.32  - x  x  i  A  u  of +  <u',v',w'>  i f  .say  then  and R(z) = 3  by  a u + pu' €  dim < x , u , u ' , v , v ' , w , w ' > 1  {f-^f }  = 5  i s a Pg-pair.  Hence  <v,w> - <v',w'> ,  then  2  for i f  f _ 1  v  A  u')  y 2 f  =  i  A  u  '  +  are contained i n  v  H .  '  A  w  '  where  Therefore, (1,1)  {f-^fg}  Let  such t h a t  2  x  are i n  and l e t  in  + U(f )]  ;  has rank one.  ff-^fg}  vectors  x  w  = X-^Au' + \ V A W  2  H e R|(U)  [U(f )  Then Proof:  Let  o f independent  some b a s i s =  a,p ;  •  0  <v,w> fl <v' , w ' > = 1 , a n d  Pg-pair  l  that  on the  o ^ \ e F ,  Corollary  f  a,p e F , then  f-^ - \~"^f2 = X ^ A ( U  dim  hand,  <v,w> 4= <v',w'>  f^ and  other  the hypothesis  <v,w> n < v ' , w ' > =)=  I f f o r some  i s independent,  On t h e o t h e r  <v,w,v',w' >  Suppose  a ' u + p ' u ' 4 <v,w,v',w'> ,  + p'u',v,w,v',w'}  1  for  <v,w> n < v ' , w ' > 4= o .  form.  be a  {x ,...,Xg} 1  <x > c W ( f , f ) 1  <u,v,w>  1  2  be and  ,  <x ,...,Xg> . 2  d i m <v,w> D <v , w ' > = 1 .  The p r o o f  i s contained i n the proof  o f Theorem  II.31.  32.  We s h a l l  o b t a i n another  c o r o l l a r y to Theorem I I . 3 1 ;  b u t f i r s t we n e e d t h e f o l l o w i n g Definition if  <f ,...,f 1  1 < i  { f - ^ . . .,f^}  11.33  < k  ] t  >  e R|(U)  k fl  1=1  U  =1  , 1  C o r o l l a r y 11.34 Pg ,  Let  i s a (1,1)  Proof:  Let  {u-^Ug}  I I . 31,  there  = u Av  2  1  is  2  + UgAWg  By d e f i n i t i o n , 1  2  5  .  there  such  dim  H e  R  2  such t h a t  basis for be a b a s i s  .  Since  k D U i=l  1  {f-^f^f^}  <f ,f ,f^>  be p a i r w i s e .  W(f  1 3  f ) 2  f-^ = u A v 1  1  U ( f ^ )3 <u u > l S  2  Then  .  2  i s a (1,1)  ,  1 .  = 2  i s a representation  {f^f^f^}  ±  1  of  ±  {Uj^Ugl^ e V(f )  U(f^) o W(f ,fg) 1  <f ,...,f^>  for  that  let  ( U ) ,  are representations  a Pg-pair,  <f ,f ,f >  and  H ,  }  f  II.27)  basis  exist  1  independent i n  {f-^f^f  and there  (see D e f i n i t i o n  dim  i s a (1,1)  .  By Theorem  + u Aw 2  and  1  , {f ,f^} 2  f^ = u ^ v ^ + u Aw^ . 2  basis  for  32.  CHAPTER  III  THE RANK TWO S U B S P A C E S WHEN  In the  r a n k two  III.38). has  a  this  the  (l,l)  we w i l l  have  r a n k two  basis  are  We a l s o subspace  show  that,  d i m e n s i o n a t most subspace  has  given  then i t  i n Theorem  show t h a t  with a  has  if  (1,1)  If  dim U = 5  three  (Theorem  the  a basis  then  ,  it  r a n k two  sub-  vectors  whose  of  III.45.  dim U = n  basis,  when  dimension two,  (Definition II.33).  dimension three,  representations  two  HAS DIMENSION 5  chapter,  subspaces  If  space has  U  then  and i f  H  is  a  rank  d i m H <_ n - 3  (Theorem I I I . 4 o ) . We w i l l p.  need  to refer  to  the  following result  (see  14 o f [3]):  Remark I I I . 3 5 that  intersect  exists  an  1 •< i  <_ k  that  V  j  Let  {V^,...,V }  be n - d i m e n s i o n a l v e c t o r  k  pairwise  in  dimension ( n - l ) .  ( n - l ) - d i m e n s i o n a l space or there  . c W ,  exists  1 <_ i  <: k  an  W  their  bases.  Y  ±  3 W  (n+1) - d i m e n s i o n a l s p a c e  W  there Q  , such  .  T h e f o l l o w i n g two r e m a r k s w i l l t h e maximum d i m e n s i o n o f  Then e i t h e r  such t h a t  Q  spaces  the  r a n k two  be  useful  subspaces,  for  finding  and f o r  finding  33. R e m a r k I I I . 36  ,f^}  {f ,. .. 1  pairwise-P^  Proof:  dim U = 5  Let  be i n d e p e n d e n t  Since  dim  W(f ,f j)  = 3 , i f j  are  pairwise-P,-  ±  i  Let  homogeneous q u a d r a t i c coefficients , then  [f-^,... , f )  k < 3  f  =  will  j  a  .  are  k  k  <  dim U = 5  ,  then  B y Lemma 1.19 ( l ) , {f^, . . . ,ffe}  Hence  .  : i = 1,...,r]  equations  in  be a s y s t e m o f  i n indeterminates closed f i e l d  a non-trivial  solution  F  .  If  f o r the system  F .  i s contained  dim U = 5 in  i n Chapter  > H e R^U)  .  11  of  [4].  Let  H .  .  [x ,...,x,-J 1  be a b a s i s o f  U .  Then each  f^ ,  has the form E af.x.Ax, l<i<J<5 3  We  {f^ = o  exists  Let  Then  Let  i  , 1 <_ i , j  be i n d e p e n d e n t  k  *  L>  H  .  above r e s u l t  T h e o r e m I I I . 38  <_ 1 <_ k  {f- . . . , f ]  j  i n an a l g e b r a i c a l l y  there  The  1  ^  homogeneous e q u a t i o n s  Proof:  Then  Let  .  Remark I I I . 3 7  of  .  H  .  2  , i = 1,...,k , a n d  j  > 3 ,  i  < n  R (ll)  =-4  dim U(f .)  ¥(f ,fj)  r  in  He  .  dim  with  ,  consider  ;  a|. e F .  3  the vector  z =  k £ (3, f , i=l 1  zero,  a n d show f i r s t  {B^}  not a l l zero.  that  R ( z ) <_ 1  1  for  ,  B, € F  not a l l  i  k > 4  a n d some  34.  Now  z =  £ 8,f,  1=1 p  ( a(l) k  , k  a(2)  and  {k.^}  From  (*)  in  s  g  n  CT  ^  * P( i> 2^  *  1  k  k  £ X a  and t h e above  statement,  the n o n - t r i v i a l equations  Theorem  1.2  , v i z : (with  0  s  1  x  2  where  2  permutation of  a  1'<_  ^  <. 5 ,  i t i s easy  a r e l i n e a r homogeneous  AX.  p(i,i )x, 1 ^ 1  i i  are a r b i t r a r y integers  {^(i-^ig)} Hence  =  = l  1  functions  {1,2}  i = 1,2  t o see  ,  .  that  {^,...,6^.} .  of  i n the sustem o f e q a t i o n s ( l )  r = 2, n =  5)  (1) p ( l , 2 ) p ( 3 , 4 ) + p ( l , 3 ) p ( 4 , 2 ) + p ( l , 4 ) p ( 2 , 3 ) = o (2) p ( l , 2 ) p ( 3 , 5 ) + p ( l , 3 ) p ( 5 , 2 ) + p ( l , 5 ) p ( 2 , 3 ) = o (**)  (3) p ( l , 2 ) p ( 4 , 5 ) + p ( l , 4 ) p ( 5 , 2 ) + p ( l , 5 ) p ( 2 , 4 ) = o  W  P(l,3)p(4,5) + p(l,4)p(5,3) + p ( l , 5 ) p ( 3 ^ ) = o  (5) p ( 2 , 3 ) p ( 4 , 5 ) + p ( 2 , 4 ) p ( 5 , 3 ) + P ( 2 , 5 ) P ( 3 , 4 ) = o  are  i n fact quadratic  B-^,...,^ pendent there  in  Hence  of i f  R (U) 2  Example  Also,  b y T h e o r e m 1.2,  a non-trivial solution 1  k  will  indeterminates (**)  f o r (**)  , not a l l zero;.  k  { f ^ , . .. , f } e H  now  , then  exhibit  has 3  k > 4 , then  i f .  R ( z ) <_ 1  inde-  For  these  by Theorem  U = <u ,u ,u x  . n u ( f ) = <u ,u ,u > . 1  2  5  2  1.2.  k < 4 .  a 3-dimensional rank 2  subspace  .  III.39 ±  i n the  H e n c e , b y Remark I I I . 3 7 ,  B ,...,P  We in  .  equations.  exists  values  F  homogeneous  ,x^,x > 5  .  H - <f- ,f ,f > L  2  5  ;  35. f , = X AU-, + U^AUU 1 4 1 3 2 )t  d  5  2  3  1  f 2 = ( x ^ + X ^ ) A U ^ + UgAU-^  For  e F ;  3 E  z = -  a^i  = u-j^Aa-j^x^ + u A ( a u ^  +• a x ^ +. a^u.^)  1  2  2  + u ^ A ( a ^ x ^ + a^x^ - a2 ]_)  •  u  If 1  any one o f  <  1  i s zero,  [a ) ±  R ( z ) = 2.  If  + o ,  < ^» °S z =  u  i  A  a  i  x  4  (  +  + uy\(ayt^ = ( *  u  l  U  +  §S  u  2  ( l 4  ) A  ( l 3  a  a  u  A  a  u  - H^ 2 3  X  u  3^ ^ l 3  +  ' 2 5 a  x  +  a  3 i' u  + a^)^)  ) A  v  <2  a  +  2  - (a  H7 3 1  +  u  a  +  a  2 5 X  +  +  a  a  3  ) u  3 l U  l-  }  }  d and  R(z) =2  Results Interest the  .  obtained later  i n t h i s paper w i l l  i n a p a r t i c u l a r k i n d o f r a n k two s u b s p a c e .  r a n k two subspace  with a  (1,1)  basis  j u s t i f y our This  is  (see D e f i n i t i o n  We now o b t a i n t h e maximum d i m e n s i o n o f a r a n k t w o s u b s p a c e such a b a s i s such a  when  We g i v e ' a l s o  an example  with  of  basis.  Theorem III.40 (1,1)  dim U = n .  basis  for  Let  dim U = n .  <f ,...,f > 1  k  Let  € R|(U) .  {f , . . . , f } k  Then  k  =  be a (n-3)  II.33).  .  36,  Proof:  Suppose l  f  f  f  =  k = n-2 .  l  U  A  l  y  +  = u Ay  2  x  n-2  *  l  u  A  +  2  A  i  z  U A Z 2  n-2  y  2  u  T h e n we c a n r e p r e s e n t  V  +  2  n - 2 >  Z  <u ,u ,y ,...,y _ 1  Now  2  1  n  { u ^ u ^ y ^ , . . . jy _2^'  u-^Av  combination of  + \VAW  U = <u ,u ,y ,...jy _ 1  2  1  n  {z^^,. . . , z _ } n  f-^,... , f  o =(= X e P  }  n  ,...,z _ > E U . n  2  has the form  2  a n d t h i s h a s r a n k <_ o n e . since  >  2  dim U = n .  This implies  Hence  {\x- y ,. . . , y _ 3 .  i s dependent on  2  Z l  m u s t b e i n d e p e n d e n t f o r , i f n o t , some  n  linear  ,  2  L>  n  1  Thus  2  n-2  Z j  *  j ^ i J J  =  a  If fj  y  = 1 =o , = u A(y 1  Hence, w i t h o u t l o s s is  +  j  P  y  -  J  s  d  j * P j  2  2  n  e  P'  e  n  d  e  n  1  i  o  n  z =  1  1  n-2 £ a, f, 1=1 1  z  f o r some  1~ n  2  1  < j  < n-2  .  [z ,...,z _ 3 n  1  2  z  •  we c a n s a y H  e  n  c  e  < y  l'•••'y _2  € P , not a l l zero,  , o f X e F ,  and  1  Q  1  j  n  = u . A y 4- u A \ 1  2  l ±> •, • • > -2^"  = y  1  J  argument t o t h e one a b o v e ,  n-2 E a.y, i-l  n-2 = \ L a.z. 1=1  i  1  }  •  T h e r e f o r e f o r some  n  F  o f g e n e r a l i t y , we c a n a s s u m e  = <z ,...,z _2>• 1  e  + u A( E ajyj)  pjU )  n  i  a  write  Using a similar y  '  { y - ^ , . . . >y _2^  dependent on  f l ' ' ' * * n-2^  l  u  d  1  y  has r a n k a t most o n e .  Hence  n  >  37.  k  i s a t most  (n-3)  •  We now e x h i b i t p  H e R ( ) ' f  f  f  f  For  a  L  u  2  U  =  < u  i> 2* 3>... , u  ^ l U  =  l  U  A  A X  X  n-2  n-l  V  +  +  U  2  X  n - l n  A x  € F , n o t a l l zero ,  i  z =  n-3 E i=l  fs 1  /n-3  R(z) = 2  z  of  b y Theorem 1,15-  {f-^, . . . , f _ ^ } n  " ^ l * ' ' * ***n-3  a  r  e  Corollary III.41  /n-3  \  V ^ V i + a )  =  and  y  x  = U-^AX^ + UgAX^  n-4 n-3  t  {f^} i n  = U-j^Ax^ + UgAX^  1  2  e  (n-3) such v e c t o r s  +  U  {  f  ±  1  i i 3 X  +  Thus any l i n e a r This implies  combination that  and g e n e r a t e a rank two subspace.  L e t dim U = 5 ,  dim H > 1 , and l e t  a  A  1  has rank 2 .  independent  2 l ?  ,  b  e  H e R|(U) , w i t h a  (1,1) b a s i s f o r H .  k = 2 .  Then  We have seen so f a r t h a t i f dim U = 5 , and H e R (U) , then 2  H  then  has a  H has dimension a t most 3 . ( l , l ) basis  (Corollary II.30).  Example I I I . 3 9 t h a t t h e r e do e x i s t 3 - d i m e n s i o n a l H € R (U) . 2  By C o r o l l a r y I I I . 4 1 , such subspaces  I f dim H = 2 , We see by subspaces cannot have  38. bases.  (1,1) when  H  We s h a l l p r o c e e d t o  has  dimension  If I I I . 36,  i  is  2  {f-j^fg,^}  are  + j , 1 < i, j < 3 .  =  1,2,3  or  U(f )  We n o t e a n d d i m U(f^) d i m f? U(f.) i - l  > 2 .  H ,  .  Thus  U(f ^)  either  w  dim W = 5 ,  ,  then But  o U  {f,,f ,f  give but  U(f. )  .  =2,3  = 3  ,  >_ 2 .  Thus  are, p a i r w i s e - P .  We s h a l l p r o v e  = 3  two t h e o r e m s  , 5  which  1  representations first  }  p  o  dim W(f^,fg)  n U(f^)]  1  since  dim U  = 1,2,3 •  i  dim[W(f ,fg)  = 3  i  i = 1,2,3-  ,  y  d i m hi i=l  dim W(f , f j )  dim U = 5 ,  since  1  then  t h e n by Remark  ±  now t h a t  = 4 ,  of  Now d i m U ( f ) = 4 ;  c  i  a basis  pairwise-P,-  Hence b y Remark I I I . 35, i  H  three.  {f ,f ,f^} 1  o b t a i n the bases of  it  will  for  be  {f ,fg,f  J  1  convenient  for  i n each o f us  to have  these  cases;  the  following  [S;  T]  definitions. Definition  III.42  <SUT>\<T>  . In  T = {x  g + 1  the  , . . . ,x } k  following  y =  that  2 a.x, i=l 1  a-, > • • •  subsets  case where ,  x  1  S,T  of  U ,  S = {x^,...,x  c U ,  1 •< i  <_ k  }  =  and  , we s h a l l u s e  the  convention? [S;  We n o t e  For  i  = [x ,...,x 1  in this ,  1  T]  a,  case,  e F ,  1  s  nonzero.  s  if  1 <: i  ; x  s  y e <. k ,  +  1  ,...,0^] [S;  T]  where  , at  • then least  one  of  ,  39.  Definition  III.45. SAT  In  F o rsubsets  = {xAy : x e S  t h e case where  SAT  = xAT .  <x^,...> regard  x  k  >  , then,  for T .  t o keep  Also,  SAT a s  L e t dim U = 5 .  for  such t h a t  H e Rg(U) ,  = 1,2,3 •  U"  o  Then  = <u ,u ,u^> 1  l  f  2  f  3  S  i  hasa basis  U  + U^U-J  =  +  X^AU  2  dim U  {u-^u^u  1  U  2  {f-^fg}  has a basis  representations:-  3  A U  1  such  i s a P^-pair,  {w ,w ,w^,y _,y^} 1  = 3 ,  Q  U-, AU^,  = - yAu^+  Since  be a b a s i s  ,x^,Xj-} ,  have  2  }  2  ,  Q  {f-^f ,f^}  = X^AU^  {f-^f ,f  x^, u ^ , u ]  2  U  we s h a l l  f c  yy € [ x ^ ; x ^ , u ^ , u ] n [ x ^ ;  Proof:  i s t h e space  [x^,...,x ]AT .  Let  U ( f ) z> U  such t h a t  2  f  i f  write  T .  Theorem I I I . 4 4  i  {x} , we s h a l l  thenotation consistent,  as a s e t and w r i t e for  U ,  a n d y e T} .  i s the singleton  Similarly  S  Similarly  S  S,T of  2  U  i l  0  2  t h e n b y Theorem 1 1 . 2 9 ,  = W(f- ,f ) L  = <w ,,w ,w^> ,  2  ]  2  that f  = y^Aw  1  f  = y  2  Now  A 5  ^  1  + WgAW^ , + w Aw^ ,  2  x  U(f,)  U  3  p  Hence t h e r e [y.V 5 y  ;  w  exists  i^ 2 w  j W  3-'  ,  dim U ( f J •  o  y'e ^(f^) "  B  u  t  ^ l f  p  such that j f  2' 3^ f  a  and dim U = 5 .  =4  r  y' 4 e  P  a  l  r  w  u  i  s  y' «  >  0  e  -  p  5  •  Hence  40.  y'  G  iy^i y y w-^w^w^My^; y^w^v^w ] .  f h a s j-  a representation  = y ' A u + vAw ,  contrary that Since  1  1  subspace  U  Suppose on t h e 1  subspaces  1  f ^ = y'A(aw  a,b e F .  o f t h e 3-  2  + bw ) + (cw + dw )Aw' .  1  2  + (by' - dw')Aw  1  ,  g  dim <w ,w > n <v,w> _> 1 .  (1,1) b a s i s f o r  H .  I I I . 4 0 , and t h e h y p o t h e s i s t h a t u ^  .  2  i . e . , u = a w + bw  , then  Q  f ^= (ay' - cw')AW a  1  a r e 2-dimensional  Hence we c a n r e p r e s e n t  2  <w ,w >  1  ;  2  2  {f ,f ,f^}  u  u e <w ,y >  <w ,w > , <v,w>  dimensional  1  <u,v,w> = U . o  J  We show now t h a t  Then  By Theorem 11.16,  1  ,  2  2  making  T h i s c o n t r a d i c t s Theorem  dim H = 3 .  <w^,Wg> , and u e [wy, w^,w ] .  Hence  Without  2  loss of  g e n e r a l i t y , we c a n s a y u = w^ + c-^w^ + CpW^ ,  c^ e F .  We have now: l  f  f  y^  =  = y  2  f  A  w  A 5  i w  w Aw^ = ( y ^ - C W ) A W  +  X  Since  least v €  Then  2  f  ,  o  u 4 <v,w> .  [w^  we  Then  f f  w  3  u] .  2  2  2  = a 1  -  Q  2  1  2  ±  ±  + u Au^ ,  1  2  2  bu  , w = cw  + du)  \xl , w = -1 a u  = ( y - C W )ACT U 2  1  A  = (y^ - c w ) A a " u 5  are 2 - d i m e n s i o n a l  L e t v = aw +  1  1  2  + WjAu ,  - b c w ) A u + Y^ ' 1  1  1  2  Hence we may suppose  2  y  ) A W  , they i n t e r s e c t p a i r w i s e i n dimension a t  = y'Au + ( a w + bu)A(cw  P or =  1  .  = (y' + adw Let  W  2  Now  [w ; u]  2  ,  2  <w ,u> , <w^,u> and <u,w>  of U  one.  VAW  5  + W AU  1  2  = (y -G  + W AW^  2  = y'Au +  subspaces  1  2  + ^ A U ^  ,  W  '  ,u =  o 4= Y e F . au^ .  1  + du .  41.  f ^ = j" Aan^ + u We h a v e x  = a  5  y  _  1  the  result  (y  -  € [x^j  5  5  , 1  2  Let  H e Rp(U)  basis  <u  dim U =  - such t h a t  2  - 1  (y^  -c-jWg)  and n o t i n g  •  5  that  Cf^fg,!"  Let'  1=1  ) = 2  .  }  Then  be  a  basis  U  has  a  1  such t h a t  g i v e n by e i t h e r  ,  .  d i m fl U ( f  5  representations  ,  x^u-^Ug]  u ,u^x^x >  1 }  x^ = a  y = ay" 5  45  i  on s e t t i n g  x ,u ,u ]fl(x ;  Theorem I I I . for  A u 2  { f ^ f ^ f ^}  (l)  or  (2)  have  below:-  3 (1) ^ 0 ^ U ( f ) ±  = <u,u^> j  f ^ = ^ -[_ +  u  fg  = Xj-AUg +  U^AU^  =  u^Ay'  x  AU  +  uAy  2  u e <u  A u  3 _0 U ( f ) 1  = <u,u'>  [x^x^  ,  = Xj^AU^ +  UgAU^  fg  = X^AUg +  U-j^AU^  f  = yuAu' +  ?  U  €  yAy' ,  has  Since  a basis  {f-^fg)  is  ,  <u- ,Ug> L  + Y a  {u^UgjU^y^yg)  representations  ,  2  W(f  u^u^u^]  y,y' O  Proof:  u >  l 5  f ) 2  = <u ,u ,u > x  2  5  3  y,y'e  (2)  1 >  e 6  u'  [x^x^ F  .  £ <u ,Ug,u^> 1  u^u^u^]  .  •  •  P ^ p a i r , by Theorem 1 1 . 2 9 , such t h a t  {f-^fg}  have  U  .  42.  2  f  y  =  5  A U  2  U  +  1  A U  3  W(f  ;  f )  1 ?  2  dim f[ U ( f ) = 2 .  By h y p o t h e s i s ,  = <u ,u ,u > . 1  Let  <u,u'> =  3-dimenslonal  space  d  L  then  2  1  J  are 2-dimensional  <u- ,u ,u^>  /? U ( f , ) . "  <u,,u_,u,> . 1  <u,u'>  5  1=1  i=l T h i s . i n t e r s e c t i o n i s contained i n  Since  2  subspaces  dim  o f the  <u,u >n<u- ,u > _> l . /  L  2  Hence, by p r o p e r t y 4.c ( I n t r o d u c t i o n ) we c a n assume u e <u^,u > . 2  Since  U ( f ^ ) => <u,u'> , then by Theorem 11.17  has  a representation either  for  some  uAv' +  f^ = UAY'+U'Awt.  f o l l o w s from Lemma I I . 1 8 t h a t {f ,f ,f^}  isa  2  III.4l.  U  A  U '  + vAw  u e <u ,u >  then i t  2  f ^ = u ^ v ' + UgAw* .  Then  4» <u- ,Ug>  <u,u'>  .  L  Corollary  Therefore, since  u' e [ u ^ ; \x u. ] .  ,  g  I f <u,u'>=<UpU >  ( l , l ) basis f o r H , contradicting  Hence 1  Y  o f Y e F • Suppose  1  or  U'AW'  f^  1}  2  How we can w i t h o u t l o s s o f g e n e r a l i t y assume u e [u, j u_] , s i n c e i t i s easy t o see t h a t a s i m i l a r case h o l d s  1  if  u € [ u ; U.J .  5  Thus, we can assume  2  Let f  2  u' = au^ + b u =  2  uy\aw';  UAV'+  a t i o n s as i n (1). a u  - 1  u A(au 2  Writing  1  u  x  5  f  ±  i  b = o ,  =1,2,3}  -  1  then have r e p r e s e n t f  ±  = y^A^ +  = y ^ U g + i ^ A u ^ = (y^ - a ~ b u ) A 1  g  + bug) .  for a  {t ;  If  2  I f b ± o , we r e w r i t e  2  + a* u A(au  a ^ Q-.  hence  + bu ) ;  5  1  2  ;  u' e [ u ^ ; u ] .  ^ ,  1  Now u  g  f  ?  = uAv' + ( a u + b U g ) A w ' .  for a  ?  - 1  u  p  ;  x^  for ay , 4  43.  x  for  5  a(y  - a~ bu ) 1  5  1  (1).  have  It  In o  i  e F  Y  and  the  •  hence  i s easy  e  any  He  have  a basis  to note types this  claim,  H  III.44 can  be  H  has  that  By  then  {f  i  i s one  of  shall  Let  by  dim  a 3-dimensional of  the  5-dimensional  dim[U(f,)0W(f  3  u^u^u^].  U  of  = 3  dim  and  types.  space  basis  ,  i s one To  of  H  Rg(U)  such  ,  {-f  of  the  ,f ,f^}  i n Theorem III.45.  dim  H  ,  .  is a  = 3  .  r? U ( f , ) = 2  dim  Since  .  1  i s a basis  2  two  substantiate  ,  that  5  that  would  1=1  U(f^) U  seem  = 5  i n Theorem  He  pairwise-P  and  ) = 2 1  It is interesting  which  those  U  <>  i f  are  some  ,  the  = 5  d  n U(f 1=1  dim  we  for  Y  ( i . e . III.45).  , f j ] > 2  1  [x^,x,_;  e  i t would  0  subspace  w'  2  theorems,  {f, , f , f , }  1,2,3}  ,  since  a basis  one  Remark I I I . 3 6 , =  H  show t h a t  replaced  + bu )  = u A u ' + vAw  f  three  has  (au^  3  two  theorem  a basis  : i  2  dim  1  Proof:  v',  (2),  case  1  in fact  Theorem III.46. Then  for  u_,u_,u_]  5  second we  that  preceding  which  i n the  see  h  such  that  to  [x ,x_;  the  Rg(U)  u^  i s Just  4  Prom  and  a l t e r n a t i v e case,  This v,w  ,  of  H  W(f- ,f ) L  2  4-dimensional  , is  subspace  then  i.e.  dim  fi U ( f 1=1  ) > 2  .  A  3 If  to  dim  show t h a t - i f  with  the  required  fl U ( f )  1=1  dim  1  = 2  , we  have  the  result.  It  remains  1  3 D U(f,) 1=1  properties.  = 3  > then there  exists a  basis  44. 3  d i m n U ( f , ) = 3 , t h e n b y T h e o r e m 111,44 ,  If a basis  1=1  U  has  1  {u^UgjU^x^x,-}  s  u  c  i  [f f , f ^ }  t h a t  l  1 5  have  2  repre sentations  -  u(g )  =  <x  4  Then  n  u(  g  x  =  +  l  f  X  R  + UgAu^ ,  AU.  5  3  f  Si  =  2  f  Let  = X^AU^  l  f  +  U , A U ,  2  y € [x^j x ^ U - ^ U g l n [X JX^,U- ,U ]  yAu^ + UgAl^  f  =  2  (x^  -  ) = <u, , u » >  U ^ ) A U  is  2  ,  and  a basis  for  H  The f o l l o w i n g a r e d i m e n s i o n 3 , when T h e o r e m III.45. H e Rg(U)  ,  Note  -  v 5  U , ) A U ,  3  g  :  2  = f  2  g-j = f j .  ;  2  Hence  =  l  f  f  2  f  3  e F  ;  the  properties.  rank  subspaces  two  two t y p e s  dim H = 3 >  j  Q  where  -dim U  U = <u ,Ug,u^,x^,x > 5  1  o  ;  given  in  an example the  = 3 ;  basis  i = 1,2,3.  f?U(f )  = <Ug,u^>  i  X^AU^ + UgAU^  5 =  of  of  required  E x a m p l e III.39 i s  that  ±  E x a m p l e I I I . 47  w i t h the  examples  U(f ) 3 U  is 3  5  ±  (x_  +  has d i m e n s i o n 2 .  dim U = 5 ,  dim U = 5 ;  {f fg,f }  a  1  1  {g-^, g , g-^}  For  L  5  Let  i=r  r  ,  1 • 3  2  +  U  UgAx^ + z =  = u. 2  AU,  N  1  3  3  5  3  A(  Zaj^f  a i  u  5  4- a,u,Ax,-  3 3  5  i  - a x 2  5  + a x ) +..u A(a x ?  4  1  1  4  +  agU^)  of  of  2  45. If  =  o  ;  z  and If  a _= o  z  and If  a  i  4  °  + -a^x^)  + agU^Au^ +  = u  + a  +  2  A ( - a  2  R(z)  ;  2  = UgA(-a x^  5  x  5  4  )  u A(a x 5  5  -  5  a ^ )  2.  =  = x^A(a u 1  -  1  a^u ) +  U ^ A ( - ^ U g 4- a ^ x ^ )  2  =2 .  R(z)  i = 1,2,3;  >  z - a u A(^i 3  x  2  a^u^Ax;-  then  - ^  2  x  +x ) +o ^ u ^  5  a  =  (a^u  2  + a  u  1  1  )  A  ( x  + a /a  4  2  u )  ±  + c^iyNf —  ?  u^) + 0 ^ X 5  + ^  4  2  a  x  5  a  l  + ( —  2  - — ) u  ?  ]  3 3 5 a, Let  Y'  a  =  -  a  2  /  l )  a  3  '  I  = o  Y'  F  ;  + ( 2 2 3 3) 5 _a  U  +  a  u  Ax  z a n d a  If  Y  '  + o  ;  z  =  (a^u +  and .  (a  R(z)  2  + a u ) A ( x 1  1  3 2 - § U  =  U  3  )  =  A  (  4  R  2  + —  4  - ^  2 3  •  ( ) = 2.  1  z  u ) ?  X  5  +  Y U  3  }  2. 3  Example III.48  U = <u ,u ,u ;x^,x > . 1  x  = X AU  f  2  = X AUg  f  4  X  5  5  + UgAUj +  5  = U A(U 2  2  :  f  X  \  (a^UgH-^^)A(x +^-u )  U-JAUJ  + U^) +  ,  X AX 4  5  W(f ) ±  = <u ; u + g  1  > .  46. z = Ea^f^ = x ^ A ( a u 1  (a^Ug + a u 2  If  z =  0^=0;  and If  a  2  = o ,  a  ±  2  2  2  - ayc^)  2  2  R(z) i =  = 2  .  1 , 2 , 3  ;  z - (a x 3  (a u 1  - (,a^ [(a and  4  x  R(z)  2  + a^u^) + x ^ ( a u  1  + u^)  1  - a u )A(-l 2  2  cXgU-^ +  U  2  + a )u 5  = 2  .  5  2  a  +  a^Ug)Au^  1  2  1  a., + (+a u )A(^  + x )  i  - a u )A(a /a 2  + aye,-)  A  a.  4  +  2  1  = 2 . 1  + o ;  ay^)  2  + ( a u + a^u )A(u  z = UgA(a u^ +  and If  + a^u )Au^ + U A ( -a x^ +  1  X,-A(CI U  R(z)  + a^x^) +  1  5  ^  + a u ]A[u 2  1  + x^) +  3  -  + i)u ]; 2  a. -A)^ +  CHAPTER  THE RANK TWO  In E € R (U)  this  it  3  its  we  dim H < 3  (1,1)  has a  SUBSPACES WHEN  chapter,  then  2  IV  basis.  If  show t h a t  HAS DIMENSION  dim H = 3 ,  l e t  first  case,  we  case  i s contained  show t h a t  H  (Theorem  IV.6l).  There  (Theorem  IV.62).  Representations  given  ff-^f^f^}  Our if  f ^ , f , f-3,  of  dimension  first  at least  three,  In t h e second vectors  basis vectors are  TV. 57, 5 9 -  vectors and  .  f o r such a b a s i s  f o r these  aim i s to obtain  are independent  0  III.  are 3 p o s s i b i l i t i e s  major  be  1  of pairwise~Pg  IV.53 a n d Theorems  i n remark  ,  d i m [ ^ U ( f . ) ] = 5,6  i n Chapter  has a b a s i s  6  dim H = 2  If  1=1 The  6  dim U -  i f  (Theorem IV.72).  T h e o r e m 11.26 shows t h a t  basis.  U  the result  i n a r a n k two  that  subspace  dim[ i U ( f . ) ] = 6 ,  then  1=1 <f ,fg,f^>  contains  1  addition extended This  {f ,f } 1  2  to a basis  result  a basis  is a  of pairwise-Pg  Pg-pair,  then  of pairwise-Pg  i s contained  i n Theorem  this  vectors IV.6l.  vectors. pair for  Ifin  c a n be <f ,f ,fj> 1  2  .  To a c h i e v e - t h i s of  vectors  }  {f^f ,f 2  a i m , we c o n s i d e r a l l p o s s i b l e  s a t i s f y i n g t h e above  Theorem I I . 26 t e l l s us t h a t e a c h p a i r must be e i t h e r a is  pairwise-P^  or at l e a s t  t h e c a s e when bases f o r However, that  P^- o r a P g - p a i r .  { f , f^,f^}  <f f ,f^> 1; 2  i  1 < i < J < J  one o f w h i c h i s a  {f-^f^f^)  Pg-pair.  For  we e x h i b i t  3  (1,1)  two  basis.  t h i s c a s e i s n o t o f g r e a t i m p o r t a n c e as we can show  <f^_, f  2 >  contains at l e a s t  fy>  case i s reduced t o the second  one • P g - p a i r .  [ fj U ( f . ) ] ,  Thus  this  case.  In t r e a t i n g the l a t t e r case,  i»l  ,  Thus e i t h e r  is pairwise-P^  n  conditions.  {t^t^  one p a i r i s a  triples  p l a y s a prominent r o l e .  the  intersection  I t has d i m e n s i o n a t  1  most two_, s i n c e we have a t o show t h a t basis  [fj, f ) 2  {f^f^g-j}  necessary  for  P g - p a i r say  {f^f  } .  In order  can be e x t e n d e d t o a p a i r w i s e - P g Kt^f^,  f^> ,  we have found it-  t o c o n s i d e r t h e t h r e e cases?  dim[ 3 U ( f . ) ] = o , 1 , 2 .  1=1 Theorem IV. 49. pairwise-P,Then  Let  and i n d e p e n d e n t i n  [ 5 U(f )] ±  f , f, 0  H e Rg(U) ,  have  has a b a s i s  representations  and l e t H  {f-^f^f^}  such t h a t  be  dim | U ( f ) = 6 .  (u^u^u^x^x^Xg)  ±  such t h a t  ^ 6  3  f  x  a  U  +  v  A U  3  ^  u 4 <u >  < u , v > = <u ,Ug> , 1  Proof;  Since  intersect i  pairwise  = 1,2,3 .  hypothesis i  {f^f^  f^}  are pairwise-P^ ,  .  fTj(f ):  i = 1,2  ±  Now d i m U( f ) = 4 ,  i n dimension 3 .  ±  f r o m r e m a r k 111.35 a n d t h e  d i m ^ U( f . ) = 6  that  U( f . )  U  dim U  ,  - 1,2,3 • By t h e o r e m 11.29,  f  u 4 <Ug>  3  x  Hence i t f o l l o w s that  where  l  =  y  4  A  U  i  +  U  2  A V  3  >  {u^,Ug,v-j,y ,y^}  f  2  y  5  A U  2  +  u  exist i  A  i s independent.  4  f-j = y^AW-L + WgAW^ , and  =  there  where  V  3  1  5  w_ = a u , + b u  0  1  4  1  1  Z - u and that  R( Z) = 3 <f  l 3  Q  l A  .  Let  (y -v 4  ; 5  5  s  Wg e <u^,Ug  a , b e F , and w^ € <u ,Ug> . 1  w^ 4 <u^,Ug> ; i . e . ,  Z = -(f^+fg+f-j) .  Then  - a w ) 4- U g ( y - v - b w ) 5  Rg(U) .  Q  Since  6  We now show t h a t  A  b y Theorem 1.13,  fg,f >  we h a v e  >  1  we c a n WLG assume  on the contrary:that  <u ,Ug,w > = U 1  5  f o r some  = ygAW-j_ + (au +bUg)AW^ . Suppose,  where  <w ,Wg,w^> = < u , U g v ^ > = T]  d i m <u^,Ug> n <Wg,w^> > 1 , Thus  >  B y Theorem I I . 1 6 ,  | U ( f ) = <u ,Ug,v_ ,y ,y ,y >. i  representations  5  5  5  + w  l A  y  6  contradicting the hypothesis  Hence  w  1  e <u ,u.g> ]  .  Prom t h e  proof  of  T h e o r e m 1 1 . 2 9 , we s e e  Since <u ,u 1  3  2 J )  Rewriting:  l  f  2  =  the  y  5  ,  for  =  yjjAU  ^5  =  -  £3 5  1  A U  y6  +  U  U  1  U  and  2  Q  e [v^;  w^  X € F  A ( V  2  + U )  3  IA(V +U)  and  4 <u > .  w  2  1  = <w w ,w^> = 1>  2  u-^u.g] . u e  < u  Thus  i> 2 u  •  >  l  +  X W  2  ,  ,  3  for  A( 3 ^) v  •  +  u =  We s e t  and  f^,f  ,  2  H € Rg(U)  and i n d e p e n d e n t  v = Xw  = y^ ,  4  ,  2  a n d we  is a  (1,1)  By T h e o r e m IV.49,  have  .  and l e t  in  x  {f^f^f^}  such t h a t  H  he  basis for  [J U ( f . ) J 1  Kf^f^f^  has a  ] = 6  d i m [ | U(f,) I  f f - p f g , ?j)  Proof:  +  x  Let  17.50  pairwise-Pp. 5 Then  ±  u >  then o 4  + u ,  r e q u i r e d forms  Corollary  ,  some  2  A W  u^ =  3  1 ?  2  and  4 <u >  w  -f  X  <w ,w > = < u  w - j > = "Cu-pUg, v^>  = X(v^+u)  w  that  X  €  R (U) 2  basis  X  {u ,u u^,x ,Xp,Xg} 1  4  2 J  such  that  f j_ •= X A 2 _ + o A - j 4  f  u 4  <^>  2  -  X  «  XgAtt  and  both nonzero  u  u  AUg  +  U  + v  A  in  F ;  where  <u,v> =  , How  u = a ^  v = cu-^ + d u  f ^ = X g ( a u + b u ) + (cu-j^+dUg) u ^ . A  1  2  representations  ,  5  A ^ - J  4 <u > . 2  U  have  >  u  L  f-^f^f^  A  2  ;  + bu  c,d  2  e F .  <u ,u > 1  for  2  some  a, b  Thus  On r e a r r a n g e m e n t  we  have:  51.  ?  = (-•a ) u  2  3  3  f  A  2  ,  5  = ( & ^ - c u ) u + u A(du -bx ) 6  3  {f^i',.,1%}  and hence  + U A(-X )  1  A  is  1  2  5  6  ,  ( 1 , 1 ) basis f o r  a  <f^ , f , f  > € R|(U)  2  Theorem 17.49 a n d i t s c o r o l l a r y give t h e form of a basis  of  a r a n k two subspace  theorem.  We now e x h i b i t  the  kind.  above  Example  o  =  1  =  f  f  2  ?  3  <UpU  x  a n example of a rank two subspace  of  a. e F  1  +  = x^ u  2  + u^ (u +u )  A  =  ,  u^>  2 )  u  ] | A  U A(U 5  +U ) 2  1  +  2  6  B  X  A  X. A(U1+3U )  o +  e F ,  2  U A(^ +U ) 5  3  1  + 1  ,  2  .  not a l l z e r o ,  Z = -' 1 a f i=l ±  i  = u [a x +a x +(a +a +a- )u ] lA  +U A 2  and  the c o n d i t i o n s of t h e  IT.51 U  For  satisfying  R(Z) = 2 .  Hence  1  lj  5  6  1  2  3  3  [ a 2 x 5 + a 5 6 x 6 + ( a ^ a ^ a ^ ) u^ ]  j  generate a rank two subspace  {fpfg,^}  o f d i m e n s i o n three. We s h a l l now show t h a t , of pairwise-P,- vectors then  H  has  a basis  for  if  {f^f^f^ } -  H € R (U) 0  of pairwise-Pg  and  vectors.  is  a  basis  dim £ U(I\ ) = 6 We n e e d  the  52. following.  Lemma IV. 52 <f  1 ?  f ,f >  .,  2  Let  ( f ^ f ^ }  € R (U)  3  dim \ U ( f j ) = 6  (2)  {f-^fg}  3  g  isa  Pg-pair  fy  e  }  2  (1,1)  basis f o r  Proof:  Since  {f^fg]  Since  x  x  = u  f  g  = tu^x^ + ^ A 5  assume  Pg-pair,  +  U  2  A X ^  isa  \ U(f ) ±  (Theorem II.31)  U(f j) c [ I J ( f )+U( f ) ] , p  2  y ' - u ' + a'x^ + b'x^ +  choose  such t h a t  2  c'Xp.+  , d'xg ,  {u,u'} e <u >  both polynomials  X - , , Xp 6 F  u-^y + u A y ' .  . Let  y = u + ax-j + bx^ + c x ^ + dxg  where  froB.  and by Lemma 11.16, we can  {y^y'} e <u ,x^,...,xg>  Since  .  ( l , l ) basis, i t follows  f ^ has a r e p r e s e n t a t i o n 1  has a b a s i s  ,  X  {f^, f , f^}  and  have r e p r e s e n t a t i o n s : -  2  Theorem I I . 18 t h a t Since  isa  f  5  are pairwise-Pg  ^ ^ f ^ f > .  ^ {f-^^fg] l A  basis f o r  .  } {f-p f , g^}  form a  (u-^Ug^x-j, . . . , X g ]  (1,1)  satisfying:  2  <1)  Then  be a  g  above a r e n o n - t r i v i a l , we c a n  a+X  b  1  c+X  4o a'  Then  '  b  +  4  and e'  l  X  d  2  d'+X  =  X f _ + ]  X f  =  u  [  a + X ) x + b x + ( X + c )x^+dxg ]  1  l  A  £  o  2  + f^  2  1  3  4  2  + U g A t u ' + a ' x ^ X-j+b' ) x + c x + ( x + d ' ) x g ] v  4  is  such that  Hence  V(g^)  Let  for  l3  has a b a s i s  =  ?2 ~  U  U  1  Proof: (1,1)  A X  a  l  X  A  3  +  U  2  A X  5  +  u  2  a X  y  4 6  (1,1)  such that  basis  i s proved.  of pairwise-Pg  dim[J U ( f ) ] ±  such  = 6 .  that  3  + Ug y'  j  A  <y,y'>  n u(f ) » o ,  2  ±  The r e s u l t  follows  and t h e f a c t .  a n d t h e lemma  3  c < u , x ^ .. . ,xg>  pairwise-Pg  5  {u^Ug^x.^,.. . ,Xg]  <y,y'>  basis  tl( g^) n <x ,Xg> = o .  representations  I  £3 = u  a  and  be a  <r Rg(U)  ?J  have  3  "^1  where  { f ^ f ^ f ^ }  <f f f^>  U( f ^) ]  {f ,fg,f J  = o  h  are pairwise-Pg  2  Then  1  >>  [ f ^ f ,g^}  Remark IV.55  vectors  n <x^ yi >  5  ,  i = 1,2  .  d i r e c t l y from the d e f i n i t i o n that  { f : i = 1,2,3} ±  are  of a  54. I T . 5^  Theorem  independent  H e R (TJ)  Let  .  2  vectors  i n  H  Let  { f ^ f ^ f ^ }  such that  be p a i r w i s e - P  \ U(f.) = 6 i=l  dim  .  5  ,  Then  1  l> 2>  <f  f  f  3  Proof:  >  h  a  s  C*)  a  1  'basis o f p a i r w i s e - P g  1  B y C o r o l l a r y IT.50 a n d i t s p r o o f ,  vectors.  { f ^ f ^ f ^ }  have  repre sentat ions  f  f  form a  S  3  Then  x  is a  (1,1)  C  b y Lemma  p  I T . 55  basis  5  A  1  + u A(du -bx ) ,  1  2  gives  the r e s u l t . f }  1  1  5  6  +  U  2  A ( - X  5  a + o ,  )  ,  b + o ,  3  '  ^  C l' 2' 3 f  f  f  }  .  + m^u^+du^-bxg) .  Pg-pair,  ^ b a s i s  Let  ¥ ( f , g.^) «= < u , u > 2  1  for <  ;  { f ^ f ^ f ^ }  dim  f  f 1 J  2  {f^fg,^} Otherwise  form a  be a  ? U(f ) =6  of pairwise-Pg  Suppose  {f  = (-u ) u  >  f  3  >  ,  2  *  T  n  and result  e  TV.52.  e Rp(U)  Proof:  say  2  }  5  i sa  <f,,f ,f,>  f  4-  2  Corollary  A  ;  ?  f o r <t^ f^,f^>  basis  4  >f  A U  - <V V V**' ^ ^  (x +axg-cu )AU  2 ^  2  5  {g-^f )  follows  u  3  - f =  X  = (axg-cu ) u  (1,1)  l  +  X  J|  where  Let  ! = 4AU  ±  vectors  (1,1)  .  Then t h e r e  for <f ,f ,f > 1  are pairwise-P at least  Pg-pair.  basis f o r  Lemma  2  .  5  two of IT.52  5  exists  a  .  Then Theorem { f ^ f ^ f ^ } ,  gives  the result.  I V . 54  55.  Remark 17.56 in  H  Let H e R (U) .  Let  2  { f ^ f ^ f  }  be  independent  such t h a t (i)  dim  \ u(f,) 1  3 U(f.)  (ii)  i=l  .  .  are pairwise-Pg .  ff-pfg^f^}  I f any p a i r ,  U ( f ^ ) n TJ(fg) U(f^)  = o  1  Then  Proof:  =6  1  say  isa  f f ^ f g ]  has d i m e n s i o n  T h e r e f o r e no p a i r  P^-pair,  t h r e e and hence must  i sa  P^-pair,  and  then intersect  {f ,f g,f^} 1  must be p a i r w i s e - P g .  Theorem 1 7 . 5 7  L e t H e R ( U ) , dim H > 3 •  be  in H  2  independent  (i)  dim  such t h a t  \ U(f ) = 6 . i=*l ±  3U(f  (ii)  Let { f ^ f ^ f ^ }  ±  ) = o  .  i=l Then  {u,,u J.  }  o f W(f,,.fJ ,  d  {VL^U^XJ,  ( f ^ f g , f-^}  X  d  . . .,xg]  f  and f o r any b a s i s  [ \ U ( f ) ] has a b a s i s 1=1 ±  such t h a t  = u  l A  x  3  +  fg = U  l A  X  5  + Ug Xg  x  are pairwise-Pg  Ug X^ A  A  {f-^fg/f^}  ,  ,  have r e p r e s e n t a t i o n s :  56.  F  =  5  X  5  A  I  W  +  X  A W  4  ,  2  < i. 2 w  w  >  < , 6 x  =  x  >  >  5  r  Proof:  By Remark IT. 56,  f f ^, f ]  isa  2  f  U AX-  =  x  1  +  5  {f-^f^f^}  Pg-pair,  U  2  A X  ,  4  [f^,f}  f  ^ ^ U ( f ^ ) = <u^,u , x^, .  =  2  U  1  A X  . »,Xg>  2  u AXg  l  =  u  i  A  X  3  +  u  2  4.  a X  c h a n g i n g u-^,u ( i i ) holds,  .  2  .  1  s  l  x^ e [ x ^ ; u , u ] 1  a 4  0  •  Then  1  that  n  x  4  U ^ A X ^ +  z > .  € [x ;u ,u ] 4  1  U  2  = u ^ x ^+  A X ^  1  { i ^ £5} f  1  2  Hence  f^ = x^ w A  x  +  XJJAW  2  l  2  s  a  1 (  2  a = 1 .  <w-_, v,w > = L  g  P g - p a i r , by C o r o l l a r y  <w ,w ,x |> 1  L  Let this  J  11.34,  i n t e r s e c t i o n be x  = u^x^ +  = <x^, z  2  By  _+CU- )  where  and we have: - f  where  {w ,w } e [ z , z ; x ^ ] . 1  2  there  L e t x^ - ax^ + bu^ + c u  U A(X  A  2  x ^ e [x^+cu-^Ug]  2  *  2  €  and  1  f ^ = X ^ A W + v Wg , Since  2  4  1  •  <x^, y^> = <y ,y^>  dim < x + c u , u > n <v,w > = 1 . Then  2  and  {y^y^  where  € [x^,Xgju ,u 3  3  c  ,  2  Without l o s s o f g e n e r a l i t y , we s h a l l take  f^=  15  u  1  i . e . , without  are pairwise-Pg  2  2  j  A  [f-^, f , f^}  [z^ z^}  ,  2  Theorem I I . 16, <y£,z  Since  and  {x-j, Y 2}  exist  A  1  2  x ^ , x ^ so t h a t  f ^ « x ^ u + x^ w  U(f^) = <y ,y ,z ,z >  [x^,x JU ,U ] 4  and  Since  ;  2  We now show we can choose f  .  have r e p r e s e n t a t i o n s +  5  are pairwise-Pg  ± }  z> 2  U  2  .  A X ^  <x^> . ,  Thus  57. I n a s i m i l a r f a s h i o n , without a l t e r i n g f x ' x'} o °  we can choose  <  and  f  Thus  ^ 6  x  x  2  =  =  y  i_  u  <  A X  i. 2  z  z  5 + ^  2  > 6  5  x  5  -3 = X^AW  + x^ w A  2  =  x  .  and  3  u  x  3  f  2  x  ^ 5^ i^ 2  6  {v v } e [x-^x^jxg] l 5  '5  u  A  v  i  >  j 1  +  ^  x  6  X  >  v 2  X'AX/D  o  p  .  2  i s zero i n the f i r s t .  appears i n <x^Xg> x^  for  . and  3  x^ ,  Therefore  X'AX' ,  X ^ A X ^ i s c l e a r l y zero i n  A v  x  x^  for  x^'  XgAv  and the c o e f f i c i e n t  3  -1 X ^ A W  X^AW-^  < v 2  x_  3  f^  I t f o l l o w s t h a t n e i t h e r term  f^ =  f^ = <5 i +  U  v r h e r e  [w w | e [x^x^x^]  the second e x p r e s s i o n obtained above f o r of  3U  From above, we a l s o have  the c o e f f i c i e n t of  3  u^  [ 6 1^ 2^  e  x  With respect to the independent set 3 _< i < j < 6  ox-  so t h a t  >  A X  u^  5  i^ 2 v  >  f o r x^  <w^,w > =  3  2  2  yi  =  <x  and  x  •  >  Xg  for  Writing Xg  5  we have the r e s u l t .  Lemma IV, 56 in  Let  H e R ( ) u  2  •  Let  { i \ , f , f ~} 2  3 Ii s a t i s f y i n g ( i ) dim E U ( f ) = 6  be independent  ( i i ) {f-^fg}  i  is a  3 P g - p a i r ( i i i ) dim_fi {f^f'^g^} and  U(f ) = I . ±  i s a b a s i s of p a i r w i s e - P g  u ( f 5 ) n W C f ^ f g ) - u(g^)  e <f t ,fj>  Then j  vectors f o r  n Wtf^fg)  -)  l3 2  .  <f ,f^f^> 1  58. Proof: l  f  =  Since i  u  A  + u  3  X  ff-^f^  2  A X  4  is  f  »  =  2  a  i  u  Pg-pair,  5  A X  +  U  A X  2  they  6  have  1  3  •  |  U(f )  u  € <u ,u >  *  1  <u ,u 1  1 ;  Uj.  = u  3  f  =  .  l  u  A  1  W  +  1  e  =  W(f 3  f  i  2 >  "  Let  A  1  v'  +  W  x  A W  X  +  {f.,,f^}  u  A V  .  loss  '  w  Then  +  5  2  Choose  5  5  [x ,x ;u ]  .  Then  {f-pgj]  a  .  Then  f ^ ^ S j }  6  2  Pg-pair  u(g ) ?  n v(? ,? ) 1  2  = <^x>  •  is  f.^ e u ^ x ^  v'  e  a a  w'  e  + = x^  and  Let  .  Therefore  Y ^ 0 4 y + C ^ 0 .  ] ;  4  z«e  Pg-pair r  a  [x^XgjUg] .  + X^A( V-Yiig)  [ x ^ ; x , X g , U g , x ^ ] 0 [ x , X g j u g, x , x  5  ;  is  (Introduction)  [x^XgjUg]  4  assume  If  0  2  e  .  ff , f^}  < u , x ^ , . . . .xg> ,  y,y'  = u-^w+yx^)  5  and  generality,  [x ,xg;u ,x ]  £  we c a n  result.  Now  of  then  16,  and P r o p e r t y 4 . C  <w,x^,v'> c .  that  2  Pg-pair  | U(f^)  = a x ^ + b X g + GUg . = f^  a  j  i  <u ,x^3...,Xg>  we h a v e t h e  is  2  _n U(f )  B y Lemma I I .  <w,w%v> c  Hence w i t h o u t  4  .  1  since  <u> =  f r o m Theorem II.18  pairwise-Pg  4Av'  Let  <u >  where  = -Oi^y y'>  Then z e  .  By C o r o l l a r y I I . 3 4  f^) U  follows  >  V  [x^Ug] ,  u  . . . ,Xg>  J )  U( f ^)  Suppose  UgAtx^jUg] . 3  A  are  P^-palr. w'  '  W  x^  It  Thus  2  Case  2 >  .  2  ff , f , f - j }  f  representations  t  '  h e  = U-^z + X ^ A Z' ;  [u ;x,-,x ]  and  required  2  6  fl  ff ,g ] 2  5  vectors,  is  Case 2  {f-^f^}, {f^f^}  Suppose  f ) = 3 ,  dim  W(f ,f )  « 3 ;  dim  ¥(f ,f )  n ¥(f , f^) = 1 .  1  3  1  3  + W(f , f  dim W ( f  5  But t h i s  2  )] « 5 ,  2  2 >  P -pairs.  are  am  by  (iii),  itT5jli.es  dim U ( f ^ ) ,  w h i c h exceeds  Then  dim ^ ( f - ^ f ^ )  Hence  this  case i s not p o s s i b l e . In order  If  where  bases //hen t h e  o f Theorem I V . 5 8 h o l d , we need fchs f o l l o w i n g lemma.  conditions  Lemma,  to exhibit desirable  f e C g ( U ) and  f € x-^x^x^x^] +  » <x^,. •.,x^> ,  U ( f )  f  then  €  X-^ACX.-J  [ x ^ ; X  ] A [ X ^ ; X  2  2  ]  +  [x^ j X ^ ^ X g ] A[X-jjX^,Xg3 ,  Proof:  Since  <x^,x,„>  f  t h e n , by Theorem I I . 1 7 , e i t h e r X-^AV'  +  X A 2  w  "  °r  for  some n o n z e r o  the  coefficient  f  of  in  f €  f  ,  <x  x  1  ,x^,x.,,x,, > ,  n  f ,  in  (i)  l  5  f ^ f  dim  x  i  A [ X  2  zero  , X ^ , X  o 4 Y € P ,  Since  ]  Let  4  Y -J_A O X  in  f ,  X  and  +  V  form,  A  W  then  c o n t r a d i c t i n g the  ] + [x^XpjAfx^jx,,] ,  Hence  <x ,x,- , v,w> = ]L  of  1  x,, A X ,  I L nonzero  V A W S [xj^x^x^JAtx^x-^Xg]  H e Rg(U) with  \ U(f ) = 6  f -  has t h e f i r s t  the c o e f f i c i e n t  be p a i r w i s e - P g  t  f  If  i t follows e a s i l y that  Theorem I V . 59 { f  P .  X ^ A X ^ is  assumption t h a t = y x A p + VAW  has a representation  has a r e p r e s e n t a t i o n  y  U(f) ,  2 - d i m e n s i o n a l subspace o f  is a  dim H > 3 * L e t  and independent  ( i i ) <iim ^  in H  U ^ )  satisfying  = 1  ,  ,  60. Then  for  = S  <u,>  that  <u  x >  u >  - W(f ,f )  2  x  fu.^, u , x^_,...,  }  2  f  »  2  U  ]_A 5 +  = u-, AS  f^  '+  A 6  X ^ A  X  2  f  =  3  ,  6  2_ =  X  fo ~  A ^ +  S  l A  x  = tL  5  Proof;  f  l 3  =  u  i  V ' A  u  X  3  +  u  W'  4 <u 3  2  A X  3 5  exists  . •., g> x  2  x >  g^  4 U(f )  , y  5  €  ±  <.t'^, f _ , f-<> p  , , i =  1,  such  that  6 ,  Y X ^ A  <?A ]p  £  x  Since  x  1  <~x ,Xy  x  X  = 2  U  5  ,  1  A X  , <: w, v> r  }  {f-^f-j} >  Corollary  e  [x^ju^Up]  W €  [XgJU-^Up]  v  o  is a  2  A  hy  +  has i s  pA ij.  {f-,,f }  = u^A7 + w v  holds,  U  such  and  3  + V A W  2  Since  A  u  +  U-^A -J  =  f  2  y e  where  there  <f ,f ,f > x  0  '  Furthermore,  x  has a  ±  y  <f ,f ,g >  u  that  X  2  any r e c t o r  .  1  ii_V(f )}  ,  2  such  U  X  U(f.)  1=1  x  11.34,  4  Y  v'  €  S  i  F  [ u - ^ x ^ X g ] , w'  5  +  c  <u ,x _,.,.,x^-> »  U  2  A X  .f  f  ?  *  B  j  T  h  e  o  r  e  [u ;x ,Xg] 2  4  II*  m  Pg-pairs  a r e  = u-^y  +  zz  f  A  .  representations  16,  3  { 2^ 3} f  6  e  they have  Pg-pair,  2  ,  and ( i i )  , z e  [x^jiipj  ,  _  z'  € [xgjttg]  '  z  =  Y  2 6 X  Then  f  .  a J i a  =  1  Without  f  3  =  u  l  y = 1 ,  take  = 1,2,  i  one.  since  We  [f v  f  2 )  fj}  e U(g-j)  3  Y i  =  u  A  U  2  +  V  A  >  W  w s [x ;u ,u ] . 6  is  1  independent,  Now g^ = ation is  X^AV  v  7  i n  .  A Z  Q  +  or  2  g^ ,  s [u^x^Xg]  .  of  If  {f-^, f , f ^ }  has  2  y 4 u(f ) , ±  , z  g  Q  2  v e [x^u-^Ug] € < f f , f> 2  2  Hence  hence  .  ff , f , Y ±  By Theorem .  A U  L  2  +  U ^ A U  2  g ) 5  V  11.17,  The f i r s t of  _ 1  2  g^ = u -  A  W  that  w' e [ u j x , x g ] .  of  representi n i t  u-^u^ i s  we c a n t a k e  *  either  g ^ = yx^AXg + V ' A W ' ,  Since the coefficient  4  ,  4  and  and  5  the coefficient  2  ,  e <u ,x^,Xg>  Q  proposition,  i t follows easily and  as r e q u i r e d .  combination o f  g ^ « Yx^AXg + V'AW' since  we  <z ,v,v'> = < u , x , X g >  y = 1 ;  i s not so.  4  of generality,  and  VAV'  Since  y " ^  can take  e [u ,u jx ,xg]  nonzero  1  ° 4 Y e P ,  1  and t h i s  1  U  linear  U( g^) « < u , u , x ^ , X g >  + XgAW  loss  combination  b y t h e above  i s not possible  zero,  v',w'  we  = Y .  2  2  without  a suitable  Since  2  a  are pairwise-Pg .  2  2  Let 1  1  g^ =  ,  + c f u Aoacg ,  y | <u ,x^,x^>  {f ^ , f , f ^ }  Hence,  .  u  6  .  z =  f ^ = ] _ A y + x^AXg ,  Hence  can take  =  ,  Hence,  , v' e [ x ; u ]  2  2  .  to obtain  € [x^;u ]  u s  l f  A C G C ^  some l i n e a r  1  a t most  of generality,  Yx^AXg , o + y € P  +  and  y € <u ,x^,x^> , rank  y  + cf^  U^AX^  f ^ = u^Ay + ax^Aaxg can  A  loss  62. Lemma 17.60  in  Let  H such  P -pair  that  H e Rg(U) .  U(f.)  {f-^f^f  6  ( i ) dim ? U ( i \ ) =  ( i i i ) dim  6  Let  = 2 .  }  be  [f f ]  ( i i )  l a  is a  2  there  Then  independent  exists  g, €  a  1=1 l> 2> f  <f  3  f  >  s  u  c  vectors  for  Proof;  Since  ations  |  f  U(f )  If  < u  either  at  h  { ^ f ^ g ^ }  <f ,f _ f^> 1  2  {  f 1  >  f 2  u  U  A  s  ,  f  x  or  =  2  >  g,  {f^, f^}  f-j i s e i t h e r  f^  = u  + V'A*'  *  2  x  *  Pg-pair,  In the f i r s t  v  + Ug w  { f , f , f-^} 1  basis  and t h e r e s u l t  f o l l o w s b y Lemma  second  case,  we c a n a s s u m e  v ' e IJ(f-^  €  [ X  , X  3  4  J U  generality, w'  = dx  5  +  1  , U  2  ]  ,  w' e  v ' = x^ + a  a u _ + b u 2  Let  ]  g  2  Z « | f  +  ±  ^  b.jU  =  u  l A  1  (x +x +u ) 5  5  Without  In the .  2  i  € P .  +  U  2  A ( X  4  + X  6  Thus  loss of  and  CgXg,d,a ,b ,c 1  IT.52. 2  +  2  i sa  2  w' e U ( f )  [x^Xgju^Ug] . +  P^-pairs.  or  A  ,  2  basisf o r  are both  case,  x  IV.52.  (1,1)  v'  <u ,u > .  then  l)  Lemma  {fg,f^}  l A  U ( f ^ ) r>  (1,  represent-  ,  2  i sa  u  have  2  + U AXg  5  111  and  11.17,  u  x  f o l l o w s from  Theorem  A  A  {f ,f }  C )  f^}  By  l  l  ,  i s a  2  {f-^, f  u  B y  ff ,f^}  The r e s u l t  Suppose  Pg-pair  a  •»6 *  and  A W  2  .  2  l  UgAX^  {f-^f^}  <f ,f ,fy>  3  x  i s a basis of pairwise-Pg  .  >  i* 2> 3* * *  f-j = u ^ v +  1  t  = u-^x-^ +  x  =  ±  n  ^  )  +  63. ( 3  +  x  a  i i u  +  b  i  Consider!  = - c^d + c If  c  v'  f  +  i 2 f ) A( dx +a u +b u +c x )  c  x  2  5  0  0  0  0  1  1  0  1  0  0  1  0  0  0  0  0  0  0  1  0  1  a  l  a  2  A  u  1  l  b  b  0  2  l  c  + c  R(z) _< 2 .  + c x,. , 1 4 '  i 4^ x  ( 2" l l^  +  c  u  c  u  u  6  A X  1  w i t h o u t l o s s o f g e n e r a l i t y , we =  + a u 2  + b u  1  2  (x +a u +b u ) . 5  2  1  2  Let  2  (x^+b u + ( a + l ) u ) + 2  2  ±  2  ,  2  u  2 ( A  w' 4  u  A  X  = x  u  "  i  u  i  5  A  X  u  i  b  U  u  +  5 +  U  2  u  v ' = x^ + a^u^ + b ^ u (x^+a^ i  +  u  + b  2  1  5  1  2  1  u A[xg+(b -l-b )u -x ] 2  2  1  Then  1  l ( 5 A  X  + B  2 2' U  <f ,f ,g >  Theorem I V « 6 l (£ f ,fj} lt  2  Let  2  5  2  2  2  1  are pairwise-Pg  and  « <f ,f ,f > .  3  1  H e R (U) 2  be independent  2  with  elements  3  dim H > 3 . of  u  +  1  .  4  {f-^f^g^} 1  1  i 2)  A  «= ( x y - a - ^ + b ^ U g )  2  H  c,x,16  Hence  2  2  + b^u. + 22  u A(xg+b u ) = [x +b u +(a +l)u ]A[x +b u +(a +l)u ] 2  .  4 >  A X  + u AXg .  + f  i~ 4" i i) x  A  + a_u_ 21  c  3  can assume  = f^ - ^  ?  _  =  f^ =  and g  A X  R(Z) = 3  then  Hence  and  n  0  u  w'  g  0  then  + b,u 12 x  1  2  0  0  = u A(x^+c Xg) + ( 2 ~ i i )  2  2  I f t h i s d e t e r m i n a n t i s nonzero,  = i ( 3  1  2  0  1  f  1  0  - c d = o ,  2  2  1  .  2  = x-z + a , u , 5 11  Now  u  Let  f o r which  +  A  2  ,  64. diia[  | U ^ ) ] = 6 .  pairwise-Pg then t h i s  pair <f  Proof;  The f i r s t  1 5  If,  i n addition,  f , f^>  result If  U(f, ) = o 1  is  pairwise-P^ o  ff-^f^f^}  (b)  dim  2 .  Pg-pair  ,  pairwise-Pg  f r o m R e m a r k I Y . 56, Pg-pair,  result  then-  follows  from  H  such that  is a  dim  n\, i " , f^) 2  dim ^ U ( f . ) = 6 .  1  basis  be  independent,  Then  either  1  <f^ , f , f^>  for  2  or  .  f\ U ( f . ) < 2 .  1=1 2  for  dim  3 U(f. ) = 2 , i=l  and by C o r o l l a r y 11.34,  ff^f^f^}.  <f ,fg,f^>  follows.  Note:  For representations ; Theorems  If  1  basis  The  Let  1  =3 W ( f , f ) 1  .  (1,1)  U(f.) = o,l  f\  17,58,60.  (1,1)  that  is a  The s e c o n d  H e RJ(U)  1=1  (a)  immediate  Let in  (a)  for  of  is a  of  1  Theorem IT.62  U(f^)  {f^fg}  to a basis  {f-^fg}  T h e o r e m I Y . 5 7 a n d Lemmas  Proof:  a basis  .  2  Lemmas I Y . 5 8 , 6 0 .  contains  2  c a n be e x t e n d e d  for  1=1  <?•]_, f , f ^ >  vectors.  vectors  dim  Then  1  IV,57,  .  The r e s u l t  of  ff-^fg,^}  IV.59  for (b).  f o l l o w i n g are three  are pairwise-Pg  in  examples  H e R (U) , 2  then  is  a  s e e R e m a r k I V . 53  of  ± U(f ) ±  ff-^fg, f y has d i m e n s i o n  6,  65. each example i l l u s t r a t i n g Examples  one o f t h e t h r e e  cases  i n Theorem I V . 6 2 .  IV.65  Example I Y . 6 5 . I  dim  3 U(f.) = 1 1=1  .  1  1 U ( f ) = <u u ,x ,...,x > ±  f  = U  1  X  l A  1 3  + U AX 2  3  2  >  x  3  These if  =  U  1  ( 2  A  U  vectors  a  + X  Z = \ a  f  1  a  i  5^  + X  +  generate  € F ,  1  3  4  A X  6  3  of dimension 3 since  + ( a g + a ^ ^ + ayig]  3  2  + (— u +x^)A(a x -a u ) 3 2  5  and and  •  then  = u A[a +a )x J  X  a rank 2 subspace  4 o ,  1  .  6  ,  4  f g = U-^AX^ + U g A g f  5  if  = o  6  1  R(Z) = 2 V  ;  2  ;  Z = a f 1  1  and  + ctgfg  R(Z) = 2 .  a c t g + o) 1 3  Example  I V . 65.2  I U(f.) = <u ,u ,x ,...,x > 1  f  f f  l 2  =  u  2  i* 2 x  +  5  U  2  6  A X  4  = u  l A  2  ( 4+2x ) x  6  x  ;  (1,1)  basis  ,  + U A( 2  x  for  F = Complex n o s .  '  = u x<- + u A g l A  a  ff^fg,^}  3  +3X ) 5  •  <f ,fg,f > 1  5  66. For  a  i  € F  not  = ? a.f. 1  all  zero  = u A(a 3 2 5 3 4 x  1  x  + a  + a  x  + 2 a  1  + U2 ^ l 4 2 6 3 3 A  x  a  x  + a  and  Example  1  f a  x  3 6^ x  + 5 a  3 5^ x  =2 .  R(Z)  ^ U(f. ) = o  IV.63.3  1=1 =  J u(fi)  •1 fg  For  a.  1  u  <u ,u ,x 1  A  AX^ + UgAX^  =  U-^X  =  X^  e F  any o f  .,Xg>  3  i "5  A  X  5  ,  *  +  UgAXg  +  X^AXg  not  all  +  If  2  Z « ( u  +  Z = y a.f.  '  £  UgA(a x +a x ) 1  is  {0^}  zero,  4  2  zero,  1  a  + a  3  ( 2 aT 4 ( l 4 u  +  2  x  and  ) A  a  x  R(Z)  + a  =  We c a n now show t h a t  = u . Av ( a x - , + a . x _ )  1  1  + ayc^Ax^ +  6  R(Z) = 2 .  a. ^ ) A ( a x  r  1  2  x  2 6 x  5  )  If  1 3  25  n  a^x^AXg  +  .  a. | o ,  )  >  2 if  dim U = 6  and  H e Rg(U)  >  66. then  H _< 3 .  dim  that  i f  We a l r e a d y know from C h a p t e r s I I and I I I  f f_, • • • f ] 1  }  in H ,  a r e independent  k  and  k  dim r U ( f ) < 5 ,  then  ±  result in  3 .  k  i f we now show t h a t when  {f^, . . ., f } k. < 3 .  then  ±  show t h a t any f o u r independent v e c t o r s such t h a t  | U(f.) = 6  dim  a r e independent  fc  and . dim \ U ( f ) = 6 ,  H ,  We s h a l l have the d e s i r e d  In f a c t ,  f f ^ . . », f ^ ]  we  in  Cg(U)  cannot g e n e r a t e a rank two  subspace. To do t h i s , we take two 3 - d i m e n s i o n a l rank two subspaces  <f ,f , f 1  g  >  and  <f  i s n o t a r a n k two subspace. pairwise-Pg 0  1  dim  0  U(g,) = o , l o r 2 .  1=1,2,4  2  ,  and show t h e i r  By Theorems 1 7 . 6 1 , 6 2 ,  r e s p e c t i v e l y ; and  *  d  f,f^>  f g ^ g g , g^l ,  bases  <f, , f , fi,>  1 ?  dim  fg^g^gj+l 0  1=1,2,3  for  there e x i s t <f ,fg,f > ,  U(g,) = o 1  We must show  sum  x  3  or 2  <g,,...,gu>  1  and R (U)  1  2  d  f o r a l l s i x p o s s i b l e cases. We n o t e t h a t t h e f o l l o w i n g r e s u l t s a r e t r u e f o r any dimension  n  Lemma IY. 64  of  Let  pairwise-P^ i n D  (ii)  U ,  H e Rg.(U) .  H  h ( f j= 0 .  i=l  1  n > 6 ,  satisfying  unless otherwise s p e c i f i e d .  Let  f f - ^ f g , f-j]  ( i ) dim  he independent,  ? U(f.) = 6 i=l x  67. If  f  satisfying  e Cg(U)  4  (a)  dim  ,  independent  | U( f , ) = 6 1=1  of  (b)  (c)  f i  dim  d  ,  =1  4  then  f  u 1  j  * 3**•*  u  x  2  f  f  f  Let  By Theorem IV.57,  1  =  =  2  3  =  i  u  A  U  X  5  l  3  X  A  .  +  =  ^ch  s  U  A X 2  X  Az  <u> -  6^  } X  4  +  X  i  n  we c a n t a k e  6 2  j  = u  Since  basis  , < Z , Z '>  *  U(f.)  4  a  .  Then  <X^,X^>  =  u e <u,,u^> I3 2  •  .  By T h e o r e m  dim  fl U( f . ) i=l,2,4  and  w'  exists g^  s  =  dim  £ U(f. ) = 6 1=1,2,4  and  +  7  Hence  Now  Y^AXg  .  ,  o  f-j =  X^AZ  is  for  s u c h an  rank 3 .  z, z ' }  some a  ,  o |  is  Z = g^  Hence  -  af  ,  L  v'  where  e  [u^x  since  k Rg(U)  1  ,x ]  ,  g  we w i l l  In  take  fact  <z,z'>  is  =  <x^,x > . h  independent.  = v ' w ' H-.OZAX^ + ( x 4 - a z ' ) X g A  <f ,fg,f  <z, z '> = <x^,x^> .  independent.  {x^+az'^z}  <f ,...,f^> 1  ,  independent  a 6 F ,  F  <f- , f g , g^> =  of generality,  + XgAZ'  4  e  y  Without loss  fv',w',x ,Xg,x^,x^}  {v',w',Xg,x,-,  ^  = 1  1  such that  L  [Ugjx^,Xg]  Y = 1 .  Also  g^ e < f - , f g , f^>  V'AW  IV.57,  .  1  there  d  *  '  A Z  has  k  that  4- U g A X g  5  1  [ i U(f,)] 1=1  <f ,...,f >4Ro(U) ^ 1  1  Proof:  are  0  1  n U(f.) 1=1,2,4  }  ff,,f ,f,.i  1  pairwise-P  f f ^ f ^ f  4  A  For has  J +  >  68. Lemma I V . 6 5  Let  pairwise-Pg  (i)  in  H € Rg(U)  H  .  that  and  (a)  (c)  ( i i )  such  be i n d e p e n d e n t ,  dim 3 u(f. ) = 1 . 1=1  f }  ,  independent  are pairwise-P  k  dim £ U(f. ) = 6 , i=l  then  < f  By Theorem I V . 5 9 ,  ( f - ^ f g , f^}  (b)  A  dim  n  i> 2 3 f  of  j f  >  1  f  2  = u  l  A  x  +  5  U g  X ^  A  U ( f . )= 1  k  ( iJ 2 3^  has a basis  f  f  5g  =  U-^X  +  =  u  + v w  l  A  U g  U g  A  ,  X g  A  ,  where  3  = [ ? U(f ) ] , v € [x ,u ±  Let  4  <u> =  Cu-^Ug^x^, . . . ,xg>  1 5  u ]  and W e [xgjU-^Ug]  2  f) U( f . ) . 1=1,2,4  Then  u e <u  1  <u> = <u- >  or  we c a n t a k e  u  L  1.  u =  {u,u- } L  g  = u  another  basis  e [x^ju^Ug]  u > .  1  i s .independent.  .  Either  d  In t h e l a t t e r  case,  (Theorem IV.59)=  .  B y T h e o r e m I V . 59,  v'  n  that f  Case  such  4 Ro(U) • 2  <f , . , ., f > x ^  1  Proof:  }  1  f^ s Cg(U)  (f, , f  f f ^ f ^ f  satisfying  dim ^ U ( f . ) = 6 1=1 If  Let  we c a n r e p l a c e such that  { f ^ f g ^ }  ,  w' e [x ,u ,u ] 6  1  2  .  i  f 1  ,  f  > 4 3 f  2  g^ =  u ^ U g +  Thus  u(g ) 4  b  V'AW'  =  < u  1  y  ,  , U g , x  4  , X g >  69. Hence  = U(g^) .  fg^g^}  is a  P^-pair.  B y T h e o r e m 11.20,  <e>^ ^> i s n o t a r a n k 2 s u b s p a c e a n d h e n c e Case  2.  <£ Rg(U) .  <f , ...,f^> 1  }  u = u^ . F r o m t h e p r o o f o f T h e o r e m I V . 5 9 , we s e e t h a t we c a n  replace g\  =  ff^f^f^}  u  A  U  2  Then  i  +  y  by a basis  AZ ,  y € [x^u-^Ug]  Z = g^ + yg'^  for  ( 1 - Y )+ o ,  which contains the  ff  = (1-Y)U AU 1  and  ,  4  2  has  yyAZ  Hence  i s not a  that  z e [x^u-^ u ] .  + VAW+  2  such  g ^} 7  2>  and  o 4 Y s F .  <g-j,g' >  f"  x >  rank  <f^, . . . , f ^ >  rank 2 subspace.  3  , We h a v e  result.  Lemma T V . 6 6  Let  vectors  <f ,f ,f-,> x d. y  If  for  1  f^ e Cg(U)  (i)  dim  be a '(1,1)  ?  € R (U) , of  are pairwise-Pg  0 U(f. ) = 1 , 1=1,2,4  basis  such that  2  2  c  i s independent  f f , , f ,fh}  (iii)  ff^,f ,f.j}  of pairwise-Pg  dim | U(f, ) = 6 . 1  f f ^ f g , f^} (ii)  then  satisfying  dim \ U ( f ) = 6 1 i  < f f  4 Rp(U) .  >  h  1  Proof:  From Remark IV.53, 1 U ( f . ) 1  has a b a s i s  x  f u , u , x , . . .,Xg] x  f  l  2  =  u  where  such that  3  i  A  X  3  +  U  A X 2  4  »  f  2  =  U  1  A X  5  +  <z,z'> c <u ,x^,..,Xg> 2  z' \ < u , x , x > 2  if  6  and  f f - ^ f g , f^} U  ,  2  A X  6  >  f  3  =  have U  1  A  Z +  z 4 ^-^x^xy  representations U  2  A Z  '  1  ;  < z , z ' > n U ( f ) = o , i = 1,2. L e t i  70. n U ( f . ) = <u> . 1=1,2,4  I t follows  easily  f r o m Lemma I I . 1 8 t h a t  1  that  we c a n a s s u m e  without  loss  Lemma I V . 59 a n d i t s p r o o f , y 4 <u ,x ,x > x  3  ,  5  f'  3  = u ^ v +  Similarly, f  4  =  U  2  where  a suitable  U ^ A  W  ,  W'AW"  +  w" € [ x g ; u J  .  2  i  a  f  ' i  6  and  2  1  ,  x  3  w e  <Ug,x^,Xg>  F  i s independent.  f ^ € c|(U) , dim | U(f ) ±  n 1=1,2,4  2  ,f ,f^  4  gives  2  2  a l l nonzero Let  =  i )  w* e [ x g j x ^ U g ] .  Hence  f  i  WAW*  such  that  > ^'g  =  ,  f  where  R(z) = 3 ,  R^ ) * U  <f ,f ,f^> 1  3  2  vectors  (iii)  v'4<u ,x ,Xg>  2  o r ^ , . . . i n  Let  (i)  and  v € <u ,x^,Xg>  and w ' e [ x ^ ; u ]  Lemma I V . 6 7  If  gives  2  where  and  < f , . . .,f^> 4  for  f^,f ,  ~ u A ( a x + a x ^ + a v + a _ w ) + a^UgAv' +  w € [x ;x ,u ] 4  .  f  Choose  1  ?  l i n e a r combination of  [a-^Xj+agX^+OjV+a^w, v ' }  Z = |  i = 1,2  l i n e a r c o m b i n a t i o n -of  ,  From  2  ±  A V '  u = u^ .  f ^ = u ^ y + x^AXg , y € < u , x , . . . , X g >  y 4 U(f ) ,  A suitable  of generality  {f ,f ,f^} 1  2  be a  2  6 R (U) , 2  independent = 6  ( i i )  U(f.)= o ,  of  ( l , l ) basis such that  ff-^f^f^}  ff-^fg,^} then  of pairwise-Pg  dimJ U ( f ) 1  such  that  are pairwise-Pg  <f*...,f > 4  4 R§(U) *  = 6 .  ,  71., Proof:  B y R e m a r k X V . 53,  such t h a t f  = u  2  l  A  <z,z'> c z 4 <  u  ,  1  we s e e  + Ug x  5  A  ,  g  = u  <Ug,x , . . . x > 3  x  3  ,x >  ,  5  that  3  ,  6  and  has  ±  have  [ t ^ t p f j }  x  ?U(f )  g  = u-^x^ + U g x  f  + Ug z'  where  z  l A  ,  A  ±  A  n U(f ) = o ,  f  1 =  i  4 <Ug,x ,Xg> 4  we c a n a s s u m e  fu^u^x-^ . . . ,x }  representations  <z z'>  z'  a basis  .  ,  4  1,2;  Prom Theorem IV.57,  f^ = x ^ v + x ^ v ' A  ,  A  < v , v ' > = <x,_,Xg>  = x,_ w .+ XgAw' ; <w,w'> = < x ^ , x > A  We  shall  write  f  We  shall  Case  1.  = x  4  3 A  + x^ ( b ^ y b g X g )  (a x +agXg) 1  5  following  a ,b^ = o j  a ^ =f  2  a^ = 1  fjj. =  X 3  A  bg 4 °  y'  = ulAy  + b x  X  2  5  u  4  A  x  x  •  x  4  + axg + b u  (u +x ) (x +x )  Then  1  5  A  3  R(Z) =3  5  ,  a, A  4  some  y'  f  p 2> 3  of  b e F .  + Ug (x +x +a for  Without loss  of  .  6  Without loss 2  •  f  f  y e [x^x^Ug]  where  A  2  y = x «  + Ug y' ,  4 < , 4, 5>  e F .  ±  .  A suitable l i n e a r combination of 5  i  cases:  ,  0  a ,b  ,  A  c o n s i d e r the  generality,  f'  4  6  Let  Z =  y'-ab^)  4 <Ug, , 6> • x  4  ,  i v e s  and  generality,  o 4 a s F  x  §  + f  g  + aX'^  +  + (a^-bgXg) (x +ax )  since  A  4  6  .  .  72. C  a  s  e  2  -  a  i^ 2 b  generality,  a  ° >  =  4 °  2  a  i n case  and  y'  1 ,  4  + o  \  f'  2  + f  +  ,  A  f  f  5  Suppose  2  5  1  for  +  3  4  +  5  6  &  take  f  2  }lo }  i  2  = X A X '  6  3  .  1  since  s  without loss  4  2  a € F  '±> 2 l ^'2  a  + x Ah x  6  .  l f  some n o n z e r o  = u-^x'p. + ( u - a u ) A X g  will  2  f^  5  A  b^ = o ; 2  y = x^ + axg + b u  (x +x +ax, +aax )  one o f  = x A(x +a x )  k  where  2  R(Z) = 3  say  .  U A ( x ^ + X g + a y '-abu-j^)  6  non-zero,  of  Let  + (x +b x ) (x +ax )  Case 5-  Without loss  b ^ A X j -  + u^y'  + af  2  = (u^Xg) +  X^AXg  .  4  1  =  = u^y  ?  <u ,x ,Xg>  Z » f  Then  .  = 1 .  g  f^  As  ,  z  of  ,  r  o  *  811(1  e  n  <u ,x^,Xg>. 2  ^ ^  e  es  generality,  + x^bgXg .  5  As i n t h e  = u-^y + u ^ y '  e  y'  above  Now  2 cases,  y s <x _,Xg>  ,  1 )  4 <u  y'  Let Z = f +f +af' 1  2  +  5  = u-^x^+x'^)  f  4  + u A(x +ay ) +  + au Ay + X ^ A X ' x  +  5  + au  l A  3  y  A  +  3  2  l f  2  2  + u A(x +ay'+x )  3  (h x  (u -a u ) Xg  x^AbpXg  = (u +x ) (x +x' ) 1  /  4  2  4  2  -a u ) x 2  1  A  6  a^ = 1  6  1  A  we  2  ,x  J +  ,Xg>  .  = (u +x ) (x +x 1  5  A  + ay y ;  ) + u (x +ay'+x ) 2 A  6  y e [xgjx^,^] R(Z) = 3  5  4  y e [^;x ,u ]  A  Then  /  5  since  6  ;  1  y € <x ,x > . 4  6  o4aeP  f o r some  since  y' 4 < u x x g > . 2 >  Case 4:  Suppose  a  i ?  loss of g e n e r a l i t y ,  f  b  4 >  are a l l nonzero,  i  a^ = 1 .  4 = A ( x + a x ) +^ ( b ^ + b p X g ) x  5  5  2  6  » (x +h x ) (x +a x ) 5  Y  Now  =  1  b  1  2  l A  2  +  6  X  Hence  L  4  2  = u. /\(x^-b x^) 1  /  1  ±  +  l A  A Y X  U' A1 2  1  1  4  ;  6  ,  + (u -b u )AXg  ,  )+ (u -b u ) z'  .  1  2  1  2  f - u^x'^ +  = u "i  4  a  3  1  5  5  A  V l 2 '  f^ = u A(z+b z  f  4  f = u (x +b- x ) + (u -b u )AX f  i = 1,2 . Without  U '  2  A X  and f  4  4  1  1  ,  =  1  f g  X '  5  A  =  A ( X '  xi^x/^ 5  + u'^Xg ,  + ( ag-b-^Xg)  +  Y X  4  A X  a - b^ = o , we have case 1 . I f a - b ^ 4 ° ,  If  g  w  6  .  e  2  have case 3. Combining the four cases, we have the r e s u l t ; v i z . , <f  1  ,....,f  4  >  4 R^  U )  '  74. Lemma I V . 6 8 .  H  in  Let  H e R (U)  .  2  such that  (i)  f\  Let  f f f, f } p  U(f,) = o  be  2  (ii)  independent  ?U(f.) = 6 .  dim  1=1  1=1  1  p If  that  <f  1 ?  (a)  € C (U)  4  i=l  B y Remark I T . 56, has  i  ±  combination  f  i.e.,  fl  U(f ) ?  can f i n d  x  2 A  [u (ax +bx ) are  n U(f)  4  a.x.  4  3  U( f )  x = |  (ax +bxg) 1  L  W(f^,f )  5  in  4  +  f  4o  and  y = fr  B x  •  (ax +bxg)j  indeterminates  4  in  = x  g  F .  l  •  A  x  f , f x  + x  5  ±  1=3  Now  ,  .  Thus have  2  2 A  Xg  ,  Also  find  a  W(f^,f^)  linear  n U(f)  + o ;  >2,  we  >  af  Consider the v e c t o r U 2  then  (b).  c U(f^)  and  W ^ , ^ )  f  such that  2  1  .  such that  t h a t we c a n a l w a y s  and  f-  Since  independent  = o ,  such  v  (Theorem IV.57)  show f i r s t  of  1=3  3 ,  ff-^f^f  are p a i r w i s e - P g  1  by h y p o t h e s i s  2  We w i l l  }  = x^x^ + x^x^,  f  n <x- ,x > = o L  U(f,)  n  1=1,2,4"  fx , . . . , X g ]  U(f^) = <x^,...,xg>  U( f ^ )  (b)  ff-^f^f  a basis  representations  of  .  2  [ ^U(f ) ]  and  ) = 6  independent  1  4 R (U)  4  ,  2  d i m | U(f  ...,f >  Proof:  f  where  Then the  x  dim W ( f , f ) ?  1  ,  + bf  a , ±  4  p. s F' ,  = x^lax^+bx^)  g  v = X-jX + X y g  X-^ X , a , b 2  i U l  coefficients  , u of  2  +  75x 3>  •*• 6 x  a r e  >  • +  l3  X a  X ^  + X 3 2  + x P  a  h  2  +  a  determinant  °3  a  a  -  4  u  2  u  2  a  (**)  -  5  5  h6 The  -  26 " P  b  of the c o e f f i c i e n t s  &  5  X  P  3  5  -a  o  o  -a  -b  o  of  X , 2  o  ,  U  is  2  = a (3 a -P a ) 2  6  6  U ; L  -b  5  5  6  + ab(P a -P a +a 3 -a P ) 3  6  6  5  4  5  5  4  + b (p a -a 3 ) 2  1|  which Since of  a'  i s a homogeneous x and y ab  are independent,  and b  there  exist  i.e.,  the vector  polynomial  J+  o f degree at l e a s t  v  a,b  in  P  3  2 in  a and b .  one o f t h e  i n t h e above p o l y n o m i a l  non-trivial  Now  5  making  coefficients  i s nonzero. (**)  Hence  a l l zero;  i s zero.  U  cannot  2  then  +  X,  both  be z e r o ,  f o r i f they  were,  for  nontrivial  X,\ x  2  ,  and n o n t r i v i a l  c o n t r a d i c t i n g the independence exist  \  1 }  X ,u ,u ,a,b 2  1  of  x and y .  such t h a t  2  ( u , u ) + (°,°) »  and  zero.  f^) n u(f) 4 o .  2  Thus  W(  1  2  U(f )  = <f ,f,f- > .  ?  1  5  n <x ,x > = o ,  ±  <f  1  2  Hence t h e r e  (X , X ) 4 (0,0) , ±  (a,b) 4 ( 0 , 0 ) ,  We now have <f ,f ,f >  f e u ] , fp^} •  2  and t h e v e c t o r  f > - <f ,f> g  x  i = 3,4 .  ( s a y ) , and  x  U ( f ) ~> <x ,x >  Now  Hence  2  v is  but  dim[W(f ,f )^U( f)] 3  4  1 or 2 . In  the f i r s t  (lemma I V . 5 8 ) ,  pairwise-Pg  Lemmas I V . 6 4 , 6 5 , 6 7 In basis  3  of  this  ,  4  <f, f ^ , f >  has a pairwise-Pg,  4  (Lemma I V . 6 0 ) ,  and from t h e p r o o f o f  U(g) n U(g^) n U ( g ) 4  •  0  4  I f the dimension  i n t e r s e c t i o n i s one, we have the r e s u l t  {g,g ,g } 4  isa  (1,1) b a s i s f o r < f , f , f > .  independent  Let  in H  by a s i m i l a r  I f i t i s two, C o r o l l a r y 11.34 shows ?  f o l l o w s by Theorem 111.40 and Lemmas Theorem I V . 6 9  are  .  p r o o f t o t h e one above. 3  ff,f-^,f )  and t h e r e s u l t f o l l o w s by  t h e second case,  f{g,g ,gij.}  Lemma IV.60  case, we can assume  H € Rg(U) . such t h a t  Let  4  The r e s u l t  IV.66,67. ff^f^f^}  be  dim \ U ( f ) = 6 . I f 1  =  77. f  ]\  e  C (U) ,  independent  2  dim p S U(f ) = 6 , 1  Again,  f f f f } such t h a t •1' 2> 3'  < f ^ . . .,f > 4 R (U)  then  i  Proof :  of  2  4  By Theorem I V . 6 l , we can assume  ff^f^f^}  by Theorem I V . 6 l , we can f u r t h e r assume  pairwise-Pg  .  Now  dim  r\ U ( f , ) = 0,1,2, 1=1  are pairwise-Pg  f f ^ f g , f^}  are  and  1  dim  o U( f , ) = o , l , 2 . 1=1,2,4  dim  ^ U(f.) = 2 1=1,2,4  I n t h e case where  1  we have  <f ,f ,f > 1  2  Hence  ,  5  1  2  isa  4  is a  1  is a  2  ff ,f ,f ]  ff ,...,f^}  i n Lemmas IV.64 _> 68.  H ;  ent i n f  <t  f  ,f  <f ,...,f > 1  Proof: Hence  4  >  a  n  d  i  m  - ^  T  u  f  i ^  4 R^(u) .  L  £ 12  4  Let  2  d  i f  <f ,...,f^> 1  with  We have t h e d e s i r e d r e s u l t .  j  >  2  The o t h e r cases a r e d e a l t  dim U(f- ) + U ( f ) = 5 • dim  1  dim \ U(f _) = 5 .  4 \> 2 y  4  (1,1) b a s i s f o r < f , f , f > .  H € R (U) .  Let  (1,1) b a s i s f o r  (1,1) b a s i s f o r  c o n t r a d i c t i n g Theorem 111.40.  Theorem IV. 70.  By  1  f f , f , f^} 1  1  dim ft U ( f . ) = 2 . 1=1  1  C o r o l l a r y 11.34,  dim ? U( f . ) = 2 1=1  2  U(f.) = 6 .  If  =  6  f  ;  f f _ , f , f-^} ]  2  e C (U) ,  4  2  t  h  e  n  he  independ-  7SV Similarly  dim  <f ,...,f > 1  j  R2(U)  1  4  T h e o r e m I V . 71 ent  in  Let  such  H  U(f. ) = 6 .  By Theorem  .  H e Rg(U)  that  dim  .  Let  Proof:  <f ,fg,f^>  This is  Theorem I V . 7 2 : dim H < Proof:  In the  H e Rg(U)  .  Then  If  be  independ-  k < 3 .  For  =  of pairwise-Pg  immediate from'Theorems  Let  ]  .  1  a basis  f e  IV.69,70  vectors.  and  dim U = 6 ,  6l.  then  3 . Let  dim I U ( f . )  1  has  1  ff.^ . . . . f  = 6 .  r U(f.)  1=1 k = 3 ,  17.69,  ff ^,. . ., f ] k  = 4,5,6  .  be  In the  Independent first  in  case;  H .  Now  k = 1 (Corollary  II.21).  1  second  k < 3 (Theorem  c a s e ; k _< 3 ( T h e o r e m I I I , 38). IV.71).  In the  third  case,  79CHAPTER  THE  RANK TWO SUBSPACES WHEN U  In dim  U = 7 ,  basis.  this  an example  dim  H = 2 ,  has  first L  2  are  a  (1,1) ,  if  H e R|(U)  dim H = 4 , i t  see  ,  If  (1,1)  V.85.  Theorem 111.40.  basis.  and  has a  c o n t a i n e d i n Theorem  such a b a s i s ,  I n the  (Theorem I V . 61) give  If  If  dim H = 3 ,  and  3  then  d i m TU(f )  = 5,6,7  ±  Theorem I I I . 4 5 g i v e s .  3  results  [f-^fg,^]  ease,  {f- ,f ,f }  of  i t has  a basis  we show t h a t  dim H < 4 .  Both these  For  HAS DIMENSION J  chapter,  then  V  representations  second case,  {f-^f^f^}  .  In  H the  for are  pairwise-Pg  a n d T h e o r e m s I V . 62,.'57, 59 a n d R e m a r k I V . 53  representations  for  T h e o r e m s V . 74,  75 g i v e  Theorem- V . 73  Let  {f-^fg,^}  ,  I n the t h i r d  representations  H € Rg(U)  with  for  case,  {f-^fg,^}  dim H > 3 .  .  Let  3 {f ,fg,f^}  be i n d e p e n d e n t  1  Then  {f-^fg,^}  be e x t e n d e d  1  ?  Proof:  or  say  1  1  {f-^fg^} 2  that  dim  , we h a v e  3  {f-^fg,^}  dim[lu(f.)]  Since  6  1  1  (*)  x  {f-^fg}  [f ,fg,g^}  dim[u(f )nu(f )nu(g )i  We show f i r s t  U(f.)  either  is  ?  + x  , which  of  can  <f ,fg,f^> 1  a ( l , l ) -basis  contains  <?dim!U(f "1=1 + X g  = 7 .  .  for  = 1 .  1  ±2  dlm[su(f^)]  such t h a t  to a pairwise-Pg basis  P.-pairs. • 1=1  H  contains a Pg-pair,  Furthermore, <f ,fg,f >  in  ) -  at l e a s t 2  dimW(f.,f  l<i<j<3 - 5 < x  two  where  3  80  ij  x  =  £  2  y  d  m  ^  W  1  f  '  ^  a  1 < i  If  n  d  x  =  d  i  < j < .3  + Xg^+ x  2  zero.  If  I *  f  i j 1^,  x  = x  be  i  ^  m  and  , we h a v e  y <_ J  B  Theorem 1 1 . 2 6 ,  y  o < x < 3 .  6 < y £ 9 .  x = 1 , we h a v e  x = 2 , we h a v e  •  y = 6  so t h a t  Also,  putting  dearly  and hence  at least  x  cannot  each  x^j = 2 .  two o f t h e  x  's X J  are  2 .  If  be 9 •  x = 3  then each  s  x  . = 5  B u t b y e q u a t i o n (*) above,  and hence  y <_ 8 .  We h a v e  c o n t r a d i c t i o n and hence t h i s case i s n o t p o s s i b l e . d i m ?i U ( f . ) = 1 , 2 , . I n the f i r s t case, [f, ,f_,f_}  1=1  pairwise-Pg least  .  I n the second c a s e ,  {f-^f^f^]  a  Therefore are  3  1 2  1  must  y  contains at  two P g - p a i r s .  3  We now c o n s i d e r t h e c a s e Let By  {f ^ } 1  2  [ £. U ( f . ) ]  =  u  i  now  A  5  x  +  U  2  be the P g - p a i r s and  2>  3  Hence  A X  31,  there  such that  6  {f^fg,:^)  '  a  i  s  n  d  f  ff  1  ?  f  2  ,g-j}  <f ,f ,f 1  2  In {w,w'}  say  exists f  3  =  u  1  is a  there (1,1)  1=1 3  = 2 .  1  PI U ( f ) = < u _ , u > ±  a basis  ]  2  {u^u^x^,...,x } y  = U , A X , + u.Ax, ,  U* )  a  r e m a i n s t o show t h a t  that for  {? fj}  Theorems I I . 1 8 ,  for  f  ,  2  dim[ n U ( f . ) ]  when  l  A  w  b  a  +  s  l  U  2  s  f  o  exists basis  A W  '  w  i  t  h  <f ^  r  1  2  ,  u  ( 3^  f  3  f  >  3  *  g^ e < f , f , f ^ > 1  2  of pairwise-Pg  < x  I  7  >  t  such  vectors  > . t h e above  representation of  w e [Xj, u , x ^ , . . . , X g ] 2  combination  aw  +  Bw'  e  Without loss  of generality,  ,  we s h a l l  at least  T h u s some l i n e a r  <u ,x^,„..,Xg> 2  f^ ,  ,  take  with  B =)= °  6 = 1 .  •  Then  one o f  8i.  Then  3  Renaming ax^  2  =  f± f  - au )Aw + u A(aw +  =  f  2  (u  ^ l  =  f  If,  say,  ,  ±  x  3  i  v e U( =  fg  +  f l  f  Then for  u 2  +  u  +  5  = U^Ax|  v 4 U(f )  vectors  A  U ^ A X  =  If  g^  a  )  A  + u A(ax  3  2  +  x 5  u  A 2  (  a x  ax,_ 4- X g = X g ,  u  f^  2  2  A  x  5  x  aw + w' have  we  UgAXg  + UgAv  = 1,2,  ,  , i.e.  =  U ^ A ( X  1  V  then  )  <f ,fg,f^>  T h e o r e m V . 74.  5  €  <Ug,X  4-  .  Pg ,  independent  in  X ^ )  4-  U  form a  -  2  (i)  {f^fg,^}  is  (ii)  dim  Then  \ U(f,) 1=1  [ £ U(f,)]  a  s  } 5  4-  (1,1)  ,X >  p a l r  v)  pairwise-Pg  ,  for  t  h  e  n  l  e  t  of  pairwise-Pg  proved. the  2 kinds  of  V.73. .  Let  {f-^fg,^}  be  pairwise-  satisfying (1,1)  basis  for  <f ,fg,f > x  3  ;  = 7 * has  a basis  .  .  basis is  .  g  are  p _  a  A ( X g  The t h e o r e m  H e R^(U) .H  ,X^,X  {f-^f^f  ( f ^ j  i n Theorem  Let  = v ,  i  {f-^fgjg,}  bases  x^)  + g) •  We now g i v e r e p r e s e n t a t i o n s pairwise-Pg  +  ?  >  2  = i  2  au )Ax  - au ) = uj , w = xj  + x^ = x^ ,  l  "  u  (i^  f  -  x  w')  g  [u-^u^x...,x  }  such  that  82  {f-j^fgjf  }  have  f  l  f^  =  representations:  l  u  A  x  3  +  .  U  2  A X  4 >  ,  = u^Ax-, + UgAv  where  2j)  i  The p r o o f  T h e o r e m V . 75 in  H  (i)  u  l  1|  and  J(  v 4 U(f ) i  Let  = 1,2 .  H e R|(U)  „  Let  {f-^f^f^}  be i n d e p e n d e n t  satisfying  dim[ 1 U ( f . ) ] 1=1  > = ~j\ u ( f )  = 7  ;  (ii)  any b a s i s  ,  i  there  dim n U ( f , ) = 1 . 1=1 of  <u-^,Ug>  i s a basis  ¥(f^,fg)  such  {u^u^x^,... ,x-J  that  of  3 such  [ £-U(f.)] 1=1  that  1  f  l  f  2  f  3  =  U,AX  1 =  U  l  Ax  7  3  U,Ax  1 =  r  +  U A  X j +  ,  +  U AXg  ,  +  X^AXg  2  2  5 7  Furthermore, representations  any three  (*) g e n e r a t e  vectors  {f^f^f  a r a n k two s u b s p a c e  }  with '  of dimension  three. Proof:  ,  i s c o n t a i n e d i n t h e p r o o f o f Theorem V . 7 3 .  Then f o r <  ,  2  4 <u x _ Xg>  v  Proof:  v € <u ,x^,...,x^>  B y Theorem V . 7 3 ,  i  f  1  >  f  2  } f  ^  a  r  e  P i a  r  w  i  s  e  -  p  6 •  Thus,  83.  by  Theorems  11.16,  11.31,  for  <u > =  3 n U(f ) , 1=1  and any v e c t o r  x  u  f o r which  2  <u ,u > 1  {u-j^u^x-^,... ,Xj)  l  f  =  U  = Wff-^f 3  2  for  1  A X  3  u  [ E U(f ) ] i  2  4  A x  fg  = U  f^  = u^Au' + V A W  X  A X  +  +  5  ) ,  there  such  ,  2  ,  where and  <v,w>c<u ,x ,...,x_,> 2  U(f- )3<x > 5  Since  {f-^f } ,  {fg,f }  are,Pg-pairs,  <v,w>  intersects  each  <Ug,x^>  1  .  If  u  i s i n either  0  that  >  U AXg  of  i s a basis  hypothesis  by C o r o l l a r y 11.32,  and  <Ug,Xg>  i n t e r s e c t i o n ; then  (ii).  Hence  .  7  d  contradicting  3  VAW e  i n dimension  d i m fl U ( f . ) >_ 2 , 1=1  UgAXg [ x ^ ; U€ g]u A [Ax [g X ; g ;U gU g] ] .  Ther e f o r e etU hg eA r ef x ^e ;x i sUtg ] xa'n d € [x^ ; • Now UgAx^  and  such  2  Xg € [x ; Ug] 6  l  = U, A x , 1 3  + UgAxj^  ,  2  = U, Ax,_ 1 5  + UgAXg  ,  = U^Au'  + yx^AXg ,  f  f  that  f  f o r some  o |= y e F .  3  2 Let  a  l  f  Since  Then  = Y  f  2  f  3  = u., Ax-, 1 3  + a  =  + a UgACiXg  U-.  1  c  = u Au' x  U(f )3<x > 5  Ax D  7  ,  UgAcxx^  _1  , ,  + bx^AaXg u  e  [x-,; x ^ , X g ]  .  Renaming  a  _  1  u  2  84.  ax^  by  x^ ,  axg  by  Xg  and  (*)  , with  desired representations  u'  by  x  , we h a v e  ?  i U(f.) 1=1  the  = <u , u , x  ,...,"x  We now  consider  z = 2 j_£j_ >  e F ,  a  Then  z =  ^  ( ]_ ^ +  A  a  2  z  = u-^A ( a - ^ X j  and  If  ctgX^  x  + a^x )  not a l l  zero.  + u A (a-^x^ + a ^ X g ) +  7  ayc^AXg  2  = o ,  a  If  U  >  '  1  ctg  4=  R(z)  +  = 2  a  2  x  ^  +  aye-, ) + ( a  1  u  2  -  ayCg)AX  4  .  o , a-, z = u A(a x 1  and  1  R(z)  + a x  5  g  =2  Thus e v e r y  non-trivial  is  a rank 2 vector,  independent  and generate  We w i l l  use  is  a basis  then  a  (1,1)  Pg  H  has  (Theorem V . 8 l ) .  dim J U ( f 1=1 5  case  ) = 5,6,  and use  it  g  + ^ x ^ A ^ x ^  + c^Xg)  and hence  {f^,f ,f 2  ?  H € R (U)  basis',  three  There are or 7 .  three  We p r o v e the  }  are  subspace.  f o l l o w i n g lemmas t o  for  to prove  l i n e a r combination of  a rank 2  the  {f, , . . . , f , }  + (u  .  (f^,f ,f^} 2  + a-f-j)  5  others.  and  show t h a t ft  dim'2 U ( f . )  o f w h o s e members a r e cases, the  according  result  for  if = J  pairwise-  as  the  ,  third  85.  Lemma V . 7 6 .  Let  {f ,.. ., fJ  H e R (U)  with  2  be independent i n  x  dimH > 4 .  H  Let 4 dim2 U ( f ) = 7 , i=l  such t h a t  x  and l e t  {f,,f ,f  Then two o f dim  } ;?  2  be p a i r w i s e - P ^ o  {f-^f^f  } , say  3  such t h a t  dim 2 U ( f . ) 1=1  = 7  1  {f-^fg}  , a r e such  that  2 U(f.)= 7 . 1=1,2,4 1  Proof;  Since  {f-^f ,f  dlm[U(f  ) + U(f )] . case,  are pairwise-Pg ,  = 6 .  p  the f i r s t  }  2  Then  dim  2 U ( f . ) = 6,7 . i=l,2,4  In  1  dim  v, U ( f . ) = 7 . 1=1,2,4  I n the second  case,  1  we h a v e  the r e s u l t .  Lemma V . 7 7 . <f,,f ,f^> 0  1  f  2  3  1  , independent of  2  4 dim2 U ( f )  (i)  <f ,...,f |_>  has a  J  pairwise-Pg Proof: Pg  = 7 •  ±  1  .  2  1  e C (U)  4  Let {f ,f ,f,} be a ( l , l ) basis f o r p 3 e Ro(U) , such t h a t dim2 U ( f , ) = 7 . If 2 1=1 {f^f^f^}  (ii)  such  <f ,...,f > x  ( l , l ) basis,  three  that  e R (U)  ,  2  4  then  o f w h o s e members a r e  .  By T h e o r e m V.73> we c a n assume By T h e o r e m V . 7 4 , f  f  ,  x  2  = U AX X  = u-^x^ +<U AV  ,  A X  5  +  U  2  A X  2  v  6 <u ,x ,...,Xg> 2  ?  have  ]  are pairwise  representations  2  '  U  2  + u Ax^ ,  5  6  =  1  {f-^f^f^}  {f-^f ,f  i<y(f^} and  v  1  = < u , u , x , . . . ,x > 1  2  <u ,x^,Xg> 2  3  .  ?  ,  86.  B y Lemma V . 7 6 a n d T h e o r e m V . 7 3 , we c a n assume dim  E U(f ) = 7 1=1,2,4  For  dim  and  {fy',f ,f,,}  are pairwise-P^  0  D  E U ( f , ) = 7 , Theorem ¥ . 7 3 1=1,2,4  implies  1  either  {f^fg,?^}  dim  (  a  x  ^)  basis  for  <f ,f ,f^> 1  or  2  .  1  In a  s  U(f,)= 1  n  1=1,2,4  is  i  the f i r s t  (1,1)  basis  In  case,  for  i t i s easy  <f ,...,f^> case,  { f ^ . . . , f ^}  , and the r e s u l t  1  the second  t o see t h a t  let-  <u> =  follows.  U(f^) .  n  Since  i. —1 ^ 2 $ ^ =  ¥(f ,f ) n  s  0  1  D U(f.) , 1=1,2,4  d  then  u € <u,,u >  we c a n w i t h o u t  loss  .  0  1  1  of generality  B y Theorem  II.18,  2  assume  = u .  From  Theorem V . 7 5 a n d i t s p r o o f , f Let  =  4  Since  4  4  ,  f o r some  + aw e [ x ; x ^ , X g ] , <f ,...,f >  Lemma V . 7 8 . H  such t h a t  6 Rg(U)  4  Let  o  + a w ) + UgAv +  ?  a e F  R(z) = 3 .  ?  1  ?  g  4  i f  [x , x ^ , X g ] .  and w e  + x ) .  v | <Ug,x ,Xg>  possible  in  Y e F  1  Y  ?  4=  o  z = f ^ + f g + f - j + a f ^ = u A ( x ^ + x,~ + x (Ug + a x ) A ( x  x  f o r some  + yx^Axg  U^J^AW  H e Rg(U)  .  .  Hence t h i s .The result  Let  dim1 U ( f ) 1=1  If  f  4  that  case i s n o t follows.  }  {f^f^f  = 7 •  ±  such  be  independent  e c|(U) , 4  independent  of  { * W 3 f  }  s  u  c  h  t h a t  (  l  ) d  i  m  5 ^ i U  f  )  =  7  '  87.  (ii)  <f ,...,f > 1  basis,  three  Proof:  e R|(U)  4  ,  o f w h o s e members  By Theorem V . 7 3 ,  Pg  .  <f  ,f ,f  Moreover, >  0  then  or  <f ,...,f > x  are p a i r w i s e - P g  we c a n assume  either  {f-^f^f  dim n U ( f , ) 1=1  = 1  }  .  is  has  4  (l,l)  The f i r s t  (1,1)  are  pairwise-  .  {f-^f^f^} a  a  basis  case i s  for  contained  1  i n Lemma V . 7 7 -  We n e e d c o n s i d e r  B y Lemma V . 7 6 ,  then only the  we c a n assume  dim  second E  U(f.)  1=1,2,4 and by Theorem V . 7 3 , Pg . <f, , f 1  is  {f^f^f }  c a n be  4  As above, e i t h e r {f-^f^f^} is ,f > or dim H U(f ) = 1 . i=l,2,4 k  d  4  = 7  pairwise-  (1,1) basis for A g a i n , the f i r s t case  1  c o n t a i n e d i n Theorem  V.77-  3 By Theorem V . 7 5 , {u, , u , x ^ , . . . , x } 0  1  f  2  f  3  <u> =  +  UgAXg  ,  = U-^AX^  +  X^AXg  .  1  H  1  2  Case  1.  basis  3  5  1=1,2,4  <u >  <u> = < u >  a  that  = U, A X „  By Theorem V . 7 5 , either  has  = U_ AX^ 4- UgAx^ ,  l  f  Let  such  7  [ E U(f.)] 1=1  U(f,)  .  n  Then  u e W(f , f  1 2  ) = <u ,u_> n  1 2  .  we n e e d c o n s i d e r o n l y two p o s s i b i l i t i e s :  = <u>"  u = u  .  1  .  or  {u,^}  is  ,  1  t a k e n t o be a  case.  independent  i.e.,  88.  As i n the y € [Xj,  u ,x ,x ] , 2  without a^,b^ If  4  loss  = o  2  rank  Theorem V . 7 5 ,  [x ; u ]  (b f 1  , w e  g  4  generality,  b ^ =(= o ,  of  v e  6  of  e F ,  a  proof  f^  4 =  f  [x ;  i  u  A  +  .  Ug]  g  y  v  A  >  w  Thus,  = u ^ A y 4- ( x ^ + a ^ U g ^ b ^ X g  + h u ) 2  2  .  •- b f  5  2  - f^)  1  = u A(b x 1  1  - b x  7  2  has  - y)  3  one.  If  b  = o ,  has  rank  2  (a b _f ]  2  + b f  2  1  - f )  ?  = u A(a b x  4  1  2  1  + b ^  5  - y)  one. 2  Hence  in this  Case 2. y e  case,  <f- .  u A s= ui n t. h e  proof  2  fx,,; u ^ x ^ x ^  {f ,f } 3  L>  is  4  a  >  . . ,f >  of  P -pair, k  R  2  ( )  .  u  Theorem V . 7 5 ,  e  v  i  4  \] k > 7  we  ,  •  f^  = UgAy + V A W  [x,-j u.^  By Theorem  ,  Then  .  II.26,  2 z = af^  + pf^  4 C (U)  ,  2  a,p  both nonzero  in  F  .  Again  2 <f- ,...,f^> Lemma V . 7 9 -  4 R (U) . T h e lemma i s Let H e Rg(U) . Let  in  that  L  of  H  such  { f , ,f_,f_} 1  (ii)  2  basis, Proof:  three  such t h a t  3  i )  of  e R (U) 2  whose  ) = 6 .  ..,  ,  E U(f.) 1=1,2,4 ,  Lemma V . 8 0 .  Let  = 7  1  <f ,...,f^>  are  •  has  1  pairwise-Pg  assume  that  +U(f )]  L  e  t  t  f  (1,1)  ,  .  .  '  are  Then  b y Lemma V . 7 8 . • ^  t ±> 2> 2^ t  a  {f^f^fij  =6  2  The r e s u l t f o l l o w s •'• v •  H e R§(U)  independent  2  i=l  dim[U(f ) .  independent  dim E U ( f . ) = 7  then  1  be  f. e C ( U ) 4 d  4  (i)  members  Then  If  1  B y T h e o r e m V . 6 l , we may  pairwise-Pg dim  dim E U ( f 1=1  <f ,...,f _> 1  proved. {f^f^f^}  2  b  e  i n d e  P ndent e  ,  89. H  In  dim £ U(f\, ) = 5 .  such t h a t  I f f ^ e C^(U)  inde-  4 pendent o f (il)  {f^f^f^}  <f ,...,f > 1  4  ( i ) dim £ U ( f ) = 7  such t h a t  e R^U)  ±  then  <f" ,...,f^>  (1,1)  has a  1  o a s i s , t h r e e o f whose members a r e p a i r w i s e - P g . Proof;  ———  3  dim E U ( f ) = 5  and t h e r e f o r e  1  (Theorem 1 1 . 2 6 ) .  d i m [ U ( f ) +- U ( f )] = 5 3 0  dim £ U ( f . ) = 7 = dim E U ( f ) .  Hence  1=2  1=1  1  The  1  r e s u l t f o l l o w s i m m e d i a t e l y from Theorem V . 7 8 . Theorem V . 8 l . H  in  4  dim E V(± )  such t h a t  then i t has a P  Let HeRp(U).  be  independent P  I f <f . . . ,f^> e R g ( U ) ,  = 7 ,  ±  (1,1)  L e t {f ,. . . , f j j 13  b a s i s , t h r e e o f whose members a r e p a i r w i s e -  6 •  Proof;  By Theorem 1 1 . 2 6 ,  dim £ U ( f \ . )  1  = 5*6,7  •  The r e s u l t  1  f o l l o w s i m m e d i a t e l y from Lemmas V . 7 8 , 7 9 , 80 . P  Lemma V . 8 2 . in is  H a  L e t H e R^(U) •  such t h a t (1,1)  ( i ) dim £  basis f o r  are p a i r w i s e - P g . are such t h a t  Let [ f f . , }  4  2  1  lUf ) ±  <f ,...,f > 1  lf  Then two o f E  dim  =7  ,  be independent  4  ( i l ) {^,...^4}  , and  (ill) {f^f^f  {f^fg) ,  { f ^ f g , ^ } , say  U ( f )= 7 . i  i, ™1 2 $ ^  Proof:  If  Otherwise  dim 1 U ( f , ) = 6 ,  1  1  dim \ U ( f , ) = 7 .  1  1  then  dim  E  1=1,2,4  By Theorem  U(f.)= 7 .  III.40,  }  x  90.  4  dim  S'U(f )  =  1  6,7  •  I n the f i r s t  case,  dim  £  2 In  1=1,3,4  the  second  case,  we h a v e  The p r e c e d i n g next  dim U = 7  ,  R|(U) ,  it  (1,1)  has  a  To show  H e  the  Theorem V . 8 3  We s h a l l  in  we p r o v e  H s R?(U)  H  that  and l e t  dim  £ U(f  Proof: basis Pg  ,  2  independent  ) = 7  .  then  of  .  <f-^,...,f > 4  B y Lemma V . 8 2 ,  e R  2  ,  ( U )  we may f u r t h e r  pairwise-Pg  .  . be  4  =7  .  If  such  > 4 R ^ ( U )  and  ,  1  {f ^ , . . . , f ^ }  <f,,.. . ,f  are  k  dim £ U ( f . )  if  dim H = 4  x  B y T h e o r e m V . 8 l , we may a s s u m e for  If  {f, ,. . . , f )  ^  such t h a t  following  now show t h a t  dim H + 5  1=1 e C (U)  i n the  dim H < 4 .  d independent  be u s e f u l  t h r e e o f whose members  result,  Let  will  then  1  result.  chapter.  basis,  first  the  result  lemma, a n d i n t h e  f,_  U(f. ) = 7  that  .  {f-^,. . . , f ^ } {f-^fg,^} assume  is  are  dim  a  (1,1)  pairwise-  £  U(f^)  = 7.  5 Hence  dim £ U ( f . )  = 7 ,  and  [f  1=2 for pair, f^  3  by Theorem V . 81.  3  = 2 .  dim W ( f g , f ) 3  has  a representation  U(f ) ±  =  <u ,Ug> 1  ,  and  bases f o r  their respective  a  basis  (1,1)  Therefore  is  a  (l,l)  basis  5  <f ,...,f > 2  ,. .. ,f }  for  <u ,Ug> 1  u^Av^ 4- UgAw^ {f.^...,^} subspaces.  <f^,...,f^>  <f ,...,f^> x  Let  4  Since  ,  [fg,f^}  = W(fg,f )  Pg-  Then  <_ i _< 5 ,  since  ( f g , . . . fJ t  Hence  are  {f^, . . . , f ^ }  c o n t r a d i c t i n g Theorem '  a  .  5  1  is  (1,1) is  III.40.  91.  Theorem V.84 independent  Let in  H  H e R|(U) such t h a t  .  Let {f ...,f ) k d i m L U(f. ) = J . 1=1 i s  be  k  1  k < 4 .  Then  k = 4 ,  If  o f w h o s e members Proof:  <f ,.„„,f 1  >  1 |  are pairwise-Pg  B y Theorem 11.26,  k-1 , dim 2 U ( f ) = 5 , 1=1  has a  (1,1)  basis  three  .  d i m ^ U ^ )  = 5,6,7 .  If  w  k < 4 .  III.38,  then  k-1 < 3  k-1 d i m 2 U(f, ) = 6 1=1  If  and hence  ,  b y Theorem I I I . 3 8 , then  k-1 < 3 ~~  and hence  b y Theorem  k-1 dim 2 U ( f , ) = 6 , k-1 < 3 1=1 ~ k < 4 . Finally, for  k < 4 .  If  b y Theorem I V . 7 1 , and a g a i n k-1 k d i m 2 U ( f ) = 7 a n d d i m 2 U(f.) = 7 , T h e o r e m V . 8 3 shows 1=1 1=1 1  k  1  than 4 .  cannot be g r e a t e r  result  k <_ 4 .  The second  i s c o n t a i n e d i n Theorem V „ 8 l .  T h e o r e m V . 85 When  Hence  Let  H e R|(U) , d i m U = 7 .  dimH = 4 , H  has a  members  are pairwise-Pg  Proof:  Let  ff-^...,f  Theorem 11.26,  f c  }  (1,1)  three  dimH < 4 .  o f whose  . be independent  d i m 2 U(f,) = 5 , 6 , 7 1=1  f o l l o w s b y Theorems  basis,  Then  III.38,  IV.71,  in  H .  i f k > 2 . . . . and V.84.  By The r e s u l t  92.  CHAPTER V I  THE M A I N RESULTS  In this and  n > 6,  it  has  a  chapter,  then  (l,l)  we show t h a t ' i f  dim H < n-3basis.  These  H e R (U),  Furthermore, i f results  dim U = n  2  are  d i m H > 4,  c o n t a i n e d i n Theorem  VI.100. Let  H = <f^,...,f >•  c o n s i d e r e d the  I n Chapters  k  cases  dim £ U ( f . )  consider  and t h a t k <_ 5  the  case  (f ,...,f )  is  3  k  (Theorem V I . 9 2 ) .  preceding chapters  J.  6,  I V , V , we  In this  chapter,  Ic  1  we w i l l  = 5,  III,  dim 2 U ( f ) i  a  (1,1)  = 8,  basis  for  We s h a l l t h e n u s e  and above  show t h a t  k <_ 5 ,  <f ,...,f >, 1  k  the r e s u l t s  from  to o b t a i n the main r e s u l t s  the  stated  i n Theorem V I . 1 0 0 . Theorem V I . 8 6 . in  H  H e R^U).  Let  such that  dim | U(f )  Let  = 8,  ±  {f^fg,^}  Then  be  { f ^ f ^ f ^  independent is  a  (l,l)  2  basis  of p a i r w i s e - P g vectors  Proof;  By T h e o r e m 1 1 , 2 6 ,  ^f ,f )  is  1  2  Pg-pairs.  Hence  dim | U ( f ) 1  dim  3  a Pg-pair.  = 8  R (U). 2  and  Similarly  1  2  Thus  3  3  e R (U) . 2  + U(f )]  it^t^),  2  = 6,  3  are  then  and hence basis  hence  Since  are p a i r w i s e - P g ,  f o r m a (1,1)  and  {f-^f^  are p a i r w i s e - P g .  U(f-) => W f f ^ f g )  (f ,f ,f ) 2  2  1  (f ,f ,f )  1  <f^f ,f >  dim[U(f )  [f^f^f^}  U ( f ) = 2.  C o r o l l a r y 11-34,  for  for  by <f ,f ,f > 1  2  3  e  93.  Theorem V I . 8 7 . in  Let  such that  H  f  2  l =  u  A x  Proof:  +  2  u  A x  (f-^f^f  }  +  f  such that  4'  f  - i u  2  Ax  V*6'  5  B y Theorem VI.86, for  representations  = ^ A x ^ + UgAx^,  f 2 =  u  1  A  since  x  f  1  + 2 u  7  8'  dim i U ( f )  pendent  in  Let  H  = u.jAx  2  ...,x >.  have g  such that  a  (1,1)  basis  for  i  = 1,  2,  3;  f  3  so t h a t  1  2  take  2  1  u = u^.  have  independent  If  be  inde-  e R (U),  <f  2  2  13  .,.,t^> e R ( U ) ,  a  (1,1)  2  is  1  2  I  U(f )  > 1,  1  3  are  2  1  i  for  then  d i m U(f^) = 4.  u e <u, , u _ > .  <u ,u ,x^,  > 2;  1  2  =  ±  3 > dim ¥ ( f ^ , f )  < u , u ' > = <u ,u >-  of  f ^ = u-^Ax^ + UgAx^,  = u Ax^, + u A X g j  that  basis  By Theorem V I . 8 7 ,  3  X  (u ,u ,v ,vj_,. . .,v ,v^) 1  ( f ^ f ^ f ^  <f^...,f^>  2  = 1,2,3if  this  d i m U ( f ^ ) = 6, Let  B y Lemma I I . 1 8 , e a c h  of  d.  c a n be r e p r e s e n t e d i n t h e f o r m  3  of  e C (U),  [f^f^f^]  <f ,f ,f >.  which c o n t r a d i c t s the f a c t  2  (l,l) basis  are  were n o t s o a n d t h e i n t e r s e c t i o n were empty,  1  8-  A x  f^  for  L  (f ,f ,f )  2  g  = 8.  ±  representations  + u Axg,  <u> c $ TJ(f. ) . 1=1 1 '  u  = u-^x,. + u A x g ,  2  d i m J ^ U ^ ) = d i m [ U ( f ^ ) n <u- ,u >]  Hence  + 7  { f ^ f ^ f ^  dim | U ( f )  B y Theorem 11.26,  g  Let  2  By Theorem VI.86,  pairwise-Pg vectors  f  f  Ax  are p a i r w i s e - P g .  2  2  ,is a  basis  representations  l  = u  3 > . . . ,Xg]  H e R (U).  is  1  (f^,f ,f^}  j X  have  independent  has a  ±  Hence  3  u  2  {f^,f ,f^)  Proof;  u  { f . ^ f , f ^3  {f ,...,f ^} 1  2  t i' 2  d  such that  independent of then  n  1  be  = 8.  ±  Theorem V I . 8 8 .  and  a  <f ,f ,f >.  )  i U(f )  3  [f-^fg,^}  pairwise-Pg vectors  A x  [ f ^ f ^ f  Then  ±  3  l 3  Let  2  d i m i u ( f ) = 8.  (u ,u ,x ,...,XQ] 1  H e R (U).  It  is  f  ±  easy to  independent.  = uAv  ±  + u»Av j ;  see  Hence  WLG,  we c a n  94. 4 dim n U(f ) = 1 ;  Suppose has  dimension  dim  U(f )  = 1,  i  1.  Since  =• 4 ,  4  2, 3,  f  1  a n d thus  ij. = ] _ u  A  2  r> <  11-32,  U j  >,  then  (f , . . . , f ^}  u  n <u ,x^>  d i m <w,w'>  2  Since  4  W(f^,f ) =  arepairwise-Pg.  2,  Therefore,  II.16,  <v,w,w'> c < u , x , . . '. ,xg>. 2  1  £ U(f ).  2  dim  and b y Theorem  3  + wAw',  v  then  L  Uff^)  and  n <u ,u >  i f  u- e U ( f ^ ) ,  U ( f ) c < u , u , x , . . . ,xg>; 4  i . e . ,U(f )  ±  By C o r o l l a r y  3  = 1,  2  d i m <w,w'> n <u ,Xg> = 1 ,  (*)  2  d i m <w,w'> n <u ,Xg> = 1. 2  If  u  ^ <w,w'>,  2  contradiction. <u ,u >. 1  forms  Hence  Now  2  Hence  u  2  with  e <w,w'>  = 4  d i m U ( f ^)  a Pg-pair  d i m <w,w'> = 3 ,  (*) i m p l i e s  1  a t least  one o f  Theorem V I , 8 9 .  .$ U ( f ^ ) =  ( f ^ f ^ f ^ } , 2  L e t H e R (TJ).  Hence  2  u ^ A v ^ + u Av^_  i s a (1,1)  (f^,...,f^)  and t h e r e f o r e  a n d U ( f 4) r> < u , u > .  f ^ has a r e p r e s e n t a t i o n  Therefore  which i s a  basis f o r  say f . 1  II.31).  (Theorem  <f ,...,f >• 1  4  L e t ( f , . . . ,f^}  2  f^  be i n d e -  1  4 pendent  i n H  such  a (1,1) basis  has  Corollaries are  pairwise-Pg.  If  dimi U(f.)  1  either  x  I f dim = 7,  L  (ii)  = u Ax ]  j L  dim  3  JU(f  3U f f ^ 3  3  + u Ax , 2  4  f  2  ) = 6;  1  4  i  =  6,  8.  7,  By  we c a n assume  4  d i m Jyjff^)  b y Theorem V . 7 3 , t h e r e  then,  2  ±  <f , . . . , f >  dim| U ( g ) = 8.  that  dim | U(f )  [f.,f ,f_} p  =  8.  a r e two c a s e s ;  i s a (.1,1) b a s i s = 1.  B y Theorem V . 7 4 ,  (i)  such  V - 7 3 , Theorem VI.86,  ( i ) (f- ,f ,f }  or  f  (g-^-.-jg^)  IV.6l,  Then  1  B y Theorem 11.26,  Proof:  Case  d i m S TJ(f ) = 8 .  that  (f-^f^,^}  = u Ax ]  5  + u Ax ; 2  6  have r e p r e s e n t a t i o n s :  f  3  = u Ax 1  7  + u Av, 2  95.  3  = <u ,u ,x  2 U(f )  1  1  v £ U(f ),  i  1  EU(f )  3 >  . . .,xy;  = 1, 2.,  v e < u , u , x , . . . ,x >, l i  since  = <u n ,x^,...,XQ>  1  p  x , . ..,x >.  2 )  3  H <u ,u > 1  2  dim W ( f , f ) 4  > 2,  ±  for  0,  T  if  1 = 1 , 2 ,  are p a i r w i s e - P g .  U(f^) = <x >.  Let  Let  g  not,  and  g  3  d i m U ( f ^ ) H ¥*  Then  ?  2  {f^f^f^} and  2  1>  W* = < u u U(f^)  2  = 3;  and  b y T h e o r e m 11.26, and t h i s  3,  implies  d i m U ( f _ ) fl ¥* > 3ll  Suppose <xg,u ,u ,y>; 1  dim  i = 1  and  2  4  U(f )  = 8.  i  3 <u ,u >.  4  y e W*.  2  E  U(f )  1  If  d i m U ( f ^ ) fl W** = 3.  It of  .  Clearly  (f^  - y~ f^)  6x^  4  + w').  + w'}  is  independent.  follows  easily  11.26, t h e a b o v e 3  [x^^xg; For and  u^Ug]  that  and  U(f ) 4  contradicts  g = f_ ]  n U(g) = < u > x  1  e F, f^  6,  (f ,f ,g) 2  + U(g)]  + Yf*  [f^gj  Theorem I I . 2 6 and hence  this  for  3  + w)  +  + w, are  3  = 8. By Theorem  U ( f ^ ) r> <u^>. that  3  II.18,  Theorem  d i m [ U ( f _ ) H W*] = i)  for  = t^A (x  3  e  4  7  e [vju^Ug]  making  (w,w'}  = U^A(6X  y e [x^x^j u^Ug];  + jSy  Y  Similarly  we c a n WLG t a k e  ay  0 4=  A X ^ ,  asumption, and the f a c t  0 4= Y e P j  some  2  0 <u^,Ug> = 1.  4  U(f^) 3 <y,y'>;  imply  Y U  a  Hence  2  2  3  dim U ( f )  f ^.  = u^Aw + u Aw';  For such a  2  forms  [u^,u ,x,_,Xg, v , x , 6 x ^  dim[XJ(f ) + U ( f )  Suppose  say  f^  g = 5 f ^ + f^  Let  6 e F,  F o r some  p a i r w i s e - P g and  it  4= u-^Aw +  0 4= Y e P .  2  f^  then x  2  then that  has r a n k o n e .  x  yu^Ax^ + u A w ' , UgA(5x  f^  1  2  [xg^u^Ug]  otherwise  =  4  e ¥** = < u , u , x ^ , . . . > g > ,  {f ^ , f , f ^},  are p a i r w i s e - P g and  3  y  follows  [f ,f ,f^} 2  U(f )  y e [x^jU-^u^x^x^x^x^Xg],  Otherwise,  P p . - p a i r w i t h a t m o s t one  Then  2  + Y  y'  e  some X 7  )  a , 0 e P. +  UgA^+yv),  a P -pair. ?  case does n o t  This occur.  96. (ii).  Case f  l  =  l  u  A  3  x  B y Theorem V . 7 5 ,  +  2  u  A  V  x  2  f  =  i  u  A  x  5  +  |  u  i U ( f ) = < u , u , x , . . .,x >. ±  1  2  U ( f ^ ) 3 <x >Since  6  A x  1  2  that  most  f.^  pairwise-Pg. dim  ¥(f ,f )  dim  U(f )  i f  2  ,f  i  > 2,  0 W* = 3 ,  1 |  3  ),  Suppose  ±  say U(f^)  this  and  is  (f ,f ,f 1  of  2  i f  }  again. T  and therefore f^  0 ' and  basis  for  z'  for  case a g a i n .  2  h  Finally, a basis  ;  . . . ,x >, g  d i m U(f )0W*= 4  fact  , f ^}  3  are  with the fact  [f^f^}  is a  that  u e  i f  J U  2  ];  Theorem  = 2, 3  < z , z ' > fl < u , u > ={= 0, ] [  U ( f ^) o < u > .  1  2  i f  x  for  Again,  As i n t h e p r o o f  < z , z ' > fi < u ^ , u > = 0 , 2  i ]  3  We h a v e C a s e  J  i = 2, 3-  t h e n as a b o v e ,  ] L  <f ,f ,f ^>. Then  as a b o v e , (i)  we h a v e s h o w n t h a t  w h i c h we s h a l l s i m p l y c a l l  < z , z ' > fi  ff^f^f^}  (i)  <u ,Xg> 1  i s a (1,1)  again. <f ,...,f 1  (g ,...,g> ) 1  Then  2  z e [x _;u ],  a n d we h a v e C a s e  2  are pairwise-Pg.  2  1  [ f , f ^}  u^Aw + u A w ' .  >.  (f ,f ,f^1  2  Since  2  <f ,f ,f  If  ±  since  [ U  Pg-pair,  contradicting  Otherwise  z £ U(f ),  e [xgj^J.  <f ,f-^,f >  2  has a r e p r e s e n t a t i o n  i s a (1,1) b a s i s  Otherwise,  6  U ( f ^ ) o <u,v,w>;  Thus,  WLG, z e [x^x^jU-^Ug] .  (f^f^f^)  Then  [f , f  Together  T h e o r e m V . 7 3 , we c a n a s s u m e  and  A x  1  i s a (1,1) b a s i s  We h a v e t h e f i r s t  4  a n d U ( f ^ ) = < u ^ , X g , z , z ' >,  1  then  x  Now d i m <z,z > n U(f.j_) = 1, i  3  a Pg-pair,  +  T h e o r e m 11.26 i m p l i e s  B u t then  4  < z , z ' > => <u >  7  'forms a P ^ - p a i r w i t h a t  Hence  implies  d i m < z , z ' > fi U ( f ) >. 1.  then  x  by the preceding  f^  1  U ( f ) -D < U ^ > ,  <z,z'> c < u , x , — , x ^ > . 2  A  ±  be w r i t t e n i n ( l , l ) f o r m ,  Hence  11.31.  l  u  representations:  T, U ( f ) = KvL^u^Xy  <u >.  = 1, 2, 3 -  v e [x^;u^3, w e [xgju^] . which cannot  =  3  a n d T h e o r e m 11.26, i t f o l l o w s 1  3  are pairwise-Pg,  2  (f , f  f  W* = < u , u , x , . . . , x >.  [ f . ^ f , f ^}  one o f  J  Let  7  Let  g  3-  3  have  { f - , f g,^}  4  such  J +  >  has  that  97.  dim is  3  2 U(g ) = 8 ,  i t follows from Theorem VI.88 that  1  {g  • • . ,g^}  a (1,1) b a s i s .  Theorem V I . 90.  Let  H e £ ( U ) . Let  [f, , . . . , f j  pendent i n  H  dim t U(f ) = 8 .  such that  be inde-  D  1  d  Then  <  f  1  has a (1,1) basis  : 1=1,...,5}  {g  B y Theorems I I . 2 6 ,  > 5  dim 2 U(g ) = 8 .  such that •  Proof:  f  i 1  1  IV.71,  4 E U(f, ) = 7 , 8 . 1  dim  1  By  Theorems I V . 8 l ,  VI89,  <f ,...,f^> 1  of whose members are pairwise-Pg.  has a ( l , l )  We can assume that  are i n f a c t these basis members, and that  that  5  dim t U ( f ^ ) = 8 ,  If  dim E U ( f . ) = 8 .  1  (f ^ , . . . , f 4 }  [f , f , f _ ) are  4  pairwise-Pg.  b a s i s , three  VI.89  we can assume by Theorem  The r e s u l t follows by Theorem V I . 8 8 .  3-  4 dim E U ( f ) = 7 ,  If that  dim  ±  =  we can assume that by Theorem V . 8 2 ,  1  1  \  = 7-  ^{? ) ±  Then  dim  i = 1  ^  _U(f ) = 8  and  1  the r e s u l t follows from Theorems V I . 8 9 , 8 8 . Theorem VI. 91. pendent i n  H  H e R^U) •  Let  ±  [f ,...,f 1  },  1  B y Theorem V I . 9 0 ,  (1,1) basis f o r By  Theorem V I . 8 8 ,  By  Theorem III.40,  {f ,...,fg} 1  such that  {f ^ . . . ,f,_}  then  2  is a  dim | U(f ) = 8 . jL  i s a (1,1) basis f o r  <f ,...,fg> £ R ( U ) . 1  f g e c|(U), 1  we can assume  x  be inde-  Re-  6  <f ,... ,f^>  }  dim | U ( f ) = 8 ,  such that  <f ,...,f > k proof:  {t^...,?  dim | U ( f ) = 8 . I f  such that  independent of  Let  <f ,...,fg>. 1  98.  Theorem ¥ 1 . 9 2 . if-[_,••  •, )  Then  k < 5,  {g  1  f  b  k  L e t H e Rg( J)  © independent i n H and  : 1 < i < k)  Proof:  <f ,...,f > 2  = 8.  jL  1  1  By Theorem 11.26,  1=1  6, 7 , 8 .  =  1  t h e n by Theorem I V . 7 1 ,  d i m ^ U ( f ) = 6, 1  k - 1 <_ 3, and  k <_ K.  thus  ) = 7,  dim | lT(f i  1  t h e n by Theorem V.84-, k - 1 <_ 4 ,  k <5. If  d i m ^ U ^ )  dim l U(g ) = 8.  such that  k-1  If  such t h a t  has a (1,1) b a s i s  fc  k-1 dim £ U ( f , )  If  dim H > 5 . L e t  with  T  k-1 , dim E U(f ) 1  1  = 8  1  k dim | U(f )  and  1  = 8,  i  then Theorem VI.91  k <_ 5 .  shows  k <_ 5.  Hence  By Theorem 11.26, k > 3-  f o l l o w s b y Theorems V I . 8 8 ,  Let H e R|(U).  pendent  such t h a t  in H  independent of  such that then  i  ±  4  [f^f^f^} ±  [f^f^f.^]  {g ,.••,g )  Let  dim 1 U ( f ) = 7 -  ( i i ) dim 2 I T ( f ) = 9 ,  basis  The r e s u l t  8 9 , 90.  Lemma V I . 9 3 .  and  1  and  be i n d e -  I f f ^ e C (U), 2  ( i ) <f , .. . ,f^> e Rg(U) x  <f ,...,f > 1  has a ( l , l )  if  such t h a t dim ^ U ( g ) = 8 . 1  Proof:  By C o r o l l a r y V . 7 3 , we can assume  pairwise-Pg.  Then  dim  1 = 1  E  2  2  ^U(f ) = 8.  [f-^f^f^} The r e s u l t  1  are follows  from Theorem V I . 8 8 . Lemma VI.94.  Let H e R | ( U ) .  Let  {f ,...,f ) 1  i f  be independent  99. 4 in  H  such that  d i m J U ( f ) = 7-  of  (ii)  d i n | U ( f ) = 9,  {f ,...,f ) 1  then  ±  5  for  pairwise-P^-  ( i ) < ? ^ . . . , f > e R ( U ) and 2  <f ,...,f  >  1  Theorem V I . 95-  is a  1  f f ^ f ^ f  dim E U ( f . ) = 8. 1=1,2,5 i  }  are  The r e s u l t  y  in  H  Let  H e R ^ U ) . L e t {f _, . • . , f ) k . . _ such that d i m E ^ U ^ f ^ ) = n , n _> 6 .  follows  <f ,...,f >  has a (1,1)  k  For  be i n d e -  k  ]  Then  has a ( l , l ) b a s i s .  x  Proof:  f f , • • . , f ^}  such that  n = 7, k = 4 , < f , . . . , f ^ >  1  basis  VI.88.  by Theorem  pendent  has a (1,1)  1  <f ^ . . . , f ^>  Hence  6  2  dim ^ U ( g ) = 8.  that  basis  indepen-  f" € C ( U ) ,  B y T h e o r e m V . 8 l , we c a n assume  Proof:  For  k <_ n n > 8,  basis.  n = 6 , 7, 8 ,  the r e s u l t  i s c o n t a i n e d i n Theorems  V . 8 4 , VI.92•  V.71,  For The r e s u l t and  such that  4  i&i,-'-such  For  If  ±  dent  (1,1)  0  n > 9,  follows  d i m £ U ( f , ) _> 7 , 1  8  by Theorem 11.26.  b y i n d u c t i o n f r o m Theorems V I . 9 3 ,  94, 92, 8 8 ,  III.40. The f o l l o w i n g i s a n e x a m p l e  H e R (U),  dimH = n-3,  2  Example V I . 9 6 .  Let  f  1  =  f  2  =  f  3  =  x X ]  X  x  l  A x^ + x  A x  be a b a s i s  n  4  5  basis f o r  dimU = n .  [x^...,x )  _ A x  of a ( 1 , 1 )  + x + x  2  2  2  A x^ A x A  5  Xg  for  U.  100.  f f  X-,  ii  —  ~  =  n-4 n-3  1  _ -j- x ^ A x 2  A x  n-2  -i  n-1  x, A x , + x„ A x„ 1 n-1 2 n  n-3  a  For  ±  and  e F;  z = 2  =  x1.  +  x  [  f  < l 3  A  a  x  A (a x  £  ± ±  a  1  a  2  4  x  •••  +  +  a  n-3 n-l X  }  + a x ^ + ... + a . _ 3 ) > x  I |  2  n  n  R(z) = 2 . It  i s p o s s i b l e f o r H e R (lT) v e c t o r s when  s h a l l show t h i s  pendent i n  H  Let  H e R|,(U).  U(f ) 3 U  (ii)  dim 2 U ( f . ) = m,  ±  Q  ,  dim U  o  We  [ f , .. . , f )  Let  k  be i n d e -  1 < 1 < k;  = 3,  dim U .  6 < i <  k < m-3  Then  (f. ,...,f  and  1  1  1  i s a (1,1)  basis  P^ v e c t o r s f o r < f ^ . . . , f > . f c  Since  [f-^, •.' ,^ )  ( i ) holds,  k  ( i ) and ( i i ) , k > 3 j  1  f  2  f^  =  1  l  x  A  =  X  =  X^  2  u  +  l  W  A  U  A  ( ^  2  A  +  ^  ±  l -  V  A  Wg  +  dim i U ( f ) = 6 .  have r e p r e s e n t a t i o n s :  k  l  are p a i r w i s e -P,..  and f o r k = 3 ,  [f ,•..,f )  By Theorem I V . 4 9 ,  f  dim U = n > 7.  such t h a t  (i)  fo pairwise  dim H = n - 3 ;  i n the f o l l o w i n g theorems and example.  Theorem VI.97-  Proof:  t o have a ( l , l ) b a s i s  2  of pairwise-P.-  By  +  3  U  2  Vg  )  + W  3  A  V^  101.  k  f  with  =  £3* • • • i  1 < i  Q  <\i^,u ,u^,x^ 2  ±  hence  < k,  expressed  a  Let  1  < i  <_ k .  (1,1)  form  (1,1)  Theorem VI.98.  V  k> F,  u" = < u , u , u ^ > ;  of  < k,  i  =  < k,  1  k  Let  U ( f ^) 3 U ,  dim  Q  k = n-3,  {f  ±  i  .  u  ..  {f  , . . .,f  ,f _ 3 n  3  . is  a  basis  k < n-3  1  .. • .  follows  by Theorem  1 • •  result  VI.97.  The f o l l o w i n g i s  an example  of  a (l,l)  basis  of  2 pairwise-Pj- vectors  for  H e R (U),  d i m H = n-3,  2  dim U = n;  <f^,..•,f _ >• n  3  Example V I . 9 9 -  U =  <u ,u ,u ,x ,..., 1  =  U f  i  =  x  i  A  2  0  ^  U  l  1  n  e F,  3  not a l l  - S  z =  u,A 1  ( £ a,x, 1=1  U  (£g  2  A  a  u  +  3  A  i  u  non-zero  z =  +  x  1  n  _ >i 3  Q  ^ i ^  +  6, ,...,6 „ ^1' "n-3 a ,...,a _  3  <u,,u ,u >. 1 d' 3  0  For  }  = 3,  0  for < f f ,> . k ^ ' Proof: d i m I u ( f . ) <_ n , a n d h e n c e b y T h e o r e m V I . 95, 1 k : ) • • • When k = n-3 , d i m £ U ^ T J ) = n , 7 < n = d i m U . The  H =  and  < f , . . . , f > . , By  d i m U = n>7-  For  I U(f )  1 < i  2  for  2  c o r o l l a r y IV.50, each  + v'Au ,  1  1  o  1 < i  2  basis  such that  k <_ n-3 •  Then  A  vAu  H e R|(U);  H  k  1  ±  k <_ m~3 •  in  W  As i n t h e p r o o f  is  k  +  e <u ,u >,  v  i n the  ff ^ . . . , f )  independent  U  and d i s t i n c t on  Theorem I I I . 4 0 ,  be  ^k 2'  +  . . . ,x^>.  }  c a n be  f  K  A  non-zero  <w ,v > c U , 1  k  X  j  zero;.  ±  x  + £ a, u )  1=1  a ^ ) .  1  1  ^- " 3 n  and d i s t i n c t i n  f , x  1  ^  F.  102.  R(z) < 2  Now  implies  n-3  n-3  2 a 0 x 1=1 1  for then  some n-3  X  4=  1  \[  1=1  1  1  n-3 a x , + E a,u_ 1 i=l 1  P. Since  in  0  T,  1  3  4 0.  j3.  \  ! < _ ! < . n-3,  4  0,  1  a  = 0  ±  and  a ^  1 <_ 1 < n - 3 .  = Xa ; ±  f o r two o r more values o f i ,  then  ^6 " = X ±  c o n t r a d i c t i n g the d i s t i n c t n e s s o f  If  a  4 0  ±  for two o r more  . But a  ±  1,  4 0 for  n-3 exactly  one  implies  i  which i s a c o n t r a d i c t i o n .  E ( z ) = 2.  Hence  T h e o r e m VI.100. dim  0 ,  H < n-3-  Proof: k  dim^E^U(f )  If (l,l) tions?-  = xAx  i  + yAy^,  2-dimensional  combination  z = af  i  z  + /3f j |  subspace  ( f ^, .. . ,f^.}  in  U.  where  We n o t e  members  Then  1 <_ I , J <_ k.  Thus  representation  xAx  the main r e s u l t s  f  + yAy ,; ±  (i.e.,  x , y f  f  k  have  has a representais a  <x_,y> that  any l i n e a r  f i , f j is  Hence a n y n o n - z e r o v e c t o r  f  e U.  z = xA(ax in  H  i  + 0Xj) +  has a  We c a n t h u s  restate  T h e o r e m VI.100) u s i n g t h e f o l l o w i n g  of H .  D e f i n i t i o n VI.101. subspace  H.  of d i m e n s i o n  1 £ i < k,  of a n y 2 b a s i s  a,j3 e F ,  H  y A ( a y . + )3y . ) . 1 J  definition  in  f o l l o w s b y c o r o l l a r y 11,21, a n d  then  k  1  then  R ^ ( U ) ,  TI.95.  ( f ^, • . . , f ) ,  f  H e  h a s a (1,1) b a s i s .  a r a n k two subspace  basis  fixed  The r e s u l t  III.38,  H  If  be i n d e p e n d e n t  {f^, . . . , f ^ }  <_ n .  1  Theorems  d i m H > 4,  If  Let . ,  d i m U = n > 6.  Let  A r a n k two subspace  i f any non-zero  vector  f e H  H has a  is a  fl,1)-type  representation  103, f  = xAx  + yAy  f  Cg(Cr),  in  f  2 - d i m e n s i o n a l s u b s p a c e of  where  is a fixed  <x j> 9  U.  Theorem. V I . 102.  Let  d i m H < n-3-  dim H > 4,  then  convenience  of the r e a d e r ,  If  d i m U = n > 6. E  If  H e R|('C),  then  is  a (1,1)-type  sub-  space.  F o r the a summary o f  the r e s u l t s  obtained i n t h i s  A r a n k two v e c t o r f  = x^AXg + XgAx^;  where  f  f,  if  Yj, • •> •  to t h i s for  y  A  1  y  <x^  t  + 7^ Y^ A  2  e I , T  l  then  paper.  representation are independent  x , , . ..  The 4 - d i m e n s i o n a l s u b s p a c e sense that  has a  we s h a l l now g i v e  ...,x^> i s w e l l - d e f i n e d  s  <x^,...,x^>  i s  dimension one. when  d i m U = 4,  Thus  dim U >. 4. of  Thus the i n t e r e s t i n g r a n k 2 s u b s p a c e s a r e  t h i s maximum d i m e n s i o n i s  are as f o l l o w s .  Let  dim U »  attained.  f  e H  has the f o r m  f  c a l l s u c h a s u b s p a c e a (1,1)-type f o r s u c h a subspace i s  and  When a r a n k two subspace  = xAx  a f i x e d 2 - d i m e n s i o n a l s u b e p a s e of  n>6.  (n-3),  has d i m e n s i o n h i g h e r t h a n o r e q u a l t o 4, t h e n e v e r y  basis  those  dim. U >. 5 •  Then the r a n k two s u b s p a c e s have d i m e n s i o n a t most  vector  of We r e f e r  U(f).  the r a n k two s u b s p a c e s a r e  The main r e s u l t s  H  i n the  .. . ,y^>.  as  t h e r a n k two s u b s p a c e s t o be n o n - t r i v i a l , When  IT.  any o t h e r r e p r e s e n t a t i o n  s  <x_, . . . . x^> = < y  4 - d i m e n s i o n a l subspace  in  f  II,  + yAy ,  where  f  and  subspace.  the' f o l l o w i n g ; -  x .,y f  f  non-zero <x,y>  e U.  An example  We of a  is  104'.  dim H = k;  Let  dim J = n;  H = <f ,...,f >; 1  i  f  =  X  A  x  i + 2  Other r e s u l t s subspaces  X  2  A  x  i  ; +  ^  1  3  < "  1  k  obtained characterize  t h e r a n k two  o f d i m e n s i o n two and t h r e e . The  r a n k two subspaces  o f d i m e n s i o n two e x i s t when  E v e r y 2 - d i m e n s i o n a l r a n k two subspace  dim U >_ 5. type  +  n  k < 3-  k  l  U = <x , . . . ,x >.  T  is a  (l,1)-  subspace. The  3 - d i m e n s i o n a l r a n k two subspaces  a r e t h e most  varied. First, are  t h e r a n k two subspaces  anomalous i n t h e sense  dim H < n - 3 j  [ H ] when  t h a t , whereas when  we have t h a t when  dim U = 5,  dim U = 5  dim U > 5, t h e r e do e x i s t  r a n k two s u b s p a c e s o f d i m e n s i o n 3• However, we have n o t f o u n d i t c o n v e n i e n t t o c h a r a c t e r ize  t h e r a n k two s u s p a c e s by c o n s i d e r i n g t h e u n d e r l y i n g  IT.  We have t r e a t e d Let  We c o n s i d e r e d dim  U(f )] ±  space  then i n t h e f o l l o w i n g way i n s t e a d .  [f^,f ,f^} 2  he a b a s i s o f t h e r a n k 2 subspace  the v e c t o r space  U'f^)] .  We showed  H.  that  i s e i t h e r 5, 6, 7 o r 8.  3 Case 1.  dim Then  following  E U ( f ) = 5* 1=1 1  H  has a b a s i s  two t y p e s : -  (g ,g ,g ) 1  2  3  which i s one o f the  105, 3  x^ l  =  x,_ A  S3  =  U  2  „  1(b)  •u + u  A  2  A  U.  A  1  U2  A  +  y  ?  + u  y<  A  3  3  [2  U 'f ^ ) }  =  <U, ,U ,U-,X  c.  j  p  ,S>  i l  U *  g  x  =  X^  A  U., +  U  g  2  =  X,-  A  U  U  s  u  A  u' 4-  g  3  2  +  2  1  y  A  u  A  U  ,X >|  V  K  ^  €  J 6  <U,  J_  ,U >: 0  ^  "••^2^3" •  3  A y '  3 Qui a_ j .  dim S U { f ) = 6, f h e n E has a b a s i s ,  1  (g_^g^g,} ' ""'4 * 3  whieh i s ont of th©  f o l l o w i n g t h r e e types?»  2{a,)  i s a ( l / I ) b a s i s and go  (g^gg^gg}  H  is a  t y p e gribBpaoe.  3 2(b)  t E t'(f )3  «  t  , u , x - , .. g  g  g  .  u  1  A  + Ug  A x  g  3  .  X  3  A y  + X  A yt  4  .,xg>.  g  j  <y,J«> «  3 2(c)  [ 2 Ii'(f )j m  1=1  1  - < u , ' a , x , •. . , x > . 1  2  3  g  <Xg,Xg>,  *  106. g  =  1  gg  =  63  =  U-  t  A  x  A  Xp. + Ug  3  Ay  + u  2  A  x^  A  Xg  + x^ A x g j  ye  <u ,x ,...,Xg>. 2  3  3  dim [ 2 T . \ > . )1 i=l  = 7.  A  K  has a b a s i a  (g^,g ,g^J 2  which I s one o f two types  3(a) {&•• pi -" 3^ g  i s  (~?~) M i l s i n d so H  s  2  i§ s (£,1)-  3  «1  s  t3  ss  2  A  x  A  + U  3  g  x^  A  X^ + X|  dim E tr{f ) - 8 i-l subspace.  If  1  A  P  Xg. th#n  H  i s a (1,1)-fcyp«  107.  BIBLIOGRAPHY  [1]  Bourbakl,  N.,  Algebre  [2]  Elements  H o d g e , W.V.D. a n d  B e r t i n i , E.,  Vol. I*  Van  Der  [5]  Halmos, 2nd  W a e r d e n , B.  P.  R.,  Edition,  i n die Projective  Modern A l g e b r a ,  Dimensional  1958.  Algebraic  1947.  Raume, V i e n n a ,  L.,  Finite  Methods o f  Cambridge  Einfuehrung  Chap. I l l  1948.  P e d o e , D.,  Mehr-Dimensionaler  [4]  Mathematique, A l g e b r e ,  Multlllneare,  Geometry,  [3]  de  Vector  Geometrie  1924.  Vol..I I .  Spaces,  

Cite

Citation Scheme:

        

Citations by CSL (citeproc-js)

Usage Statistics

Share

Embed

Customize your widget with the following options, then copy and paste the code below into the HTML of your page to embed this item in your website.
                        
                            <div id="ubcOpenCollectionsWidgetDisplay">
                            <script id="ubcOpenCollectionsWidget"
                            src="{[{embed.src}]}"
                            data-item="{[{embed.item}]}"
                            data-collection="{[{embed.collection}]}"
                            data-metadata="{[{embed.showMetadata}]}"
                            data-width="{[{embed.width}]}"
                            async >
                            </script>
                            </div>
                        
                    
IIIF logo Our image viewer uses the IIIF 2.0 standard. To load this item in other compatible viewers, use this url:
http://iiif.library.ubc.ca/presentation/dsp.831.1-0080623/manifest

Comment

Related Items