UBC Theses and Dissertations

UBC Theses Logo

UBC Theses and Dissertations

Characterization of transformations preserving rank two tensors of a tensor product space Moore, Carolyn Fay 1966

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

Item Metadata

Download

Media
831-UBC_1966_A8 M6.pdf [ 1.3MB ]
Metadata
JSON: 831-1.0080595.json
JSON-LD: 831-1.0080595-ld.json
RDF/XML (Pretty): 831-1.0080595-rdf.xml
RDF/JSON: 831-1.0080595-rdf.json
Turtle: 831-1.0080595-turtle.txt
N-Triples: 831-1.0080595-rdf-ntriples.txt
Original Record: 831-1.0080595-source.json
Full Text
831-1.0080595-fulltext.txt
Citation
831-1.0080595.ris

Full Text

CHARACTERIZATION OP TRANSFORMATIONS PRESERVING RANK TWO TENSORS OF A TENSOR PRODUCT SPACE by CAROLYN FAY MOORE B.Sc.j U n i v e r s i t y of B r i t i s h Columbia, 1964 A THESIS SUBMITTED IN PARTIAL FULFILMENT OF THE REQUIREMENTS FOR THE DEGREE OF MASTER OF ARTS i n the Department of Mathematics We accept t h i s t h e s i s as conforming t o the r e q u i r e d standard THE UNIVERSITY OF BRITISH COLUMBIA September, 1966 In presenting t h i s thesis i n p a r t i a l f u l f i l m e n t of the requirements f o r an advanced degree at the University of B r i t i s h Columbia a I agree that the Library s h a l l make i t f r e e l y available for reference and study,, 1 further agree that permission f o r extensive copying of t h i s thesis f o r scholarly purposes may be granted by the Head of my Department or by his representatives. I t i s understood that copying or publication of t h i s thesis f o r f i n a n c i a l gain s h a l l not be allowed without my written permission. Department of MATTTCMATTHS The University of B r i t i s h Columbia Vancouver 8 , Canada ABSTRACT Let U ® V be a tensor product space over an a l g e b r a i c a l l y closed f i e l d P ; l e t dim U = m and dim V = n ; l e t T be a l i n e a r t r a n s f o r m a t i o n on U ® V such t h a t T preserves rank two tensors. We show that T preserves rank one tensors and t h i s enables us t o c h a r a c t e r i z e T f o r a l l values of m and n . i i i . TABLE OP CONTENTS page CHAPTER ONE 1 1. " I n t r o d u c t i o n 1 2. Some P r o p e r t i e s of Rank Two Subspaces " 2 CHAPTER TWO 9 T(R 1) c R-j_ unless m = n = j5 CHAPTER THREE 13 T(R 1) c R 1 when m = n = 3 CHAPTER FOUR l 6 C h a r a c t e r i z a t i o n of Transformations P r e s e r v i n g Rank Two Tensors BIBLIOGRAPHY 17 i v . ACKNOWLEDGEMENT I would l i k e t o thank Dr. Roy Westwick f o r the many hours and v a l u a b l e a s s i s t a n c e he gave t o the w r i t i n g of t h i s t h e s i s . A l s o , t o the U n i v e r s i t y of B r i t i s h Columbia and the N a t i o n a l Research C o u n c i l , I give thanks f o r t h e i r f i n a n c i a l a s s i s t a n c e . 1. CHAPTER ONE 1. INTRODUCTION. Let U and V be m-dimensional and n-dimensional vector spaces over an a l g e b r a i c a l l y closed f i e l d P . The tensor product of U and V denoted by U<8>V i s the dual space of the space of a l l m u l t i l i n e a r functions mapping U x V into F . An element x e U ® V has rank k i f x = S x. ® y. 1=1 1 1 and x^,...,x k are l i n e a r l y independent and y-j^'-'^y^ a r © l i n e a r l y independent. We define R k(U®V) to be •[ x e Ucav rank of x = k ] . I f x e R, (U0V) and x = 2 x. ® y. 1=1 1 1 then, by d e f i n i t i o n , U(x) = <x1,...,xfc> and V(x) = <y-L,...,yk> . U(x) and V(x) are w e l l defined by Lemma 1.2 of [ 1 ] . * I f k x = E x.<g>y. and x-,j...x, are l i n e a r l y independent, then 1=1 1 1 i k the rank of x = dim <y-^ ,... ,y^> by Lemma 1.1 of [ l ] . The subspaces i n Rg(U®V) are of four types, by Theorem 2 A of [ 1 ] , , Type 1: U(x) i s constant as x ranges of H . (H i s a subspace i n R 2(U®V)) * Numbers, i n square brackets r e f e r to the bibliography. 2. Type 2: V(x) i s constant as x ranges over H . Type 3: There e x i s t s u e U and v € V such t h a t each element of the subspace has a r e p r e s e n t -a t i o n of the form x ® u + v ® y where x e U and y e V . Type 4: Those subspaces which are r e f e r r e d t o as " s p e c i a l type 3 subspaces" i n [1]. In Lemma 1.1, we show t h a t e very type 4 space has a b a s i s of the form: u ® x^ + y^<g> v u # x 2 + y 2 ® v y±® x 2 - y 2 ® x 2 where dim <u,y 1,y 2> = 3 and dim<v,x-pX 2> = 3 . The maximum dimensions of type 1, type 2 and type 3 spaces are m-1, n - l and the minimum of \ m-1, n - l } r e s p e c t i v e l y ; by Theorem 2.5 i n [1]. These maximum dimens-i o n s can always be achieved. Except f o r a p a i r o f subspaces of types 3 and 4, the i n t e r s e c t i o n of two d i f f e r e n t types of subspaces i s a t most one-dimensional, by Theorem 2.6 i n [1]. 2. SOME PROPERTIES OP RANK TWO SUBSPACES Lemma 1.1; Let H be a rank 2 subspace of dimension t h r e e . A/ I f H i s not of types 1, 2, or 3 then H has a a b a s i s o f the form: 3 . X-^  = u <£>x-^ + y^ <g> v X 2 = u a Xg + y 2 ® v = y-j® x 2 - y 2 ® x 1 , and u , y 1 5 y 2 are l i n e a r l y independent and v,x-^,x2 are l i n e a r l y independent. B/ I f H i s a type 4 subspace and X-^  e H, X 2 e H then H = <X-L,X2,X^> . Proof: (A) From the proof of Theorem 2.4 i n [ 1 ] , we can w r i t e H = <X1,Y-^,Z1> where X-^  = u<g> x^ + y-^o v Y-j^  = u<a x 2 + y 2 ® v Z l = zl<g> z 2 + 2 3 ® z 4 xx9Y^»y2 a r e l i n e a r l y independent and v,x-^,x2 are l i n e a r l y independent. Now, u i U ( Z 1 ) . I f u e U ^ ) then Z^ = ug>z| + z 2 ® z ^ . This i m p l i e s t h a t z^ € <x1,v> D <x2,v> = <v> , by Lemma 2 .1 of [ 1 ] . Then, H i s a type 3 space. This c o n t r a d i c t s the assumption t h a t H i s not of the types 1, 2, or 3. We use, u / U(Z-L) to show tha t U(Z 1) c <u,y^,y2> . F i r s t , we show dim(<Zj L,z^> H <u,y1>) = 1 . Suppose, dim(<z 1,z-j> fl <u,y1>) = 0 . tThen, f o r X-^ + Z-^ t o be rank 2, <x-L,v> = <z 2,z^> . By . Lemma 2 . 3 of [ 1 ] , H i s type 2. This i s a c o n t r a d i c t i o n . Suppose dim(<z 1,z-j> n <u,y^>) = 2 % By Lemma 2 .2 of [ 1 ] , H i s type 1. Therefore, dim(<z1,z-^> 0 <u,y-L> = 1 and " s i m i l a r l y dim(<z 1,z^> n <u,y2>) = 1 . This i m p l i e s , as 4 . u i U(Z 1), that U(Z 1) e <u,y1,y2> and s i m i l a r l y V(Z 1) c <v ,Xj,x 2> .f Now, U(Z^) and <y;j_*y2 > a r e D°th 2-dimensional subspace of <u,y^,y2> and therefore they i n t e r s e c t i n at lea s t one dimension, say <y> . Let ay^ + Py^ = y * Then, we can form X 2 = u ® + y-^® v Y 2 = u ® x 2 + y ® v (Y 2 = dX1 + ^Y±) Z 2 = zj& z 2 + y « . By Lemma 2.1 of [ 1 ] , <z£> = <x£> . Let x£ = x, 2 z 2 = X 2x, x^ = \xlt yj_ = A . ^ , z j = X 2 z^ , z^ = X 2 z J . Let = Y 2 = u ® x + y<g> v Z 3 = X 2 Z 2 = z i ® x + v ® z 5 • Now, z£ e <u,y,yp . Let z£ = au + By + yy^ . We can assume 6 = 0 since Z^ = (au + yy^)# x,+ y®(z^+^x). Let = , Y^ = Y 5 , and Z^ = Z^ - aY 3 = yy|® x + y # z ^ , where z^' = z^ - av . Let z = y^ , z^ = 1/yz^ and we ar r i v e at a basis of the form = u ® x^ + z®v ' Z 5 = z ® x + y® z 4 ( Z 5 = v"" 3^) ' By Lemma 2.1 of [ 1 ] , z^ = \x| f o r some X e P . Therefore, l e t t i n g x4-= w , the basis i s 5. X = ug> w + z <g>v Y = u® x + y«>v Z = z <a x + y & Xw To prove ( A ) , i t remains t o show tha t X = -1 . Consider X + Y + Z = u<8(w+x) + (z+y)®v + z ® x + y<g?Xw Now, w+x,v,w are independent. Therefore, dim <u+z, z+y, Xy-z> = 2 . This i m p l i e s t h a t z+y = ia(Xy-z) f o r some \i e F . • Therefore, X = -1 . (B) To prove B, Me assume X-^,X2 are i n the b a s i s of a type 4 subspace, H , and show tha t H = <X1,X2,X-^> . From the proof of A, H = <X,Y,Z> . Al s o from A, = u®(w+x) + zg>(w+x) - z® w + (z+y)<#v + y® Xw = (u+z)®(w+x) + (z+y)@v + (Xy-z)®w . z = y£ = x , y x x = 1/X 2 z 2 y = dy1 + BV2 W = X-! = XnX Therefore, , Z = z & x - y © w = X-jy-j® X ^ Z g - = y,® ( X 1 x J 1 z 2 Therefore, H has an element of the form, y 2 0 x 1 . To show w1 = x 2 ' f i r s t consider 6. S = a ( u 8 x 2 + y 2 ® v ) + P(y1<s w - y ^ X - J = au<ax 2 + Py-^w' + y 2 ® ( ^ v " Now, S e H . Therefore, S i s rank 2. This i m p l i e s , s i n c e u , y 1 , y 2 are l i n e a r l y independent and x 2 , x l s v are l i n e a r l y independent; t h a t , f o r every bT,P ^ 0 , there e x i s t s Y,Y' such t h a t w' = yxg+y* (av - px-^). Obviously Y' = 0 and iv * = 6 x 2 f o r some 6 e F . To show 6 = 1 , l e t Xi. = y^® 6 x 2 - y 2 ® x 1 . Now l e t VI - y^Q x 2 -5 _ 1 y 2 ® x • U s e x i + X 2 + X 3 * a s w e u s e d X + Y + Z i n the proof of A , t o show -6~ 1= -1. Therefore, 6=1 and H — X^-^ ^ X^^ X^^1 • C o r o l l a r y 1.1: D i s t i n c t spaces of type 4 i n t e r s e c t i n at most one dimension. Lemma 1. 2; Let V"1 = <{u'<£>x1 + v'®y 1} i = 1, ...,m-l> be a type 1 subspace. Let V"2 = <{u"g)z i + v"® w i 3 i = l,...,m-l> be a type 1 subspace. I f dim<V 1 n V"2> >. 1 then <u»,V> = <u",v"> . Proof: Assume without l o s s of g e n e r a l i t y t h a t u'(»x 1 + v ' ^ y ^ = u"® + v" ® ' This means that u» <g> x 1 + v'$ y 1 - u"<& - v'^w-j^ = 0 . I f dim<u' ,v' ,u",v"> = 4 then x 1 = y-^  = = = 0 which i s a c o n t r a d i c t i o n . . Suppose dim<u!,v»,u",v"> = 3. We may assume t h a t v" = a u 1 + 8v' + Y u" • Therefore u , S ) ( x 1 - aw 1) + v « ® ( y 1 - Bw 1) + u " ^ ) ( - z 1 - y ^ ) = 0 7. This i m p l i e s x 1 = aw 1,y 1 = pw, , and t h e r e f o r e x and y.^  are dependent. This i s a c o n t r a d i c t i o n . Therefore, dim<u»,v',u",v"> = 2 o r , i n other words, <u',v'> = <u",v"> . The next Lemma i s analogous t o Lemma 1 .2 f o r type 2 subspaces. Lemma 1 . 3 : Let = <{xi(g)u» + y i ® v ! } I = l , . . . , n - l > be a type 2 subspace. Let V"2 = <{z^®u" + w^C^v"} i = 1, ...,n-l> be a type 2 subspace. I f dirrKv-^ n v 2> >. 1 then <u',V> = <u",v"> . Lemma 1 . 4 : Suppose n = 4 and m 4 . Let X = <(u®x i + yi(g>v} i = 1 ,...,min(m-l,n-l)> be a type 3 subspace. Let Y = <{u'(g)z i + w ^ v ' } i = 1,... ,min(m-l,n-l)> be a type 3 subspace. I f dim<X n Y> > 2 then <u> = <u'> and <v> = <v'> . Proof: Suppose <u> ^ <u'> . without l o s s of g e n e r a l i t y , assume X 1 = u(g>x1 + y-jg>v = u'® z 1 + w ^ v ' -X 2 = u ® x 2 + y 2 ® v = u'® z 2 + w2(g)v« Then <u,y1> = <u',w-L> = <u,u»> and <u,y2> = <u»,w2> = <u,u'> . Let y 1 = au + 8u« and y = a'u + B'u' . I t i s e s s e n t i a l t h a t 3 £ 0, B' £ 0 ; , otherwise, X 1 and X 2 are rank* one. Consider, B'B""-1^ - X 2 = B , B ~ 1 ( u ® x 1 + y 1 © v ) - ( u © x 2 + y 2 ® v ) = uS>(B'B" 1x 1 - x 2) + { p ' p _ 1 ( a u + Bu') -a'u - Bu» } ®v = u s ^ B ' B " 3 ^ - x 2 ) + ( B ' B ^ a - a') u $N This i s rank 1 which c o n t r a d i c t s the assumption t h a t and X 2 form a type 3 subspace. Therefore <u> = <u'> . S i m i l a r l y <v> = <v'> . CHAPTER TWO In t h i s chapter, we assume T i s a l i n e a r t r a n s -f o r m a t i o n and T(R 2) c R 2 . We show that T(R 1) c R 1 f o r a l l cases except m = n = 3 . The l a t t e r case i s d e a l t w i t h i n the next chapter. Lemma 2.1; (a) I f dim V > 4 then, f o r a l l u € U, v e V, T(ug>v) has rank _< 2 . (b) I f dim U _> 4 then, f o r a l l u e U, v € V, T(u®v) has rank _< 2. Proof: Assume dim V _> 4. Let u ® v , be any rank 1 tensor. We can express u ® v as u a ( a ' x 1 - x 2 ) where dim( <x^,x2>) = and a' 0, a' ^  1 . Extend x ^ , x 2 t o a set of f o u r inde-pendent v e c t o r s x-^,x 2,x-j,x^ . Consider the f o l l o w i n g two spaces: S-^  = <u<Sx1 + v®x^, u^(x- L+x 2) + v&x ^ * u <g>x^  + v(g>x1> and S 2 = < u ® x 2 + v&a'x^, u®(x 1+x 2) «*» v®x-j, u ®x^ + v®x.j> . Any l i n e a r combination of tensors i n S-^  i s rank two. Consider X = O ^ U & X - L + v ^ x ^ ) + p(u®(x 1+x 2) = v@x^) + y(u®x^ + v ® x 1 ) . = u(g)(ax 1 + 6 x 1 + 8 x 2 + yXj) + v®(ax l f + px-^ + yx^) I f X i s rank 1' or 0, then e i t h e r ct = B = Y = 0 or <ax-^  + px-j^ + B x 2 + yXy> = <ax^ + Bx^ + yXj> . The l a t t e r i m p l i e s a = 0 s i n c e x 1,x 2,x-^,x i + are l i n e a r l y independent and axjj occurs only on the ri g h t h a n d s i d e . T h i s i m p l i e s 6 = Y = 0 f o r s i m i l a r reasons. Therefore S-, and s i m i l a r l y 10. S 2 are rank 2 subspaces. Extend S-^  and S 2 to (m-l)-dlmensIonal rank 2 subspaces. Now, T maps S and S 2 i n t o (m-1)-dimensional subspaces. Also, dim(S 1 n S 2) j> 2. Therefore, T maps S-^  and S 2 into subspaces of the same type. (When dim V=4, S 1 and S 2 cannot be mapped into type 3 and type 4 subspaces. This i s proven at the end.) Now, T(u(g>(a'x1 - x 2 ) ) = T(a ,(u(g>x 1 + v ^ x ^ ) -( u 0 x 2 + V(g)a'x^)). By Lemmas 1.2, 1.3 and 1.4 we know T(u(g)(a'x^ - x 2 ) ) can have rank no greater than two. It remains to show that i f dim V=4, and S 2 cannot be mapped into type 4 and type 3 spaces. Suppose T(S 1) i s a type 3 subspace and T(S 2) i s a type 4 subspace. Then, T(u(g>(x 1+x 2) + v(g>x^) = u'&xj^ + x 2 & v « T(u<g>x^ + v<g)X1) = u'®y«^ + y 2 © v ' T ( u g ) x 1 + ,vg>xi|) = u'(g>z^ + z 2 ® v » By Theorem 1.1, there ex i s t s a ' , 6 j Y such that X = T(a« ( u ® ( x 1 + x 2 ) + v ® x 5 ) + B ( u ® x 5 + v ® x 1 ) , + Y ( u ® x 2 + v ^ a x ^ ) ) = x2<g)y^ - y£<2>x-[ • Obviously, y ^ 0 . Let 'X' = T(a'(u®(x 1+x 2) + v ^ x ^ ) + p( u ® x ^ + v ^ ) + Y( u<& x! + v<£>x^)) . = u'<g>(a'x.j_ + By£ + Y z | ) + (ax 2 + By 2 + yz^)®v* .'Consider, X« - X = yT(u£)(x 1 - x 2 ) + £V®(l-a)x^) • = u'<&(a'x^ + ByJ|_ + yz^) + (ax 2+By 2+yz 2)&v 1  J^ 1 - x 2 &> yj^ + y 2 ® x£ . 11. Since ^ / 0, X 1 - X i s a rank 4 tensor. Therefore, T maps a rank two tensor i n t o a rank four tensor. This i s a c o n t r a d i c t i o n . By a s i m i l a r p r o o f , T(S-^) i s a type 4 subspace i m p l i e s T ( S 2 ) cannot be a type 3 subspace. By a p p l i c a t i o n of Lemma 2.1, we have proved the f o l l o w i n g Lemma. Lemma 2.2: T maps rank 1 tensor i n t o tensors of rank _< 2. Theorem 1: Except p o s s i b l y when m = n = 3> T(R^) c R-^  . Proo f : From Lemma 2.2, T(R 1) c {0} IJ R 1 u R 2 . Now, i f T(x®y) = 0 then; i f m > 1, n > 1; there i s a rank 1 tensor mapped i n t o a rank 2 tensor. Therefore, i t i s s u f f i c i e n t t o show that no rank 1 tensor can be mapped i n t o a rank 2 tensor. Suppose T ( u i ® v m ) i s r a n k 2. Extend to a b a s i s of U ; say, (u^, ...,u ); and extend v m t o a b a s i s of V; say, ( v 2 / ' * * 3 Y n ) ' Consider the space S = <Sn,...,S > where 1 m 5 1 = U l ^ v m 5 2 = u 1 © v 1 + u 2 ® v m = ul(g> v 2 + u 2 ® v 1 Sm = u l ® v m - l + u2&>vm-2 • T(S) i s a rank two, m-dimensional subspace. This i s e s t a b l i s h e d 12. i f every l i n e a r combination. ctSn + ouS 0 + a-,S, +. ..+a n 8 , ^ ? 1 l j> p m-1 m = 0 . Consider i s rank two unless ct-^  = a 2 = a m - l m-1 X = a u 1 ® v m + a 1 ( u 1 ( g ) v 1 + u 2 ® v m ) + 5 ^ a ± (u ],®v i + u 2 ® v i - l ) m-1 m-1, , = u l (g>(av m +z a j_v i) + U g o M a ^ + E * o ^ v ^ ) I f ,X i s not rank 2, then m-1 m-1 < a vm + ± l ± a i V = < a l v m + . ^ 2 a i v i - l > ' Now, v m 2_ does not appear on the righ t h a n d s i d e . This i m p l i e s c t m _ 1 = 0 . This means v m _ 2 does not appear on the r i g h t h a n d side and a 2 = 0 • By t h i s method i t i s shown tha t a-, = ct 0 = ...= a n = 0, . Therefore, no l i n e a r 1 d m-1 combination of si***«-» s m i s r a j i k 1 except a ( u 1 ( g ) v m ) , a € F . Since by assumption T ( u ] _ ® v m ) h a s rank 2, T(S) i s a rank two m-dimensional subspace. Unless m = 3, there are no such subspaces. This i s a c o n t r a d i c t i o n . S i m i l a r l y , consider the subspace, S S spanned by ul<S>vm u 2<5D vl + ul<S>vm - ' U n ® V l + U n - l ® v m - .',. 1 - ' T(S') i s a n-dimensional rank 2 subspace. This i s a.j.-' ' c o n t r a d i c t i o n u nless n = 3 • ^  Therefore T maps no rank 1% tensor i n t o a rank 2 t e n s o r , w i t h the p o s s i b l e e x c e p t i o n when n = m = 3 . 13. CHAPTER THREE In t h i s chapter we assume m = n = 3 and show i n t h i s case also that T(R 2) c R 2 implies T(R 1) c . Lemma 3 . 1 : I f m = n = 3 then T(R 2) c R 2 implies T(R]_) c R U R 3 . Proof: F i r s t , we show that no rank 1 tensor i s mapped into 0 . Suppose T(u(£>v) = 0 . Extend u and v to bases of U and V re s p e c t i v e l y ; say, U = <u,x2,y2> and "V = <v,x^,y^> . Choose any a fi 0 and consider the family of subspaces, S(a), with the fol l o w i n g b a s i s : + x 2 ® a x 1 y 2 ® y l -+ x 2 ® v  x 2 ® y l ' T(S(a)) i s a 3-dimensional, rank 2 subspace i f every tensor i n S(a) and not i n <x 2 © y ^ > has rank 2. Suppose a ' ( u ® y 1 + XgC&ax^ + ^{y2®^i + x 2 ® v ) + y'(x2®yl^ ' 'A^f = (a'u + p»y 2 + v lx 2)<g>y 1 + x2g>(a'ax-L + 0'v) i s not rank 2. Then, a 1 = 8 ' = 0 and we have v ^ x ^ y - ^ . But ' T ( y ' x 2 ® y 1 ) i s rank two as T ( u ® v ) = 0 . Therefore, every l i n e a r combination of tensors i n the basis of S(a) i s mapped into a rank 2 tensor. This implies T(S(a)) i s a type 4 subspace. Since a l l S(a) int e r s e c t i n 2 dimens-ions, T(S(a)) i s the same space for every a fi 0 . Choose a fi 1 . Now T ( S ( l ) ) c T ( 8 ( a ) ) . This i m p l i e s , f o r some a,b,c, e F , T ( u ® y 1 + x2^)x1) = T((au + b y 2 + cx 2)<&y 1 + x 2 g ) ( a a x 1 + bv)) This i m p l i e s , T ( { ( a - l ) u + b y 2 + cx2)Qy1 + x2Q{{axx-l)x1 + bv]) = 0 Therefore, { ( a - l ) u -I- b y p + cx0}(R>y-j + x 2 @ { ( a a - l ) x 1 + bv] i s not rank 2. Therefore, one of the f o l l o w i n g three cases must h o l d : Case 1: ( a a - l ) x ^ +bv = 0. This i m p l i e s aa = 1, b = 0 and T ( { ( a " 1 - l ) u + c x ^ s ^ ) = 0 Case 2: ( a - l ) u + b y 2 + c x 2 = 0. This i m p l i e s a = 1, b = 0, c = 0 and T(x 2 ( x > ( a - l ) x 1 ) = 0 . Case 3: <x2> = <(a-l)u + b y 2 + cx 2>. This i m p l i e s a = 1, b = 0 and T(x 2®(cy 1 f ( a - l ) x ] L ) ) = 0. Now, T(u®v) = 0 . Therefore, Case 1 must h o l d w i t h c = 0 and T(u<&y^) = 0 . Since the ex t e n s i o n of v to a b a s i s v,x^,y^ i s a r b i t r a r y , we h'ave t h a t T ( u & y ) = 0 f o r a l l y € V. Now, the problem i s symmetric w i t h respect t o u and v. Therefore, T(x®v) = 0 f o r a l l x € U . Now choose y independent of v and x independent of u and we have a c o n t r a d i c t i o n ; namely, T(u®y + x & v ) = 0 . This shows that no rank one vec t o r i s mapped i n t o 0 . Suppose a rank one t e n s o r , c a l l i t XgQcy-^ , i s -| r mapped i n t o a rank 2 tensor. Extend x 2 and to bases of U and V . Let U = <u,x 2,y 2> and V = <v,x-L,y1> . Then, by c o n s i d e r i n g the spaces, s(a) as defined above, we a r r i v e at cases 1,2 or 3 as above. Therefore, we have the contra-d i c t i o n t h a t a rank 1 v e c t o r i s mapped i n t o 0. Therefore T ^ ) c R 2 U • Lemma 3. 2: Let m = n = 3 . I f T ( R 2 ) c R 2 then T(R 1) c R 1 or T ( R 1 ) c R^ . Proof: Assume T ( R 1 ) fi. R 1 and T ( R 1 ) fi R^ - Then, T(R 1) c R 1 U R^ by Lemma 3.1. Now, we can f i n d x ® y and x'<S>y' such t h a t x,x' are l i n e a r l y independent; y,y' are l i n e a r l y independent; T ( x & y ) i s rank 1 and T(x r®y') i s rank 3. I f t h i s i s not the case then T(R 1) c R^ or T(R^) c R^ and we are f i n i s h e d . Let T(x®y) = and T(x,<g>y«) = x1^z1 + x 2 5 o z 2 + x ^ g i z ^ . Consider, T(ax®y + x«®y») = x 1(g>(ay ] L + z x ) + x 2 & z 2 + X J < $ Z J . This must be rank 2, f o r a l l a fi 0 . Therefore, there e x i s t 8 , Y e P such "Chat ay-^ + z^ = P z 2 + y z - j °r ay-j_ = P z 2 •+ Y z^ ~ z j _ • Let a 1 € F , a 2 e F and a-^  fi a 2 . There e x i s t 8-^,Y-J_,32,Y2 € P such t h a t °l yl = P1 Z2 + V 5 " 2 i a 2 y l = $2Z2 + ^2Z"*> " z l This i m p l i e s I ^ l a l Z2 + Y l a l " Z v i ~ a l z l = $2° 2 Z 2 f Y 2 a P z 3 ~ a 2 z ] ' Nov, d i u ( < z 1 , 2 2 , z ^ > ) = 3. Therefore, a ^ 1 = a^ 1 . This i s a c o n t r a d i c t i o n . Therefore, T ( R ] _ ) c R i o r T ( R ] _ ) c 1 Theorem 2: Let m = n = 3 . T(R 2) c R 2 i m p l i e s T(R X) c R 1 . Proof: From Lemma 3 -2, i t i s s u f f i c i e n t t o show t h a t T(R 1) jL . Assume T(R ] L) c R^ . Let T(u 1(g)V 1) = u 1 g , y 1 + u ? ^ ) y 2 + u^foy^ and T ( u 1 0 v 2 ) = u 1(g,z 1 i u2(£)z2 + u^0z^ where v-^ v 2 are l i n e a r l y independent. Let A : V - V such that Ay^ = z. i = 1,2,3. A has an eigenvalue, x . Then T ( u 1 ® ( v 2 - = u x ® ( A - \ l ) y 1 + u 2 ® ( A - \ l ) y 2 + u_, (A - Xl)y-j; • There e x i s t s a,B,y not a l l 0 , such t h a t ay-^ + By 2 + yy-j i s the eigenvector corresponding t o X. Therefore, (A - X l ) ( a y 1 + By 2 + yy^) = 0 and (A - X l ) y 1 , (A - U ) y 2 . (A - Xl)y-^ are dependent. This means T(u-^<g>(v2 - Xv-^)) i s not rank. 3 which c o n t r a d i c t s the assumption t h a t T(R^) Therefore, T(R n ) c R-, . 1 7 . CHAPTER FOUR Theorem 3 : I f F i s a l g e b r a i c a l l y c l osed and T(R 2) c R 2 then T(R 1) c R 1 . Proof: The r e s u l t f o l l o w s immediately from Theorems 1 and 2. For a l g e b r a i c a l l y c l osed f i e l d s of c h a r a c t e r i s t i c 0 , the s t r u c t u r e of T i s given by the f o l l o w i n g theorem, which i s quoted from [ 2 ] . Theorem 4: Let T(R 1) c R-j_ . "Let T^ be the l i n e a r t r a n s f o r m a t i o n of V&U i n t o U&V xvhich maps y® x onto x ® y . I f m = n , l e t cp be any non-singular l i n e a r t r a n s f o r m a t i o n of u onto V . Then i f m fi n , there e x i s t n o n - s i n g u l a r l i n e a r t r a n s f o r m a t i o n s A and B on U and V , r e s p e c t i v e l y , such t h a t T = A ® B . I f m = n, there e x i s t non-singular A and B such t h a t e i t h e r T = A® B or T = T 1 (cpA$>rp~ 1 B ) . " For a l g e b r a i c a l l y c l o s e d f i e l d s of a l l c h a r a c t e r -i s t i c s , Theorem 4 h o l d s ; but the proof i s , as y e t , unpublished. 13. BIBLIOGRAPHY G. F. Iwatai C h a r a c t e r i z a t i o n of Rank Two Subspaces of a Tensor Product Space, U n i v e r s i t y of B r i t i s h Columbia, 1955". Marvin Marcus and B. N. Moyls, Transformations on T e n s o r P r o d u c t Spaces, P a c i f i c J o u r n a l of Mathematics, V o l . 9/No." C, fi959)j 1215-1221. 

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.831.1-0080595/manifest

Comment

Related Items