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) = and V(x) = . U(x) and V(x) are w e l l defined by Lemma 1.2 of [ 1 ] . * I f k x = E x.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 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^ v u # x 2 + y 2 ® v y±® x 2 - y 2 ® x 2 where dim = 3 and dim = 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^ 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 = . Proof: (A) From the proof of Theorem 2.4 i n [ 1 ] , we can w r i t e H = where X-^ = u x^ + y-^o v Y-j^ = u 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^ € D = , 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 . F i r s t , we show dim( H ) = 1 . Suppose, dim( fl ) = 0 . tThen, f o r X-^ + Z-^ t o be rank 2, = . 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( n ) = 2 % By Lemma 2 .2 of [ 1 ] , H i s type 1. Therefore, dim( 0 = 1 and " s i m i l a r l y dim( n ) = 1 . This i m p l i e s , as 4 . u i U(Z 1), that U(Z 1) e and s i m i l a r l y V(Z 1) c .f Now, U(Z^) and a r e D°th 2-dimensional subspace of and therefore they i n t e r s e c t i n at lea s t one dimension, say . 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 ] , = . 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 v Z 3 = X 2 Z 2 = z i ® x + v ® z 5 • Now, z£ e w + z v Y = u® x + y«>v Z = 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 = . From the proof of A, H = . 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(y1x1 + 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 >. 1 then = . 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» x 1 + v'$ y 1 - u"<& - v'^w-j^ = 0 . I f dim = 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 = 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 = 2 o r , i n other words, = . 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 = . 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 > 2 then = and = . Proof: Suppose ^ . 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 = = and = = . 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 = . S i m i l a r l y = . 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( ) = 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-^ = 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 = . 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(ux^ + vxi|) = 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 ^ ) ) = x2x-[ • Obviously, y ^ 0 . Let 'X' = T(a'(u®(x 1+x 2) + v ^ x ^ ) + p( u ® x ^ + v ^ ) + Y( u<& x! + v<£>x^)) . = u'(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 = 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 ulvm u 2<5D vl + ulvm - ' 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 = and "V = . 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 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)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: = <(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 = and V = . 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'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,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 = $2° 2 Z 2 f Y 2 a P z 3 ~ a 2 z ] ' Nov, d i u ( < z 1 , 2 2 , z ^ > ) = 3. Therefore, a ^ 1 = a^ 1 . This i s a c o n t r a d i c t i o n . Therefore, T ( R ] _ ) c R i o r T ( R ] _ ) c 1 Theorem 2: Let m = n = 3 . T(R 2) c R 2 i m p l i e s T(R X) c R 1 . Proof: From Lemma 3 -2, i t i s s u f f i c i e n t t o show t h a t T(R 1) jL . Assume T(R ] L) c R^ . Let T(u 1(g)V 1) = u 1 g , y 1 + u ? ^ ) y 2 + u^foy^ and T ( u 1 0 v 2 ) = u 1(g,z 1 i u2(£)z2 + u^0z^ where v-^ v 2 are l i n e a r l y independent. Let A : V - V such that Ay^ = z. i = 1,2,3. A has an eigenvalue, x . Then T ( u 1 ® ( v 2 - = u x ® ( A - \ l ) y 1 + u 2 ® ( A - \ l ) y 2 + u_, (A - Xl)y-j; • There e x i s t s a,B,y not a l l 0 , such t h a t ay-^ + By 2 + yy-j i s the eigenvector corresponding t o X. Therefore, (A - X l ) ( a y 1 + By 2 + yy^) = 0 and (A - X l ) y 1 , (A - U ) y 2 . (A - Xl)y-^ are dependent. This means T(u-^(v2 - Xv-^)) i s not rank. 3 which c o n t r a d i c t s the assumption t h a t T(R^) Therefore, T(R n ) c R-, . 1 7 . CHAPTER FOUR Theorem 3 : I f F i s a l g e b r a i c a l l y c l osed and T(R 2) c R 2 then T(R 1) c R 1 . Proof: The r e s u l t f o l l o w s immediately from Theorems 1 and 2. For a l g e b r a i c a l l y c l osed f i e l d s of c h a r a c t e r i s t i c 0 , the s t r u c t u r e of T i s given by the f o l l o w i n g theorem, which i s quoted from [ 2 ] . Theorem 4: Let T(R 1) c R-j_ . "Let T^ be the l i n e a r t r a n s f o r m a t i o n of V&U i n t o U&V xvhich maps y® x onto x ® y . I f m = n , l e t cp be any non-singular l i n e a r t r a n s f o r m a t i o n of u onto V . Then i f m fi n , there e x i s t n o n - s i n g u l a r l i n e a r t r a n s f o r m a t i o n s A and B on U and V , r e s p e c t i v e l y , such t h a t T = A ® B . I f m = n, there e x i s t non-singular A and B such t h a t e i t h e r T = A® B or T = T 1 (cpA$>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.