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Rank preservers on certain symmetry classes of tensors Lim, Ming-Huat 1971

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RANK PRESERVERS ON CERTAIN SYMMETRY CLASSES OF TENSORS by MING-HUAT LIM B.Sc, Nanyang University, Singapore, 1965 A THESIS SUBMITTED IN PARTIAL FULFILMENT OF THE REQUIREMENTS FOR THE DEGREE OF DOCTOR OF PHILOSOPHY in the Department of MATHEMATICS We accept this thesis as conforming to the required standard. THE UNIVERSITY OF BRITISH COLUMBIA July 1971 In present ing t h i s thes is in p a r t i a l fu l f i lment o f the requirements fo r an advanced degree at the Un ivers i ty of B r i t i s h Columbia, I agree that the L ibrary sha l l make i t f r e e l y ava i l ab le for reference and study. I fu r ther agree tha permission for extensive copying of th is thes is fo r s c h o l a r l y purposes may be granted by the Head of my Department or by h is representa t ives . It is understood that copying or p u b l i c a t i o n of th is thes is f o r f i n a n c i a l gain sha l l not be allowed without my wr i t ten permiss ion. Department of The Un ivers i ty of B r i t i s h Columbia Vancouver 8, Canada i i . Supervisor: R. Westwick ABSTRACT Let U denote a finite dimensional vector space over an alge-braically closed field F . In this thesis, we are concerned with rank one th r preservers on the r symmetric product spaces VTJ and rank k preser-2 vers on the 2nd Grassmann product spaces AU The main results are as follows: r r (i) Let T : VU -> VU be a rank one preserver. (a) If dim U > r + 1 , then T is induced by a non-singular linear transformation on U . (This was proved by L.J. Cummings in his Ph.D. Thesis under the assumption that dim U > r + 1 and the characteristic of F is zero or greater than r .) (b) If 2 < dim U < r + 1 and the characteristic of F is zero or greater than r , then either T is induced by a non-singular r r linear transformation on U or T(VU) = VW for some two dimensional sub-space W . of U r s (ii) Let T : VU •+ VU be a rank one preserver where r < s If dim U > s + 1 and the characteristic of F is zero or greater than g — , then T is induced by s - r non-zero vectors of U and a non-singu-lar linear transformation on U i i i . 2 2 (i i i ) Let T : AU -»• AU be a rank k preserver and char F ={= 2 . If T is non-singular or dim U = 2k or k = 2 , then T is a compound, except when dim U = 4 , in which case T may be the composite of a compound and a linear transformation induced by a correlation of the two dimensional subspaces of U i v . TABLE OF CONTENTS PAGE INTRODUCTION 1 CHAPTER I. RANK k VECTORS IN SYMMETRY CLASSES 4 1. Definitions and Remarks 4 2. Properties of Rank k Vectors 8 3. Applications 22 CHAPTER II. RANK ONE PRESERVERS ON SYMMETRIC SPACES 26 1. Maximal Pure Subspaces of Symmetric Spaces 26 2. Intersections of Maximal Pure Subspaces 38 r s 3. Rank One Preservers from VU- to VU , r < s 53 r 4. Rank One Preservers on VU 70 CHAPTER III. RANK k PRESERVERS ON GRASSMANN SPACES 92 BIBLIOGRAPHY 105 V . ACKNOWLEDGEMENTS I am indebted to my supervisor, Professor R. Westwick, for his generous and valuable assistance during the research and writing of this thesis. I would also like to thank Professors B.N. Moyls and G. Maxwell for their helpful suggestions and criticisms. I am grateful to the University of British Columbia and the National Research Council of Canada for their financial assistance. Last, but not least, I wish to thank Miss Eve Hamilton for typing this thesis. INTRODUCTION Let U be a finite dimensional vector space over an arbitrary field F . Let G be a subgroup of the symmetric group and x be a character of degree one on G . Denote by U^G) the X symmetry class of tensors over U associated with G and x • A. non-zero element of U^G) is said to have rank k i f i t can be expressed X as the sum of k but not less than k non-zero decomposable tensors (pure products) in Um(G) . A linear transformation of U^ CG) is said . X X to be a rank k preserver i f i t maps the set of a l l rank k vectors into itself. Recently, there have been investigations concerning the structure of rank k preservers on classical symmetry classes of tensors by Beasley [1], Cummings [4], Djokovic [6], Marcus and Moyls [12], Moore [16] and Westwick [18;19;20]. The purpose of this thesis is to continue this investigation: th mainly on rank one preservers on r symmetric product spaces and partly on rank k preservers on 2nd Grassmann product spaces. We begin by studying some basic properties of rank k vectors in general symmetry classes of tensors (Chapter I); Most of the theorems are generalizations of well-known results about the classical spaces. For i example, we show that (i) the rank of a vector in U^G) is unchanged X in 11 i f we extend U ; (ii) for each rank k vector in U (G) and each orbit X of G , we associate a unique subspace of U ; ( i i i ) z,.* •••*z, + ••• + z, * • ••*z, is of rank k i f for each 11 1m kl Km orbit 0 of G , the dimension of the subspace spanned by the vectors z..^  where d e 0 , j = 1 , • • •, k is k.10j where j0J denotes the cardinality of 0 . From our results on rank k vectors, we obtain an application on intersections of symmetry classes of tensors and an application on equalities of two associated transformations (induced transformations). In Chapter II, we consider rank one preservers on the r*"*1 r symmetric product spaces, VU , where U is a finite dimensional vector space over an algebraically closed field F . We first classify the r maximal pure subspaces of VU and study their intersection properties. We are able to determine the structure of an infinite family of certain maximal pure subspaces such that any two of them have a non-zero intersection. With the help of the results on maximal pure subspaces, we prove the following main theorems of this Chapter. (i) If dim U ^  r + 1 , then a rank one preserver on VU is induced by a non-singular transformation on U . (ii) Let T be a rank one preserver on VU . Let 2 < dim U < r + 1 and the characteristic of F be zero or greater than r . Then either r r T is induced by a non-singular transformation on U or T(VU) •= VW for some two dimensional subspace W of U . r s ( i i i ) Let T be a rank one preserver from VU to VU where r < s . Let dim U >_ s + 1 and the characteristic of F be zero or greater than g — . Then T is induced by s - r vectors z, , ••• , z of U and a r J 1 s-r non-singular transformation f on U in the sense that T(xn x ) = z z -fCx.) -f(x ) , x e U , i = 1 , • • • , r . L r l s—r . i r i (i) and (ii) partially answer a question raised by Marcus and , 3. Newman [14, p. 62]. (i) was first proved by Cummings [4] under the assumption that dim U > r + 1 and the characteristic of F is not a prime p <^  r . An example is given to show that there is another type of rank one preserver from $XS to tHj , r < s , i f F is of prime characteristic p such that rp <_ s . I n Chapter III, we consider rank k preservers on 2nd Grassmann 2 product spaces, Au , where U is an n-dimensional vector space over an algebraically closed field F with characteristic not equal to two. We 2 show that i f T is a rank k preserver on AfJ and either (i) T is non-singular; or (ii) n = 2k ; or ( i i i ) k = 2 ; then T is a compound of a non-singular transformation on U , except when n = 4 , in which case T may be the composite of a compound and a linear transformation induced by a correlation of the two dimensional subspaces of U . These results lead to corresponding theorems on rank 2k preservers on the space of a l l n-square skew-symmetric matrices over F . The result on rank two preservers 2 on /^ U was also obtained independently by M.J.S. Lim [10]. CHAPTER I RANK k VECTORS IN SYMMETRY CLASSES §1. Definitions and Ramarks. Let F be an arbitrary field. Let G be a subgroup of the symmetric group S^ of degree m . Let x be a character of degree * * one on G ; i.e., x : 0 -*• F is a homomorphism where F is the multiplicative group of F . Let V- , V_ , • • • , V be finite dimensional vector spaces I I m over F such that V. = V for i = 1,2, •••,m and a l l a e G . l a(i) * Let W be the cartesian product of the V. , W = V, x ••• x V i . i m 1.1. Definition. Let U be any vector space over F . A multilinear function f : W -> U is said to be symmetric with respect to G and x i f f(X • • ,X " N) = x(a) f(X l S • • • ,X ) for any a e G and arbitrary X^  e V^ , i = 1,2,• • •,m . 1.2. Definition. A pair (p,u) consisting of a vector space P over. F and a multilinear function u : W -»• P , symmetric with respect to G and x > is a symmetry class of tensors over V,,V„,...,V associated — I m  with G and x i f the following universal factorization property is satisfied: For any vector space U over F and any multilinear function f : W -*• U , symmetric with respect to G and x > there exists a unique linear transformation h : P ->• U such that f = hy , i.e., such that the diagram 5. W : > U 7[ commutes. Given a p a i r G and ,x » a symmetry class of tensors over V, , V„ , ••• , V associated with G and x exists and is unique to 1 2 m within canonical isomorphism (see [13], [17]). We shall denote such a space by (V , . • . , V ) (G) . v . . . . . . . . v = v , then A 1 m such a space is usually denoted by Vm(G) (see [13]). The vectors X u(X,,'••• • ,X ) are denoted by X, * • • ' • * X and are called 1 m 1 m decomposable elements (pure products). The notion of a symmetry class of tensors generalizes the classical tensor, Grassmann and symmetric spaces, for an appropriate choice of G and x » i«e»> 1. If G = {e} , where e is the identity permutation in S , X = 1 , then (V,,V-,--- ,V ) (G) is the tensor product of vector spaces 1 2 m x m V, V , ®V. . In this case, the decomposable element X. * X 1 m . , l 1 m 1=1 is denoted by X.® • • • ® X , 1 m 2. If G = S and x = "sign of the permutation" character, m then V^G) is the Grassmann space Av • I n t n i s case, the decomposable element X, X is denoted by X-A- • * • * * Ax I m . l m 6. m 3. If G = S m and x = 1 » then V ^ ( G ) is the symmetric space W . The decomposable element X, * •••* X is denoted by 1 m X, X . 1 m 1.3. Remark. Since u is multilinear and symmetric with respect to G and x » w e have: (i) X, * (aX.+BX'.) * ...*X = aX.*...*x.*."-.*X +BX,*...*Xj*...*X 1 i i m l l m l i m for each X. , Xf. z V. , each ct,t3 e F and each i = 1,2, .. . ,m . (ii) X a ( i ) * '" *Xa(m) = X ^ X l * ' * " *Xm ' f ° r 6 a c h a e G , X. e V. , i = l,...,m . i i 1.4. Remark. It can be shown from the universal factorization property that the decomposable elements span (V,, ••••,V ) (G) . 1 m x 1.5. Remark. Let IL , ••• , U be subspaces of V. , • • • , V 1 m 1 m respectively such that U. = U ... for i = 1,2,...,m and for a l l a e G . l a(i) Let u, be the restriction of the map u to U, x... x u Then 1 1 m it can be shown that (<range u^ > , y^) is a symmetry class of tensors over t, IL, , • • • , U , associated with G and x where < range y. > denotes i m 1 the linear closure of the range of y. . Therefore we identify (U_,•••,U ) (G) 1 1 m x with a subspace of (V,.•••,V ) (G) . 1 m x Let T. : V. V. be linear transformations such that T. = T ... i i i I a(i) for i = l,2,...,m and for a l l a e G . Define a mapping (j) : V x...x V• + (V ,---,V ) (G) by setting <}) ( X X ) = T X * ... *T X . 1 m l mx 1 m i l mm It is easily seen that <{) is multilinear and symmetric with respect to G and x • Hence by the universal factorization property of (V..,«'-,V ) (G) , there exists a unique linear transformation on ^ X '(V,,«*»,V ) (G) , denoted by K(T.,---,T ) , such that 1 m x 1 m K(T,,«",T ) x. *•••* x = T . .X . . *•••* T x . l m l m 11 mm When = ••• = T m = T , we shall denote K(T ,.••,Tm) simply by K(T) 1.6. Definition. The above transformation K(T_, ):(V. ,«",V ) (G)+ 1 m l m x (V„ •••,Y ) (G) is called the associated transformation (induced transformation) 1 m x of T.,---,T . 1 m Our definition of associated transformation generalizes the one in [11,13]. The associated transformations in the classical tensor, ttl Grassmann and symmetric spaces are the m tensor product T ^ ® • • • ^ m •  the m^i compound C ( T ) and the m*"*1 induced power of T , P ( T ) , m m respectively. 1.7. Remark. If S. : V. -*• V. are linear transformations such that : i i i S. = S ... for i = 1 , ••• , m and a e G , then we have K(T.. S-, • • • ,T S ) i a(i) ' ' • ' 1 1 m m = K ( T T ) K ( S S ) 1 m l m 8. 1.8. Definition. A non-zero vector in (V-,«««,V ) (G) is said to 1 m x have rank k i f i t can be written as a sum of k but not less than k non-zero decomposable elements in (V, ,".,V ) (G) . The set of a l l 1 m x rank k vectors in (V...-.,V ) (G) is denoted by R, ((V. ,---,V ) (G)) 1 m Y k. 1 m Y 1.9. Definition. A subspace MC (V, ••••,V ) (G) is called a rank k r — 1 m x subspace i f every non-zero vector in M is of rank k . A rank k subspace M is said to be maximal i f no other rank k subspaces properly contain M . A rank one subspace is usually called a pure subspace. 1.10. Definition. A linear transformation A : (V, .•••,V ) (G)-»-(V,, • • • , V ) 1 m x 1 m x is called a rank k preserver i f A maps the set of rank k vectors into itself. 1.11. Remark. Some knowledge of the structure of rank k subspaces will be useful in characterizing rank k preservers. By definition, i t is evident that a rank k preserver A maps a rank k subspace M into a rank k subspace A(M) . Moreover, A|M is a monomorphism and hence dim M = dim A(M) '. §2. Properties of Rank k Vectors. Throughout this section, let ^ i ' " " * ' ^ m ^ ^ ^ denote a symmetry class of tensors over V-,,«->,V associated with a subgroup G of S 1 m m and a character x o n G • We also let 0^  , ••• , 0fc be a l l the orbits of 9 . 2.1. Theorem. Let x, + • • • + x. = y, + • • »+y e R. ((V., • • • ,V ) (G)) 1 'K 1 q . K l m x where x. = x. *. ..* x. y = y *...* y for each j = 1 , , k 2 j l jm ' -n •'nl n^m and n = 1 , ••• , q . Then for each orbit <X , we have k q Z <x., : d £ 0.> C E <y , : d e 0.> j-1 3 d 1 ""n-l n d where <x, : d e 0.> denotes t h e subspace spanned by the v e c t o r s jd 1 x., , d e 0. . Jd l q Proof: Suppose that for some j » i £ j £ k , < p c . d : d £ C\> <j: E <ynd : d e 0±--1 n=l Then for some s e O . , x. i E < y , : d s O . > . J n=l Consider the associated transformation. K(T.,»««,T ) : (V.,««»,V ) (G) 1 m 1 m x + (V,,---,V ) (G) where T» = T , D s for a l l a e G Z = 1 , ••• , m 1 m x -Co\JL) and T, , ••. , T are defined as follows: 1 m If Z e C\ , T^ : -> is a linear transformation such that ToCx. ) = 0 a n d T„ Z j s Z I <y , : d £ 0.> is the identity mapping. , nd l n=l If Z £ 0 i , T^ : -> is the identity mapping. We have KC^, • • • ,Tffl) (X;L + • .. + x f c) = K(T^, • •-.T^ ) (y 1 + . ., + y ) , 10. Since K(T 1 }...,T m)(y n) = y n for n = 1 , • • • , q and KC^, • . . ,Tm) ) = T - x . , *•••* T x. = 0 , i t follows that 1 j l m jm K ( Tr'--'V ( xi +--- + x j-i +Vi +V-- + xk ) = y i + " - + y q e V(V--'>Vx(G)) ' This is a contradiction since the left hand side is a vector of rank less q than k or the zero vector. Therefore < x, . : d e 0. > C E < y , : d e 0 > id x — .. •'nd l n=l k q for each j = 1 , ••• , k . Hence Z <x : d e O . >C E < y „ : d e 0. > . • i jd l — , Jnd x J=l J n=l 2.2. Corollary. Let x + y = Z where x = x *...^x y = y y a n < j Z = Z.. *•••* z are non-zero decomposable elements of (V, .••••V ) (G) . 1 m r 1' ' m x Then for each orbit 0. , we have < Z : d e 0. > C < x : d e 0.> + < y, : d e 0.> i d l — d I 7 d l This Corollary generalizes a Theorem of Cummings [4, p. 17]. 2.3. Theorem. Let x + .. . + x = y + . . . + y e" R. ((V.,•••,V ) (G) where 1 k 1 k k 1 m x x, = x.. *• • • * x. and y. = y._ *...* y. for each i = 1 , • • • , k . j J l jm 3 J ' j l 'jm k k Then for each orbit 0. , we have £ <x. , : d £ 0. > = Z < y. , : d'e 0. > . i j = 1 Jd x j = 1 'jd Proof: This follows immediately from Theorem 2.1. 2.4. Corollary. Suppose that x *•••* x = y. *...* y e vm(G) and x TO 1 m x x *...* x ^ 0 . Then < x ... -x > = < y, ,..«,y > . 1 TO I ' m ' l '"'m 11. Corollary 2.4 generalizes a lemma of Marcus and Mine [13]. 2.5. Definition. Let x = x, *•••* x be a non-zero decomposable element 1 m of (V., •••,V ) (G) . For each orbit 0. of G , define < 0.(x) > to be I m x l l the subspace spanned by the vectors x, where d e 0. . By Theorem 2.3, d i this definition is well-defined. 2.6. Corollary. Let x 1 + ••• + xfc e R k((V 1, • • • , v m ) x ( G ) where x_. is a decomposable element for each j = 1 , ••• , k . Let x^+j = '*•••* Z^  4 0 . If for some orbit 0. , there is a s e 0. such that Z i < 0.(x.) > +...+< 0.(x.)> l i s T l 1 l k then x^ +• • •+ has rank greater than or equal to k . k+1 k Z x. = -x, ,.. . This implies that k = 1 . J- l J Proof. If Z x. = 0 , then J-l J By Theorem 2.3, we have Z e < 0.(xj > which contradicts the hypothesis. s 1 I k+1 n If Z x. = Z y. e R ((V ,•••,V ) (G)) where | < n < k , then k n k Z x, = Z y. - x, . This implies that n = k - 1 since Z x. • i 1 *. i 3 k+1 r . i 3 3=1 J j=l J 3=1 is of rank k . By Theorem 2.3, we have < C^Cy^) > +••••+ < ^ (y^,-^ > + < 0. (x ) > = < 0. (x, )> +• • •+ < 0. (x, ) > . Hence Z e < 0. (x, )>+••• •+ <0. (x, )> , i k+1 l 1 l k s l 1 i V a contradiction to the hypothesis. k+1 Therefore Z x. is of rank greater than or equal to k . 3=1 3 12. Corollary 2.6 will be used in Chapter III. The following corollary of Theorem 2.4 is well-known. k k 2.7. Corollary. (a) Suppose that E x. ® y. = E u. ® v. e R, (V «>V ) 1=1 1=1 Then < x1,...,xk > = < u1,-..--,uk > , < y r---»y k > = < V " >vk > (b) Suppose that x^® • • • ® x^ = y^® • • • ® y^ ^ 0 in V^® • • • Then < x ± > = < y ± > , . i = 1 , , k.. 2.8. Theorem. Let , • * • , b e supspaces of ^i»""*>^m respectively such that U. =U/.> , i = 1 , ••• ,m, for a l l a e G . Then I a(i) V ( u r - ' V v ^ ^ V ( v r - ' V v ( G ) ) where y. is a decomposable Proof: Let y = E y. e ^((1^, • • • » u m) x( G)) wh re y element for each j = 1 , ••• , k . By definition, i f L e CL , then < u^(yj) > S£ ^ £ f° r e a c n j = 1 » > k . Suppose that n y = E Z. e R ((V1,«',,V ) (G)) where Z. is a decomposable element for j = 1 J n 1 m x J each j = 1 , ••• , n . Then n '_< k . According to Theorem 2.1, we have < 0. (Z.,) >+•••+< 0. (Z ) > <r < 0. (y,) >+-••• + < 0,, (y, ) > cr U„ i f I e 0. l 1 l n — i w l .i'k — £ l 13. If n < k , we conclude that the rank of y is less than k in (U,,'«',U ) (G) , 1 m X which is a contradiction. Therefore n = k and y e R.,((V. ,.«««V ) (G)) . K 1 m X 2.9. Lemma. Let f. : V. -»• F be linear transformations, i = 1 , ••• , m i i where V. = V ... for a l l i and a l l a e G . Let f : V1 x-..x V -*• F l o(i) 1 m be defined as follows: . m f(w ,---,w ) = E x(o) n f .(w ) 1 m aeG i=l 1 Then f is multilinear and symmetric with respect to G and x Proof: Note that the function f is well-defined since V. = V ... , l a(i) i = l , ••• , m , a e G .' For each a , 3 e F and wi , w^  e V i , we have f(waw.+0w'...,w) = E x(o)( n f (w ))-f (aw +Bw!) 1 i i m a(j) j a(x) i i = E X(a)( nf ( W j)).af (w.) aeG j f i J J m + V ( 0 ) ( - I - f < ' ( J ) ( w J ) ) , e f ^ i ) ( w i ) aeG j f i J J = af (wn, • • • ,w , • •-,w ) + fBf (w ,• • • ,w' , • • • ,w ) ± 1 m 1 i m Hence f is multilinear. Also for each x e G , 14. m f ( wx(l)'' * * 'Wx(m)> = ^ ^ ^ a d ) (WT(i)> m = Z x(o) H f 1 (w.) where we let x(i) aeG j=l aT - J-(j) 3 -1 m = 2 X(O)X(T)X(T A) n f _ l f (Wj) aeG j=l ax (j) -1 m = X(x) I x(o-x ) n f (w ) aeG j=l ax J = x(T)f(w ,•••,w ) l m Hence f is also symmetric with respect to G and x 2.10. Lemma. Let x = x, *• ••* x e (V,,•••,V ) (G) . Then x = 0 1 m l m x implies that for some i , dim < o^(x) > < |o_^| , where 10_J denotes the number of elements in 0. . l Proof; Suppose that dim < O^x) > = 1 | for a l l i = 1 , ••• , t . For each j , let f. : V. -*- F be a linear transformation such that f.(x.) = 1 • . 3 3 3 3 f j (x d) = 0 for a l l d such that j ={= d and j , d are in the same orbit-of G . Consider the multilinear function f defined in Lemma 2.9. By the universal factorization property of (V^,•••»Vm)^(G) > there exists a linear mapping h : (V-.'-'.V ) (G) -»• F such that 1 m x 15. h(w *...*w ) = f(w,,***,w ) . 1 m 1 m Since f a ^ ( x . . ) = 1 i f and only i f a ( j ) = j , i t follows that m n f / . . ( x . ) = o i f a 4= 1 . Hence we have j = 1 3 T m f ( x . , . . - , x ) = x ( D n f,(x.) = l . 1 m j = i J J Therefore h(x *...*x ) = f(x-,»-«,x ) = 1 which is a contradiction 1 m 1 m since x, *•••* x = 0 . Hence dim < 0.(x) > < lo.l for some i . 1 m l 1 I ' 2.11. Theorem. Let x. = x., *•••* x. ,1=1., ••• , k . Suppose that j j l jm v v for each i = 1 , ••• , t , dim (< C ^ C x ^ > +• • •+ < ° i ( x k ) > ) = 1 0 ^ | k . Then k I x. e R, ((V , • • •,V ) (G)) . j = 1 J K 1 m x Proof: We proceed by induction with respect to k . By Lemma 2.10, the proposition is true for k = 1 . Assume now i t is true for k - 1 where k k _> 2 . -By Corollary 2.6, Ex. has rank equal to k or k - 1 . 3=1 1 k k-1 Suppose that E x. = I y. e R. , ((V-, ' * *,V ) (G)) where , .( V-.'-'.V ) ( j = 1 J j = 1 J K-1 1 m x y^ = y ^ * . . . * , j = 1 , ... , k - 1 . We divide the proof into two parts: 1 6 . (i) Suppose G is transitive. Let V = V, = ••• = V . 1 m k k - 1 Then m _> 2 . Suppose that for some i , I < i < k , E < 0,(x.)> c: E < 0 , (y.)> J + i J J=l J k k - 1 k Then we have E <01(x.)> = E <0,(y.)> since dim ( £ <0.(x.)>) = (k-l)m . J + 1 " i i 1 J J - 1 1 J J + 1 1 J ' Now x. = -(x,+--'+x. ,+x.,,+••'+x, ) . + y, + ••• + y, , . From Theorem 2 . 1 , l 1 l - l i+l Ic J l Jk-1 k k - 1 k we have <0-(x.)> cz £ <CL(x.)> + E <0.(y.)> = E <0,(x.)> , which contradicts J- 1 — . i . J- J . -i J- J J- i J f i J J=l J J+i • the hypothesis. Therefore for a l l i = 1 , ••• , k , k k - 1 E <0 n(x.)>4z E <01(y.)> . k - 1 Hence ther are two vectors x , x 3 i E <0n(y.)> and n + d . n t l D T 2 J - 1 1 I k - 1 If <x , x, > + ( E <CL (v )>) is not a direct sum, then we can choose n t l D T 2 j=l 1 "J" k - 1 a vector x such that x i <x ^  ,x > + E <0.(y.)> since s t 3 s t 3 n t l dt 2 j = 1 1 J k - 1 dim (<0.(x,)>+•••+<(),(x. )>) = mk and dim (<x _ ,x. > + E <0.(y.)>)< mk . 1 1 I K . nt^ at 2 j=l 2 k - 1 By the choice of x , we have both <x ,x > + ( E <0 (y.)>) and s t 3 s t 3 nt 1 j = 1 l J k - 1 <x ,x > + ( I <0 (y.)>) are direct. Either s + d or s + n . Therefore S T 3 D T 2 j=l 1 2 in any case, there are two vectors, say, x ^ , x ^  such that k - 1 <x ,x > + ( £ <0 (y.)>) is direct and Z + q 1 Q 2 i=l 1 J J 17. Consider the associated transformation K(f) : V^G) V^G) X X where f : V -»• V is a linear transformation such that k-1 f(x^ d ) = 0 and f| <xqd > + E ^ ( y )> is the identity mapping. k k k-1 k-1 We then have K(f)( E x.) = K(f)( E x.) = K(f)( E y.) = E y. . Since j=l 3 j f e J j = i J j=i 3 k-1 E y. is of rank k - 1 , i t follows from Theorem 2.3 that j - l 3 k-1 E <f(x ) , ••• , f(x. )> = E <0 (y )> .-M 31 J m J - l 1 However, f(x , ) = x , and qd 2 qd 2 k-1 hence x , e E <CL(y.)> , a contradiction. Therefore Ex. is of rank k q d2 3=1 1 3 3=1 3 (ii) Suppose G has at least two orbits. Consider the associated transformation K(f.,'*',f ) : (V,, •••V ) (G) ->• (V., • • • V ) (G) where 1 m l m x l m X f : V •> V is a linear mapping such that n n n r r 6 f I <0, (x.)> = 0 , f I <0,(x„)> +•••+ <01(x.)> = identity i f n e 0, n i l n 1 I l k . 1 and f : V -»• V is the identity mapping i f n I 0, . f . n n n T l k-1 k-1 Let K ( f f ) ( E y ) = E y' where K(f f ) y . = yl , j = 1 1 m j = 1 1 j = 1 3 1 m 3 3 k k k-1 Then K(f ,--«,f )( E x.) = E x. = E y! . By the induction hypothesis, m j=l J j=2 3 j=l 3 k k k-1 Ex. is of rank k - 1 , hence by Theorem 2.3, E <0 (x.)> = E <0 (y'.)> J-2 J j=2 3 j=l 1 3 18. k-1 k-1 £ « U y . ) : j=l 2 k k-1 k k-1 Since f = identity i f n e 0 , i t follows that Z <CL(y. > = Z <0.(y!)> n l 2 j=l 3 Hence Z <0 (x.)> = Z <0 (y.)> . Now x = -( £ x.) + Z y. , by Theorem 2.1, 5=2 1 2 j=l ^ J 1 j=2 J j=l 2 we have k-1 <0 (x )> c Z <0 (x.)> + Z <0 (y.)> = Z <0.(x.)> . X j=2 Z 2 j=l Z 3 j=2 2 2 k This contradicts the hypothesis.' Therefore Z x. is of rank k j=l 2 From Theorem 2.11, we have the following known result. 2.12. Corollary. If x^ , ••• , x^ are linearly independent and k y. , ••• , y, are linearly independent, then Z x.® y. is of rank k J=l J 2.13. Remark. Theorem 2.3, Theorem 2.8 and Theorem 2.11 generalize Theorem 3, Theorem 5 and Theorem 6 in [8] respectively. 2.14. Theorem. Let x + y = z where x , y , z are non-zero decomposable elements of (V^, ',V^) (G) . Then for a l l i , <0i(x)> = <0±(y)> , except possibly for one value j of i , in which case dim <Oj(x)> <_ dim (<Oj(x)> fj <0_.(y)>) + 1 and dim <0_.(y)> <_ dim (<0^ .(x)> 0 <0^(,y)>) + 1 19. Proof; Suppose that there exist distinct s and q such that <0g(x)> f <0g(y)> and <0q(x)> =f <0q(y)> Without loss of generality, we may assume <0g(x)> ^  <0 (y)> . Choose d e 0 such that x, I <0 (y)> where x = x, * ••• * x s d T s 1 m Let T, : V, V, be a linear transformation such that T,(x,) = 0 a d d a a and T d | <0g(y)> is the identity mapping. Let T n : v n v n b e t n e identity mapping i f n i 0 and T = T, i f n e 0 . Consider the associated r r " r s n d s transformation K(T..,•••,T ) on (V.,•••,V ) (G) , we have 1 m 1 m x K ( Tl'*"' Tm ) (x+ y ) = K ^ . - . y z = 7=1=0 Since T is the identity mapping i f n £ 0 , by Theorem 2.3, <0.(y)> = <0.(z)> n s for a l l i , i =f s . In particular, <0^(y)> = <0q(z)> =(= <0q(x)> . By Corollary. 2.2, we have <0^(x)> cr <0^(y)> + <0^(z)> = <0^(y)> . Therefore <0 (z)> ci: <0 (x)> . Let z = z, *••••* z . Choose r e 0 such that q ~r q 1 m q z i <0 (x)> . Let f : V• ->- V be a. linear transformation such that r T q r r r f (z ) = 0 and f I <0 (x)> is the identity mapping. Let f : V •*• V r r r ' q ; r r ° n n n be the identity mapping i f n £ 0 q and f n = f r *f n £ °q * Consider the associated transformation K(f.,,''',f ) , we have 1 m 20. K(f ,---,f )(x+y) = x + K(f ,--..f )y = K(f ,--.,f )z = 0 . 1 m 1 m 1 m Therefore K(f..,«'«,f )y = -x 4= 0 . Since f is the identity mapping for 1 m n n e 0 , i t follows from Theorem 2.3. that <0 (x)> = <0 (y)> . This yields s s s a contradiction. Therefore there is possibly only one j such that <0 (x)> + <0j(y)> . Now, assume that such 'a j exists and dim <0j(x)> > 1 + dim (<0^ . (x)> f~\ <0^  (y)>) Then i t is not hard to see that there are two independent vectors x, , x , , d p where d , p e 0. such that <0.(y)> + <x,,x > is a direct sum. Note that J 3 d P i f <0j(z)> C<0^(y)> , then <0.. (x)> cj: <0.. (y) > + <0^(z)> , a contradiction to Corollary 2.2. Hence there is a r e 0. such that z . r <o.(y)> • We have either < x d > + (<zr>+<0_. (y)>) is direct or <xp> + (<zr>+<0^  (y)>) is direct. We may assume <x > + (<z >+<0.(y)>) is direct. Let g : V -> \ *. d r j r r be a linear transformation such that (x^) = 0 ,'g (z r) = 0 and 8 r I < 0 j ( y ) > is the identity mapping. Let § n : v n v n b e the identity mapping i f n { 0. and = i f n e 0_. . Consider the associated transformation K(g^,''',gm) , we have K( g ; L, ---.g^Cx+y) = y = K(g 1,--- >g m)z = 0 , 21. a contradiction since y is assumed to be non-zero. Therefore dim <C\(x)> <_ 1 + dim (<CL (x)>f\<C\. (y)>) , and similarly dim <0^(y)> <_ 1 + dim (<CL(x)> f\ <0^(y)>) The above theorem contains the known facts in tensor, Grassmann and symmetric spaces as special cases. See Lemma 3.1 [19], Lemma 5 [3] and Theorem 1.14 [4]. 2.15. Theorem. Let KO^, • • • ,Tm> : O^t *' ' i V^) (G) •* • • •,Vffl) (G) be an associated transformation such that T. : V. ->• V. is non-singular for i i x each i = 1 , . ••• , m . Then K(T, ,-",T ) is a rank k preserver for 1 m al l possible k . Proof: Since K(T,,--.,T )K(T~1,-..,T 1) = K(I,••',!) is the identity . 1 m l m . mapping on (V^ • • •, v m) xC G) • KC^,'**,^) is non-singular. Let z = z, + ••• • + z. c R, ((V , • • •, V ) (G)) where z 1 , • • • , z. 1 K t c l m x 1 fc are decomposable elements. Then K(T^, • • • ,T )z =)= 0 . Let K(T 1,..-,T m)( E z.) - Z y . £ Rj'((V1,---,Vm) (G)) 1=1 1=1 where y^ , ••• , y are decomposable elements. Clearly j _< k 22. Now, we have K(T7\• • • ,T 1 ) K ( T 1 , - - - , T )(z) = z 1 m 1 m = KCT^.-.^T^Cy^.-.+y ) z being of rank k implies that j > k . Hence j = k . Therefore K(T,,«'«,T ) is a rank k preserver. X m The problem of rank k preservers is concerned with the converse of Theorem 2.15. §3. Applications. Let , u"2 be supspaces of U and , V"2 be subspaces of V . It is well-known that (i) ( I L j ® f l (U 2®V 2> = (t^f l l ^ ) ® ( V ^ V ^ [7, p. 20] A u i A A u 2 = A ( u i n u 2 } * We now apply Theorem 2.3 and Theorem 2.8 to obtain a result on symmetry classes of tensors generalizing (i) and ( i i ) . 3.1. Theorem. Let U. and W. be subspaces of V. where U. = U , x x x x a (x) ' W. = W , V. = V ,.s for i =• 1 m and for a l l a e G . Then x a (x ) x a (x) we have ( U r'- ,' Um ) x ( G ) ^ ( W l'"' , Wm ) x ( G ) = ( U i n W l » - " » W x ( G ) * Proof: It is obvious that (IL fl W., • • •, U fl W ) (G) C (U. , • • • ,U ) (G)f\(W1 , • • • ,W ) ( G ) 1 1 m m x — I M X - L M X 23. Let z be a non-zero vector of (U, , •",U ) (G) A'(W,, • • • ,W ) (G) . 1 . ' m x 1 . m x Then z e R, ((Vt,•••,V ) (G)) for some positive integer k . By K 1 m X Theorem 2.8, z z R. ((U., • • • ,U ) (G)) A R. ((W_, • • •,W ) (G)) . Therefore - K l m x ' K l m x z = x, + • • • + x, = y,+•••+ y, for some x. e R, ((U., , • .., U ) (G)) 1 Tc 1 k j 1 1 m x and s ome y e R^ ((W^ ,•••,W ) (G))•where j = l , ... , k . By Theorem 2.3, for each orbit 0. of G , we have l <0. (x. )>+••• + <0. (x. )> = <0. (y-)> + • • • + <0. (y. )> CT W f\ U , q e 0. . I l I Tc I Jr I k — q q i Therefore z e (U.. A W.., • • • ,U AW ) (G) . This completes the proof. 1 1 m m x 3 . 2 . Corollary. Let and U2 be two subspaces of U . Then (vu^ n (vu2> = v(u xnu 2) . As an application of Corollary 2.4, we prove the following 3.3. Theorem. Let K(T) , K(S) : ^ (G) -»- Vm(G) be two associated X X transformations. Suppose (i) p(T) , the rank of T , is greater than m or (ii) x = 1 • Then K(T) = K(S) i f and only i f T = AS for some A e F with Am = 1 . Proof: The sufficiency of theorem is tr i v i a l . We proceed to prove the necessity. 24. Case 1. Let v.^  e V be such that TCv^) = =f= 0 . Extend to a basis z i » *"" > z n °f ^ he range space of T . By hypothesis, n > Let T(v±) = z± , i = 2 , ••• , n . Since K(T) ='K(S) , we have K(T) (v..*•••*v.*•••*y L l) = K(S) (v .*v.*.--*v ,,) , i = 2 , 1 l m+1 1 I m+1 m Hence 2 *...*£;*...*z = Sy *• • .*Sv.*" -*Sv f 1 , i = 2 , 1 i m+1 . • 1 I m+1 Since z^ , ••• , z ^ ^ are linearly independent, z^*- • • *z\*• • • * z m + ^ t 0 by Lemma 2.10. Therefore by Corollary 2.4, we obtain <Zl>'"'h'"',Zmrl> = <Sv1,...,S^i,...,Svn+1> , i = 2 , . . . , m + l . Therefore m+1 m+1 m+1 (1) H <z 1, • • •,z i, • • ' . z ^ ^ = p\ <Sv ,••«,Sv ,••«,Sv >^ i=2 i=2 Since z^ , ••• , z ^ ^ are linearly independent, the left hand side of (1) is < z 1 > • Therefore the right hand side is a one dimensional m+1 ^ subspace of V . But O <Sv1,•••,Sv.,•••,Sv-> 23 <Sv > . Hence J. i m+j. — ± <Sv^ > = < z j > = <Tv^ > since Sv^ is obviously non-zero. By symmetry, S(u) =f 0 implies that <S(u)> = <T(u)> . It follows that <T(v)> = <S.(v)> for a l l v e V . This implies that T = AS for some X e F . Clearly 25. Case 2. Suppose x = 1 • Let v e V . Then we have K(T) v *...* v = K(S) v *...* v = Tv *• •••* Tv = Sv *•••* Sv . If Tv = 0 , then we have Sv = 0 . If Tv =|= 0 , then Tv *• • • * Tv + 0 and hence by Corollary 2.4, <Tv> = <Sv> . Therefore we have T = AS for some X e F . Clearly Am. = 1 . 3.4. Remark. The above theorem (Case 1) is proved by Marcus in [11] under the assumption that the underlying field is the complex numbers. CHAPTER II 26. RANK ONE PRESERVERS ON SYMMETRIC SPACES §1. Maximal Pure Subspaces of Symmetric Spaces. Throughout this section, let U denote a finite dimensional r vector space over an algebraically closed field ' F . Let VU denote the r-fold symmetric product space over U where r >_ 2 . 1.1. Definition. Let x, , • • • , x , be non-zero vectors of U . Then the pure subspace {x, x -u : u e U} , denoted by x, x ' «U , 1 r - l 1 r - l r is called a type one pure subspace of VU . 1.2. Definition. Let S be a two dimensional subspace of U . Then the pure subspace {s^ s^ : s^ e S , i = 1 , • • • , r} , denoted by S ^ , r is called a type r pure subspace of VU . 1.3. Definition. Let S be a two dimensional subspace of U . Let x, , ••• , x , be vectors of U such that x. i S , i = 1 , ••• , r - k , 1 r-k l T where 1 < k < r . Then the pure subspace {x, x , *s, s, : s. e S , i = 1 , ••• , k} , denoted by x, x , -S,, . , 1 r-k 1 k I 1 r-k (k) is called a type k pure subspace of VU . The following result is proved by L.J. Cummings in [4] r 1.4. Theorem. (i) If dim U > 2 , then every type i pure subspace of VU 27. is a maximal pure subspace where 1 < i < r . (ii) If dim U > 2 and the characteristic of F is zero or greater than r , then every maximal pure subspace of VU is of type i for some i = 1 , • • • , r . In this section, we shall show that i f the characteristic of F is a prime p _< r , then there is one more type of maximal pure subspace besides the above mentioned ones. 1.5. Definition. For each u , y- , • • • , y , e U , k < r , we denote r r the pure vector u u e VU by u and the pure vector r k u u-y, y , e VU by u -y, y , . 1 r-k 1 r-k 1.6. Theorem. Let char F = prime p and r > pfc for some positive integer t . Let x^ , •*• , x be r - p t non-zero vectors of U . Then the r-p t . r set M = {x^ x ^-uF : u e U} is a pure subspace of VU . Moreover r-p t. dim M = dim U . Proof: Let k = pfc . For any x , y e U , we have (x+y)k = x k + (^ ) x k + • • • + (k) x k i«y i + • • • + y k = x^ + yk since P | (k) for a l l i = 1 , ••• , k - 1 Hence x, x •, *xk + x. x , -yk = x, x • (x+y)k . 1 r-k 1 r-k  J' 1 r-k 28. 1 1 k k k k For any non-zero a in F , we have a x = (a x) where a JL is the k*"*1 root of a . Hence a x. x , «xk = x, x , *(a kx) k . 1 r-k 1 r-k Therefore M is a pure subspace of VU . Let u, , ••• , u be a basis of U . Let a. , ••• , a e F I n 1 n Then . 1 n k n k k £ a. x, x , 'u. = £ x. x , '(a. u.) . , i 1 r-k l » i 1 r-k l l i=l i=l 1 n k k = x .•(£ a. u.) = 0 1 r-k . - l I i=l 1 n k implies that £ a. u. = 0 . Therefore a. = 0 , i = 1 , ••• , n . i=l 1 1 1 n Also for any y in U , we have y = £ 6. u. for some i=l 3^ in F , i = 1 , ••• , n . Therefore k n k x- x «y = x1 x . • ( £ 3. u.) 1 r-k 1 r-k . .. I I i=l n k = £ x, x , • ( 3 . u.) i=l 1 r-k ^ i i n k k = £ 3. x, x *u.. . . , I 1 r-k i i=l Hence x, ••••• x , «u. , i = 1 , •" , n , is a basis of M . 1 r-k l 29. The pure subspace in Theorem 1.6 will be denoted by t x^'-.-x t-U P r-p r 1.7. Definition. Let x = x, x be a non-zero pure vector of VU . 1 r We shall use U(x) to denote <x, , ••• , x > . The one dimensional 1 r subspaces < x :£ > » i = 1 » ""* » r a r e called the factors of x . A one dimensional subspace <u> is said to be a factor of x of multiplicity k i f x = u^. y y . where <y.> 4= <u> , i = 1 , ••• , r - k . 1 J r-k Ji ' The above definition is well-defined since X l x = z.. z f 0 1 r 1 r 1 implies that <x.> = <z ,.\> for some a e S , i = 1 , ••• , r [4, Corollary 1.9] r ± a(i) r Let u, , ••• , u be a basis of U . Let VU denote the symmetric I n algebra over U and ^t^i»"""»^n] denote the polynomial algebra in n determinates £ , > • • • » £ over F . Then there exists a unique isomorphism 1 n d> : VU -*• F[£,»* • • ,£ ] such that 1 n * u ± = £ ± (i = 1 » * *' » n) r r The restriction of if to VU is an isomorphism of VU onto the vector space of homogeneous polynomials of. degree r (see [7, pp. 202-203]). 3 0 . r Clearly x^ is a non-zero pure vector of VU i f and only i f < f > ( x ^ x ) is a product of r linear homogeneous polynomials in F [ £ , ] • In the Gaussian domain F[g-,''»,£ ] , linear homogeneous I n 1 n polynomials are irreducible elements. 1.8. Lemma. Let x^ , •* * , x^ be k non-zero vectors of U . Let r-k r > k + 1 and x^ xk" A ~ z i z r ^ 0 w h e r e A e V U and z± e U . Then < x i > » •** » < x i l > a r e k factors of z^ z^_ and A is a pure r-k vector of V U . Proof; Consider the isomorphism <j> from VU onto FfC^,•••,£] . We have • (x, x^A) - ty(z± z r) 4 0 . Therefore tyix^ • (j)^) ty(A) = ty(z±) ••• ^(z^) =j= 0 . Since F i ^ , • • •, ^] is a Gaussian domain and since <KX-^ ) > • ••• > ^Cx^) > ^(.z^) <Kzr) are linear homogeneous polynomials, i t follows that for each i = 1 , ••• , k , <<f>(x.)> = <<(>(z. )> for some j . where 1 _< j.<_ r and j =f j i f t =}= s , 1 3^  1 1 t s and '('(A) is a product of r - k linear homogeneous polynomials. Hence <x^ > , ••• » < x j c > are k factors of z^ z^ and A is r-k a pure vector of V U . 31. The following Lemma is equivalent to Theorem 1.17 and Theorem 1.18 in [4] . The proof given here is rather simple. 1.9. Lemma. Let x and y be non-zero pure vectors of VU such that dim (U(x) + U(y)) _> 3 and <x , y> is a pure subspace. Then x and y have a common factor. Proof: If x and y are linearly dependent, then clearly x and y have a common factor. Hence we assume x and y are linearly independent. Let x = x^ x^_ and y = y^ y^ . Since dim (U(x) + U(y)) >^  3 , either dim U(x) >_ 2 or dim U(y) _> 2 . We may assume that dim U(y) >_ 2 . Then for some i , j , k , x . ,y. , y, are linearly independent. Without loss of 1 J K generality, we may assume i = l , j = l , k = 2 . Extend y^ , y~ t to a basis u, , ••• , u of U where y. = u, , y~ = u_ and x, = u„ . 1 - n 1 1 / / i j For each non-zero A e F , x + Xy is a pure vector by hypothesis. n Let x + Ay = z(A) = z.(A) z (A) . Also let x. = £ a.^ u and n z.(A) = E a. (A) u where i = 1 r . Let f : U -> U be the l i t t linear transformation such that f(u.) = 0 and f(u.) = u. , i > 2 . 1 i i r Consider the induced transformation P (f) on VU , we have Pr(f)(x+Ay) = P r(f)(z(A)) 32. It follows that x, • f(x_) f(x ) = f(z n(A)) f(z (A)) . I f l / r 1 r f ( X j ) = 0 for some j , then x has a factor < y i > a n d w e a r e done. Hence we may assume f ( x j ) 4 0 f° r a H j = 1 » *** > r . Without loss of generality, we may choose vectors z^(A) , '*• , zr(A) such that Corollary 1.9 of [4] implies that x^ = f(z^(A)) . Hence we have z^(A) = x^ + a^^(A) u^ Now, let g : U -»• U be the linear transformation such that g(u2) = 0 and g(u^) = u^ , i = 1 , 3 , ••• , n . Consider the induced r transformation P (g) on V U , we have P r ( g ) (x + Ay) = Pr(g) (z(A)) It follows that g(x1)-----g(xr) = g(z1(X)) g(zr(A)) . If 8( xj) = 0. f° r s o m e j » then x has a factor <y£ > a n d w e a r e done. Hence we may assume that s( xj) T" 0 ^ o r a H J = 1 » Therefore g(z^(A)) = n g(x^ ) for some non-zero n e F and for some where A i 1 <_ k^ £ r . It then follows that n 8 l ( X ) - x 1 + a 1 1 ( A ) « 1 - n n ^ u t ) Hence z^(X) = n (a^ i u i + ^ 3 x]_) * This shows that z(A) has a factor A A 33. < a k Al u l + \ 3 V ' S i n c e { < a 1 3 x l + a l lV ' '• < a r 3 x l + a r l u l > } is a finite family and F is an infinite field, we see that there are distinct non-zero X^  , X„ e F ' such that z(X^) and z(X2) have a common factor. But <x , y> = <z(X1) , z(X,,)> , we conclude from Lemma 1.8 that x and y have a common factor. 1.10. Theorem. Le t - dim-U-.-:> 2--_ Vand: char F = prime j? .." Let x^ , ••• x r be non-zero vectors of U where r > pfc , t a positive integer. Then t r M = x^ x is a maximal pure subspace of VU . r-p r Proof: Let y = y^ y^ be a non-zero pure vector in VU such that M U{y} spans a pure subspace. We shall show that y e M . Let U.. = <x. , x.> and W.. = <x. , y. > , i , j = 1 , ••• , r 13 I j lk. - l tc k = 1 , • • • , r . Let V = < y ± , y > , i + t , i , t = 1 , • • • , r . Since F is algebraically closed, F is an infinite field. Since dim U >_ 3 , U =f (UU. .) U(UW., ) \J (UV )• Choose a vector u e U such that ( i j lk i t u {. (UU. .) (J (UW. ) (J (UV. ) . Then we consider the pure vectors i j lk i t t p x = x^ x t*u and y . r-p If for a l l i and j , <x^ > = <y > , then y e M and we are done. Hence we assume now <x. > f <y. > for some i and j l 1 1 o o o Jo 34. Since u , x. , y. are linearly independent, i t follows from Lemma 1.9 o Jo that x and y have a common factor. Let < u i > > ,* *' > < u i e > ^e a"H t* i e common factors (counting multiplicities) of x and y . Let x = u1u^'x' , y = u± uj^ -y* r-k where x' and y* are non-zero pure vectors of V U . Then x' and y' have no common factors. Since <u> is not a factor of y , we see that <u> is a factor of x' of multiplicity p1" . Since <x , y> is a pure subspace, i t follows from Lemma 1.8 that <x' , y'> is a pure subspace r—lc t of V U . Suppose for some d , <x > is a factor of x' , l < d < r - p d — — Let <y. > be a factor of y' . Then u , x, , y. are linearly independent. J l ' • . d J l This implies that dim (U(x') + U(y')) >_ 3 and x' and y 1 have a common t factor, a contradiction. Therefore x' = Xu*5 for some X in F and u^ u^ = n x^ x^ for some n e F . Since dim (U(x') + U(y')) _< 2 , t by our choice of u , we see that y' = z^ for some z in U . Consequently y e M . Hence M is a maximal pure subspace. t 1.11. Theorem. Let dim U > 2 and char F = prime p . Then M = {x^  : x e U} is a maximal pure subspace of V U and dim M = dim U . Proof: That M is a pure subspace and dim M = dim U can be seen from the proof of Theorem 1.6. Suppose M is properly contained in a 35. t pure subspace N . Let x be a non-zero vector of U . Then x*U^ is properly contained in the pure subspace {x«A : A e N} , contradicting the fact that x'U*5 is a maximal pure subspace of *>V+"4j ... Therefore M is a maximal pure subspace. t t If char F = p , we shall denote {x*5 : x £ U} by t p U and the pure subspace in Theorem 1.6 are called power type pure  subspaces of degree t . The following elementary fact will be needed. 1.12. Lemma. Let V be a k-dimensional vector space over an infinite field. Then there is an infinite subset W of V such that any k elements in W are linearly independent. 1.13. Theorem. Let U be an n-dimensional vector space over an algebraically r closed field F . Let n >_ 3 and M be a maximal pure subspace of VU . Then M is of type i for some i ,' 1 <_ i <_ r , or M is of power type of degree t for some positive integer t . Proof: Case 1. Suppose that for a l l non-zero pure vectors x and y in M , dim (U(x) + U(y)) <_ 2 . We have two subcases: (a) For some non-zero x e M , dim U(x) = 2 . Let y be any non-zero element in M . Since dim (U(x) + U(y)) <_ 2 , i t follows that U(x) Z) U(y) . Let S = U(x) . Then M C S ^ . Since both M and S ^ 36. are maximal pure subspaces of VU , we conclude that M = S, ,. . (r) (b) For a l l non-zero x e M , dim U(x) = 1 . Then M cr (u r : u e U} Since M is a maximal pure subspace, dim M =f 1 ; for otherwise M is property contained in a type one pure subspace. Hence there are two r r independent pure vectors y and z in M . Clearly y and z are r r r linearly independent. Note that we have y + z = u for some non-zero u c U . In view of Corollary 1.2.2, <y , z> ^  <u> . Therefore u = az + by for some a , b e F . It then follows that (az+by)r = a rz r + ••• + (£)(az)r k-(by) k + ••• + b ry r r r = z + y IT IT TO IT-Ic It Hence a = b =1 and ( k)a ..b = 0 , k = l , - " , r - l since T f—1 IT—Ic k TC z , z -y , ••• , z «y , *'• , y are linearly independent vectors of VU . Thus (£) = 0 for a l l k = 1 , ••• , r - 1 . This implies that characteristic F = prime p , r = p C for some positive integer t and t hence M = UP in this case. Case 2. There exists a pair of non-zero pure vectors x and y of M such that dim (U(x)+U(y)) ^ . 3 . If dim U(x) = 2 and dim U(y) = 1 , then by Theorem 1.2.14, U(x) DD(y) , a contradicition. Similarly i t is impossible that dim U(y) = 2 and dim U(x) = 1 . Let z be a non-zero vector in M . We shall show that 37. z has a common factor with x or y . If dim (U(z)+U(x)) ^ 3 or dim (U(z)+U(y)) >_ 3 , then by Lemma 1.9, we are done. Now, assume dim (U(z)+U(y)) < 3 and dim (U(z)+U(x)) < 3 . Then dim U(x) = dim U(y) = and U(x) 4 U(y) . It follows that U(x) D U(z) and U(y) ^  U(z) j for otherwise dim (U(x)+U(z)) _> 3 or dim (U(y)+U(z)) JL 3 .. Hence dim U(z) = 1 and U(x) f\ U(y) = U(z) . Since x and y have a common factor, say <f> , (Lemma 1.9), we see that <f> = U(z) . Therefore U(z) is a common factor of x , y and z . Assume now dim M = m . By Lemma 1.12, let z, , ••• , z , ••• be an 1 n infinite subset of M such that any m z^  forms a basis of M . By the above argument, for each i , z^ has a common factor with x or y . Since x and y have at most 2r distinct factors, i t follows that there are infinitely many z^ with a common factor, say < x 2 > • Hence there is a basis z. , ••• , z. of M with a common factor <x > . By Lemma 1.8, i , l 1 1 m we see that <xj^ 1 S a factor of any non-zero pure vector in M . Let <x^ > , ... , < x j > be a l l the common factors of non-zero elements of M . If j = r, - 1 , then clearly M = x^ xr_^'U If j < r - l , let M' = {v, v . : x, x.-v, v , e M} 1 r-j 1 j 1 r-j By Lemma 1.8, we see that M1 is a pure subspace of IV~'u . Since M is a maximal pure subspace, i t follows that M' is also a maximal pure 38. subspace. Now dim (U(x')+U(y')) <_ 2 for a l l non-zero x' , y' e M' ; for otherwise, by the same argument as above, there is some non-zero v E U such that <v> is a common factor for a l l A e M' , A =f 0 . Hence by the same argument as in case 1 , M' = S. .x for some two dimensional (r-j) s TJ S subspace S of U or M' = U where r - j = p for some positive insteger s and char F = prime p . Thus s M = x, -x. - S, . v or M = x, x -UP 1 J (r-j) 1 r_ ps In case M=x, x.-S. . v , then we must have x, , ••• « x. A S ; for 1 J (r-j) 1 J i f some x^ e S , then M is properly contained in the pure subspace X- x". x.*S, . , - * , a contradiction to the fact that M is maximal. 1 I 2 (r-j+1) This completes the proof. §2 . Intersections of Maximal Pure Subspaces. In this section, we study the intersection properties of maximal pure subspaces and determine the form of an infinite family of maximal pure subspaces of.type one or power type such that any two members of them have a non-zero intersection. These results will be used in a latter section. Unless otherwise stated, throughout this section, U is assumed to be a finite dimensional vector space over an algebraically closed field F such that dim U > 2 . 39. t 2.1. Theorem. Two power type pure subspaces M = u^ u «UP r-p and N = v t«U P (r > p"1") of VU are equal i f and only i f r-p u t = Av^ v ' for some A e F . r-p r-p Proof: The sufficiency is clear. We prove the necessity. Choose a vector z e U such that z £ <v 1 >U-"U<v > . Then M = N implies that r-p t t . ± -P - - - p r-p r-p u .. ...u «z = v, v *w 4= 0 t 1 t 1 for some w e U . Since < z> + <v^> for a l l i = 1 , ••• , r - pfc we have <z> = <w> . Therefore u^ u t = Av^  v t for some A e r-p r-p t JL 2.2. Theorem. Let M = x^ x t«U P and N = y^ ^ y ^-UP be r-p r-p two distinct power type pure subspaces of VU where t >_ Z and r >_ p*" Then dim (MQN) = 1 i f f either (i) t > I and y^ y ^ = Ax^ x t«f P p for some f e U , A e F ; or r-p r-p t Z (ii) r _> p + p and x 1 x t = z i z t Z ' r-p r-(p +p ) t y i ' - - " y z = z i z , t A z ' f V r-p r-(p +p ) for some a , f , z. e U . Otherwise MHN = 0 . l 4 0 . Proof: Suppose that MAN ={= 0 . Then there are non-zero vectors t I f and a e TJ such that x^ x fc.f = y £ , A P T" 0 • r-p r-p Either <f> = <a> or <f> =j= <a> . If <f> = <a> , then t ={= I , otherwise by Theorem 2.1, M = N , a contradiction to the hypothesis. Hence t > £ and t I u . f F " r-p - ~ r-p" y ^ = Xx^ x t«f P P for some X e F . t £ If <f> ^  <a> , then clearly r >_ p + p and for some z± , i = 1 , ••• , r - (p +p ), I x l x t = z i z t i "aP r-p r-(p +p ) t y l y l = Z l Z , t + * ' f P r-p r-(p +p ) t Conversely, i f (i) holds, then MAN = <x^  x t*^ P > a n d I • t r~ P i f (ii) holds, then MAN = <z^  z ^ -a P - f P > . r-(p +p ) t 2.3. Theorem. Let M = x, x 'UP and N = y.. y , be two 1 t J1 "'r-l r-p r t maximal pure subspaces of VU where r _> p and t is a positive integer. t _ 1 Then dim (MAN) = 1 i f and only i f either (i) y r - l = * X 1 X t * a P r r-p t ~ p t for some a e U , X £ F or (ii) r > p and y, y , = x, "x. x .a 1 r - l 1 i t r-p for some a e U and some i where 1 < i < r - pfc . Otherwise MAN = 0 . 41. Proof: The argument is the same as that given in the proof of Theorem 2.2. t 2.4. Theorem. Let M = x ^«UP be a power type subspace where r-p r > p^ and N = S ^  be a type r subspace of VU where S is a two dimensional subspace of U . Then dim (MAN) = 2 i f and only i f x ^ x e S r-p Otherwise M 0 N = 0 Proof: If MAN ={= 0 , then let A e MAN and A =f 0 . We have t A = x, x _ «uP = s, s 1 t 1 r r-p for some u e U , s ^ e S , i = l , ••• , r . Clearly this implies that Xj , ••• , x t e S . Conversely, i f x^ , ••• , x t £ S , then r-p r-p t t MAN = <x^  x t " V l » x i x t"V2 > w ^ e r e S = <v^  , Vy> . . r-p r-p Hence dim (MAN) = 2 . 2.5. Theorem. Let M = UP be a power type pure subspace and let r t N = y^ y t (k) ^ e a tY'Pe K pure subspace of VU where p >_ k > 1 p -k and S is a two dimensional subspace of U . Then (i) MAN = 0 i f p1" > k ; 42. t and (ii) MAN = <sP , s P > i f p = k where S = <s^  , s^> Proof: (i) Suppose that p > k and MAN + 0 . Then for some non-zero t u e U and non-zero v 1 , ••• , v, e S , u P = y 1 • • • • «y •v1 v, ={= 0 . p -k This implies <u> = <y1> = <v^ > , contradicting the fact that y^ \. S . Hence M AN = 0 i f p t > k . (ii) is obvious. 2.6. Definition. Let x = x. x be a non-zero pure vector of VU . 1 m Suppose that x = u, u -z z where t < m . Then we denote I t t+1 m the pure vector z z in U by x, x u, u . The v t+1 m J 1 m 1 1 t notion x|d products will stand for x|v^ v^ for some v^ , ••• , v^ e U where <v > , ... , <v > are d factors of x . This definition is well-defined 1 d since u, u • z,. - • • •; .z = u. u -w^,, w =}= 0 implies that 1 t t+1 m l t t+1 m 1 z_,, z = w w . t+1 m t+1 m t 2.7. Theorem. Let M = x^.»--«x b e a P o w e r tyP e pure subspace of r-p r VU . Let N = y, y , -S,, . be a type k pure subspace where 1 r-k (k) r > k >^  p* and S is a two dimensional subspace of U . Let x = x1 x and y = y. y , . Then dim (MAN) <_ 2 . Moreover, we have r-p (i) dim (MAN) = 2 i f and only i f x and y have exactly r - k 43. common factors while the remaining k - p t factors of x are contained in S . (ii) dim ( M O N ) = 1 i f and only i f r >_ pfc + k and x , y have exactly r - k - p1" common factors while the remaining k factors of x are contained in S and the remaining p* factors of y are the same. Proof: (i) Suppose dim (Mi"\N) >_ 2 . Let A , B be two independent vectors in M O N . Then there are two. independent vectors u , v of U and « , s^ , s i '» "* •' sk e ^ such that t A = x i x t ' u P = y i W s i sk ( 1 ) r-p t p , , B = x, x -vr = y. y , -s' s, 1 t J l J x-k 1 k r-p If both u and v { S , then t t x i x t = ^•••••y r-k' u P } ' s i s k = ^ • • • • • y r - k l v P ) , S I s k r-p Since y^ £ S , the above equality implies that , |uP = Ay- y , |vP for some A e F . r-k' J1 J r-k1 Clearly this is a contradiction. Therefore either u e S or v e S 44. We may assume that u e S . Since <u> f <y^ > » i = 1 > * * * » r ~ k , i t follows from (1) that x and y have exactly r - k common factors while the remaining k - p*" factors of x are contained in S . -Conversely i f x and y have r - k common factors, say, x^ x^ k = *" *'^r-k » a n d t^ i e r e m a i n i n 8 k - pfc factors <x . > , • • • , <x > are contained in S , then by Theorem 2.4 r-k+1 t r-p (or Theorem 2.5 (ii) i f pfc = k) , we have t dim (x , , x -Up A S„ .) = 2 r-k+1 t 1 ' (k) r-p Therefore dim (MHN) = 2 . (ii) Suppose M(*\N = <A> where A =f 0 . Then we have t A = x l x t * u P " y l y r - k - S l " . " S k ( 2 ) r-p where s^ , ••• , s^ e S . If u fc S , by the argument in (i) , dim (MON) = 2 a contradiction. Therefore u ^  S . From (2), we see that x and y have exactly r - k - p*" common factors while the remaining k factors of x are contained in S and the remaining pfc factors of y are equal. The converse of the statement is easily checked. 45. t 2.8 Theorem. Let M = x. x -UP and N = y, y ,.S„. 1 t . • 1 r-k (k) r-p v ' be two maximal pure subspaces where r > p*" > k > 1 and S is a two dimensional subspace of U . Let x = x, x and y = y, y , . 1 t J Jl •'r-k r-p Then dim (MAN) = 1 i f f r >_ k + ' , x and y have exactly r - k - pfc common factors while the remaining k factors of x are contained in S and the remaining p*" factors of y are equal. Otherwise MAN = 0 . Proof: Suppose MAN + 0 . Let A be a non-zero element of MAN . Then t A = X l X t - u P = y l y r - k ' S l S k r-p for some u in U and s_^  in S . Since p C > k and y^ £ S , i t follows that u r S . Hence y has pfc factors of <u> . Hence x and y have exactly r - k - p*" common factors while the remaining k factors of x are contained in S and the remaining p1" factors of y are equal. The converse of the theorem is easily checked. i Combining Theorem 2.1 - Theorem 2.5, Theorems 2.7, 2.8 together with the results in section 2 of Chapter II of [4], we have: 2.9. Theorem. Let U be a finite dimensional vector space over an algebraically closed field where dim U > 2 . Then the maximal dimension of 46. the intersections of any two distinct maximal pure subspaces is 2 . 2.10. Remark. Let n be a positive integer. Let , ••• , be a collection of sets such that |A^| = n , i = l , ' " , k and |A^H A I = n - 1 i f i=j=j. If k > n + 1 , then theres exists a set W k such that W C f\ A. and |w| = n - 1 . i=l 1 r 2.11. Definition. Let C^(VU) denote the collection of a l l type one r r pure subspaces of VU . For each positive integer t , let P (VU) denote r the collection of a l l power type pure subspaces of degree t in VU . r 2.12. Theorem. Let C C P (VU) be an infinite family such that , M2 £ C implies that M^ f\ ^  =f 0 . Then r >^  2pfc and there exist r - 2pt non-zero vectors y , ••• , y with the property that M e C 1 r-2pC implies that M = y i y , t - a 5 t - D p t r-2p for some a^ £ U . (y^ , ••• , y fc are deleted i f r = 2pfc) r-2p Proof: It follows from Theorem 2.2 that r >_ 2pt . If r = 2pt , the assertion is clear from Theorem 2.2. Hence we assume r > 2pfc . 47. For each M = x t'UP E C , define a set r-p * r i M = t<x^ > , ••• , <x t>} . This is well-defined because of Theorem 2 r-p Let C* - {M* : M e C} . We claim that C is infinite. For i f C is finite and & - & & C = {M. , ••• , M. } where M. e C , i = l , ••• » 3 » then for any 1 j l pt 3 * J * M = z^ z t" ' w e n a v e < z - > E U . But U is a finite r-p i=l i=l set, i t follows that C is finite, a contradiction to the hypothesis. „ * Let N e C such that N has the maximal cardinality n * . in C . Let M e C . Since MDN f 0 , by Theorem 2.2, t t •b1 r-2p' M = Z ; L " - --z t-b P .UP (1) t t N = z z «aP -UP r-2?Z for some z. in U , i = l , ' " ' , r - 2 p t and some a , b in U . Since |N | = n , we have |{<z > , • • • , <z >}| ^ _ri — 1 . 1 r-2pC Observe that IM I > n - 1 and (i) i f |M | = n - 1 , then M = {<z >,•••> <z >} an 1 . r-2pt hence N D M : 4 8 . (ii) i f M contains an element <c> such that <c> { N , then we have <b> = <c> ftom (1), and hence | M | = n .' /•>* Since C is infinite, we observe from (ii) that there are infinitely * * & many M e C such that |M | = n . Note that i f M^  , M,, e C are such that | | = |M2| = n and M =(= M2 , then ^ A M | = n - 1 . By Remark 2.10 and (i), we conclude that MeC Now let N x , N 2 e C such that |Nn| = |N2| = n , N =j= N2 and t t - y x y t - c p -up r-2p N 2 = ?i y , t - d p t - u p t r-2p for some y , c , d e U . Let M e C . Then M ={= N^ or M =f N2 . We may assume * - * M f . From Theorem 2.2, we have 49. t t M = w w «fP -UP r-2pt t t P P 1 1 r-2p for some f , g , w in U . It then follows that M*H N* = {<w1> , ••• , <w >} , |M*HN*| = n - 1 r-2p But we have N ^ f i N 2 = (<y^ > > ••• , <y t>) • Therefore r-2p {<w > , ••• , <w >} = {<y1> , ... , <y t>> r-2p rr2p since M AN^ = N^H . It then follows that <g> = <c> . Consequently w i w t = x ? i y t 1 r-2pt 1 r-2pC t t for some j A in F . Hence M = y 1 y . f P «UP . This completes r-2pt the proof. The following result is due to Cummings [-4]. 2.13. Theorem. Let dim U >_ 2 . Let C = {M : i e 1} C (^(VU) such that |C| > r > 2 . If for every M , M e C , M. f\ M. =f= 0 , then there exist 1 2 x l • 12 5 0 . non-zero vectors y, , • • • , y n such that M. = y, y _.a.»U 1 r-2 1 ' 1 . r-2 x i . e I , a. e U . I 2.14. Theorem. Let C be an infinite collection of maximal pure subspaces r of type one or of power type in VU such that for every , in C , M l ^ M2 ^ ° ' T h e n ^ — ^ i ^ V U ^ o r *~ — P t^ v u) f o r s o m e Positive integer t , except possibly when char F = 2 , in which case, there exist non-zero vectors x , ••• , x - in U for some non-negative integer s and a subset 1 r-2 S 1 W of U such that C has the following form 2s+l 2s s C = {M : M = x, x ...IT or M = x. x .w -U 1 -s+1 1 _s+l r-2 r-2 where w e W} Proof; Let P = C H C ^ V U ) and = C f\P (VU) . By hypothesis, either is infinite or Q.^  is infinite for some positive integer t 'Case 1. Suppose () is infinite for some positive integer t . Then by Theorem 2.12, there exist non-zero vectors x.. , • • • , x in 1 r-2pC U and a subset W of U such that t t Q. = {x -x -a p .UP : a e W) . r-2pC Suppose that some type one subspace y •y ,.U e C . r - l Since 1 S infinite, we are able to choose b e W such that <b> £ ( <y 1 > , ••• , <y r_ 1 >^ • By Theorem 2.3, we have y l y r - l t t . x. x .bP - f P - 1 t (x.. x t-'bP / o n e product) «f r-2pt for some f in U . In both cases, <b> e {<y > , t.> , <y >} 1 ' lr-1 contradiction. Hence no type one subspace is in. C . Suppose that some power type subspace y 1 y t r-p I UP e where Z. < t . Choose c e W such that <c> £ {<y^ > , • • • , <y r-p By Theorem 2.2, we have r-p t t I x_ x -c P . f P _ P 1 r-2pC t ^ •x .c^ /P products)'fP r-2p t t for some f in U . Since p > p , in both cases, <c> e {<y^ > a contradiction. Hence P^ CVU) C\ C = 0 for £ < t P^ -£ Suppose now some y^ y £' e ^ where i. > t and r >_ p r-p 52. 0 Choose d e W such that <d> =)= <y^ > for a l l i = 1 , • • • , r - p and <d> =f= <x^ > for a l l j = 1 , ••• , r - 2v~ . Then from Theorem 2.2, we obtain t x 1 x • d P = r-2p t Z t r-p" Z (y1 y J p products)-f P (2) r-p t Z for some f in U . If (2) holds, then <d> = <f> and p = p because of our choice of d . This yields a contradiction. Hence (1) holds and t Z t t Z <d> = <f> , p = p - p . This implies that 2p = p and hence p is Z divisible by 2. Therefore p = 2 and t + 1 = Z . Also i f r > p , we have x n x = Ay. y p for some A in F . Therefore r-2p r-p 2 t + l 2 t 2 t C = {M : M = x- x t - j . ! * 0 " o r M = x. x - -w -U where w e W} 1 r-2 t 1 1 r-2 t 1 Case 2. Suppose that P 1 is i n f i n i t e . By Theorem 2.13, there exist non-zero vectors x, , ••• , x „ in U and W C U such that 1 r-2 — = {x- x _.a.U : a e W} . 1 1 r-2 t Assume that some power type pure subspace y^ y G ^ where r-p 53. r 21 P*" and t is a positive integer. Choose b e W such that < b > =}= < y i > for a l l i = 1 , • • • , r - p * and < b > =j= < X j > for a l l j = l , * ' ' , r - 2 . In view of Theorem 2.3, we have x _-b = 1 r-2 t y ± y t - f p (3) r-p t y l ' * ' " y i " y t ' f P ^ r-p for some f in U . If (4) holds," then <b> = <f> and pfc = 1 by our choice of b . This yields a contradiction since pfc =j= 1 . Hence (3) holds. This implies that <b> = <f> and pfc - 1 = 1 . Hence p = 2 and t = 1 . Also i f r > 2 , we have x, x _ =. n y, y „ for some n in F . 1 r-2 J1 3 r-2 Therefore 2 C = {M : M = xn x _»U or M = x. • • • • «x „'a-U where a e W} 1 r-2 1 r-2 Hence the proof is complete. r S §3. Rank One Preservers From VU to VU , r < s Let U and W be finite dimensional vector spaces over a field F . Let f : U W be a linear transformation. Let w., • • • ,w be 1 s-r s - r non-zero vectors of W where r < s . Then the mapping r s ty : XU -> VW defined by 5 4 . <f>(x1 , , x r ) = w 1 w 8 - r * f ( x i > f ^ x r ) is clearly multilinear and symmetric. Hence by the universal factorization r * r s property of VU , there exists a unique linear transformation f : VU -* VW such that * f (x x ) = w w -fCx^ f(x ) 1 r 1 s-r 1 r * for each x- , • • • , x e U . f is called the linear transformation 1 r induced by w, , ••• , w and the linear transformation f : U W . 1 s-r It is easy to derive the following two properties: * (i) f is a monomorphism i f and only i f f is a monomorphism. ft (ii) Let g be induced by y.. , • • • , y and g : U -> W . J- s r ft ft Let the rank of f be _> 2 and F be an infinite field. Then f = g i f and only i f f = ng and w. •••. . .w = Xy, y for some n , A e F ° 1 s-r J1 J s-r such that An = 1 . r s 3.1. Definition. A linear transformation T : VU ->• VW is a rank one preserver r i f every non-zero pure vector of VU is mapped to a non-zero pure vector of VW . r s 3.2. Definition. A rank one preserver from VU to VW is said to be a r type one mapping i f every type one subspace of VU is mapped into a type one s subspace of VW . 55. Throughout the rest of this section, U is assumed to be a vector space over an algebraically closed field F . We shall show that r s every type one mapping from VU to VU , r < s , is induced by some s - r vectors of U and a non-singular linear transformation on U provided that dim U + 2 . If dim U = 1 , i t is not hard to see that every rank one r s preserver from VU to VU , r < s , is induced by s - r non-zero vectors of U and a scalar multiple of the identity mapping on U . Hence, from now on, we assume dim U >. 2 . 3.3. Definition. Let r > 2 . Two type one pure subspaces y^ ^r-1*^ , r and z , z of VU are said to be adjacent i f y, y .. 1 r - l — 5 1 1 ' r - l and z^ Zr-1 n a v e exactly r - 2 common factors counting multiplicities. 2 If r = 2 , any two distinct type one pure subspaces of VU are said to be adjacent. It is easy to show that two distinct type one pure subspaces M and N are adjacent i f and only i f MO N =j= 0 [4, Theorem 2.18]. r s 3.4. Lemma. Let T : VU VU be a rank one preserver. Then the images r of two adjacent type one pure subspaces of. VU are distinct. Proof: Let = x^ X r - 2 y i , i = 1 » 2 , be two adjacent type one pure r subspaces of VU . Let f. , i = 1 , 2 , be two linear transformations of U 5 6 . onto TCM^ defined by f^u) = T(x 1 x r - 2 ' y i ' U ^ f o r e a c h u e u • Since T is a rank one preserver, i t follows that f i is a monomorphism. Assume that T ^ ) = T(M2) . Then i s a non-singular linear mapping of U . Since F is algebraically closed, ^2^^1 h a S a non-zero eigenvector, say v . Let X be the corresponding eigenvalue. Then f^v) = Af 2(v) . Hence T( X ; L x r - 2 ' y l ' V ^ = A T ^ x i X r - 2 ' y 2 ' V ^ ' It follows that y^ = Xy2 . Therefore = M2 , contradicting our hypothesis. This proves that TCM^ ) =f T(M2> . 3.5. Remark. The above proof is exactly the same as the case r = s proved in [4, Theorem 3.3] r 3.6. Definition. Let M = x„.....x ,.U be a type one subspace of VU. . 1 r - l i Then the factors of x, x . are defined to be the factors of M . 1 r - l r s 3.7. Lemma. Let T : VU -> VU , be a type one mapping where r < s . r Then the images of any two type one subspaces of VU have at least s - r common factors counting multiplicities. Proof: Let 'M, = x , x .U and M = y, y ->U be any two 1 1 r - l r J l r - l type one subspaces. Consider the pure subspaces 5 7 . Mi = y 1 y i - l * x i X r - l ' U » i = 2 » **• » r - 1 . For each i - 1 , ••• , r - l , -T(M ) A T(M ± + 1) =f 0 since M ±n M + 0 . Hence T(M^) and T(JL+^) have at least s - 2 common factors. Consequently, T(M^) and T(M ) have at least s - r common factors. r s 3.8. Lemma. Let T : VU VU be a type one mapping. For any non-zero vectors y l ' " * " ' y r 2 ° f U. , we have {T(y y _,-u-U) : u e U , u =j= 0} = {z z _9-w-U : w e W} . for some non-zero vectors z, , ••• , z „ e U and some W C U . 1 s-l — Proof: In view of Lemma 3.4, C = (T(y^ y^ ~' n'^ : u e U , u =j= 0} is s an infinite family of type one pure subspaces of VU . Since any two members of C have a non-zero intersection, i t follows from Theorem 2.13 that C = {z, z „'W«U : w £ W} 1 s-Z for some non-zero vectors z, , ••• , z • of U and some W C U 1 s-l — r s 3.9. Lemma. Let T : VU ->• VU be a rank one preserver where s >_ r > 2 . Let x, , ••• , x , be r - l non-zero vectors in U such that 1 r - l <x > 4= <x _> . If {T(x " x «-u-U) : u E U , u 4= 0} r - l 1 r-2 1 r-2 1 C {y y 0«u*.U : u £ U , u =f 0} for some non-zero vectors y. , • • • » y _ 58. of U , then {T(x1« • • • •x^_2.xr_1.u-U): u e U , u f 0} <£ {y y^-u-U : u e U , u =j= 0} . Proof; Suppose to the contrary that {T( X ] L \ - 2 ' X r - l ' U ' V ^ : u e U , u =}= 0} C {y yg_2'u.U : u e U , u f 0} . (1 For each non-zero g e U , let M = x, x 0'g'U and N = x.",«'-x -x , *g'U g 1 r - 2 & g 1 r - 2 r - l & Since M e .{x, x „-u-U:ueU,u :}=0} , we have x , 1 r - 2 1 r - l r - l for some non-zero f e U . Now, choose a non-zero vector y such that <y> f <xr* i> > < y > f < x r 2  > ' -^en by assumption T(X, x r_ 2-y- u) = y± y s - 2 * y ' ' u ( 3 ) for some y 1 e U and from (1), T(x. x"" _-x '•y-U) = y, y 0*y , , ,U (4) 1 r - 2 r - l 1 s - 2 for some y' ' e U . Since <x > 4= <x > , i t follows that . M and N are r - l 1 r - 2 y y adjacent. Hence by Lemma 3.4, T(M^) =f= T(Ny) . Therefore <y'> =f <y'*> . 59. Consider the following equalities: T(x^ X r - l " y ^ = " y s - 2 ' f * y f ° r s o m e y (from (2)) - y_.i...y . y'. a for some a (from ( 3 ) ) = y- y o * y ' ' " b for some b (from ( 4 ) ) We have <y'> = <b> , <y"> = <a> since <y'> =(= <y"> It then follows that either <f> = <a> = <y"> or <f> = <b> = <y'> If <f> = <y"> , then T(M ) = T(N ) contradicting Lemma 3.4 since x r - l y M and N are adjacent. Similarly, <f> = <y'> implies that T(M ) = T(M ) , X r - 1 y x r - l y a contradiction. Hence the Lemma is proved. 3.10. Definition. Let V be a vector space of dimension k . Then a set of vectors z, , •.. , z is said to be in general position i f any 1 m k vectors from z^ , ••• , z^ are linearly independent. i We shall need the following Theorem in order to prove Theorem 3.14. 3. 1 1 . Theorem. Let V^G) be a symmetry class of tensors over V X associated with a subgroup G of S and a character x o n G . Let m dim V = k . If k ^  m , then V^G) has a basis of tensors of the form X x.,*'««*x in which x, , ••• , x are linearly independent. If k < m ± m 1 m 60. and F is an infinite field, then V^G) has a basis of tensors of the form X x * ... * x such that x, , ••• , x are in general position. 1 m 1 m e r Proof: Consider the mapping ty : XV -»- V^ CG) defined by 1 X <Kv, , • • , V ) = V * • • • * V 1 m l m. for each v, , ••• , v in V . Clearly ty is multilinear, hence by the 1 m • J m universal factorization property of ®V , there is a unique linear transformation ty : ®V -> V^VG) such that X <J>(V-i ® • • • ®v ) = V ' * • • • * V i m 1 m for each v. , ••• , v in V . Since V^(G) is spanned by the set of a l l 1 m X decomposable elements, i t follows that ty is an onto mapping. Therefore m i t suffices to prove the theorem for the case of ®V . , The theorem is trivial in case of dim V = 1 . Hence we may assume k >_ 2 . We proceed by induction on m . Suppose first m = 2 . Let u be any non-zero vector of V ." ' Let v be linearly independent of u . Then u g u = u®v - u<g)(v-u) . Hence the set of a l l tensors of the form x^® where x^ and are linearly independent, spans 2 ®V . 61. Assume now the theorem is true for m = n ', n >_ 2 We have the following two cases: (i) k _> n + 1 . Let u^® "' ® u n + ^ e "(g^  v • By t n e induction hypothesis, u, ® ••• ® u = Z a.y... ® •••COy. 1 ^  n . .. i l l ^ J m i=l for some t , some a^ in F where y ^ , • • • , y^ are linearly independent for each i . If y , ••• , y^ n , u^+^ are linearly dependent, choose in V such that y ^ , ••• , y^ n , z^ are linearly independent. Then y i l * ® y i n ® U n + l = yil® *•• ® y i n ® Z i " y i l ® ''' ® y i n ® ^i^M' Note that y., , ••• , y. , z. - u are linearly independent. Since i i in l n+1 t u l ® ••• ® u n + l = . E n a i ( yil® ® y i n ® V l ) ' i=l we conclude from (1) that I^"V has a basis of the form x^(g) • • • CS> xn4.^ such that x^ , ••• , are linearly independent. (ii) k < n + 1 . Let u^ (g) ••• ® u n + ^ e "(g)^ • BY t n e induction hypothesis, 62. t u-L® • *' ® \ = ^ b i v i l ® * " ® v d for some t and some b. in F where v., , ••• , v. are in general 1 i l m. position for each i . Since F is infinite, we have for each i , V + U <v.. , ••• , v > d = ( j l ' "' ' Jk-1> where i n denotes the totality of strictly increasing sequences of k - 1 integers chosen from 1 , ••• , n . Choose z^ e V such that J £ Q k - l , n 1 ^ Note that the vectors z. - Au ,, . v.. , ••• > v.. are linearly dependent for n+1 x j x i ] w at most one value of A in F . Hence we are able to choose a non-zero X. in F such that z. - A. u^., , v. . , ••• , v. . are linearly ~ V l - - . . \ . U . - , , V . . » » v . . i i x n+1 1 3 x i J k - ] independent for a l l i in Q, , .We have r J xk-l,n V i l ® ® V i n ® U n + l = X i 1 ( v i l ® ® V i n ® Z i - V i l ® ® V ^ i V l " ' Hence t _ i U l ® ' " ® U n + l = \ V i < v l l ® ' - - ® v i n ® z i - V i l ® , , , ® V l n ® ( z r X l V l ) ) 1=1 6 3 . Since v ^ , — } , z^ are in general position and v., , ••• , v. , z. - X. u are in general position for each i , i t IJ. m I 1 n+l n+1 follows that (5?, V has a basis of tensors of the form x„®*"(8 x •cy 1^ m such that x^ , ••• , are in general position. 3.12. Remark. It is clear that i f V is over a finite field, then whenever m is chosen to be greater than the number of vectors in V , two of the vectors from x, , • ••• , x of U must be equal and so the 1 m above theorem is false. The following Theorem is due to Cummings [4] : r r , 3.13. Theorem. Let T : VU -»- VU be a type one mapping. If dim U f 2 , then T is induced by a nonsingular transformation on U . r s 3.14. Theorem. Let T : VU VU , r < s , be a type one mapping. If dim U > 2 , then T is induced by s - r vectors of U and a nonsingular linear transformation on U . Proof: Case 1: r > 2 . Let x, , ••• » x' be r - l fixed non-zero vectors 1 r - l of U such that <x.> + <x.> i f i 4 j • let 1 ' J T T(xn x - .U) = z, z n -U 1 r - l 1 s - l 64. From Lemma 3.8, we see that for each i = l , ••• , r - 1 , {T(x x*. x -'u-U) : u e U , u 4= 0} = {y._ v., ..-vU : v e U. CT U} 1 l r - l ' i l •'i(s-2) i — for some vectors y., > ••• , y./ e U and for some U. d U i l  Jx(s-2) l — Since z, z , • U E ( y . n y. , 0^•v•U:ve'U. C u } , i t 1 s-1 i l i(s-2) l — follows that z, z - = X; y., y., o\*v. for some v. e U. and 1 s-1 3 • ' i l •/i(s-2) 1 1 1 some X. e F where i = l , ••• , r - l . I According to Lemma 3.9, we have for each pair of distinct k and j , j l y j ( s - 2 ) > * < y k l y k ( s - 2 ) > <y Therefore we may assume without loss of generality that for each i = l , • • • , r - 1 y., y., O N = n . z, z z i • ' i l •7i(s-2) I 1 s-i s-1 for some n. e F and that any two vectors from z ,>•••> z / ,\ are i J s-1 s-(r-l) linearly independent. 65. r Our task is to show that T(VU) C <z, z *s....-..s : s. E U,i=l, — 1 s-r 1 r l Now, let u^ , ••• , u r - i b e a n y r - l non-zero vectors of U such that <u±> =f <u.> i f i =(= j . Let T(u.. u 1 -U) = w. w . -U 1 r - l 1 s - l We shall show that <z > , ••• , <z > are s - r factors of w, w 1 _ s-r 1 s-l To do this, we first show that there exists a type one pure subspace B r in VU with r - l distinct factors such that the following hold: a) T ( B ) = 8 l g s_ r- ui u ; _ x - u ; (ii) <g, > , • • • , <g > are s - r factors of w, w , ; &1 .' &s-r 1 s-l (i i i ) <u!> | {<z > , ••• , <z _->} where i = l , « " , r - l . Consider the family C = {T(u1 u^.u-U) : u £ U , u =(= 0 } Since wis w «U £ C , i t follows from Lemma 3.8 that 1 s - l C = {w. w . w ..u.U : u E V,CU} 1 s - 1 l s - l 1 — for some i ^ and some V^CU . We may assume i ^ = 1 . Since C is an infinite family, we are able to choose non-zero vectors f £ U and 66. f e V 1 such that T(u x ur_2-f.U) = v1 w g_ 2.f .U , <f> £ {<u^ > , ••• , < u r _ 2 > ^ and < f , > i i<Zj> , •••* , < Z S _ 1 > J • Note that the factors of u, u „'f«U are distinct. 1 r-2 Assume now we have shown that for some t , l < _ t < r - l , there exists a type one pure subspace d^ d r - l W l t n r - l distinct factors such that <a> T < d l d r - l ' U ) " h l h s - l ' U 5 (b) <h1> , ••• , < n s _ ( t + i ) > a r e s ~ ( t + 1) factors of W l W s - 1 5 ( c ) < h s - t * " ' ' ^ s - l " ^  { < z l > » ' < Z s - l > } " 1 Consider the families C. = {T(d, d ,-u-U : u + 0 , u e U} l 1 r - i r - l 1 where i = 1 , ••• , t + 1. Since h, h , «U e C. for each 1 s-l 1 i = 1 , ••• , t + 1 , i t follows from Lemma 3.8 that C =' {h, h , h . . u . U : u e W. C U} i 1 s-k. s-l l — l 67. for some k. , I < k. < s - 1 and for some W. C U . Since a l l 1 — i — l — factors of d, d ..-U are distinct, by Lemma 3 . 9 , we must have 1 r - l ' k f k for m f n where m , n = 1 , •• • , t + 1 . Therefore there is n 1 m 1 ' ' ' a j such that k. > t + 1 . We may assume k. = t + 1 . Now C. J - J J is an infinite family implies that there are non-zero vectors d e U and d' e W. such that 3 T(d, iT . d ,-d.U) = h h /t.0x'd'.h ^ h -U , 1 r-j r - l 1 s-(t+2) s-t s-1 <d> A {<d > , • • • , <d ..>} and <d'> A {<z,> , • • • , <z >} . Note T 1 r - l r 1 s-1 that the factors of d, d d ,.d»U are distinct. Therefore our 1 r-j r - l inductive argument shows that there exists a type one pure subspace B r of VU with r - l distinct factors satisfying conditions (i), (ii) and ( i i i ) . Now, we shall show that <g, g > =.<z, z > . For &1 6s-r 1 s-r each i = 1 , ••• , r - l , choose a vector f. e U. such that <f±> | {<g1> , ••• , < 8 s _ r > » ••* i » ••• » < u r _ i > } • From Lemma 3 .7 , i t follows that g, g -u,' u' , and 1 s-r 1 r - l z, z z ,.f. have at least s - r common factors for each 1 s-i s-1 1 i = l , ••• , r - l . Since <u'> , ••• , <u' > A {<z > , ••• , <z >} 1 r - l r 1 s-1 68. and k { <8 1 > » ••• » <8 g_ r > » < u { > » * * * •» < u r _ i > ^ » i t : f o l l o w s t h a t < g - i > » * * • > <g > are s - r factors of z, z z , for 1 s-r 1 s-i s - l each i = l , ' " , r - l . Consider the isomorphism $ from the symmetric algebra VU onto the polynomial algebra F[£, , ••• , E, ] described in §1. The polynomial 1 n . i > ( & ± g s _ r ) l s a factor of c f>( Z l zg*_i z g_ 1) in F f ^ , , £ ] for each i = 1 , ••• , r - 1 . Since <<|>(z _, )> ={= <<J>(z )> for k =f= s ic s ~' in m where k , m = 1 , ••• , r - 1 , i t follows that <j>(z and <J)(z ) are relatively prime in the Gaussian domain F[£, > • • • » £ ] » and therefore I n <Kz.. z ) is a greatest common divisor of i>(z, • • • • •'zS z .) , 1 s-r 1 s-i s-l i = 1 , •*• , r - l . This implies that 4>(g1 g _ ) is a factor of ^(z, z ) . Hence <<b(g, g )> = <4(z, z )> since they •T 1 s-r T "1 es-r 1 s-r J are both products of s - r linear homogeneous polynomials in , ••• , £ Hence <g, g > = <z, z > . This shows that <z,> , ••• , <z > 61 6s-r 1 s-r 1 s-r are s - r factors of w^  Ws-1 * Consequently T(u, u , -U) = z z «D.U 1 r - l 1 s-r r - l for some non-zero pure vector D in V U . Since dim U > 2 , Theorem 3.11 r - l implies that V U has a basis of pure vectors of the form u, u , r 1 r - l where < u ^ > k < u j > i =f= J Therefore we have 69. r T(VU) C <z- z • s, s : s. E U , i = 1 , • • • , r> . — 1 s-r 1 r i * r r Define a mapping T : VU ^  VU as follows: T*(C) = C* i f T(C) = z, z .C , C* e VU 1 s-r T is well-defined since z, z -C' = z, •••••z •C' will imply 1 s-r 1 s-r C' = C'' . Since T is a rank one preserver, i t follows from Lemma 1.8 that * T is also a rank one preserver. Moreover, T is a type one mapping implies * * that T is a type one mapping. By Theorem 3.13, T is induced by a non-singular linear transformation f on U . Consequently, T(d. d ) = z. z .f(d_). f(d ) 1 r 1 s-r 1 r for each d. in U . Case 2: r = 2 . By Lemma 3.8, we have MT(u-U) : u e U , u =)= 0} C ^  a^-u-U : u e U , u =}= 0} for some a. in U , i = l , , , , , s - 2 . Hence l T<™> ^ < a l a s - 2 , S l ' S 2 5 S l ' S2 £ U > 70. By the same argument as in case 1, we see that T is induced by a^ , ••• , ag_2 °f U and a non-singular transformation on U . Hence the proof is complete. s 3.15. Remark. Theorem 3.14 is no longer valid i f VU is replaced by VW with dim W > dim U . For example, i f dim W >^  dim VU and z, , ••• , z are non-zero vectors of W , then there1 is a monomorphism 1 s-r r s T : VU -> VW such that T(VU) = z, z . -W 1 s-1 o r where W C W and dim W = dim VU . Clearly T is a type one o — o mapping which is not induced by s - r vectors of W and a linear transformation from U to W . r §4. Rank One Preservers On VU . Let U be a finite dimensional vector space over an algebraically closed field F . In this section we show that: i f dim U >. r + 1 , then r every rank one preserver on VU is induced by a non-singular linear transformation on U ; i f 2 < dim U < r + 1 and the characteristic of F is greater than r or equal to zero, then every rank one preserver on r VU is either induced by a non-singular, linear transformation on U , or the r image of VU under the rank one preserver is a type r pure subspace. 71. We first establish the following Lemmas. r s 4.1. Lemma. Let T : VU -*• VU be a rank one preserver where r _< s . Assume that dim U > 2 and char F = 2 . Let x, , •• • , x „ be 1 r-2 non-zero vectors of U . Then i t is impossible that „k+l { T< X1x r_ 2-yU):y £U,y=f0} = {z± z -U } \J {z^ z . s-2 s-2 for some non-zero vectors z, , • •• , z , ,, of U and some W C U 1 s-2 k + 1 where k is a non-negative integer. Proof: Suppose to the contrary that ~k*t~l 9k k^ {T(M ) :yeU,y+0} = {z x z ^ . U } (J i z \ 2 k + 1«w -U :w£W} y s-2 s-2 where we denote x, x „'y-U by M 1 r-2 J y 2k+l Assume that T(M ) = z, z . ,.'U . Let u_ e U be u, 1 0k+l 2 1 s-2 independent of u^ . Let v^ = u^ - u 2 and v£ = u2 ' ^ ^ e m m a 3-4, T(M ) + T(M ) and T(M ) =f= T(M ) . Therefore U l V l U l V2 T ^ v > = Z l 2 9 k + l - V l 2 k ' u 2 k 1 s-2 k k T ( M ) = Z l.....z .v'2 .U2 (2) 2 s-2 72. for some , in W . Let z and y be two independent vectors of TJ k+1 such that z , y { <v] , v'> . Since T(M ) = z z , .. -TJ , J. Z U. x — K T X x s-z i t follows that 2k+l T ( x l x r - 2 - V f ) = z l z Jc+l* z s-z 2k+l T(x1.....xr_2.ui.g) = Z i z y s-z for some f , g in U . Clearly <z> =f= <y> implies that <f> =f <g> . We have either <f> f <u^ > or <g> =}= < u i > • W e m a y assume that <f> =j= <u^ > . Let A = x „.u-'f . In view of (1) and (2), we have 1 r-2 1 2 k 2 k B 5 X l X r - 2 ' V l ' f - z l Z 0 k + l * f l , VI s-2 C E X l X r - 2 ' V f " Z l Z 9k+l' f2 ' V2 2 s-2 for some f^ , f 2 in U . On the other hand, since <f> =)= <u^ > , by Lemma 3.4(, we have T ( v • z i z _ 9 k + i - f , 2 k - u 2 k «k „k for some f in W . Since T(A) , T(B) , T(C) e z z l r. 1«f •TT , 1 s-2 R + 1 i t follows that <f'> = <z> = <f^ > = < f > by our choice of z . 73. Now, let f^ = az and = 3z where a , 3 are non-zero elements of F . We obtain k k k k T ( B + C ) = z i 2 9 k + r f i - v i 2 +'«!•••:•" 9 k + i - f 2 - v 2 2 s-2 s-2 2^ 2^ 2^  2^ = z i 2 jc+r ( a z ) - VI + V " Z 0k+r ( e z ) - v 2 s-2 S-2 2^  2^  2^  2^ = z i 2 ,k+rz • • ( o v i > + z i - ' - " z ' k+i-z - ( p vP s—Z s—Z 2 k 2 k = z^ z k+l"Z * Cctv^ H-gv^ ) since char F = 2 . s-2 However, since u i = v i + v 2 ' w e a i S O n a v e 2k+l T(B+C) = T(A) = z^ z -z This implies that av^ + 3v2 and z are linearly dependent, contradicting our choice of z . This completes the proof. r s 4.2. Lemma. Let T : VU -> VU be a rank one preserver where s >_ r > 2 . Suppose dim U > 2 and char F = prime p . Let x i > ' ' ' > x r i D e r-2' non-zero vectors of U such that <x > =(= <x „> • If r - l 1 t t {T(x1 x 2.yU):ycU,y+0} C {z z t-u P .UP : u e U , u =j= 0} r s-2p 74. for some non-zero z. , • • • , z of U where t is a positive integer, 1 s-2pt then t t {T( X ; L V - 2 ' x r - l ' y ' U ) : 7 6 U ' 7 ^ ° } 4= { z l z t ' u P , l j P : u £ U ' u + 0 } s-2p Proof: The argument is similar to that used in the proof of Lemma 3.9. r s 4.3. Theorem. Let T : VU ->• VU be a rank one preserver. Let dim U > s + 1 . Then (i) i f r = s , T is induced by a non-singular transformation on U ; (ii) i f r < s and the characteristic F is either zero or equal to g a prime p > — , T is induced by s - r vectors of U and a non-singular transformation on U . Proof: We shall show that under these conditions, T preserves type one subspaces and then the theorem will follow from Theorem 3.13 and Theorem 3.14. r Let M be a type one subspace of VU . Since T is a rank < s one preserver, T(M) is a pure subspace of VU . Moreover, dim M = dim T(M) = dim U > s + 1 . Let T(M) C N where N is a maximal pure s subspace of VU . If N is of type k where 1 < k <^  s , then dim N = k + 1 <_ s + 1 , contradicting the fact that dim T(M) > s + 1 . Hence N is of type one or power type pure subspace. Moreover, T(M) = N . s Suppose that T(x1 x -, *U) e P (VU) for some non-zero 75. x2_ ' '" ' x r - l '*"n ^ a n d s o m e positive integer t . Then char F = a prime p If r = 2 , then u ^ U ^ u^U =f 0 for a l l , u~ =j= 0 . Hence T(U;L-U) f l T(u2-U) =(= 0 for a l l u± ., u~ =}= 0 . Also T(U;L-U) =(= T(u2'U) for a l l =(= u 2 by Lemma 3.4. In view of Theorem 2.14 and Lemma 4.1, we have {T(u-U) : u e U , u =(= 0} C P^VU) . Hence by Theorem 2.12, s >_ 2p^~ , a contradiction to the hypothesis on s . Now consider r > 2 . Let y, , ••• , y - be non-zero vectors of U '1 r - l such that <y^ > + ^ o r 1 + J • Consider the following type one r pure subspaces of VU : M l = x l X r - l * U » M2 = X l X r - 2 ' y l ' U > *" ' M r = y l y r - l ' U 5 that is, M. = x, x . »y. y. . 'U , i = 1 , r . Since I 1 r - i Jl J i - i x r_ 2. U l-U)n T(x x x r_2* U2' U ) + ° for a l l non-zero u^ , u 2 in U , by Theorem 2.14 and Lemma 4.1, s {T(x xr_2-u'u> : u e U , u =(= 0} cr P (VU) Therefore. T(M2) e P (VU) . Similarly T(M2) e P (VU) implies that 76. s s T(M_) e P_ (VU) and so on. It then follows that T(M ) e.P (VU) . J t r t t Let T(Mr) = z 1 z t«U P . In view of Theorem 2.12, Theorem 2.14 r s-p and Lemma 4.1, we have for each i = l , # , , , r - l , t t {T(y y. y -u-U) :ueU,u=f=0} C {w w -uP -UP :ueU, u=j=0} i(s-2p ) for some non-zero w w' of U . Since 1 1 K s ^ p 1 1 ) :(Mr) e {T(y1 y± y^-u-U) : u e U , u =f= 0} , it follows that t t t P P P Z , Z -«U = W . - « « r " W ^ 'U. -U 1 t xl . ( _ t N X s-p x(s-2p ) for some non-zero u. in U where i = 1 , ••• , r .- 1 . Hence x z, , • •• , z ^ has at least pfc factors of <u.> for each i by Theorem 2.1. 1 \ t r x ^ 4 • s-p Since <y, > =f <y.> for k =)= j , according to Lemma 4.2, <w#1 w > =}= <w w > 2 1 j(s-2p t) k l k(s-2pt) for k =}= j . This shows that < u j > T" < x \ > ^ o r J f k . Consequently 77. z^ z ' has at least (r-ljp 1" factors. This implies that s-p s - p t 21 (r-l)?*" and thus s i> rp1" , a contradiction to the hypothesis r on s . Therefore no type one pure subspace of VU is mapped onto s a power type subspace of VU . This proves that T is a type one mapping. Our result thus follows from Theorem 3.13 and Theorem 3.14. That Theorem 4.3 is not true i f s >^  rp where char F = p is shown by the following: 4.4. Example. Let char F = p and s >_ rp f c for some positive integer t . Let k = p*" . Let u, , • • • , u be a basis of U and w.. , ••• , w I n I n be another basis o| U . Let f : U U be a mapping defined by n n k "k th f( E X.u.) = £ X. ;w. for any A. in .F where X. is the k root of . , 1 X . - 1 1 . J x l . *-1=1 1=1 Then i t is easily checked that f has the following property: f(Xv + nw) = Xk f(v) + n k f(w) for any X , n in F , v , w in U and f is one to one. r s Define a mapping <J> : XU -» VU by 1 * ( v l > » V " Z l Z s - r k ' f ( v l ) k f ( v r ) k where z, , ••• , z , are fixed non-zero vectors of U . We have 1 s-rk 78. k k k • (Vj, •••.,Xv i+Tiw i,. ..,vr) = z± ^ g - r k ^ ^ i ^ .f(Avi+nwi) f(vr> 1 1 = z x z s _ r k ' f ( v l ) k ( A k f ( v 1 ) + n k f ( w . ) ) k f(v r) 1 - z......z , . f ( v . ) k (X kf(v.)) k f(v ) k 1 s-rk 1 l r 1 + z. z ,-f( v , ) k (n kf(w.)) k f(v ) k 1 s-rk 1 I r = Az 1.....z s_ r k.f(v/.....f(v i) k.....f(v r) k +nz 1 z s _ r k . f ( v 1 ) k f(w.)k... = X<J)(v1 , '•• , v ± , ••• , v r) + n<j>(v1 , ••• , w± , ••• , v r) . Hence ty is multilinear. Clearly ty is symmetric. Hence there — r s exists a unique linear transformation ty : VU -*• VU such that * ( V 1 v r > = Z l Z s - r k , f ( v l ) k f ( v r ) k for each v j » * * * » v r 1 1 1 ^ • Clearly ^ is a rank one preserver from r s VU to VU and is not induced by s - r vectors of U and a linear transformation on U . «. r s 4.5. Theorem. Let T : VU -> VU be a rank one preserver. Let dim U = s + 1 We have (i) i f r = s , then T is induced by a non-singular transformation on U ; (ii) i f r < s and the characteristic of F is either zero or equal s to a prime p p> — , then T is induced by s - r vectors of U and a non-singular linear transformation on U . 79. Proof; Since dim U = s + 1 , every pure subspace of type one or of power s type or of type s in VU has dimension s + 1 . Also every type one pure r r subspace of VU has dimension s + 1 . If a type one subspace of VU were s mapped into a type k pure subspace of VU , 1 < k < s , then the image would have dimension j < k + l < s + l . This would imply that some non-zero pure vector is mapped into 0 , which is not the case. Hence a type r one pure subspace of VU is mapped only onto a type one pure subspace or s a power type pure subspace or a type s subspace of VU . Suppose that some type one subspace x^ X r - 2 * y ^ S m aPP e d s onto a type s subspace ^( s) °^ ^ » w n e r e V is a two dimensional subspace of U . We shall show that this leads to a contradiction. Let C = (T(x, x „.u.U) : u e U , u 4= 0} . By Lemma 3.4, C 1 r-2 is an infinite family. We shall show that V ^ is the only type s subspace in C . Suppose that C !D (V. N , V, N} where V. N is another type s — (s) (s) (s) subspace of VU . Then V' f\ V* is 1-dimensional since V ^ f\ =j= 0 . Choose a non-zero z in U such that T < x l x r - 2 ' y ' Z ) = V * ' V s i * I where dim <v , • • • ,v > = 2 , <y> f <z> and VHV f <v > for a l l i = 1 , •• x s X ^ Note that T( X ; L x r_2* Z " U ) ^ V(s) + ° a n d T ( x l x r _ 2 * z . ' U ^ V(s) + ° ' 80. If T(x, x 'z-U) = z, z , .IJ for some z. in U , then 1 r-2 1 s-1 I ft ft z , • • • , z _ e V fl V ; hence <z > = • • • = <z ,> = V f| V . But * v «v e z, z , «U , thus for some i , <v.> = V 0. V , a contradiction. 1 s 1 s-1 1 Hence T(x^ x^ ^ 'z'^) cannot be a type one subspace. Similarly, i f char F = p , s > p1" , then T( X ; L xr_2.z-U) i Pt(VU) , 2 k where t is a positive integer. Also i f char F = p , s = p for some k positive integer k , T ( X l x 0«z«U) =j= UP since dim <vn , ••• , v > = j_ r—2. X s and vn v e T(x, x „.z.U) . Hence 1 s 1 r-2 T(x1.....xr_2.z.U) = W(s) for some type s subspace W(s) . Since Xr-2' y' Z £ x l ' x r _ 2 ' z ' U H x ± x r _ 2 ' y u » i , i t follows that v, v e V, N A W, N . This implies that 1 s (s) (s) <v, , ••• ,v>=V=W, a contradiction to Lemma 3.4 since x, x „-y-U I s 1 r-2 and x ^ x r _ 2 ' Z ' U are adjacent pure subspaces. Therefore ^( s) is the only type s subspace in C . 81. Let C be the collection of a l l type one subspaces in C . We shall show that C is finite. . Suppose that C is infinite. Since M^n M2 + 0 for every , in C , i t follows from Theorem 2.13 that C = {z, z 0-u . U : u E W e u} 1 s-z — for some non-zero vectors z.. , • • • , z in U and some WCU . According to Theorem 2.14, C = C U'{V / Y> or C = C U {V, N } U (z, • • • • • z 0-U2} . (s; (s) 1 s-2 In the latter case, char F = 2 . 2 If A = z.. z „'U £ C , we let f £ U such that 1 s-2 T(x, x *f.U) = A . Let v be a fixed non-zero vector of U such 1 r-2 that <v> =(= <f> and <v> =j= <y> . Then by Lemma 3.4, T(x 1 xr_2-v.U) = z± zs_2.v'-U (1) for some non-zero vector v' of W . Now, for any non-zero x in U such that. <x> =)= <y> , <x> =}= <v> and <x> =j= <f > , let T( X l x ..x-U) = z, z 0-x'.U . (2) l r-2 1 s-z Since z. z „.x'-U f\ V, N =|= 0 , we have x1 £ V . From (2), we h 1 s-2 (s) 1 ave 8 2 . for some v • e U . Since z. z 0«x'-v e z. z „.v'-U (from (1)) x 1 s-2 x 1 s-2 and <x'> 4= <v'> (Lemma 3.4), i t follows that <v > = <v'> . Hence x T ( x l " , , , V 2 ' v , D ) - Z l Z S _ 2 * V ' ' V U < T ( x l x r - 2 * V ' y ) > ^ <T(x1 xr_2-f-v)> U ^(x^^ x 2-v.y)> . This is impossible since dim T(x, x 0«v'U) = dim U > 2 . 1 r-2 If A | C , by the same argument as above, we have K x ^ - . - . x ^ . v.U) C Z l Z s _ 2 , v ' - V U < T ( x 1 x r_ 2-vy)> (J <T(X;L x r _ 2 ' v ' v ) > » which is also impossible. Hence C is finite. Since C is infinite, i t follows that char F = a prime p and for some positive integer t , the collection V of a l l power type subspaces of degree t in C is infinite. In view of Theorem 2.12, we have s > 2pt and t t A V = { y n y -\P -Up : w £ W } 1 s-2p * for some y^ in U and some W C U . By Theorem 2.14, we see that there is no other type one subspace or power type subspace in C except 83. 2 t + l possibly y 1 y _ -U in which case char F = p = 2 . s-2 -Since T(x^ x^^'y-U) = v^ s^ > w e maY choose a non-zero vector u in U such that <u> =f <y> and T ( x l x r - 2 , y * u ) = V l V s where < v ^ > + < v j > i f 1 + j • ' BY Lemma 3.4, we see that 9t+l T(x 1 x r_2' U' U ) e Vt ° r T ( x l x r _ 2 ' U , U ) = y l y t+l' U s-2 Hence v^ v g has a factor of multiplicity at least pfc , a contradiction. r Therefore no type one pure subspace of VU is mapped onto a type s r s subspace of VU . This means that every type one subspace of VU is mapped s onto a type one subspace or a power type subspace of VU . By the argument used in the proof of Theorem 4.3, we conclude that T is a type one mapping. Our result thus follows from Theorem 3.13 and Theorem 3.14. • 4.6. Lemma. Let W be a finite dimensional vector space over a field F of characteristic 0 or of characteristic greater than a positive integer t . Then every non-zero pure vector of the form xfc -'.y^* , 1 <_ j <_ t , in VW is in the span of M = {ufc : u e W} . Proof: Let A^  , 84. , ^ t + j _ ^e t + 1 distinct elements of F . Then (x+X^y)*' = xt + • • • + (£) x t~ k.(X iy) k + ••• + X^yt e M for each i = l , ••• , t + 1 . Consider the following system of non-homogeneous equations in t + 1 variables , ••• , ^ t +^ ' t+1 z x m c . i - i 1 1 t+i . Z A? C. i - i 1 1 = 0 i f m + j , 1 <_ m _< t , t+1 1 , E £. = 0 . i-1 1 ' (1) Since the determinant of the coefficients of (1) is det 1 A, t+1 t+1 i t follows that (1) has a solution, say £. = d. , i = 1 , » "^ t+1 _. . Hence Z d.(x+A.y) = (.) x «y^  is in the span of M i=l 1 1 3 (^ ) =j= 0 , we conclude that x*" 3 *y2 is in the span of M .. , t + 1 Since 4.7 . Theorem. Let W be a finite dimensional vector space over a field F of characteristic 0 or of characteristic greater than r . Then r r VW is spanned by M = {u : u e W} . 85. Proof: We proceed by induction and assume that any pure vector having less than k + l distinct factors is in [M] , the span of M , where r 2 _< k < r . Let A e VW have exactly k + l distinct factors. It is clear that we may write A = z, z -z z in such a way I t t+1 r that z has exactly 2 distinct factors and z ,, z has I t t+1 r exactly k - 1 distinct factors.. By Lemma 4.6, we have t + 1 t Z l Z t = }. d l V i . • i = l for some d. in F and some v. in W . Hence l l t+1 t t+1 t A - ( I d V Z t + l Z r = .\ d i ( v i ' Z t + l « r ) i=l i=l t t By the induction hypothesis, v.-z ,.. z e [Ml since v.-z ..•••••z J J V l t+1 r I t+1 r has at most- k distinct factors. Therefore A e [M] . This proves that r VW has M as a spanning set. 'Theorem 4.7 is well-known (see [2], p. 131). We have included a proof for the sake of completeness. From now on, we assume that U is an n-dimensional vector space over an algebraically closed field F with char F = 0 or , r r char F > r . We also assume that 3 < n < r + 1 . Let T : VU -> VU be a 86. • r • rank one preserver. Since every type k subspace of VU has dimension r k + 1 < r + 1 where 1 < k < r , and every type one subspace of VU has r dimension n < r + 1 , we see that every type r subspace of VU is mapped r onto a type r subspace of VU under T . 4.8. Lemma. If there are two distinct type r pure subspaces M and N r . r of VU such that MAN f 0 and T(M) = T(N) , then T(VU) = T(M) . Proof: Let M = S , N = S and T(M) = T(N) = S. . where S , S , S l ( r ) ^( r) ^ are two dimensional subspaces of U . By hypothesis, M f\ N = S 1 n s = (sns) 1 o 1(r) ^(r) 1 2 ( r ) Hence S2 . =f= 0 . Let S± = <y1 , y2> , S2 = <y1 , yg> . Consider S , = <Y9 , Yo> • Then S f\ S = <yT> , S n S = <y2> 3 2 3 3(r) 2(r) 3 3(r) X(r) preserver Hence T(S ) f | s/ \ 2 <T(y^) , T(y,)> . Since T is a rank one J(r) and <y2 , y^ > is' a two dimensional pure subspace, i t follows that <T(y*) ,'T(y^)> is two dimensional. Hence T(S ) = S, . because any r two distinct type r subspaces of VU have at most one dimension in common. Let z = ay^ + 3y2 + yy 3 where a , 3 , y are a l l non-zero scalars. Consider S^  = <y^  , z> = <y^ , 3y2 + Yy 3 > • Since 87. SA 0 S o 3<(6y+Yy )r> , S n S ^ <yh > V ) 3 ( r ) 2 3 4 ( r ) V ) 1 we have T(S^ ) (~\ S ^ 3 <T(y1) , T((3y2+yy3) )> which is two dimensional. Hence T(S^ ) = . Consequently, T( ,y3>) = since r r V<y^  , y 2 , y^> Is spanned by a l l pure products y where y e <y^  , y 2 , y^ > (Theorem 4 . 7 ) . Now, let w e U such that w £ <y^  , y 2 , y^ > • Let W = <y^  , w> Consider the type one subspace y^ yn"U . Let P = y^ y^'U • Since r dim (P fl V<y1 , y 2 , y3> ) = 3 , we have dim (T(P)O. S ( r ) ) L 3 • Since the maximal dimension of the intersections of two distinct maximal pure subspaces is 2 , we conclude that T(P) C s ^ . We observe that T ( W ( r ) ) ^ S(r) - < T ( y l ) ' T ( y l y 1 ' w ) > • S i n c e » y i _ 1 - w > " a two dimensional pure subspace, <T(y^) , T(y^ yn*w)> is also two dimensional. Hence T(W/ .)= S, s . By Theorem 4 . 7 , we conclude that ( r r (r) r T(VTJ) = S ( r j • This completes the proof. 4 . 9 . Lemma. Suppose that for any two distinct type r subspaces M , N such that MON =f 0 , we have T(M) =f T(N) . Then T is induced by a non-singular transformation on TJ . Proof: Let y be any non-zero vector of U . Let y , y^ , y 2 be linearly independent vectors. Let S^  = <y , y^ > , S2 = <y , y2> . Let 88. T(S± ) = S* of . 1, x ' T ( s ? ) _ S9 where Si , S' are two dimensional subspaces <r> (r> (r) ^(r) 1 2 U . Since S =f= S , S f l S ' +' 0 , by hypothesis, S* + S* (r) Z(r) 1(r) 2(r) Therefore S' O S' = T(S n S ) = <y,r> for some y* e U . (r) \v) ±(r) ^(r) Hence T(y ) = Ay' for some X in F . We claim that T(y y-U) = y' y'-U . Since T(y y-U) is a pure subspace, i t is contained in a maximal pure subspace. If T(y y-U) is contained in a type k pure subspace g, g , -W... . where . 1 r-k (k) 2 <_ k < r , then y' e g.^  8r-k' W(k) a n d n e n c e <^i> ~ <y' > > y' e W . This implies that g^ e W , a contradiction. If T(y yU) is contained in a type r subspace W ( r ) , then dim (S n y y - U ) = 2 implies that dim (T(S ) H W ) _> 2 ; (r) 1(r) ( r ) and dim (S f\ y yU) = 2 implies that dim (T(S - ) 0 W, .) >_ 2 . (r) l{r) K X ) Since T(S ) and T(S„ ) are both type r subspaces, i t follows that '\r) 2(r) T(S ) = W. v = T(S„ ) , a contradiction to our hypothesis. Hence 1(r) U ; ^(r) r T(y y-U) is a type one pure subspace of VU since there are no power type pure subspaces under the assumption that char F = 0 or char F > r . Since y , r e T(y y-U) , i t follows that T(y y-U) = y 1 y'-U . 89. r - l r - l r - l By Theorem 4.7, let x^ , ••• » x be a basis of V U . Note that 3 <_ dim U < r + 1 implies that r >_ 3 . Clearly, i f i =(= j , then x^ and x_. are linearly independent. Consider any type one subspace fc r - l z. z , -U . Let z, z , = Z X.x. where X. e F and i = 1 , ••• , 1 r - l 1 r - l . , 1 1 I i=l We shall show that T(z^ Z r - 1 * ^ "*"S a t y P e o n e P u r e subspace. Suppose to the contrary that (i) T(z 1 z r - l ' U ^ — S(r) ° r T ^ Z 1 Z r - l " U ^ — v^ k"^(k) , 2 _< k < r , for some two dimensional subspace S of U and some v n , ••• , v . e U - S . 1 r-k Let T(x. x. 'U) = x! x!'U , i = 1 , ••• , t . Note that we l i i l have T(x.) = n. x! for some n. e F where i = 1 , ••• , t . For i l l l j r r r j f i , <x^  , x^ > is a two dimensional pure subspace of VU implies that r r _ r r T(<x. , x.>) = <x! , x'. > is a two dimensional pure subspace of VU . Therefore x^ and xj are linearly independent i f i ^  j . Consider case ( i i ) . We choose a vector v of U such that v k <v > ( J * " U < v >US U(U<x! , x!>) . This is possible since <v.> , S 1 r-k . i . I i l i+J r - l r _ 1 <xl^  , xj> are a l l proper subspaces of U . Let T(x^ -u) = x^ -v . r - l r _ 1 For each i > 2 , let T(x. «u) = x! *u. . — i l l We shall show that <u.> = <v> for i = 2 , ••• , t . Since i -i n r - l r - l r—1 r—1 <x, -u , x. 'U> is a pure subspace for i > 2 <x! .v , x! .u.> is also l i r v — » l ' i i 90. a pure subspace. By our choice of v , <xj^  , v , x_j> is three dimensional, r - l r - l hence from Lemma 1.9, x' -v and x! ' U . have a common factor, say , 1 i i <w.> . Note that we must have <u.> = <w > , i > 2 . Either <u.> = <x'> l l i — i 1 r - l r - l or <u.> = <v> . If <u.> = <x'> , then <x' «v , x! *u.> is a pure l l 1 I ' l l r-2 r - l subspace implies that <x| «v , x| > is a pure subspace (Lemma 1.8). r-2 r - l Consequently, by Lemma 1.9, xj^  -v and x!^  - have a common factor, a contradiction. Therefore <u.> = <v> , i > 2 . I — Now, we have u. = a. v for some a. e F , i > 2 and i i l — Z r - l T( Z ] L z r _ l ' u ) = T ( 1 X i X i U ) 1=1 r - l t r - l = X, x' -v.+ E X. x! -a. v 1 1 . = 2 i i r - l t r - l = (X.x' + E X. a. x! )-v 1 1 i=2 1 1 1 Let T(z, z -«u) = w. w . Then 1 r - l 1 r r - l t r - l '(X, x! + E X. a. x! ")-v = w. w 4= .0 1 1 . 0 i i i 1 r 1 i=2 In view of Lemma 1.8, <w^> = <v> for some j , 1 £ j _< r . Since w, w e v, v , "S,, N , we have <v> = <v.> for some i or v e, s . 1 r 1 r-k (k) l This contradicts our choice of v . Hence T ( Z 1 Z r - l ' U ) ^ V l Vk'S(k) Similarly T(z 1 z r - l " U ^ £ S ( r ) ' H E N C E T ^ T " r one pure subspace of VU . In view of Theorem 3.13, singular linear transformation on U . Combining the above two Lemmas, we have the following main result: r r 4.10. Theorem. Let T : VU + VU be a rank one preserver where U is a finite dimensional vector space over an algebraically closed field F with characteristic equal to 0 or greater than r . If 3 £ dim U < r + then either T is induced by a non-singular linear transformation on U r or T(VU) is a type r subspace. In particular, i f T is non-singular, then T is induced by a non-singular transformation on U . We have so far not been able to determine whether there does r in fact exist a rank one preserver on VU such that its image is a type r subspace when 3 <_ dim U < r + 1 . 91. * Z R 1*^ 0 is a type T is induced by a non-92. CHAPTER III RANK K PRESERVERS ON GRASSMANN SPACES Let U be an n-dimensional vector space over an algebraically closed field F of characteristic not equal to two. Let AU be the r*"*1 r r Grassmann product space of U . If T : /\U -*• AU is a rank one preserver, then the structure of T is known: T is an r ^ compound of a non-singular linear transformation of U , except possibly when n = 2r , in which case T may be the composite of a compound and a linear transformation induced by a correlation of the r - dimensional subspaces of U [18]. In this chapter, we shall study the structure of rank k preservers 2 on AU for a fixed positive integer k . We show that a rank k preserver 2 2 T on AU is also a rank one preserver on AU > provided that T is non-singular or n = 2k or k = 2 . Denote by H^  the set of a l l n-square skew-symmetric matrices over F . A linear transformation on H is called a rank 2k preserver n c  i f every rank 2k matrix in H n is mapped into another rank 2k matrix. 2 Let u, , • • • , u be a basis of U . Let ty : AU •+ H I n , x n be a mapping defined by <J>(u.Au.) = E-- - E-- » 1 < i < j < n 2 and extend linearly to a l l AU • E denotes the n-square matrix with 1 in position i , j and 0 elsewhere. It is shown in [15], that cf) is an 2 isomorphism of AV onto H such that for each positive integer k , the 2 2 set, R^ C^ U) > of a l l rank k vectors in is mapped under ty onto the set of a l l rank 2k matrices in H . Moreover, T is a rank k preserver n 2 -1 on AU i f and only i f ty T ty is a rank 2k preserver on H n . 93. 2 1. Definition. For each z e R^ (AU) , we write R(z) = k . 2. Theorem. Let x^ , y^ , ••• , , be 2k vectors of U . Then k 2 I x j[A y i e ^(^U) i f and only i f x^ , y^ , • • • , x^ , y^ are linearly independent (see [8], Theorem 7). 2 2 3. Lemma. Let T : AU -»- AU be a rank k preserver for some positive integer 2 „ k . If z e R£ (AU) , I < k , then R(T(z)) < k . 2 Proof: Clearly we are able to choose a vector w in AU such that 2 z + Xw e Rj^ (AU) for a l l non-zero X in F . Suppose that R(T(z)) = m . Then <$>(T(z)) is of rank 2m . Since T is a rank k preserver, T(z+Xw) = T(z) + XT(w) is of rank k for a l l X =f= 0 . It follows that <j>(T(z) + XT(w)) = <j>(T(z)) + X <f>(T(w)) is a rank 2k skew-symmetric matrix for each non-zero X in F . On the other hand, since F is an infinite field, we can choose a X in F , X 4= 0 , such that o o 1 <j>(T(z)) + A q <}>(T(w)) is of rank greater than or equal to 2m . Hence 2m <^  2k .» This proves the Lemma. r r 4. Lemma. If T : AU AU is a rank one preserver, then T is also a rank k preserver for a l l possible k . 2 Proof: From [18], we know that T is a compound of a non-singular linear 2 transformation of U . In view of Theorem 1.2.15, T is a rank k preserver, k Now, let x = E x i e ^ ( M ^ ) where x^ is of rank one for each i . Let i=l 94. t r r * T(x) = E y. E R (AU) where y. e R-(AU) , 1 < i < t . Since i = 1 i t i l t k t t E y = £ T(x.) , we have t <_ k . Now T (x) = T( E y.) = E T(y.) i=l i=l 1 i=l 1 i=l 1 r 2 and T(y i) e R^(AU) , i = 1 , , t . Since T is a rank k preserver, t >_k . Consequently t = k . Therefore T is a rank k preserver. 2 2 5. Theorem. Let T : AU + AU be a rank k preserver. If T is non-singular, then T is a rank one preserver. Proof: Case 1. dim U _> 2k + 2 . By the non-singularity of T , we 2 shall show that i f there is an A e Rt(AU) such that R(T(A)) = I <_ k , 2 then there is a B e AU such that R(B) <_ t + 1 and R(T(B)) _> I + 1 . Indeed, suppose T(A) = u l A u 2 + ... + u 2 - £ _ l A u 2 £ for some independent vectors u i > * " * » u 2 £ °^ ^ where 1 <_ I <_ k Extend u^ , • • • , u ^ to a basis u^ , • • • , » " ' * • u n °^ ^ " 2 Since T is onto, there exists a vector z e AU such that 2 T(z) = u2£+i A u2£+2 * L e t z = z l + *** + zs e R S^U) where 2 z e R, (AU) , m = 1 , ••• , s . Let T(z ) = E ot u. A u. where m l m . . mil l 1 K j J J a .. £ F . Then mij T(z) = E T(z ) = £ ( E a u A u ) = « £ A u £ + . m=l i<j m=l J J 95. This implies that for some m , we have a =j= 0 . ° mo,2£+l,2-e+2 2 Hence there is a w in ^(AU) such that T(w) = ^  a.j u.Au. , a u + l t 2 l + 2 + 0 Hence the 2l+l , 21+2 entry of <b(T(w)) is a 2 ^ + 1 . For each A in F , the minor of order 2t + 2 in the upper left corner of the matrix <j>(T(A+Aw)) is of the form a2£+l,2-e+2 A* + b l * 3 + b2 ^ + ( 1 ) Since F is infinite and &2l+l 2l+2 ^ 0 ' ^  i s n o n - z e r o f o r some non-zero A in F . Hence 4(T(A+A w)) has rank > 2t + 2 . o T o — Therefore R(T(A+A w)) > I + 1 . However R(A+A w) < R(A) + 1 . o — o — Now, assume that T is not a rank one preserver. Then some rank one element A^ is mapped to a rank j element, j > 1 , since T is non-singular. By Lemma 3, j _< k . In view of the above argument, there is an 2 A 2 e AU such that R(A2> <_ 2 and R(T(A 2)) ^ j + l > 2 . If j + l < _ k , 2 continue the process as above, we see that eventually some element of AU 2 of rank less than or equal to k is mapped to an element of AU of rank greater than k . This contradicts Lemma 3. Hence T is a rank one preserver. 96. Case 2. dim U <_ 2k + 1 . We first note that the maximal 2 -1 rank of a l l elements in AU is k (Theorem 2). Assume that T is not a rank one preserver. Then some rank one element is mapped under T ^ to a rank Z element, Z > 1 . If Z =|= k , by the non-singularity of T - 1 , there exists a B e AU such that R(B) £ 2 and R(T _ 1(B)) >Z+1. If £+l=}=k, continue the process, we see that eventually there is a ? -1 2 2 z e AU such that T (z) = w e ^(AU) for some w e AU and R(z) < k . Hence T(w) = z , contradicting the fact that T is a rank k preserver. Therefore T ^ is a rank one preserver. By Lemma 4, T ^ preserves a l l ranks. Hence T is a rank one preserver. 6. Remark. The proof of Theorem 6 is analogous to that of Theorem 3.1 [1] and Theorem 1 [6]. 7. Remark. It is well-known that the maximal dimension of a rank n subspace (a subspace with every non-zero element of rank n) in n(F) , the vector space of a l l n-square matrices over an algebraically closed field F , is one. Hence the maximal dimension of a rank k subspace in 2 2 AU is one i f dim U = 2k , since A : AU H is an onto isomorphism such A. 1 1 2 that <j)(Rk(AU)) is the set of a l l rank 2k matrices in H n . 2 2 8. Theorem. Let T : AU -* AU be a rank k preserver. If dim U = 2k , then T is a rank one preserver. Proof: In view of Theorem 5, i t suffices to show that T is non-singular. 9 7 . 2 Assume that T is singular. Then some z e is such that T(z) = 0 , z + 0 . Clearly t ={= k , since T is a rank k preserver. 2 Let w E AU such that R(z+w) = k and R(w) = k - t . Hence 2 T(z+w) = T(z) + T(w) = T(w) E (AU) . Assume that w = u- A u_ + ••• + u_ . , /\ u„. where i = k - t . 1 2 2j-l 2j Extend u^ , ••• , u ^ to a basis u^ , ••• , u ^ , ••• , of U . * Let w = A + ••• + u2k A u^ . Then for a , b in F , ' a w + bw* = u 2 A (bu3-au1) + . . . + « 2 J A < b u 2 J + r a u 2 j - l ) + b u2j+2 A u2j+3 + b u2kA u i ' * * 2 Clearly i f b f 0 , R(aw+bw ) = k and hence T(aw+bw ) E Rfc(AU) . * 2 If b = 0 and a =j= 0 , then T(aw+bw ) = T(aw) E Rfc(AU) . Therefore * 2 i f a and b are not both zero, then T(aw+bw )E R^ (AU) . It follows * 2 that <T(w) , T(w )> is a rank k subspace of AU with dimension equal to two. This contradicts the fact that the maximal dimension of a rank 2 k subspace of AU is one. Therefore T is non-singular and the proof is complete. 9 • Lemma. If dim <x^  , y^ , • • • , , J^.-^ = 2k + 1 and x^ A y^ =J= 0 for a l l i = l , ••• , k + l , k > l , then k+l 2 Z x i A y i e \(AU) i=l 98. k+1 Proof: It is clear that Z x. A y. has rank less than k + 1 because . .. 1 ' N J TL 1=1 x i » ^ i '» """ » » yk+l a r e l i n e a r l v dependent. Also there is one j such that dim , y 1 , • • •; , *x , 9j . • • • » V l , y k + 1> = 2k . Hence k+1 2 t x. Ay. e ^(AU) . Since x. or y. £ ^ . y ^ - : ' v, V i » W . k+1 by Corollary 1.2.6, Z x.Ay. is of rank at least k . This •- , I ' i i=l k+1 proves that Z x . A y . is of rank k . . , i w i i=l 2 10. Example. The following is a 3-dimensional rank k subspace of AU when dim U = 2 k + l , k > 2 . Let u^ , ••• , u 2 k + i be a basis of U . Let f, = u, A 1 + U2 A uk+2 + + \ A " 2 k f2 = u l A \ + 2 + u2 A V 3 + + A A u 2k+l f 3 = \ + l A \ + 2 + u i A \ + 3 + - + V l A u2k+l 2 Clearly f^ } > e ^(AU) • Let a^ , a^ , a^ be non-zero elements in F . We have ^ V i = 33\+l A \ +2 + U l ^ (Vk+1 + 32\+2 + a3Uk+3) + •'••+ V l ^  ( al u2k-l + a2 U2k + a3 U2k+l ) + \ * ( al U2k + a2 U2k+l ) 99. 3 3 Since dim < U f c f l > u ^ , V J^a.u^., •• • ,^._vJ^V2k-241'V V2k + a2 u2W-l > 3 b y Lemma 9, E a.f. is of rank k . i-1 1 1 Similarly, using Lemma 9, we check that both a^ f^ + a^ i n , a2 2^ + a3 3^ a r e °^ r a n k k " B y Theorem 2, a^ f^ + a~ £~ -*-s °^ r a n k k Hence , , are linearly independent and <^i»^2'^3> ^ S 2 a rank k subspace of AU . It is known that every rank m subspace of M ,. ,..(F) is J r m+l,m+l at most 4-dimensional (see [21]). We feel that every rank 2k subspace in **2k+l * S m o s t likely of dimension <_ 3 . 2 2 11. Theorem. Let T : AU ->• AU be a rank k preserver. Suppose dim U = 2k + 1 and the maximal dimension of a l l rank k subspaces of 2 AU is 3 . Then T is a rank one preserver. Proof: Assume that T is singular. Then k >_ 2 and for some 2 2 2 z e Rt(AU) , T(z) = 0 where t <_ k . Since T(Rj^(AU)) C R^AU) , t =(= k . 2 Choose a vector f in R, (AU) such that z + f is of rank k . o Tc-tv o Denote f q by u^ A \ + u 2 A \ + 1 + * * * + u j A uj+k-l w h e r e 3 = k - t . 2 Then T(z + f Q) = T(z) + T(f ) = T(f ) e Rjc(AU) . 100. Extend , , • • • ., , " j ^ . ^ t o a basis u 1 , * " , u 2 k + 1 of U . Consider f l = U l A V l + U2 A \+2 + + V * u2k f2 = U l A \+2 + U2 A \ + 3 + + V A u2k+l f3 = \+l A uk+2 + U l A \+3 + + V - l A U2k+1 Let a , a, , a„ , a„ be four non-zero scalars in F . Then o ± I 5 \ a i f i = a3 V f l A \+2+Ul A ( ao Uk + al\+l + a2 Uk+2 + a3 Uk+3 ) + * * ' 1=0 + « j A ( a o V j . 1 + a l u k + j + a 2 U k + j + l + a3Uk+j+2) + + V A ^ al U2k + a2 U2k+l^ Since 3 3 3 d i m V ' ^ ' V / A V l V . I n ° i V j - l + l ' u j + l , . E 1 0 i ^ J + l ' "• ' 1=0 • 1=0 J 1=1 \' al U2k + a2 U2k+l > = 2 k + 1 ' by Lemma 9, Z a.f. is of rank k . Similarly, using Lemma 9, we check i=0 1 1 that a f + a, f. + a, f, , a 'f + a 0 f„ + a, f_ are of rank k . o o 1 1 3 3 o o 2 2 3 3 2 2 Also we have a Q f + a.± f± e P^ CAU) and aQ f Q + a 1 f 1 + a~ f„ e R^ CAU) 101. i = 1 , 2 , 3 , by Theorem 2. 2 Since T is a rank k preserver and ^(^Q) e R^ CAU) , from the above argument, we see that 3 3 2 I c. T(f ±) = Z T( C if.) e ^(AU) i=0 i=0 3 i f one of c. + 0 . Hence £ d. T(f.) = 0 implies that a l l d. = 0 . i -0 • 1 1 Therefore T(f Q) , ••• , T(f^) generate a 4-dimensional rank k subspace 2 of AU which is a contradiction to our hypothesis. Hence T is non-singular. This proves that T is a rank one preserver by Theorem 5. 2 12. Corollary. Let T be a rank 2 preserver on AU . If dim U = 5 , then T is a rank one preserver. Proof: This follows from Theorem 11 and the fact that every rank 2 2 subspace of AU has dimension less than or equal to 3 i f dim U = 5 [9] 2 2 13. Theorem. Let T : AU ->• AU be a rank 2 preserver. Then T is a rank one preserver. Proof: Because of Theorem 8 and Corollary 12, i t suffices to consider the case when dim U _> 6 . It has been shown in [9], that every rank 2 subspace 2 of A^ n a s dimension _< n - 3 i f dim U = n >. 6 . 102. Assume that T is not a rank one preserver. By Lemma 3, some rank one vector is mapped to the zero vector or a rank 2 vector. If some rank one vector is mapped to the zero vector, i t is clear that some rank one vector is mapped to a rank 2 vector since T is a rank 2 2 preserver. Hence T(u^ A v^) e R^AU) for some independent vectors U l • V l * Extend u, , v_ to a basis u, , u„ , v, , • • • , v „ of TJ 1 1 1 / 1 n—z Consider the following vectors: f l = U 1 A v i f2 = U 1 A V 2 + U 2 A V 1 fn-2 = U l A V n - 2 + U 2 A Vn-3 ' Let a. a „ be n - 2 elements in F . . Then i n-z n-2 n-2 n-3 I a.f = u A (} V i ) + ' U 2 A a i + l V • i=l i=l i=l n-2 2 n-2 n-3 E a.f. £ R,(AU) implies that E a.v. and E a.,,v. are linearly i=l 1 1 1 i=l 1 1 i - l - 1 1 1 n-2 2 dependent and thus a n 2 = '*" = a 2 = ^ ' a l ' ® ' Hence ^ a i ^ i e ^2^ A^ i=l when some a. 4 0 for 2 < i < n - 2 . Since T is a rank 2 preserver 103. 2 n-2 and T(u A v ) e R„(AU) , i t follows that R(T( I a.f.)) = 2 whenever i=l 1 1 some a i=|=0,l_<i_<n-2. This shows that T(f^) , ••• , T(fn_2) are linearly independent and <T(f^) , ••• , T^ n-2^ > i s a r a n ^ 2 subspace 2 of /\U of dimension equal to n - 2 . This contradicts the fact that the maximal dimension of a rank 2 subspace is n - 3 . Hence T is a rank one preserver. 14. Remark. Theorem 13 is also, obtained independently by M.J.S. Lim [10]. Combining Theorem 5, Theorem 8 and Theorem 13, we have 2 2 15. Theorem. Let T : /\U ~* AU be a rank k preserver. If T is non-singular or dim U = 2k or k = 2 , then T is a compound of a non-singular transformation of U , except when n = 4 , in which case T may be the composite of a compound and a linear transformation induced by a correlation of the two dimensional subspaces of U . Using the results in [15], Theorem 15 can be stated in matrix langague as follows: 16. Theorem. Let S : H ->- H. be a rank 2k preserver. If S is non-n n singular or n = 2k or k = 2 , then there exists a non-singular n-square matrix P such that S(A) = PAP' for any A in H n , except when n = 4 , in which case S may possibly be of the form: 104. S(A) = P *34 24 23 "a34 ° "a24 "a14 "a23 "a13 l14 a13 *12 l12 0 P' where A = [a..] , a.. = -a.. i j i j J i 17. Corollary. Let S : H H' be a linear transformation such n n -that det A = det S(A) for a l l A in H where n is even. Then n S has the form as described in Theorem 16 where det P = + 1 . Proof: By hypothesis, S is a rank n preserver on H . Therefore n this Corollary follows from Theorem 16 and the fact that det 0 M2 "13 14 -a 1 2 0 a13 "a23 ~a14 ~a24 *23 °24 *34 -a 34 = det 34 24 23 -a 34 *14 "13 -a 24 a14 '12 23 13 12 BIBLIOGRAPHY 105. [I] Leroy B. Beasley, Linear Transformations on Matrices: The Invariance of Rank k Matrices, Lin. Alg. and App. 3 (1970), 407-427. [2] H. Boerner, Representations of Groups, Amesterdam, 1963. [3] Wei-Liang Chow, On the Geometry of Algebraic Homogeneous Spaces, Ann. Math. 50 (1949), 32-67. [4] Larry J. Cummings, Linear Transformations of Symmetric Tensor Spaces which Preserve Rank 1 , Ph.D. Thesis, U.B.C. 1967. [5] ' , Decomposable Symmetric Tensors, Pac. J. Math. 35 (1970), 65-77. [6] D. Z. Djokovic, Linear Transfi ations of Tensor Products Preserving a Fixed Rank, Par Math. 30 (1969), 411-414. [7] W.H.Greub, Multilinear lgebrt ringer-Verlag, N.Y., 1967. [8] M.J.S. Lim, Rank k Grassmann Fcoducts, Pac. J. Math. 29 (1969), 367-374. [9] , L - 2 Subspaces of Grassmann Spaces, Pac. J. Math. 33(1970), 167-182. [10] , Rank Preservers on Skew-Symmetric Matrices, Pac. J. Math. 35 (1970), 169-174. [II] M. Marcus, A Theorem on Rank with Applications to Mappings on Symmetry Classes of Tensors, Bull. Amer. Math. Soc. 73 (1967), 675-677. [12] M. Marcus and B.N. Moyls, Transformations on Tensor Product Spaces, Pac. J. Math. 9 (1959), 1215-1221. [13] M. Marcus and H. Mine, Permutation on Symmetry Classes, J. Algebra. 5 (1967), 59-71. [14] M. Marcus and M. Newman, Inequalities for the Permanent Function, Ann. Math. 75 (1962), 47-62. [15] M. Marcus and R. Westwick, Linear Maps on Skew-Symmetric Matrices: The Invariance of the Elementary Symmetric Functions, Pac. J. Math. 10 (1960), 917-924. [16] CF. Moore, Characterization of Transformations Preserving Rank Two Tensors of a Tensor Product Space, M.A. Thesis, U.B.C. 1966. 106. [17] K. Singh, On the Vanishing of a Pure Product in a (G , a) Space, Can. J. Math. 22 (1970), 363-371. [ 1 8 ] R. Westwick, Linear Transformations on Grassmann Spaces, Pac. J. Math. 14 (1964), 1123-1127. [ 1 9 ] , Transformations on Tensor Spaces, Pac. J. Math. 23 (1967), 613-620. [20] , Linear Transformations on Grassmann Spaces, Can. J. Math. 21 (1969), 414-417. [21] , Spaces of Linear Transformations of Equal Rank, Lin. Alg. and App. (To Appear). 

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