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 p r e s e n t i n g t h i s t h e s i s an advanced degree at the L i b r a r y I further fulfilment o f the requirements the U n i v e r s i t y of B r i t i s h Columbia, I agree s h a l l make i t agree tha in p a r t i a l freely available for thesis f o r s c h o l a r l y purposes may be granted by the Head of my Department of this thesis for It financial or i s understood that copying o r p u b l i c a t i o n gain s h a l l written permission. Department that r e f e r e n c e and s t u d y . p e r m i s s i o n f o r e x t e n s i v e copying o f t h i s by h i s r e p r e s e n t a t i v e s . for of The U n i v e r s i t y o f B r i t i s h Columbia Vancouver 8, Canada not be allowed without my ii. Supervisor: R. Westwick ABSTRACT Let U denote a finite dimensional vector space over an alge- braically closed f i e l d F . In this thesis, we are concerned with rank one th preservers on the r symmetric product spaces r vers on the 2nd Grassmann product spaces 2 VTJ and rank k preser- 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 linear transformation on U F i s zero or greater than (b) r dim U > r + 1 , then either r and the characteristic of F i s T i s induced by a non-singular r linear transformation on U or T(VU) = VW space If and the characteristic r .) If 2 < dim U < r + 1 zero or greater than i s induced by a non-singular . (This was proved by L.J. Cummings in his Ph.D. Thesis under the assumption that of T for some two dimensional sub- W . of U r s (ii) Let T : VU •+ VU be a rank one preserver where r < s dim U > s + 1 and the characteristic of F i s zero or greater than g — , then T i s induced by s - r non-zero vectors of U lar linear transformation on U and a non-singu- iii. (iii) If T 2 2 Let T : AU -»• AU be a rank i s non-singular or dim U = 2k except when dim U = 4 , in which case k or k = 2 preserver and char , then T F ={= 2 . i s a compound, T may be the composite of a compound and a linear transformation induced by a correlation of the two dimensional subspaces of U iv. 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 CHAPTER II. 22 RANK ONE PRESERVERS ON SYMMETRIC SPACES 26 1. Maximal Pure Subspaces of Symmetric Spaces 26 2. Intersections of Maximal Pure Subspaces r s Rank One Preservers from VU- to VU , r < s r 38 3. 4. Rank One Preservers on VU CHAPTER III. RANK k PRESERVERS ON GRASSMANN SPACES BIBLIOGRAPHY 53 70 92 105 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 f i e l d and F . Let G be a subgroup of the symmetric group x be a character of degree one on G . Denote by symmetry class of tensors over U associated with G U^G) X the and x • A. non- zero element of U^G) i s 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 U (G) . A linear transformation of U^CG) i s said . X X m to be a rank k preserver i f i t maps the set of a l l rank k vectors into i t s e l f . 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 on rank k symmetric product spaces and partly 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 i n U^G) i s unchanged X in i f we extend U ; ( i i ) for each rank k vector in U (G) and each orbit X of G , we associate a unique subspace of U ; 11 (iii) z,.* •••*z, + ••• + z, * • ••*z, 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 i s of rank i s k.10j where k i f for each j0J denotes the cardinality 0 . From our results on rank of 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 VU , where symmetric product spaces, U space over an algebraically closed f i e l d i s a finite dimensional vector F . We f i r s t 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 induced by a non-singular transformation on (ii) Let T U . 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 T i s induced by a non-singular transformation on for some two dimensional subspace (iii) Let T W VU is U r or T(VU) •= VW of U . be a rank one preserver from r VU s 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 1 s-r J non-singular transformation T(x n L x )= z r l z f on -fCx.) s—r .i U in the sense that -f(x ) , x r i eU ,i= 1 , • • • ,r. (i) and (ii) partially answer a question raised by Marcus and , 3. Newman [14, p. 62]. assumption that (i) was f i r s t proved by Cummings [4] under the dim U > r + 1 and the characteristic of F i s not a prime p <^ r . An example i s given to show that there i s another type of rank one preserver from p such that $XS to tHj , r < s , i f F i s of prime characteristic rp <_ s . I n Chapter III, we consider rank k preservers on 2nd Grassmann 2 product spaces, Au , where U i s an n-dimensional vector space over an algebraically closed f i e l d F with characteristic not equal to two. We 2 show that i f T k singular; or ( i i ) i s a rank preserver on AfJ and either (i) T n = 2k ; or ( i i i ) k = 2 ; then non-singular transformation on U , except when T i s non- i s a compound of a 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 §1. k VECTORS IN SYMMETRY CLASSES Definitions and Ramarks. Let F be an arbitrary f i e l d . the symmetric group one on G ; i.e., Let G be a subgroup of S^ of degree m . Let x be a character of degree * * x 0 -*• F i s a homomorphism where F i s the multiplicative : group of F . Let over Let 1.1. V. = V for i = 1,2, •••,m and a l l a e G . l a(i) * W be the cartesian product of the V. , W = V, x ••• x V i . i m Definition. Let U be any vector space over • • ,X " ) = x(a) f ( X N with lS x • • • ,X ) for any a e G and X^ e V^ , i = 1,2,• • •,m . Definition. A pair and a multilinear function and F . A multilinear f : W -> U i s said to be symmetric with respect to G and f(X arbitrary 1.2. be finite dimensional vector spaces F such that function if V- , V_ , • • • , V I I m (p,u) consisting of a vector space u : W -»• P , symmetric with respect to G x > i s symmetry class of tensors over — a G and x P over. F V,,V„,...,V I m associated i f the following universal factorization property i s 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 V, , V„ , ••• , V 1 2 m G and (V symmetry c l a s s o f t e n s o r s o v e r a associated with within canonical isomorphism space by ,x » G and x (see [13], [17]). , . • . , V ) (G) . A such a space i s usually denoted by We shall denote such a ........ v 1 V (G) m exists and i s unique to (see [13]). = , then m The vectors v v X u(X,,'••• • ,X ) 1 m are denoted by X, * • • ' • * X 1 m and are called 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 1. and x » i« »> e I f G = {e} , where e i s the identity permutation in S , (V,,V-,--- ,V ) (G) i s the tensor product of vector spaces 1 2 m x m V , ® V . . In this case, the decomposable element X. * m . , l 1 X = 1 , then V, 1 1=1 is denoted by 2. then V^G) element X,I X.® 1 If • • • ®X G = S and m , x "sign of the permutation" character, m i s the Grassmann space A • case, the decomposable = v Xm i s denoted by I n t n i s X-A. l • * • * * Axm X m 6. 3. space m W X, 1 X . m 1.3. m x = 1 » then and . The decomposable element Remark. x » and If G = S w e Since u V^( ) i s the symmetric G X, * •••* X 1 m i s denoted by i s multilinear and symmetric with respect to G have: (i) X, * (aX.+BX'.) * ...*X = aX.*...*x.*."-.*X +BX,*...*Xj*...*X 1 i i m l l m l i m X. , X. z V. , each ct,t3 e F and each i = 1,2, .. . ,m . for each f (ii) X ( i ) * '" * (m) X a = X a ^ X l*' * " *m X ' ° 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 1.5. Remark. (V,, ••••,V ) (G) . 1 m x Let IL , ••• , U be subspaces of 1 m V. , 1 • • • ,V m respectively such that U. = U ... for i = 1,2,...,m and for a l l a e G . l a(i) u, be the restriction of the map u to U, x... x u Then 1 1 m Let i t can be shown that t, IL, , • • • , U i m (<range u^> , y^) , associated with G and i s a symmetry class of tensors over x where < range y. > 1 denotes 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 (j) : V 1 and for a l l a e G . Define a mapping x...x V• + (V ,---,V ) (G) m l mx by setting <}) ( X X 1 m ) = T X * ... *T X . i l mm It i s easily seen that to G and <{) i s multilinear and symmetric with respect x • Hence by the universal factorization property of (V..,«'-,V ) (G) , there exists a unique linear transformation on ^ X '(V,,«*»,V ) (G) , denoted by 1 m x K(T.,---,T ) , such that 1 m K(T,,«",T ) x. *•••* x = l m l m When = ••• = T 1.6. Definition. (V„ •••,Y ) (G) 1 m x of T..X.. 11 = T , we shall denote m *•••* T x . mm K(T ,.••,T ) simply by m The above transformation K(T_, 1 K(T) ):(V. ,«",V ) (G)+ m l m x i s called the associated transformation (induced transformation) 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 the m^i C ( T ) and the m compound m m*"* 1 T^® ••• ^ tensor product induced power of m • T , P (T) , m respectively. 1.7. : Remark. If S. = S ... i a(i) for = m K ( 1 T T ) K i S. : V. -*• V. are linear transformations such that i i i = 1 , ••• , m and ' ' • ( l S S m ) a e G , then we have ' K(T.. S-, • • • ,T S ) 1 1 m m 8. 1.8. Definition. have rank k k A non-zero vector in (V-,«««,V ) (G) i s said to 1 mx i f i t can be written as a sum of k but not less than non-zero decomposable elements i n (V, ,".,V ) (G) . The set of a l l 1 mx rank k vectors in (V...-.,V ) (G) i s denoted by R, ((V. ,---,V ) (G)) 1 m Y k. 1 m Y 1.9. Definition. A subspace r MC (V, ••••,V ) (G) i s called a rank k — 1 m x subspace i f every non-zero vector i n M subspace contain M i s said to be maximal i f no other rank k . A rank k k subspaces properly M . A rank one subspace i s usually called a pure subspace. 1.10. Definition. is i s of rank called a rank A linear transformation A : (V, .•••,V ) (G)-»-(V,, • • • , V ) 1 m x 1 mx k preserver i f A maps the set of rank k vectors into i t s e l f . 1.11. Remark. Some knowledge of the structure of rank be useful i n characterizing rank evident that a rank k a rank A(M) . Moreover, k subspace preserver k preservers. k subspaces w i l l By definition, i t i s A maps a rank k subspace M into A|M is a monomorphism and hence dim M = dim A(M) '. §2. Properties of Rank k Vectors. Throughout this section, l e t ^ i ' " " * ' ^ ^ ^ ^ m class of tensors over V-,,«->,V associated with a subgroup G of S m m G • We also l e t 0^ , ••• , 0 be a l l the orbits of 1 and a character x o n denote a symmetry fc 9 . 2.1. Theorem. Let x, + • • • + x. = y, + • • »+y 1 'K 1 e R. ((V., • • • ,V ) (G)) q . K where x. = x. *. ..* x. y = *...* y for each 2 j l jm ' -n •'nl ^nm y and n = 1 , ••• , q . Then for each orbit l m x j = 1 , ,k <X , we have k q Z <x., : d £ 0.> C E <y , : d e 0.> j-1 3 ""n-l d where <x, : d jd e 0.> 1 1 denotes n d the subspace spanned by the vectors x., , d e 0. . Jd l Proof: q Suppose that for some j » i £ j £ k , < p c . : d £ C\> <j: E n=l <y d nd : d e 0± 1 Then for some s e O . , x. iE<y,:dsO.>. n=l J Consider the associated transformation. K(T.,»««,T ) : (V.,««»,V ) (G) 1 m 1 mx + (V,,---,V ) (G) where 1 and m T» = T , x If Z = 1 , ••• , m are defined as follows: If Z e C\ , T^ : 0 for a l l a e G -Co\JL) T, , ••. , T 1 m ToCx. ) = Z js D s and T„ Z Z £ 0 We have i -> i s a linear transformation such that I <y , : d £ 0.> , nd l n=l , T^ : -> i s the identity mapping. KC^, • • • ,T ) ( ffl is the identity mapping. X;L + • .. + x ) = K(T^, • •-.T^) (y + . ., + y ) , fc 1 10. Since K(T ...,T )(y ) = y 1} m n n for n = 1 , • • • , q and KC^, • . . ,T ) )= m T - x . , *•••* T x. = 0 , i t follows that 1 jl m jm K(T r'--'V i --(x + +x -i Vi V-- k + j + +x ) = y i + " - + y e q V V--'>Vx ( (G)) ' This is a contradiction since the l e f t hand side i s a vector of rank less q than k or the zero vector. Therefore < , . : d e 0. > C E < y , : d e 0 > id x — .. •'nd l n=l k q for each j = 1 , ••• , k . Hence Z < : d e O . > C E < y „ : d e 0. > . • i jd l — , nd x J=l n=l x x J J 2.2. Corollary. Z = Z.. *•••* z 1 m Let x + y = Z where x = x *...^x r 0. , we have i < Z: d d e 0. > C < l — y y a n < (V, .••••V ) (G) 1' ' mx are non-zero decomposable elements of Then for each orbit = y x d j . : d e 0.> + < y, : d e 0.> I d l 7 This Corollary generalizes a Theorem of Cummings [4, p. 17]. 2.3. x Theorem. Let , = .. *• • • * . j Jl jm x x Then for each orbit x + .. . + x = 1 k and y + . . .+ e" R. ((V.,•••,V ) (G) 1 k k 1 mx y y. = y._ *...* y. for each i = 1 , • • • , k . J ' j l 'jm k k 0. , we have £ < . , : d £ 0. > = Z < y. , : d'e 0. > . i Jd x 'jd 3 x j = 1 Proof: 2.4. j = 1 This follows immediately from Theorem 2.1. Corollary. where Suppose that x m x x *...* x ^ 0 . Then 1 TO *•••* x = y. *...* y e v (G) TO 1 m x < x ... -x > = < y, ,..«,y > . I ' m ' l '"'m and 11. Corollary 2.4 generalizes a lemma of Marcus and Mine [13]. 2.5. of Definition. Let x = x, *•••* x 1 m (V., •••,V ) (G) . For each orbit I mx the subspace spanned by the vectors be a non-zero decomposable element 0. of G , define l x, where < 0.(x) > to be l d e 0. . By Theorem 2.3, d i this definition i s well-defined. 2.6. Corollary. Let x + ••• + x e R ((V , • • • , ) ( ) v 1 decomposable element for each If for some orbit fc k 1 where G m x j = 1 , ••• , k . Let x^+j = x_. i s a '*•••* Z^ 4 0 . 0. , there i s a s e 0. such that Z i < 0.(x.) > +...+< 0.(x.)> l i s l 1 l k has rank greater than or equal to k . T then x^ +• • •+ Proof. If k+1 Z x. = 0 , then J-l k Z x. = -x, ,.. . This implies that J-l By Theorem 2.3, we have Z e < 0 . ( x j > which contradicts the hypothesis. s 1 I k+1 n Z x. = Z y. e R ((V ,•••,V ) (G)) where If k n Z x, = Z y. - x, . This implies that • i 1 *. i 3 k+1 3=1 j=l r J J is of rank k=1 . J J k . By Theorem 2.3, we have | < n < k , then k n = k - 1 since Z x. . i 3 3=1 < 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. Therefore k+1 Z x. is of rank greater than or equal to k . 3=1 3 12. Corollary 2.6 w i l l be used in Chapter III. The following corollary of Theorem 2.4 i s well-known. k 2.7. Corollary. (a) Suppose that k E x. ® y. = 1=1 E 1=1 u. ® v. e R, (V «>V ) Then < x ,...,x > = < u,-..--,u > , < y - - - » y 1 k (b) 1 k Suppose that < x 2.8. Theorem. that U. =U .> , i = 1 , ••• ,m, I a(i) > ,.i = 1 , ± Let ,•*•,be u Proof: Let y = element for each < ^(yj) u y = j n E = > S£ ^£ ( v supspaces of r-'Vv >k > v in V^® • • • ^i»""*>^ m respectively such for a l l a e G . Then ( G ) ) E y. e ^ ( ( 1 ^ , • • • » ) ( ) )where wherey.y i s a decomposable u G m x j = 1 , ••• , k . By definition, i f L e CL , then f° re a c n j = 1 » > k . Suppose that Z. e R ((V ,«' ,V ) (G)) where J n 1 m x , 1 1 > = < V " , k.. / V( r-'Vv^^V k x^® • • • ® x^ = y^® • • • ® y^ ^ 0 Then ± >= <y r Z. J i s a decomposable element for 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„ l 1 l n — i l .i'k — £ w i f I e 0. l 13. If n < k , we conclude that the rank of y i s less than k in (U,,'«',U ) (G) , 1 X m which i s a contradiction. Therefore n =k and y e R.,((V. ,.«««V ) (G)) . K 1 mX 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 : V x-..x V -*• F l o(i) 1 m 1 be defined as follows: .m f(w ,---,w ) = E (o) n f aeG i=l .(w ) x 1 Then m 1 f i s multilinear and symmetric with respect to G and x Proof: Note that the function i = l , ••• , m , a e G .' For each f(waw.+0w'...,w) = 1 i i m = f i s well-defined since a , 3 e F and w i E (o)( n f x V. = V ... , l a(i) , w^ e V i , we have (w ))-f (aw +Bw!) a(j) j a(x) i i E ( a ) ( nf aeG j f i X J ( )).af Wj (w.) J m + V ( 0 ) ( aeG -I- <'(J) J f ( w j f i J ) ) , e f ^i) i ( w ) J = af (w , • • • ,w , • •-,w ) + fBf (w ,• • • ,w' , • • • ,w ) ± 1 m 1 i m n Hence f i s multilinear. Also for each x e G, 14. m f (w x ( l ) ' ' * * ' x(m)> W = ^ ^ ^ a d ) = Z x(o) H f (w.) where we let x(i) aeG j=l aT -(j) (W T(i)> m 1 -J 3 -1 m = aeG 2 X ( O ) X ( T ) X ( T ) j=l n f ax _ (j)( ) A l f -1 = X(x) I x(o-x aeG Wj m )n f (w ) j=l ax J = x(T)f(w ,•••,w ) l Hence m f i s also symmetric with respect to G and x 2.10. Lemma. Let x = x, *• ••* x e (V,,•••,V ) (G) . Then 1 m l m x implies that for some i , dim < o^(x) > < |o_^| , where x =0 10_J denotes the number of elements in 0. . l Proof; each j , let •. fj (x ) = 0 d of dim < O^x) > = 1 | Suppose that for a l l i = 1 , ••• , t . For f . : V. -*- F be a linear transformation such that 3 3 for a l l d such that f.(x.) = 1 3 3 j= { = d and j , d are i n the same orbit- G. Consider the multilinear function f defined i n Lemma 2.9. By the universal factorization property of (V^,•••»V )^(G) > there exists m 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 j m nf ..(x.)=o / 3 = 1 i f and only i f a ( j ) = j , i t follows that f ^( ..) = 1 x a i f a 4= 1 . Hence we have T m f ( x . , . . - , x ) = x ( D n f,(x.) = l . 1 Therefore j=i m J h(x *...*x ) = f(x-,»-«,x ) = 1 1 m 1 m since x, *•••* x = 0 . Hence 1 m 2.11. Theorem. for each J which i s a contradiction dim < 0.(x) > < lo.l l I' for some 1 i . Let x. = x., *•••* x. , 1 = 1 . , ••• , k . Suppose that j j l jm v v i = 1 , ••• , t , dim (< C^Cx^ > +• • • + < ° i ( x k ) > ) = 1 0 ^ | k . Then k I x. e R, ((V , • • •,V ) (G)) . j J K 1 mx = 1 Proof: We proceed by induction with respect to k . By Lemma 2.10, the proposition is true for k = 1 . Assume now i t i s true for k - 1 where k k _> 2 . -By Corollary 2.6, E x . has rank equal to k or k - 1 . 3=1 Suppose that k k-1 E x. = I y. e R, . , ((V-, ' * *,V .((V-.'-'.V ) () (G)) where j J j J K-1 1 mx = y^ = y ^ * . . . * 1 1 = 1 , j = 1 , ... , k - 1 . We divide the proof into two parts: 16. (i) Suppose G i s 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=l k k-1 k Then we have E <0 (x.)> = E <0,(y.)> since dim ( £ <0.(x.)>) = (k-l)m . 1 i+i J 1 J+1 ' J+1 " J J 1 J 1 J 1 J x. = -(x,+--'+x. ,+x.,,+••'+x, ) . + y, + ••• + y, , . From Theorem 2 . 1 , l 1 l - l i+l Ic l k-1 k k-1 k we have <0-(x.)> cz £ <CL(x.)> + E <0.(y.)> = E <0,(x.)> , which contradicts J— . i . J- J . J- J J- i Jfi J=l J+i • Now J 1 -i J the hypothesis. J J Therefore for a l l i = 1 , ••• , k , k k-1 E <0 (x.)>4z E <0 (y.)> n . 1 k-1 Hence ther are two vectors x n If l D <0 (y.)> I i E 3 T and n + d . n J-1 2 1 k-1 <x n ,x t , x, > + ( E <CL (v )>) i s not a direct sum, then we can choose 2 j = l "J" l t D 1 T k-1 a vector x st such that 3 x i <x ^ ,x > + st n dt 3 t l 2 E j = 1 <0.(y.)> 1 J since k-1 dim (<0.(x,)>+•••+<(),(x. )>) = mk I K . 1 1 and dim (<x _ ,x. > + E <0.(y.)>)< mk . nt^ at 2 j=l 2 k-1 By the choice of x st , we have both 3 <x ,x > + ( E < 0 (y.)>) and s t nt l J 3 1 j = 1 k-1 < ,x x S T 3 D T 2 > + ( I < 0 (y.)>) are direct. j=l 1 Either in any case, there are two vectors, say, x ^ k-1 ,x <x 1 Q ,x ^ > + ( £ < 0 (y.)>) i s direct and Z + q 2 i J =l 1 s+ d or s + n . 2 J such that Therefore 17. Consider the associated transformation where f(x^ f : V -»• V ) = 0 d f| <x qd k-1 E ^ ( y )> >+ i s the identity mapping. k k k-1 k-1 K(f)( E x.) = K(f)( E x.) = K(f)( E y.) = E y. . Since j=l jfe j=i j=i 3 k-1 E y. j-l V^G) X i s a linear transformation such that and We then have K(f) : V^G) X i s of rank J J 3 k - 1 , i t follows from Theorem 2.3 that 3 k-1 E <f(x31 ) , ••• , f(x. )> = E <0 (y )> .- However, f(x , ) = x , and qd qd J J-l k-1 hence x , e E <CL(y.)> , a contradiction. Therefore E x . i s of rank 2 3=1 3=1 M m qd 1 (ii) 2 3 2 k 3 Suppose transformation 1 G has at least two orbits. Consider the associated K(f.,'*',f ) : (V,, •••V ) (G) ->• (V., • • • V ) (G) where 1 m l m x l X m f : V •> V n n n i s a linear mapping such that r r 6 f I <0, (x.)> = 0 , f I <0,(x„)> +•••+ <0 (x.)> = identity i f n e 0, n i l n 1 I l k . 1 1 and f : V -»• V f . n n n i s the identity mapping i f n I 0, . l T k-1 Let K ( f f ) ( E y 1 m j k k K(f ,--«,f )( E x.) = E x. j=l j=2 = Then m k Ex. J-2 J i s of rank 1 k-1 ) = E y' where 1 j 3 = 1 K(f 1 f ) y . = yl , j = 1 m 3 3 k-1 = E y! . By the induction hypothesis, j=l k k-1 k - 1 , hence by Theorem 2.3, E <0 (x.)> = E <0 (y'.)> j=2 j=l J 3 3 3 1 3 18. Since f = identity i f n e 0 n 22 k k-1 Z <0 (x.)> = Z <0 (y.)> . Now 5=2 j=l ^ Hence k-1 k-1 Z <CL(y.)> = Z <0.(y!)> £ «Uy.): j=l j=l j=l k k-1 = -( £ x.) + Z y. , by Theorem 2.1, j=2 j=l , i t follows that 1 J 2 x 1 J 3 2 we have <0 (x )> c X k-1 Z <0 (x.)> + Z <0 (y.)> = Z <0.(x.)> j=2 j=l j=2 Z Z 2 This contradicts the hypothesis.' 3 2 Therefore . 2 k Z x. j=l i s of rank k 2 From Theorem 2.11, we have the following known result. 2.12. Corollary. y. , ••• , y, If x^ , ••• , x^ are linearly independent and are linearly independent, then k Z x.® y. J=l i s of rank k 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. elements of Let x + y = z where (V^, x ,y , z are non-zero decomposable ',V^) (G) . Then for a l l i , <0 (x)> = <0 (y)> except possibly for one value i 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 <0 (x)> f <0 (y)> g g and s such that Let and T d q x, I <0 (y)> where d s T q r transformation and T = T, n d K(T..,•••,T ) 1 m K l'*"' m x+ (T T n T <0 (z)> ci: <0 (x)> q ~r q T )( y) Let on T : v n v n b e t n e n and = K^.-.yz : V• ->- V r r = 7=1=0 the associated transformation <0.(y)> = <0.(z)> <0^(y)> = <0 (z)> =(= <0 (x)> q q . r e 0 q . By Therefore such that be a. linear transformation such that is the identity mapping. ° ; be the identity mapping i f n £ 0 identity (V.,•••,V ) (G) , we have 1 mx <0^(x)> cr <0^(y)> + <0^(z)> = <0^(y)> f I <0 (x)> r ' q T,(x,) = 0 a a . Consider the associated . Let z = z, *••••* z . Choose 1 m i <0 (x)> . Let f q r f (z ) = 0 r r x = x, * ••• * x 1 m i s the identity mapping i f n £ 0 , by Theorem 2.3, s Corollary. 2.2, we have r . Choose g if n e 0 s for a l l i , i =f s . In particular, z <0 (x)> ^ <0 (y)> i s the identity mapping. g mapping i f n i 0 " s Since such that T, : V, V, be a linear transformation such that a d d | <0 (y)> r r q <0 (x)> =f <0 (y)> Without loss of generality, we may assume d e 0 s and q and f = n r r f r *f n £ K(f.,,''',f ) , we have 1 m Let f n °q * :V n Consider •*• V n 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 1 m n e 0 , i t follows from Theorem 2.3. that s f n i s the identity mapping for <0 (x)> = <0 (y)> s s . This yields a contradiction. Therefore there is possibly only one j such that <0 (x)> + <0 (y)> . j Now, assume that such 'a j exists and dim <0j(x)> > 1 + dim (<0^. (x)> f~\ <0^ (y)>) Then i t i s not hard to see that there are two independent vectors x, , x , , where if d , p e 0. such that J We have either x d <0.. (x)> cj: <0.. (y) > + <0^(z)> > d + (<z>+<0_. (y)>) i s direct. We may assume *. r r I < 0 j(y) p mapping i f n { 0. and transformation = Let g r (x^) = 0 ,'g (z ) = 0 and r Let § : v n v n b n e the identity i f n e 0_. . Consider the associated K(g^,''',g ) , we have m K( , ---.g^Cx+y) = y = K(g ,--- g )z = 0 , g;L r <x > + (<z >+<0.(y)>) i s direct. d r j i s the identity mapping. > r is direct or <x > + (<z>+<0^ (y)>) be a linear transformation such that 8 , a contradiction z . <o.(y)> • Hence there i s a r e 0. such that < p <0.(y)> + <x,,x > is a direct sum. Note that 3 P <0j(z)> C<0^(y)> , then to Corollary 2.2. d 1 > m : V -> \ r 21. a contradiction since y i s 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 i n 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^, • • • ,T > : O^t *' ' i V^) (G) •* an associated transformation such that T. : V. ->• V. i each • • •,V ) (G) be m i = 1 , . ••• , m . Then i ffl is non-singular for x K(T, ,-",T ) i s a rank 1 m k preserver for a l l possible k . Proof: Since . K(T,,--.,T )K(T~ ,-..,T ) = K(I,••',!) 1 m l m . 1 1 mapping on (V^ • • • , ) C ) • KC^,'**,^) v Let i s non-singular. G m i s the identity x z = z, + ••• • + z. c R, ((V , • • •, V ) (G)) where 1 K are decomposable elements. Then t c l m x 1 K(T^, • • • ,T )z =)= 0 . Let K(T ,..-,T )( E z.) - Z y . £ R '((V ,---,V ) (G)) 1 m j 1=1 where y^ , ••• , y 1 m 1=1 are decomposable elements. Clearly j _< k z 1 , • • • , z. fc 22. K(T7\• • • ,T ) K ( T 1 , - - - , T )(z) = z 1 m 1 m Now, we have 1 = KCT^.-.^T^Cy^.-.+y ) z being of rank k implies that K(T,,«'«,T ) i s a rank X m j > k . Hence j = k . Therefore k preserver. The problem of rank k preservers i s concerned with the converse of Theorem 2.15. §3. Applications. Let , u" be supspaces of U and 2 , V" be subspaces of V . 2 (i) ( I L j ® f l (U ®V > = ( t ^ f l l ^ ) ® ( V ^ V ^ It i s well-known that 2 A u i AA u 2 = A ( u [7, p. 20] 2 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 x W. = W x a(x) , V. = V ,. a(x)s x x for i =• 1 V. where U. = U x m and for a l l , a(x) ' x a e G . Then we have ( U Proof: r'- ' m x , U ) ( G ) ^ ( W l'"' It i s obvious that , W m x ) ( G ) = ( U i n W l»-"»Wx ( G ) * (ILflW., • • •, UflW ) (G) C (U. , • • • ,U ) (G)f\(W , • • • 1 1 mmx I X L X — M M 1 23. Let z be a non-zero vector of (U, , •",U ) (G) A'(W,, • • • ,W ) (G) . 1 . 'm x 1 . mx Then z R, ((V ,•••,V ) (G)) for some positive integer K 1 m X e t k . By Theorem 2.8, z z R. ((U., • • • ,U ) (G)) A R. ((W_, • • •,W ) (G)) . Therefore - K mx l ' K mx l z = x, + • • • + x, = y,+•••+ y, 1 Tc 1 k and s ome y each orbit e R^((W^,•••,W ) (G))•where j = l I 3.2. x. e R, ((U., , • .., U ) (G j 1 1 mx ... , k . By Theorem 2.3, for 0. of G , we have l <0. (x. )>+••• + <0. (x. )> = <0. (y-)> Therefore l , for some Tc I I + • • • + <0. (y. )> CT W f\ U r J I k — q , q e 0. . q i z e (U.. A W.., • • • ,U AW ) (G) . This completes the proof. 1 1 m m x and U2 be two subspaces of U . Then Corollary. Let ( v u ^ n (vu > = v ( u n u ) . 2 x 2 As an application of Corollary 2.4, we prove the following 3.3. Theorem. Let K(T) , K(S) : ^(G) -»- V (G) be two associated m X transformations. Suppose (i) p(T) , the rank of T , i s greater than m or (ii) x = 1 • Then with Proof: X K(T) = K(S) i f and only i f T = AS for some Ae F A = 1. m The sufficiency of theorem i s t r i v i a l . We proceed to prove the necessity. 24. Case 1. Let v.^ e V be such that basis Let i » *"" > z z n TCv^) = to a °f ^he range space of T . By hypothesis, T(v ) = z , i = 2 , ••• , n . Since ± =f= 0 . Extend ± n > m K(T) ='K(S) , we have K(T) (v..*•••*v.*•••*y ) = K(S) (v .*v.*.--*v ,,) , i = 2 , 1 l m+1 1 I m+1 L l Hence 1 2 Since *...*£;*...* = Sy *• • .*Sv.*" -*Sv i m+1 . • 1 I m+1 z ,i = 2 , f 1 z^ , ••• , z ^ ^ are linearly independent, z^*- • • * z \ * • • • * z m + ^ t0 by Lemma 2.10. Therefore by Corollary 2.4, we obtain l>'"'h'"' mrl <Z ,Z > = < Sv ,...,S^ ,...,Sv > 1 i n+1 ,i=2,...,m+l. Therefore m+1 m+1 H <z , • • •,z , • • ' . z ^ ^ = p\ <Sv ,••«,Sv ,••«,Sv m+1 ^> i=2 i=2 1 Since i (1) z^ , ••• , z ^ ^ are linearly independent, the l e f t hand side of (1) i s < z > 1 • Therefore the right hand side is a one dimensional m+1 ^ subspace of V . But O Sv ,•••,Sv.,•••,Sv-> 23 <Sv > . Hence J. i m+j. — ± < 1 <Sv^> = < z j > = <Tv^> since S(u) =f 0 implies that Sv^ is obviously non-zero. By symmetry, <S(u)> = <T(u)> . It follows that for a l l v e V . This implies that T = AS for some <T(v)> = <S.(v)> X e F . Clearly 25. Case 2. Suppose x = 1 • Let v e V . Then we have K(T) v *...* v If = K(S) v *...* v = Tv *• •••* Tv = Sv *•••* Sv . Tv = 0 , then we have Sv = 0 . If Tv =|= 0 , then and hence by Corollary 2.4, <Tv> = <Sv> T = AS 3.4. for some Tv *• • • * Tv + 0 . Therefore we have X e F . Clearly A. = 1 . m Remark. The above theorem (Case 1) i s proved by Marcus i n [11] under the assumption that the underlying f i e l d i s the complex numbers. 26. CHAPTER II 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 f i e l d ' F . Let VU the r-fold symmetric product space over 1.1. Definition. Let x, , • • • , x U where denote r >_ 2 . , be non-zero vectors of U . Then the pure subspace {x, 1 -u : u e U} , denoted by x, r-l 1 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 {x, 1 S^ , r r pure subspace of VU . Definition. Let S be a two dimensional subspace of U . Let x, , ••• , x , be vectors of U 1 r-k where x ' «U , r-l s^ : s^ e S , i = 1 , • • • , r} , denoted by {s^ is called a type 1.3. x such that x. i S , i = 1 , ••• , r - k , l T 1 < k < r . Then the pure subspace x , *s, r-k 1 is called a type s, : s. e S , i = 1 , ••• , k} , denoted by k I k x, 1 x , -S,, . , r-k (k) pure subspace of VU . The following result is proved by L.J. Cummings in [4] 1.4. Theorem. (i) If dim U > 2 , then every type r i pure subspace of VU 27. i s a maximal pure subspace where (ii) greater than i for some 1 < i < r. If dim U > 2 and the characteristic of F r , then every maximal pure subspace of VU i s zero or i s of type i= 1 , ••• ,r. In this section, we shall show that i f the characteristic of F is a prime p _< r , then there i s 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 u e VU by u r k y , e VU by u -y, r-k 1 r the pure vector u u-y, 1 u 1.6. Theorem. Let char F = prime t . Let x^ , •*• , x r-p set x ^-uF r-p M = {x^ be t and the pure vector y ,. r-k p r - p and r > p fc for some positive integer non-zero vectors of U . Then the t . i s a pure subspace of VU . Moreover r : u e U} t. dim M = dim U . Proof: Let k = p fc . For any x , y e U , we have (x+y) = x + (^) x k k = x^ + yk Hence x, 1 + • • • + ( ) x «y k k since P | ( ) k x •, *x + x. r-k 1 k k i i + • ••+ y k for a l l i = 1 , ••• , k - 1 x , -y = x, r-k ' 1 k J x • (x+y) . r-k k 28. For any non-zero a 1 1 k k k a x = (a x) where in F , we have a k JL is the k*"* 1 Therefore M Let root of a . Hence a x. 1 i s a pure subspace of u, , ••• , u I n x , «x = x, r-k 1 k be a basis of U . Let . k , 'u. = £ x. r-k l » i 1 i=l n £ a. x, . , i 1 i=l x , *(a x) . r-k k k VU . Then n x x a. , ••• , a e F 1 n 1 k k , '(a. u.) r-k l l 1 k k x .•(£ a. u.) = 0 r-k . - l I i=l n = n implies that £ i=l 1 k a. u. = 0 . Therefore 1 1 Also for any 3^ 1 y in U , we have i n F , i = 1 , ••• , n . x1 x a. = 0 , i = 1 , ••• , n . 1 y = n £ 6. u. i=l for some Therefore k «y = x r-k 1 n x 1 .• ( £ r-k . .. i=l I k 3. Iu.) k x , • ( 3 . u.) r-k ^ i i k = . £, 3. x, x *u.. . I 1 r-k i i=l n = £ x, i=l k 1 n Hence x, ••••• x , «u. , i = 1 , • " , n , i s a basis of 1 r-k l M . 29. The pure subspace in Theorem 1.6 w i l l be denoted by t x^'-.-x -U r-p P t 1.7. Definition. We shall use subspaces < x : Let x = x, x be a non-zero pure vector of 1 r to denote <x, , ••• , x > . The one dimensional 1 r U(x) £ »i > r VU . 1 » ""* » r = dimensional subspace <u> if . x = u^. y y J 1 a r e called the factors of i s said to be a factor of where x X l 1 Let <x.> = <za(i) ,.\> for some a e S r ± u, , ••• , u I n algebra over U and determinates £,>•••»£ 1 n be a basis of ^t^i»"""»^ ] n over = z.. r 1 , i = 1 , ••• , r U . Let VU The restriction of if ± = £ ± r to VU f 0 z r 1 [4, Corollary 1.9] denote the symmetric denote the polynomial algebra in n F . Then there exists a unique isomorphism d> : VU -*• F[£,»* • • ,£ ] such that 1 n * u k i ' The above definition i s well-defined since implies that r of multiplicity <y.> 4= <u> , i = 1 , ••• , r - k . J r-k x x . A one ( i = 1 » * *' » n) i s an isomorphism of space of homogeneous polynomials of. degree r r VU onto the vector (see [7, pp. 202-203]). 30. Clearly only i f in x^ <f>(x^x F [ £ , ] I n ) i s a product of Lemma. Let and Then » •** » x i > x^ x < x i k" > A i f and linear homogeneous polynomials F[g-,''»,£ ] , linear homogeneous 1 n elements. x^ , •* * , x^ r > k + 1 < r • In the Gaussian domain polynomials are irreducible 1.8. r VU i s a non-zero pure vector of ~ a r e be i z k non-zero vectors of ^ z r 0 factors of k l w h e r e U . Let r-k A e V U and z^ z^_ and A z e U . ± i s a pure r-k vector of V U . Proof; Consider the isomorphism <j> from VU onto FfC^,•••,£] . We have • (x, Therefore is • tyix^ x^A) - (j)^) ty(A) = ty(z ty(z ) ± z) 4 0 . ± r ••• ^(z^) =j= 0 . <K -^) > • ••• > ^Cx^) a Gaussian domain and since X Since > ^(.z^) linear homogeneous polynomials, i t follows that for each <<f>(x1 .)> = <(>(z. )> < 3^ and for some '('(A) i s a product of Hence a pure vector of <x^> j . where 1 r-k , ••• » r-k V U . 1 < x j and • • •, ^ ] <Kz ) r are i = 1 , ••• , k , j t =f j s i f t =}= s , linear homogeneous polynomials. > c 1 _< j.<_ r Fi^, are k factors of z^ z^ and A is 31. The following Lemma i s equivalent to Theorem 1.17 and Theorem 1.18 in [4]. The proof given here i s 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> and i s a pure subspace. Then x y have a common factor. Proof: If x and y a common factor. x = x^ are linearly dependent, then clearly Hence we assume x^_ and y = y^ x and y y^ . Since i , j , k , x . , y . , y, 1 generality, we may assume a basis u, , ••• , u 1n J dim (U(x) + U(y)) >^ 3 , either dim U(y) >_ 2 . Then are linearly independent. Without loss of K i = l , j = l , k = 2 . of U where For each non-zero and y have are linearly independent. Let dim U(x) > _ 2 or dim U(y) _> 2 . We may assume that for some x Extend y^ , y~ t y. = u, , y~ = u_ and x, = u„ . 1 1 / / i j A e F , x + Xy i s 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 l it t where i= 1 r . Let f : U -> U be the 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 P (f)(x+Ay) = P (f)(z(A)) r to r 32. It follows that f ( X j ) = 0 x, • f(x_) l / for some f( j) 4 x x has a factor f° H 0 j ra = of generality, we may choose vectors 1.9 of [4] implies that Now, l e t g(u ) = 0 2 r P (g) on V U It follows that = a > n d w e a r done. e » *** > r . Without loss z^(A) , '*• , z (A) such that Corollary r z^(A) = x^ + a^^(A) u^ g : U -»• U be the linear transformation such that , we have (x + Ay) = P (g) r x y i x^ = f(z^(A)) . Hence we have P (g) 8( j) 1 < and g(u^) = u^ , i = 1 , 3 , ••• , n . Consider the induced transformation If f(z (A)) . I f r n , then j Hence we may assume f(x ) = f(z (A)) r 1 (z(A)) r g(x )-----g(x ) = g(z (X)) 1 0. f ° r j » then rs o m e r x has a factor s ( j ) T" 0 Hence we may assume that g(z (A)) . 1 x ^ g(z^(A)) = n g(x^ ) for some non-zero A o r a H J n e F = < y£ > a n d w 1 » e a r e done. Therefore and for some where i 1 <_ k^ £ r . It then follows that n 8 Hence l (X)-x1+a z^(X) = n (a^ i i A u + 1 1 ^ (A)« -n 1 n ^ u t ) 3 ]_) * This shows that A x z(A) has a factor 33. <a k l l u V \3 + A ' S i n c e { < a 13 l x a + llV ' '• r 3 l < a x + a rl l u > } is a finite family and F i s an infinite f i e l d , we see that there are distinct non-zero X^ , X„ e F ' such that z(X^) and z(X ) have 2 a common factor. But <x , y> = <z(X ) , z(X,,)> , we conclude from Lemma 1 1.8 that 1.10. x and y have a common factor. Theorem. Le t - dim-U-.-> 2--_Vand: char F = prime j? .." Let x^ , ••• x r : be non-zero vectors of U where r > p , t a positive integer. t M = x^ x i s a maximal pure subspace of VU . r-p fc Then r Proof: M U{y} spans a pure subspace. Let k = r y^ be a non-zero pure vector in VU such that Let y = y^ U.. = < . , x.> 13 I j x 1 , • • • , r . Let Since V = We shall show that y eM. and W.. = <x. , y. > , i , j = 1 , ••• , r lk. l tc <y ± F i s algebraically closed, , y > ,i+ t ,i ,t= 1 , • • • ,r. F i s an infinite f i e l d . dim U >_ 3 , U =f (UU. .) U(UW., ) \J (UV )• Choose a vector ij lk it Since u e U such that ( u {. (UU. .) (J (UW. ) (J (UV. ) . Then we consider the pure vectors ij lk it t p x = x^ x *u and y . r-p t If for a l l i and j , <x^> = <y > , then are done. Hence we assume now y eM <x. > f <y. > for some l 1 o o 1 J i o and we and j o 34. Since u , x. , y. o o are linearly independent, i t follows from Lemma 1.9 J that x and y have a common factor. Let < u i common factors (counting multiplicities) of x x = u u^'x' 1 ,y = u > ,* *' > i > < u ^e "H * > a e t ie and y . Let uj^-y* ± r-k where y' x' and y* are non-zero pure vectors of V U . Then have no common factors. that <u> Since <u> x' and i s not a factor of y , we see i s a factor of x' of multiplicity p " . Since <x , y> 1 i s a pure subspace, i t follows from Lemma 1.8 that <x' , y'> i s a pure subspace r—lc t of V U . Suppose for some d , <x > i s a factor of x' , l < d < r - p d Let <y. > be a factor of y' . Then l ' • . J — u , x, , y. l d — are linearly independent. J This implies that dim (U(x') + U(y')) >_ 3 and x' and y have a common t factor, a contradiction. Therefore x' = Xu* for some X i n F and 1 5 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 e M . Hence M y' = z^ for some z in U . Consequently i s 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 i s a pure subspace and dim M = dim U from the proof of Theorem 1.6. Suppose M can be seen i s properly contained i n a 35. pure subspace N . Let x be a non-zero vector of U . Then properly contained i n the pure subspace fact that x'U* 5 t x*U^ i s {x«A : A e N} , contradicting the i s a maximal pure subspace of *V"4j ... > + Therefore M i s a maximal pure subspace. p If t U char F = p , we shall denote t {x* : x £ U} by t 5 and the pure subspace in Theorem 1.6 are called power type pure subspaces of degree t. The following elementary fact w i l l be needed. 1.12. Lemma. Let V be a k-dimensional vector space over an infinite field. in Then there i s an infinite subset W W of V such that any k elements are linearly independent. 1.13. Theorem. Let U be an n-dimensional vector space over an algebraically r closed f i e l d Then M i s of type of degree Proof: in F . Let n >_ 3 and M be a maximal pure subspace of VU . i for some i ,' 1 <_ i <_ r , or M t for some positive integer i s of power type t. Case 1. Suppose that for a l l non-zero pure vectors x and y M , dim (U(x) + U(y)) <_ 2 . We have two subcases: (a) For some non-zero non-zero element in M . Since x e M , dim U(x) = 2 . Let y be any dim (U(x) + U(y)) <_ 2 , i t follows that U(x) Z) U(y) . Let S = U(x) . Then MCS^ . Since both M and S^ 36. are maximal pure subspaces of VU , we conclude that (b) For a l l non-zero Since M x e M , dim U(x) = 1 . Then i s a maximal pure subspace, r y r and z M cr ( u : u e U} r dim M =f 1 ; for otherwise M i s property contained in a type one pure subspace. independent pure vectors M = S, ,. . (r) Hence there are two i n M . Clearly r r r linearly independent. Note that we have y + z = u y and z are for some non-zero u c U . In view of Corollary 1.2.2, <y , z> ^ <u> . Therefore for some u = az + by a , b e F . It then follows that (az+by) = a z + ••• + (£)(az) - ( b y ) + ••• + b y r r r r k k r r r r = z +y IT Hence T z of TO ITIc It IT a =b - =1 and f—1 ,z ( )a k IT—Ic -y , ••• , z VU . Thus characteristic t hence M = U P k «y (£) = 0 ..b = 0 , k = l , - " , r - l TC , *'• , y are linearly independent vectors for a l l k = 1 , ••• , r - 1 . This implies that F = prime p , r = p C for some positive integer t and i n this case. Case 2. There exists a pair of non-zero pure vectors such that since x and y of M dim (U(x)+U(y)) ^. 3 . I f dim U(x) = 2 and dim U(y) = 1 , then by Theorem 1.2.14, U(x) DD(y) , a contradicition. Similarly i t i s impossible that Let dim U(y) = 2 and dim U(x) = 1 . z be a non-zero vector i n 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 U(x) 4 U(y) . It follows that and U(x) D U(z) dim U(x) = dim U(y) = and U(y) ^ U(z) j dim (U(x)+U(z)) _> 3 or dim (U(y)+U(z)) JL 3 .. Hence for otherwise 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 is a common factor of x , y Assume now <f> = U(z) . Therefore U(z) and z . dim M = m . By Lemma 1.12, let z, , ••• , z , ••• n 1 infinite subset of M such that any m z^ forms a basis of M . By the above argument, for each Since be an i , z^ has a common factor with x and y have at most x or y . 2r distinct factors, i t follows that there are infinitely many z^ with a common factor, say 2 • Hence there i s a basis z. , ••• , z. of M with a common factor <x > . By Lemma 1.8, i, l 1 1 m < x j^ we see that <x <x^> , ... , If < x j 1 S > a > factor of any non-zero pure vector in M . Let be a l l the common factors of non-zero elements of M . j = r, - 1 , then clearly M' = {v, v 1 By Lemma 1.8, we see that M = x^ . : x, r-j M 1 1 x _^'U If j < r - l , l e t r x.-v, j 1 v , e M} r-j i s a pure subspace of V~'u . Since M is a maximal pure subspace, i t follows that I M' i s 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 i s some non-zero v E U such that <v> A e M' , A =f 0 . Hence i s a common factor for a l l by the same argument as in case 1 , M' = S. . (r-j) s x for some two dimensional S TJ subspace S of U or M' = U insteger s and char F = prime p . Thus M = x, 1 In case M=x, 1 i f some X1 -x. - S, . J (r-j) v r- j= p or M = x, 1 r x _ s M for some positive s -U P p x.-S. . v , then we must have J (r-j) x^ e S , then x". I where x, , ••• « x. A S ; for 1 J i s properly contained in the pure subspace x.*S, . , - * , a contradiction to the fact that 2 (r-j+1) M i s maximal. 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 w i l l be used in a latter section. Unless otherwise stated, throughout this section, U i s assumed to be a finite dimensional vector space over an algebraically closed f i e l d that dim U > 2 . F such 39. 2.1. and Theorem. Two power type pure subspaces N = u Proof: v «U r-p (r > p"") P t r-p = Av^ v ' for some r-p The sufficiency i s clear. z e U such that ± v 1 for a l l < 2.2. <z> = <w> . Therefore Theorem. u^ Let M = x^ M = N t *w - 4 = pp r-p t for some w e U . Since < z> + v^> we have are equal i f and only i f > . Then r-p t «z -P = v, r-p t t t i = , ••• , r - p 1 y^ t (ii) t Z r _> p + p x «f r-p P and x 1 x t r-p = z i y ^-U r-p P t >_ Z f e U , A e F ; or z t Z ' r-(p +p ) t y for some i ' - - " y z r-p = a , f , z. e U . Otherwise l z i z ,t z' r-(p +p ) MHN = 0 . A f V be and r >_ p*" and for some p t Ae JL and N = y^^ dim (MQN) = 1 i f f either (i) t > I y ^ = Ax^ r-p fc v for some r-p two distinct power type pure subspaces of VU where Then implies that 0 t P Choose a vector 1 u = Av^ r-p x «U r-p t P A eF. 1 u u «U r-p We prove the necessity. z £ <v >U-"U<v . .. ... u of VU 1 t M = u^ 40. Proof: f Suppose that MAN ={= 0 . Then there are non-zero vectors t I and a e TJ such that x^ x .f = y £ T" • r-p r-p , A P 0 fc 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 y t x «.ff r-p" ^ = Xx^ r-p ~ PF P t u I " for some t If z ± <f> ^ <a> , then clearly X e F. £ r >_ p + p and for some , i = 1 , ••• , r - (p +p ), x l = x t r-p i z I i" aP z t r-(p +p ) t y l y r-p l = Z l Z ,t * ' r-(p +p ) f P + t Conversely, i f (i) holds, then i f ( i i ) holds, then MAN = <x^ x I • t MAN = <z^ r z ^ -a - f > . r-(p +p ) P t ~ *^ P> a n d P P t 2.3. Theorem. Let M = x, 1 x 'U t r-p r maximal pure subspaces of VU where P and N = y.. y , 1 "'r-l be two J t r _> p and t i s a positive integer. t_ Then dim (MAN) = 1 i f and only i f either (i) y r-l = * r for some ae U ,X£ F for some a eU t or ( i i ) r > p and some i where and y, 1 1 <i < r-p fc y , = x, r-l 1 . Otherwise X 1 X t* r-p ~ "x. i MAN = 0 . 1 a P p x .a t r-p t 41. Proof: The argument i s the same as that given i n the proof of Theorem 2.2. t 2.4. Theorem. Let M = r > p^ and N = S ^ x ^«U r-p be a type be a power type subspace where P r subspace of VU where S i s a two dimensional subspace of U . Then dim (MAN) = 2 i f and only i f x ^ x eS r-p Otherwise Proof: M0N = 0 If MAN ={= 0 , then let A e MAN A = x, 1 for some x _ «u t r-p u e U , s ^ e S , i = l , t P and A =f 0 . We have = s, 1 s r ••• , r . Clearly this implies that Xj , ••• , x e S . Conversely, i f x^ , ••• , x £ S , then r-p r-p t t MAN = <x^ x t" l » i t" 2 ^ S = <v^ , Vy> . . r-p r-p t t x V Hence 2.5. x V > w e r e dim (MAN) = 2 . Theorem. Let M = U be a power type pure subspace and let P r N = y^ and y (k) ^ p -k t e a Y'P t e K pure subspace of VU where p S i s a two dimensional subspace of U . Then (i) MAN = 0 t >_ k > 1 i f p" > k ; 1 42. t and ( i i ) Proof: u eU MAN = <s This implies if p S = <s^ , s^> >k and MAN + 0 . Then for some non-zero t , ••• , v, e S , u = y • • • • «y •v v, ={= 0 . p -k P 1 1 1 y^ \. S . z t+1 v > k . ( i i ) i s obvious. t Let x = x. 1 x = u, I the pure vector notion = k where 1 Definition. Suppose that v p if p <u> = <y > = <v^> , contradicting the fact that M AN = 0 2.6. P (i) Suppose that and non-zero Hence ,s > P t x m u -z t+1 z m be a non-zero pure vector of VU . z m in U where by t < m . Then we denote x, 1 J x|d products w i l l stand for x|v^ x m 1 u, 1 u t . The v^ for some v^ , ••• , v^ e U where <v > , ... , <v > are d factors of x . This definition i s well-defined 1 d since u, 1 z_,, u • z,. - • • •; .z = u. t t+1 m l z =w m t+1 t+1 t u -w^,, t+1 w m =}= 0 implies that 1 w . 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, 1 y , -S,, . be a type k pure subspace where r-k (k) r > k >^ p* and S i s a two dimensional subspace of U . Let x = x (i) 1 x r-p and y = y. y , . Then dim (MAN) = 2 i f and only i f x dim (MAN) <_ 2 . Moreover, we have and y have exactly r - k 43. common factors while the remaining in S have exactly dim ( M O N ) = 1 i f and only i f r >_ p + k and x , y fc r - k - p " common factors while the remaining 1 are contained in S and the remaining Proof: in factors of x are contained t . (ii) x k -p (i) Suppose dim (Mi"\N) p* factors of y s s factors of are the same. 2 . Let A , B be two independent vectors >_ M O N . Then there are two. independent vectors « , s^ , i '» " * •' k k u ,v of U and ^ such that e t A = x i t' x u P= i y W s i s k ( 1 ) r-p t pr B = x, 1 If both x -v t r-p , = y. l J J y , -s' x-k 1 , s, k u and v { S , then t x i t x = ^•••••y -k' uP r } t ' s i s k = ^•••••y -kl v P ) , r r-p Since y^ £ S , the above equality implies that , |u r-k' P = Ay1 J y , |v r-k P J Clearly this i s a contradiction. for some A e F . 1 Therefore either u e S or v e S S I s k 44. We may assume that i t follows from (1) that while the remaining u e S . Since x x^ k a <x . > , • • • , r-k+1 > = 1 > *** » ~ k, r r - k common factors k - p*" factors of x are contained in S . *" *'^r-k » = < and y have exactly Conversely i f x and y have x^ <u> f y ^ » i n d t ^ r - k common factors, say, i i 8 ie r e m a n - k -p n factors fc <x > are contained in S , then by Theorem 2.4 t r-p (or Theorem 2.5 ( i i ) i f p = k) , we have fc t x -U A S„ .) = 2 t ' (k) r-p dim (x , , r-k+1 Therefore p 1 dim (MHN) = 2 . (ii) Suppose M(*\N = <A> where A =f 0 . Then we have t A where = x t* r-p u P " l y y r - k - S l " . " S k ( 2 ) s^ , ••• , s^ e S . If ufcS , by the argument in (i) , dim (MON) = 2 a contradiction. exactly l x Therefore u ^ S . From (2), we see that r - k - p*" common factors while the remaining are contained i n S and the remaining p fc k factors of y The converse of the statement i s easily checked. x and y have factors of x are equal. 45. 2.8 Theorem. Let M = x. 1 x -U t. t and N = y, y ,.S„. r-k (k) ' r > p*" > k > 1 and S i s a two P r-p be two maximal pure subspaces where • 1 v dimensional subspace of U . Let x = x, 1 Then dim (MAN) = 1 i f f r >_ k + ' , common factors while the remaining and the remaining Proof: Suppose k x and y = y, t l r-p J y ,. •'r-k J x and y have exactly r - k -p fc factors of x are contained in S p*" factors of y are equal. Otherwise MAN = 0 . MAN + 0 . Let A be a non-zero element of MAN . Then t A = l X X tr-p u P= y l y r - k ' for some u in U and s_^ i n S . Since follows that u have exactly r - k - p*" of r S . Hence x are contained in y has p fc l S S p >k C k and y^ £ S , i t factors of <u> . Hence common factors while the remaining S and the remaining k x and y factors p " factors of y are equal. 1 The converse of the theorem i s 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 f i e l d where dim U > 2 . Then the maximal dimension of 46. the intersections of any two distinct maximal pure subspaces i s 2 . 2.10. Remark. Let n be a positive integer. collection of sets such that |A^H A I = n - 1 such that and , ••• , be a |A^| = n , i = l , ' " , k and i f i=j=j. k W C f\ A. i=l Let If k > n + 1 , then theres exists a set W |w| = n - 1 . 1 r Definition. Let C^(VU) denote the collection of a l l type one r r pure subspaces of VU . For each positive integer t , l e t P (VU) denote 2.11. the collection of a l l power type pure subspaces of degree r in VU . t r 2.12. Theorem. £ C , M 2 r - 2p Let C C P (VU) be an infinite family such that implies that non-zero vectors t M^f\ ^ y =f 0 . Then , ••• , y fc and there exist with the property that r-2p 1 r >^ 2p M eC C implies that M = y i y , t- 5 a t D p t r-2p for some a^ £ U . (y^ , ••• , y are deleted i f r = 2p ) fc fc r-2p Proof: It follows from Theorem 2.2 that assertion i s clear from Theorem 2.2. r >_ 2p t . If r = 2p Hence we assume t r > 2p . fc , the 47. For each M = x 'U r-p E C , define a set P t r M* = t<x^> , ••• , <x >} i . This is well-defined because of Theorem 2 r-p t C* - {M* : M e C} . Let We claim that & - & C i s infinite. For i f C i s f i n i t e and & C = {M. , ••• , M. } where M. e C , i = l , ••• » 3 » then for any 1 j l pt w e n a v e < z > E 3 * J * M = z^ z " ' U . But U i s a finite r-p i=l i=l t set, i t follows that „ N eC Let in * C C i s finite, a contradiction to the hypothesis. * such that . Let M e C . Since M = Z ; L N . " - --z t t - •b b .U r-2p' P1 (1) P t t t «a -U z r-2 P ? |N | = n , we have P Z z. i n U , i = l , ' " ' , r - 2 p Since and some t (i) in U . >}| ^ _ r i — 1 . C IM I > n - 1 and i f |M | = n - 1 , then M = {<z >,•••> < >} an . r-2p z 1 hence a ,b |{<z > , • • • , <z r-2p 1 Observe that n MDN f 0 , by Theorem 2.2, N = z for some has the maximal cardinality N D M : t 48. (ii) we have <b> = <c> /•>* Since if M C * contains an element (1), and hence ftom <c> such that <c> { N , then | = n .' |M i s infinite, we observe from ( i i ) that there are infinitely * & many M e C are such that |M | = n . Note that i f M^ , M,, e C such that || = |M| 2 = n and M =(= M , then 2 ^AM |= n - 1 . By Remark 2.10 and ( i ) , we conclude that MeC Now l e t N ,N x 2 e C such that |N | = |N | = n , N n 2 =j= N and t - y x y t -c t -u p p r-2p N 2 = ?i y , -d -u p t t p t r-2p for some y Let M * - f * , c , d e U . MeC. Then M ={= N^ . From Theorem 2.2, we have or M =f N 2 . We may assume 2 49. t M = w w P r-2p f ,g , w for some P t P 1 t «f -U t P t r-2p 1 i n U . It then follows that M*H N* = {<w> , ••• , <w >} , |M*HN*| = n - 1 1 r-2p But we have N ^ f i N = (<y^> > ••• , <y >) • Therefore r-2p 2 t {<w > , ••• , <w since M AN^ = N^H w >} = {<y > , ... , <y >> rr2p 1 r-2p t . It then follows that i w 1 for some j A in F . Hence = x t r-2p ?i t y 1 M =y 1 t r-2p <g> = <c> . Consequently C y .f r-2p t P t «U . This completes P t the proof. The following result i s due to Cummings [-4]. 2.13. Theorem. Let dim U >_ 2 . Let C = {M |C| > r > 2 . I f for every M ,M 1 : i e 1} C (^(VU) such that e C , M. f\ M. 2 l• 2 x 1 =f= 0 , then there exist 5 0 . non-zero vectors y, , • • • , y 1 r-2 such that n M. = y, 1 '1 y _.a.»U . r-2 x i . e I , a. e U . I 2.14. Let C be an infinite collection of maximal pure subspaces Theorem. r of type one or of power type in VU l ^ 2 ^ ° ' M M ^— ^i^ T h e n except possibly when x 1 W V U ^ o such that for every *~ — ^ r P t v u ) f o r in C , , Positive integer s o m e t, char F = 2 , i n which case, there exist non-zero vectors , ••• , x - in U r-2 for some non-negative integer s and a subset S1 of U such that C has the following form 2 C = {M : M = x, 1 s+l x ...IT -s+1 r-2 s s x .w -U _s+l r-2 2 or M = x. 1 where w e W} Proof; either Let P = C H C ^ V U ) = C f \ P (VU) . By hypothesis, and i s infinite or Q.^ i s infinite for some positive integer t 'Case 1. Suppose () i s infinite for some positive integer Then by Theorem 2.12, there exist non-zero vectors x.. , • • • , x r-2p 1 U and a subset W of U such that Q. = {x t t -x - a .U : a e W) . r-2p p C P t. in C Suppose that some type one subspace y •y ,.U e C . r-l Since 1 S <b> £ ( y < > 1 infinite, we are able to choose b eW such that , ••• , y _ ^ • y Theorem 2.3, we have < > r B 1 t x. y l y x t. .b - f P P 1 t r - l (x.. x t-'b / r-2p P o n e product) «f t for some f in U . In both cases, contradiction. <b> e {<y > , t.> , <y >} 1 ' lr-1 Hence no type one subspace i s in. C . Suppose that some power type subspace where Z. < t . Choose ceW such that y 1 y r-p tU I P e <c> £ {<y^> , • • • , <y r-p By Theorem 2.2, we have x_ P r-2p 1 P_ P C t r-p for some t t I -c . f x t t f in U . Since p > p , in both cases, a contradiction. Hence Suppose now some y^ P^CVU) C\ C = 0 y £' r-p ^ •x .c^ /P r-2p P^ e products)'f <c> e {<y^> for £ < t ^ where i. > t and -£ r >_ p P 52. 0 Choose and d e W such that <d> =f= <x^> <d> =)= <y^> for a l l i = 1 , • • • , r - p for a l l j = 1 , ••• , r - 2v~ . Then from Theorem 2.2, we obtain Z r-p" t x x •d r-2p 1 t = P Z (y t y 1 r-p Jp products)-f t for some f in U . I f (2) holds, then <d> = <f> and (2) P Z p = p choice of d . This y i e l d s a contradiction. Hence (1) holds and <d> = <f> , p t d i v i s i b l e by 2. we have x Z = p - p Therefore x n t . This implies that p = 2 y p r-p = Ay. r-2p 2 C = {M : M = x- o r P 1 p is Z r > p for some in F . Therefore A M = x. x r-2 i s infinite. x, , ••• , x „ i n U 1 r-2 = {x1 and hence Also i f 1 Suppose that 1 Z t + 1 = Z . t1 exist non-zero vectors = p 2 0 Case 2. t t+l x t-j.!* " r-2 1 and 2p because of our - -w t t -U , 2 where t 1 By Theorem 2.13, there and W C U — such that x _.a.U : a e W} . r-2 t Assume that some power type pure subspace y^ y r-p G ^ where w e W} 53. r 21 P*" and t i s a positive integer. <b> =}= < y i > for all i j = l , * ' ' , r - 2 . = 1 , • • • , r Choose b e W such that p * - <b> and =j= < X j > for all In view of Theorem 2.3, we have t y x _-b = r-2 1 y - f r-p (3) p t ± t y l'*'" i y " t' y ^ f P r-p for some f i n U . If (4) holds," then <b> = <f> and p = 1 by our fc choice of b . This yields a contradiction since This implies that p =j= 1 . Hence (3) holds. fc <b> = <f> and p - 1 = 1 . Hence fc Also i f r > 2 , we have x, 1 x _ =. n y, r-2 1 J 3 p = 2 and t = 1 . y „ for some n i n F . r-2 Therefore C = {M : M = x 1 n 2 x _»U r-2 or M = x. • • • • «x „'a-U where 1 r-2 a e W} Hence the proof i s complete. §3. Rank One Preservers From Let U and W be finite dimensional vector spaces over a f i e l d F . Let f : U s-r r S VU to VU , r < s W be a linear transformation. non-zero vectors of W where r s ty : XU -> VW defined by Let w., • • • ,w 1 s-r r < s . Then the mapping be 54. <f>(x , 1 , x ) =w r w 1 8 is clearly multilinear and symmetric. - r * f ( x i > f ^ x ) r Hence by the universal factorization r property of VU , there exists a unique linear transformation * r s f : VU -* VW such that * f (x x )= w r 1 for each x- , • • • , x 1 r w -fCx^ s-r 1 1 * eU . f induced by w, , ••• , w 1 s-r f(x ) r i s called the linear transformation and the linear transformation f :U W. It i s easy to derive the following two properties: * (i) (ii) f i s a monomorphism i f and only i f f i s a monomorphism. ft Let g be induced by y.. , • • • , y Jsr and g : U -> W . ft ft Let the rank of f be _> 2 and F be an infinite f i e l d . 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 1 s-r such that An = 1 . J 3.1. Definition. A linear transformation r i f every non-zero pure vector of VU of J r s T : VU ->• VW i s a rank one preserver is mapped to a non-zero pure vector VW . r s VU to VW i s said to be a r type one mapping i f every type one subspace of VU i s mapped into a type one s subspace of VW . 3.2. Definition. A rank one preserver from 55. Throughout the rest of this section, U is assumed to be a vector space over an algebraically closed field r s every type one mapping from F . We shall show that VU to VU , r < s , i s 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 i s not hard to see that every rank one r s preserver from of VU to VU , r < s , i s induced by s - r non-zero vectors U and a scalar multiple of the identity mapping on U . Hence, from now on, we assume dim U >. 2 . 3.3. Let r > 2 . Two type one pure subspaces Definition. , and z , 1 and z^ y^ ^r-1*^ r of VU are said to be adjacent i f y, y .. r-l — 1 'r-l r-1 exactly r - 2 common factors counting multiplicities. z 51 Z n a v e 2 If r = 2 , any two distinct type one pure subspaces of VU are said to be adjacent. It i s 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 Lemma. Let T : VU s 3.4. 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 56. onto Since TCM^ defined by f ^ u ) = T(x -2' i' ^ x 1 y r U T i s a rank one preserver, i t follows that Assume that T ^ ) = T(M ) . Then f o f i i s a 2 linear mapping of U . Since r e a c h u e u • i s a monomorphism. non-singular F i s algebraically closed, ^2^^1 h a S a non-zero eigenvector, say v . Let X be the corresponding eigenvalue. Then f ^ v ) = Af (v) . Hence 2 T( x X;L It follows that hypothesis. 3.5. -2' l' ^ y r V = A T ^ i r-2' 2' ^ ' x X y y^ = Xy . Therefore =M 2 This proves that V 2 , contradicting our TCM^) =f T(M > . 2 Remark. The above proof is exactly the same as the case r=s proved in [4, Theorem 3.3] 3.6. Definition. i r Let M = x„.....x ,.U be a type one subspace of VU. . 1 r-l Then the factors of x, 1 3.7. Lemma. x . are defined to be the factors of M . r-l r s Let T : VU -> VU , be a type one mapping where r r <s. Then the images of any two type one subspaces of VU have at least s - r common factors counting multiplicities. Proof: Let 'M, = x , 1 1 type one subspaces. x .U and M = y, r-l r l J Consider the pure subspaces y ->U be any two r-l 57. M =y i For each y 1 i - l * i - 1 , ••• , r - l , x i X r - l ' T(M^) and T(JL ^) have at least T(M^) and T(M ) have at least + 3.8. for y 2 » **• » r - 1 . ) =f 0 since M n M ± s - 2 common factors. + 0. Consequently, s - r common factors. VU be a type one mapping. l ' " * " ' r 2 ° U. , we have y _,-u-U) : u e U , u =j= 0} = {z y For any non-zero f some non-zero vectors Proof: ±+1 i = s Lemma. Let T : VU vectors {T(y » -T(M ) A T(M Hence r U z _-w-U : w e W} . 9 z, , ••• , z „ e U and some W C U . 1 s-l — In view of Lemma 3.4, C = (T(y^ y^ ~' '^ : u e U , u =j= 0} i s n 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 1 for some non-zero vectors 3.9. Let z, , ••• , z • of U and some W C U 1 s-l — r s Let T : VU ->• VU be a rank one preserver where Lemma. s> _r > 2 . x, , ••• , x , be r - l non-zero vectors in U such that 1 r-l <x > 4 = <x _> . If {T(x " r-l r-2 1 1 C {y „'W«U : w £ W} s-Z x «-u-U) : u E U , u 4= 0} r-2 y «u*.U : u £ U , u =f 0} 0 1 for some non-zero vectors y. , • • • » y _ 58. of U , then {T(x « • • • •x^_ .x _ .u-U): u e U , u f 0} <£ {y 1 2 Proof; r y ^ - u - U : u e U , u =j= 0} . 1 Suppose to the contrary that {T( \-2' r-l' ' ^ X X]L For each non-zero M x , r-l y _ 'u.U : u e U , u f 0} . (1 : u e U , u =}= 0} C {y V g 2 g e U , let M = x, g 1 Since U x 'g'U and N = x." «'-x g 1 , 0 r-2 e .{x, 1 & x „-u-U:ueU,u }=0} : r-2 -x , *g'U r-l r-2 1 & , we have r-l for some non-zero <y> *i <x f r T(X, > x > y < > f e U . Now, choose a non-zero vector f < x r 2 > _ -y- ) = y u r 2 for some y ± 1 eU ' y s - 2 * y ' ' u ( x"" _-x '•y-U) = y, r-2 r - l 1 <x > 4= <x > , i t follows that . M r-l r-2 y 1 adjacent. such that -^en by assumption and from (1), T(x. 1 for some y' ' e U . Since y Hence by Lemma 3.4, T(M^) =f= T(N ) . Therefore y 3 ) y 0 *y , , , U (4) s-2 and N y <y'> =f <y'*> . are 59. Consider the following equalities: T(x^ X r-l" ^ y " s-2' * = y - f y_.i...y = y- We have and N r-1 a for some o*y''"b b since <f> = <a> = <y"> T(M a (from (3)) (from (4)) <y'> =(= <y"> or <f> = <b> = <y'> ) = T(N ) contradicting Lemma 3.4 r-l are adjacent. vectors from since y Similarly, <f> = <y'> implies that T(M ) = T(M ) , r - l x Hence the Lemma i s proved. Definition. Let V be a vector space of dimension set of vectors k (from (2)) y y a contradiction. 3.10. r s o m e <y'> = <b> , <y"> = <a> x X ° y y <f> = <y"> , then M f . ' . for some It then follows that either If y k . Then a z, , •.. , z i s said to be i n general position i f any 1 m z^ , ••• , z^ are linearly independent. i We shall need the following Theorem in order to prove Theorem 3 . 1 4 . 3.11. Theorem. Let V^G) be a symmetry class of tensors over V X associated with a subgroup G of S and a character m dim V = k . If k ^ m , then x o n G . Let V^G) has a basis of tensors of the form X x.,*'««*x in which ± m x, , ••• , x 1 m are linearly independent. If k < m y 60. and F i s an infinite field, then x * ... * x 1 m Proof: such that V^G) X x, , ••• , x 1 m has a basis of tensors of the form are in general position. e Consider the mapping ty : XV -»- V^CG) 1 X <Kv, , • • , V ) = V 1 for each v, , ••• , v 1 m m r defined by * • •• * V l m. in V . Clearly ty is multilinear, hence by the • J m universal factorization property of ® V , there i s a unique linear transformation ty : ®V -> V^VG) X <J>(V-i ® • • • ® v i for each v. , ••• , v 1 m such that ) = V m '* • • • * V 1 m in V . Since V^(G) X i s spanned by the set of a l l 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 i s t r i v i a l in case of dim V = 1 . Hence we may assume k >_ 2 . We proceed by induction on m . Suppose f i r s t Let v u be any non-zero vector of of u . Then the form 2 ®V . x^® V ." ' Let m = 2 . be linearly independent u g u = u ® v - u<g)(v-u) . Hence the set of a l l tensors of where x^ and are linearly independent, spans 61. Assume now the theorem i s true for m = n ', n >_ 2 We have the following two cases: (i) k _> n + 1 . Let u^® "' ® u n + ^ e "(g^ v • y B t n e induction hypothesis, u, ® ••• ® u = Z a.y... ® •••COy. 1^ n . .. i l l ^ m i=l J t , some for each i . If y , ••• , y ^ , u^ ^ are linearly dependent, choose such that y ^ , ••• , y ^ , z^ are linearly independent. Then in y y ^ , • • • , y^ for some V i l * a^ in F where n Note that + n ® in® n+l y are linearly independent U = y i l ® *•• ® i n ® y y., , ••• , y. , z. - u ii in l n+1 Z i " il® y ''' ® i n ® y ^i^M' are linearly independent. Since t n ® in® V l we conclude from (1) that ^"V has a basis of the form u l® ••• ® u n +l . i il® i=l = E I such that x^ , ••• , (ii) hypothesis, a ( y y ) ' x^(g) • • • CS> 4.^ x n are linearly independent. k < n + 1 . Let u^(g) ••• ® u n + ^ e "(g)^ • Y B t n e induction 62. t u-L® for some • *' ® \ = ^ i il® b *" ® d v v t and some b. in F where v., , ••• , v. i l 1 position for each i . Since U V + d where i i s i n f i n i t e , we have for each <v.. ( l j , ••• , v i, > ' "' ' k-1> J denotes the totality of s t r i c t l y increasing sequences n k - 1 of = F are i n general m. 1 , ••• , n . Choose integers chosen from J £ Q k-l,n Note that the vectors z^ e V such that ^ 1 z. - Au ,, . v.. , ••• > . . n+1 xj i ] are linearly dependent for v x w at most one value of A i n F . Hence we are able to choose a non-zero X. in F such that z. - A. u^., , v. . i x n+1 13 - - i . .\ . U independent for a l l i in Q, , k-l,n r V J il® x ® in® n+l V U = X i 1 ( v il® . -, , V . .We , ••• , v. . . » » v . are linearly . i J~ V l k x - ] have ® in® i- il® V Z ® V V ^ i V l Hence t U l®'"® n+l U = \ 1=1 _i V i < ll®'--® in® i v v z - V il® , , , ® ln® V ( z r lVl X ) ) " ' 63. Since v^ , — v., , ••• , v. IJ. m follows that such that , z. - X. u I 1 n+l x^ , ••• , m are i n general position for each n+1 (5?, V has a basis of tensors of the form •cy 3.12. Remark. whenever , z^ are i n general position and } i , it x1^ „ ® * " ( 8 xm are in general position. It i s clear that i f V i s over a finite f i e l d , then i s chosen to be greater than the number of vectors i n V , two of the vectors from x, , • ••• , x 1 m of U must be equal and so the above theorem is false. The following Theorem i s due to Cummings [4] : r Let T : VU -»- VU r 3.13. then Theorem. T , I f dim U f 2 , i s induced by a nonsingular transformation on U . r 3.14. Theorem. If be a type one mapping. s Let T : VU dim U > 2 , then T VU , r < s , be a type one mapping. i s induced by s - r vectors of U and a nonsingular linear transformation on U . Proof: of U Case 1: such that r > 2 . Let x, , ••• » r x'- l be 1 <x.> + <x.> ' 1 T(x n 1 J x i f i 4 j • let T - .U) = z, r-l 1 z -U s-l n r - l fixed non-zero vectors 64. i = l , ••• , r - 1 , From Lemma 3.8, we see that for each {T(x 1 x*. l x -'u-U) : u e U , u 4= 0} = {y._ r-l ' i l for some vectors Since e U and for some U. d U y., > ••• , y./ x(s-2) J il l — z , • U E (y . s-1 i l z, 1 v., ..-vU : v e U. CT U} •'i(s-2) i— y. , ^•v•U:ve'U. C u } i(s-2) l— n , i t 0 follows that some y., o \ * v . for some v. e U. and • i(s-2) 1 1 1 z - = X; y., s-1 3 •'il z, 1 X. e F where / i = l , ••• , r - l . I According to Lemma 3.9, we have for each pair of distinct and k j, <yj l y j ( s - 2 ) > * < y kl k(s-2) y > Therefore we may assume without loss of generality that for each y., •'il y., = n . z, • i(s-2) I 1 O N 7 z s-i for some n. e F and that any two vectors from i J linearly independent. i = l , • •• , r - 1 z i s-1 z ,>•••> z / ,\ are s-1 s-(r-l) 65. Our task is to show that r T(VU) C <z, z *s....-..s : s. E U,i=l, s-r 1 r l r - l non-zero vectors of U such that — Now, let u^ , ••• , <u > =f <u.> ± if u r -i b e a n y i =(= j . Let T(u.. u -U) = w. r-l 1 We shall show that w . -U s-l 1 1 1 <z > , ••• , <z > 1 _ s-r are s - r factors of w, w s-l 1 To do this, we f i r s t show that there exists a type one pure subspace B r in VU with r - l distinct factors such that the following hold: a) (ii) T(B)= g_-i u 8 l s <g, > , • • • , <g > 1 .' s - r w is 1 are s - r factors of w, 1 C = {T(u 1 w ,; s-l where i = l , « " , r - l . u^.u-U) : u £ U , u =(= 0 } w «U £ C , i t follows from Lemma 3.8 that s-l C = {w. w 1 for some x <u!> | {<z > , ••• , <z _->} Consider the family Since ;_ -u; & & (iii) u r i ^ and some s . - 1 l w ..u.U : u E V,CU} s-l 1 — V ^ C U . We may assume i ^ = 1 . Since is an infinite family, we are able to choose non-zero vectors C f £ U and 66. f eV 1 such that T(u u _ -f.U) = v x r <f> £ {<u^> , ••• , < u r _ the factors of u, 1 > 2 w _ .f .U , 1 2 ^ and g i i<Zj> < f , > u „'f«U r-2 2 , •••* , < Z S _ > 1 J • Note that are distinct. Assume now we have shown that for some there exists a type one pure subspace d^ t , l < _ t < r - l , r - l d r - l distinct W l t n factors such that <> a (b) W l W s - 1 where < d l d r - l ' U <h > , ••• , " l ) h < n 1 s _( t + s - l ' h i) > a r e z l U s 5 ~ ( t + 1 ) factors of 5 ( 1 T c ) < h s-t * " ' Consider the families i = 1 , ••• , t + 1 . '^s-l"^ { < > C. = {T(d, l 1 Since h, 1 » r-i ' < Z =' { h , 1 h , s-k. l h > } 1 h , «U e C. for each s-l 1 ..u.U : u s-l e W. C l— " d ,-u-U : u + 0 , u e U} r-l i = 1 , ••• , t + 1 , i t follows from Lemma 3.8 that C i s-l U} 67. for some k. , I < k. < s - 1 and for some W. C U . Since a l l 1 — i— l— d ..-U are distinct, by Lemma 3 . 9 , we must have r-l ' factors of d, 1 k n a 1 fk m for m f n where m , n = 1 , •• • , t + 1 . Therefore there i s ' ' ' 1 j such that k. > t + 1 . We may assume J - k. = t + 1 . Now C. J J is an infinite family implies that there are non-zero vectors d' e W. 3 d e U and such that T(d, 1 iT . r-j d ,-d.U) = h r-l 1 h . x'd'.h ^ s-(t+2) s-t /t <d> A {<d > , • • • , <d ..>} and <d'> A {<z,> , • • • , <z 1 r-l 1 s-1 T d r-j d ,.d»U are distinct. r-l -U , s-1 >} . Note r that the factors of d, 1 h 0 Therefore our inductive argument shows that there exists a type one pure subspace B r of VU with r - l distinct factors satisfying conditions ( i ) , ( i i ) and (iii). Now, we shall show that <g, 1 & each i = 1 , ••• , r - l , choose a vector <f > | {<g > , ••• , ± 1 From Lemma 3 . 7 , i t follows that z, 1 i = l , 6 z s-i < 8 g, 1 z , . f . have at least s-1 1 ••• , r - l . Since s _ > r g > =.<z, s-r 1 z > . For s-r f. e U. such that » ••* i g -u,' s-r 1 » ••• » < u r _i > } • u' , and r-l s - r common factors for each <u'> , ••• , <u' > A {<z > , ••• , <z >} 1 r-l 1 s-1 r 68. and < g-i k { 8 < < g * * • > <g » > » ••• » 8 _ > 1 1 i= » { < u g s _ for each r ) l s a factor of cf>( z *_ g Zl i = 1 , ••• , r - 1 . Since » i t : f o l l o w s z s-i g ic s ~' in <j>(z This implies that F[£, > • • • » £ ] » I n z ) . Hence s-r <<b(g, "1 T e 4>(g 1 g and therefore z <g, 1 6 are 6 g > = <z, s-r 1 s - r factors of w^ T(u, 1 z > . This shows that s-r W z .) , s-l )> since they J are both products of s - r linear homogeneous polynomials in Hence ) are g _ ) i s a factor of )> = <4(z, 1 s-r s-r ,£] and <J)(z S i = 1 , •*• , r - l . VU onto )> for k =f=m z ) i s a greatest common divisor of i>(z, • • • • •'z s-r 1 s-i T t z , for s-l 1 <<|>(z _, )> = { = <<J>(z relatively prime in the Gaussian domain ^(z, • 1 t h a z _ ) in F f ^ , i k , m = 1 , ••• , r - 1 , i t follows that <Kz.. 1 > r $ from the symmetric algebra s where _i ^ u F[£, , ••• , E, ] described i n §1. The polynomial 1 n . the polynomial algebra ± < l , ' " , r - l . Consider the isomorphism i > ( & » * * * •» > are s - r factors of z, 1 > s-r each > r , ••• , £ <z,> , ••• , <z > 1 s-r s - 1 * Consequently u , -U) = z r-l 1 «D.U s-r z r-l for some non-zero pure vector D in V U . Since dim U > 2 , Theorem 3.11 r-l V U has a basis of pure vectors of the form implies that r where < u ^ > k < u j > i =f= J Therefore we have u, 1 u , r-l 69. r T(VU) C <z— 1 z • s, s-r 1 s : s. E U , i = 1 , • • • , r> . r i * r r T : VU ^ VU as follows: Define a mapping T*(C) = C* i f T(C) = z, 1 T i s well-defined since C' = C'' . Since z .C , C* s-r z, 1 e VU z -C' = z, •••••z •C' w i l l imply s-r 1 s-r T i s a rank one preserver, i t follows from Lemma 1.8 that * T i s also a rank one preserver. Moreover, T is a type one mapping implies * * that T i s a type one mapping. By Theorem 3.13, T is induced by a non-singular linear transformation T(d. 1 for each d ) = z. r 1 f on U . Consequently, z .f(d_). s-r 1 f(d ) r d. in U . Case 2: r = 2 . By Lemma 3.8, we have MT(u-U) : u e U , u =)= 0} C ^ for some a. in U , i = l , l T <™> ^ < a l a , , s - 2 a^-u-U : u e U , u =}= 0} , s - 2 . , , S l ' S 2 5 Hence S l '2 S £ U > 70. By the same argument as in case 1, we see that a^ , ••• , a _2 g °f U T i s induced by and a non-singular transformation on U . Hence the proof i s complete. 3.15. Remark. by VW with s Theorem 3.14 i s no longer valid i f VU i s replaced dim W > dim U . For example, i f dim W >^ dim VU and z, , ••• , z are non-zero vectors of W , then there is a monomorphism 1 s-r r s T : VU -> VW such that 1 T(VU) = z, 1 z . -W s-1 o r and dim W = dim VU . Clearly o where W C W o— mapping which i s not induced by transformation from U T i s a type one s - r vectors of W and a linear to W . r §4. Rank One Preservers On Let VU . 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 i s 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 VU r or equal to zero, then every rank one preserver on i s either induced by a non-singular, linear transformation on U , or the r image of VU under the rank one preserver i s a type r pure subspace. 71. We f i r s t establish the following Lemmas. r Lemma. Let T : VU -*• VU be a rank one preserver where s 4.1. Assume that r _< s . F = 2 . Let x, , •• • , x „ be 1 r-2 dim U > 2 and char non-zero vectors of U . Then i t i s impossible that {T < 1 x _ - y U ) : y U , y = f 0 } = {z r 2 £ for some non-zero vectors ± z, , • •• , z 1 where Proof: k „k+l -U } \J {z^ z s-2 X , ,, s-2 of U and some W C U k+1 Suppose to the contrary that ~k*t~l z ^.U s-2 x y where we denote x, 1 x 9 } (J i z \ k ^k -U :wW} £ „'y-U by M r-2 y T(M ) = z, u, 1 1 z . ,.'U k+l s-2 U ^v > 1 = 2 k l- l s-2 V 9 u 2 ' ^ ^ e m m a 3-4, Therefore 2 k + ) = .....z 2 s-2 Zl = . Let u_ e U be 2 V l Z and £ v 2 ) + T(M ) and T(M ) =f= T(M ) . l l l 2 V k+l 0 independent of u^ . Let v^ = u^ - u T(M k + 1 «w s-2 2 T 2 J Assume that U . i s a non-negative integer. {T(M ) :yeU,y+0} = {z T(M z s-2 ' u 2 k k k .v' .U 2 2 (2) 72. for some , such that i n W . Let z and y be two independent vectors of TJ z , y { <v] , v'> . Since T(M ) = z U. x x Z J. z , .. -TJ KTX s-z k+1 , — i t follows that T ( l r - 2 - V x x f ) = l z Jc+l* z 2 k+l 2 k+l z s-z T(x .....x _ .u .g) = 1 r 2 i z Z i y s-z for some either Let f , g i n U . Clearly <f> f <u^> A = x 1 or <g> = } = i < u <z> =f= <y> • > W e m a implies that y assume that C for some 5 E X X l r - 2 ' X l l ' V r - 2 ' V X f^ , f <f> =j= <u^> . „.u-'f . In view of (1) and (2), we have r-2 1 2 B <f> =f <g> . We have 2 f f - z " Z l Z l Z k+l* l s-2 f 0 2 k I k+l' 2 ' 2 s-2 f 9 , V k V 2 i n U . On the other hand, since <f> =)= <u^> , by Lemma 3.4, we have ( T( for some f v • z i in W . Since z _ k+i9 f , 2 k - u 2 k T(A) , T(B) , T(C) e z 1 i t follows that z . «f s-2 lr 1 R+1 <f'> = <z> = <f^> = < f > by our choice of z . «k „k •TT , 73. Now, l e t f^ = az and = 3z where a ,3 are non-zero elements of F . We obtain k T ( B + C ) = i z 2 9 k k r i - i f v 2 + k +'«!•••:•" k i - 2 - 2 f 9 s-2 2^ z i 2 i 2 jc+r 2^ -I (az) V s-2 2^ = z ,k+r z 2^ V " + k+l" z 0 k+r ( o v + i> k * - (ez) 2^ •• 2 Z Z S-2 2^ v 2 2^ z s—Z = z^ k 2 + s-2 = v i-'-" 2 Cctv^H-gv^) ' k+i- z z s—Z - 2^ (pv P k since char F = 2. s-2 However, since u i = v i + v ' w 2 e a i S O n a v e k+l 2 T(B+C) = T(A) = z^ This implies that our choice of av^ + 3v and -z z are linearly dependent, contradicting z . This completes the proof. r 4.2. 2 z s Lemma. Let T : VU -> VU be a rank one preserver where Suppose dim U > 2 and char F = prime p . Let i > ' ' ' non-zero vectors of U such that <x > =(= <x „> • If r-2' r-l x > s >_ r > 2 . x r i D e 1 {T(x 1 x .yU):ycU,y+0} C {z 2 r z -u s-2p t P t .U P t : u e U , u =j= 0} 74. for some non-zero z., • • • , z s-2p 1 of U where t i s a positive integer, t then t {T( V-2' r-l' ' x X;L Proof: 4.3. y U ) : 7 6 U ' ^ ° 4= 7 } { z l t' s-2p z u P , l j P t : u £ ' + U u 0 } The argument i s similar to that used in the proof of Lemma 3.9. Theorem. r s Let T : VU ->• VU be a rank one preserver. Let dim U > s + 1 . Then (i) i f r = s , T i s induced by a non-singular transformation on U ; (ii) a prime i f r < s and the characteristic g p > — , T i s induced by F i s either zero or equal to 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 w i l l follow from Theorem 3.13 and Theorem 3.14. r M be a type one subspace of VU . Since Let < one preserver, T i s a rank s T(M) i s a pure subspace of VU . Moreover, dim T(M) = dim U > s + 1 . Let T(M) C N where N dim M = i s a maximal pure s subspace of VU . If N i s of type k where 1 < k <^ s , then <_ s + 1 , contradicting the fact that dim T(M) > s + 1 . Hence type one or power type pure subspace. Moreover, Suppose that T(x 1 x dim N = k + 1 N T(M) = N . s -, *U) e P (VU) for some non-zero i s of 75. x 2_ ' '" If ' r-l '*" ^ x n r = 2 , then a n positive integer d s o m e u ^ U ^ u^U =f 0 ± 2 for a l l =(= u by Lemma 3.4. 2 char F = a prime p , u~ =j= 0 . Hence for a l l u ., u~ =}= 0 . Also T( -U) f l T(u -U) =(= 0 for a l l U;L t . Then T( -U) =(= T(u 'U) U;L 2 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 such that < r > 2 . Let y, , ••• , y - be non-zero vectors of U '1 r-l y^ + ^ > o r + J • Consider the following type one 1 r pure subspaces of VU : M that i s , l = x l X r - l * M. = x, I 1 »2 U M = X l X r - 2 ' J r {T(x 2 u^ , u x Ul 2 _-'> u r Therefore. T(M ) e P 2 > *" ' r U M x . »y. y. . 'U , i = 1 , r-i l i - i 2 u = y l y r - l ' r . Since J x _ . - U ) n T(x for a l l non-zero l ' y x x _2* 2' U r U) +° in U , by Theorem 2.14 and Lemma 4.1, s : u e U , u =(= 0} cr P (VU) (VU) . Similarly T(M ) e P 2 (VU) implies that U 5 76. s T(M_) e P_ (VU) and so on. J t s T(M ) e.P (VU) . r t It then follows that t Let T(M ) = z r z «U s-p 1 r and Lemma 4.1, we have for each {T(y y. y for some non-zero . In view of Theorem 2.12, Theorem 2.14 P t i = l , # , , , r - l , t t w -u -U :ueU, u=j=0} i(s-2p ) -u-U) :ueU,u=f=0} C {w w 1 w' :(M ) e {T(y r y 1 P of U . Since K s ^ p 1 P 1 1 ) y^-u-U) : u e U , u =f= 0} , ± i t follows that P Z, Z 1 s-p for some non-zero t -«U t t P =W.-««r"W xl z, , • •• , z ^ has at least 1 \ t • s-p r p < y , =f <y.> > <w #1 2 1 for N X i = 1 , ••• , r .- 1 . Hence factors of <u.> x fc 4 Since t ^ 'U. -U . _ t x(s-2p ) ( u. in U where x P for each i by Theorem 2.1. ^ for k =)= j , according to Lemma 4.2, w > =}= <w j(s-2p ) k =}= j . This shows that t k < u j > w > k(s-2p ) l T" \ < x t > ^ o r Jf k . Consequently 77. z^ z ' has at least s-p s - p 21 (r-l)?*" (r-ljp " and thus t factors. 1 This implies that s i> rp " , a contradiction to the hypothesis 1 r s . Therefore no type one pure subspace of VU s on a power type subspace of VU . This proves that is mapped onto T i s a type one mapping. Our result thus follows from Theorem 3.13 and Theorem 3.14. That Theorem 4.3 i s not true i f s >^ rp where char F = p is shown by the following: 4.4. Example. Let char F = p and s >_ p r t . Let k = p*" . Let u, , • • • , u I n be another basis o| U . Let f : U n n f( E X.u.) = . , 1=1 1 X . k £ X. ;w. - 1=1 1 1 . for any A. x J Then i t i s easily checked that X ,n in F , v , w where ( v l > »V z, , ••• , z , 1 s-rk " and w.. , ••• , w I n U be a mapping defined by "k X. l in .F where th i s the k root of . *- k f(v) + n f(w) in U k and f i s one to one. r s <J> : XU -» VU by 1 Define a mapping * be a basis of U f has the following property: f(Xv + nw) = X for any for some positive integer fc Z l Z s - r k ' f ( v l ) k f ( v r ) k are fixed non-zero vectors of U . We have 78. • (Vj, ..,v ) = z •••.,Xv +Tiw ,. i r i k ^g-rk^^i^ ± k .f(Av +nw ) i 1 = z z x s _rk' f ( v l ) k k f(v > i r 1 (A f(v )+n f(w.)) k k f(v ) k 1 r 1 - z......z 1 s-rk + z. 1 = ,.f(v.) (X f(v.)) l 1 (n f(w.)) k 1 z ,-f(v,) s-rk 1 k k s r k = X<J)(v , '•• , v 1 ± r + 1 k k k f(v ) r k s , ••• , v ) + n<j>(v , ••• , w r f(v ) r z _ .f(v ) k i k I Az .....z _ .f(v/.....f(v ) .....f(v ) nz 1 k 1 r k r Clearly ty i s symmetric. — r s exists a unique linear transformation ty : VU -*• VU such that for each r v ( V 1 v r > = j » *** » v l Z 1 1 1 r s - r k Z , f ( l v ^ • Clearly ^ ) k f ( k , ••• , v ) . ± Hence ty i s multilinear. * f(w.) ... k 1 v r ) Hence there k i s a rank one preserver from s VU to VU and i s not induced by s - r vectors of U and a linear transformation on U . 4.5. «. r s Theorem. Let T : VU -> VU We have on (i) i f r = s , then T be a rank one preserver. Let dim U = s + 1 i s induced by a non-singular transformation U; (ii) if r < s and the characteristic of F i s either zero or equal s to a prime p p> — , then T i s induced by non-singular linear transformation on U . s - r vectors of U and a 79. Proof; Since dim U = s + 1 , every pure subspace of type one or of power type or of type r subspace of VU s in VU s has dimension s + 1 . Also every type one pure r s + 1 . If a type one subspace of VU were s has dimension mapped into a type k pure subspace of would have dimension j < k + l < s + l VU , 1 < k < s , then the image . This would imply that some non-zero pure vector i s mapped into 0 , which i s not the case. Hence a type r one pure subspace of VU i s 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 a PP e d s onto a type s subspace of U . We shall show that this leads to a contradiction. Let subspace z T where < l V' f\ V* x x X;L _2* " Z r -2' ' U ) V. s C subspace i s another type s N (s) V ^ f\ =j= 0 . such that Z ) = V*' s V i T( i s the only type i s 1-dimensional since y r is a two dimensional V^ N dim <v , • • • ,v > = 2 , <y> f <z> x s Note that V w n e r e „.u.U) : u e U , u 4= 0} . By Lemma 3.4, r-2 N in U x » C !D (V. , V, } where — (s) (s) VU . Then Choose a non-zero x ^ We shall show that C . Suppose that subspace of °^ s C = (T(x, 1 is an infinite family. in ^( ) ^ (s) + ° V a n d * and T ( VHV x l x r I f <v > for a l l X _ 2 * z . ' U ^ V i = 1 , •• ^ (s) + ° ' 80. If z T(x, 1 x 'z-U) = z, z , .IJ for some z. in U , then r-2 1 s-1 I ft ft , • • • , z _ e V fl V ; hence <z > = • • • = <z ,> = V f| V . But * «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^ ^' '^) cannot be a type one subspace. Similarly, i f v z char F = p , s > p " , then 1 T( x _ .z-U) i P (VU) , X;L r 2 t k where t i s a positive integer. Also i f char F = p , s = p for some k positive integer k , T( x «z«U) =j= U X l v 1 v e T(x, s 1 n r s subspace r-2' ' y , ••• , v > = s x „.z.U) . Hence r-2 1 X n X T(x .....x _ .z.U) = for some type dim <v r—2. j_ and since P 0 Z £ x W 2 (s) W( ) . Since s x l' _ ' ' z r 2 U H x x ± _2'y » u r i, i t follows that v, 1 v e V, A W, . This implies that s (s) (s) N N <v, , ••• , v > = V = W , a contradiction to Lemma 3.4 since I s and x ^ x only type r _ 2 ' Z ' U are adjacent pure subspaces. s subspace in C. Therefore x, 1 x „-y-U r-2 ^ ( ) i s the s 2 81. Let C be the collection of a l l type one subspaces i n C . We shall show that M^n M + 0 2 C i s f i n i t e . . Suppose that for every C , = {z, 1 in C 0 EWe /Y u} — z.. , • • • , z to Theorem 2.14, C = C U'{V > (s; i s i n f i n i t e . Since , i t follows from Theorem 2.13 that z -u.U : u s-z for some non-zero vectors C in U and some W C U . According or C = C U {V, } U (z, • • • • • z -U } (s) 1 s-2 2 N 0 In the latter case, char F = 2 . If T(x, 1 that A = z.. 1 2 z „'U £ C , we l e t f £ U s-2 such that x *f.U) = A . Let v be a fixed non-zero vector of U such r-2 <v> =(= <f> and <v> =j= <y> . Then by Lemma 3.4, T(x 1 x _ -v.U) = z r 2 ± z _ .v'-U s for some non-zero vector v' of W . Now, for any non-zero such that. <x> =)= <y> , <x> = } = <v> T( X l l Since z. 1 z -x'.U . s-z z „.x'-U f\ V, =|= 0 , we have s-2 (s) 1 x in U and <x> =j= <f > , l e t x ..x-U) = z, r-2 1 N (1) 2 0 x 1 (2) £ V . From (2), we h ave . 82. for some v • e U . Since x and T <x'> 4= <v'> ( x l " , , , V 2 ' v z «x'-v e z. x 1 z. 1 (Lemma 3.4), i t follows that , D ) - Z l _ 2 * Z S V U < T ( x l r <v > = <v'> . Hence x x r - 2 * V ' y x 2 dim T(x, 1 (from (1)) s-2 x _ -f-v)> U ^(x^^ 1 If ' ' V <T(x This i s impossible since z „.v'-U 0 s-2 ) > 2 ^ -v.y)> . x «v'U) = dim U > 2 . 0 r-2 A | C , by the same argument as above, we have Kx^-.-.x^.v.U) C Z Z l s _ 2 , v '- V U T(x < 1 x _ -vy)> (J <T( r 2 x X;L which is also impossible. Hence C i s finite. Since C i s infinite, i t follows that char F = a prime p and for some positive integer V of a l l power type subspaces of degree view of Theorem 2.12, we have V s > 2p t t , the collection t in C is infinite. In and t t y -\P -U : w £ W } s-2p A = { y 1 p n * 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 _ ' ' v r 2 v ) > » 83. possibly y Since vector _ -U - t+l v < v ^ + > < v j > i f x 1 w e char F = p = 2 . ma s u i n U such that ( i n which case x^^'y-U) = ^ ^ > T(x^ T where 2 y s-2 1 l x Y choose a non-zero <u> =f <y> and r - 2 , y * u ) = l V s V + j • ' Y Lemma 3.4, we see that B 9 T(x Hence s x 1 r v^ _2' ' U v g U ) t e V ° r T ( x l x r _ 2 ' U , U ) = y l t+l' s-2 y has a factor of multiplicity at least p fc t+l U , a contradiction. r Therefore no type one pure subspace of VU is mapped onto a type s r subspace of VU . This means that every type one subspace of VU i s 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 f i e l d of characteristic 0 or of characteristic greater than a positive integer t . Then every non-zero pure vector of the form x fc F -'.y^* , 1 <_ j <_ t , in VW is in the span of M = {u : u e W} . fc 84. Proof: Let A^ , (x+X^y)*' i = l , , ^ j _ ^e t+ 1 t + distinct elements of F . Then + • • • + (£) x ~ . ( X y ) + ••• + X^y = x t t k k t i e M for each ••• , t + 1 . Consider the following system of non-homogeneous equations i n t + 1 t+1 z i-i variables , ••• , ^ ^ ' t + x c . = 0 i f m + j , 1 <_ m _< t , m 1 1 t+i . Z A? C. i-i 1 1 (1) t+1 1 , E i-1 0 . ' £. = 1 Since the determinant of the coefficients of (1) i s 1 A, t+1 det t+1 i t follows that (1) has a solution, say £. = d. , i = 1 , » ^" t+1 _. . Hence Z d.(x+A.y) = (.) x «y^ i s in the span of M i=l 1 1 Theorem. Since 3 (^) =j= 0 , we conclude that 4.7. .. , t + 1 Let W x*" *y 3 2 i s i n the span of M be a finite dimensional vector space over a f i e l d F of characteristic 0 or of characteristic greater than r VW i s spanned by M = {u : u e W} . r r . Then 85. Proof: We proceed by induction and assume that any pure vector having less than k + l distinct factors i s 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 that z I exactly t k-1 A = z, I d. in F l t+1 l Z t t 1 d l i V t+1 Z t + l J V has M + }. .• i = l = t By the induction hypothesis, r VW z has r and some v. in W . Hence l A - (I d V i=l has at most- k in such a way distinct factors.. By Lemma 4.6, we have t J z r has exactly 2 distinct factors and z ,, t+1 Z for some z -z t t+1 Z r = t .\ i i=l d t v.-z ,.. l t+1 distinct factors. ( v i' t+l«r Z z e [Ml since r Therefore ) t v.-z ..•••••z I t+1 r A e [M] . This proves that 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 space over an algebraically closed f i e l d char F > r . We also assume that i s an n-dimensional vector F with char F = 0 or , r r 3 < n < r + 1 . Let T : VU -> VU be a 86. • r • 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 i s mapped r onto a type r subspace of VU under T . rank one preserver. Since every type 4.8. Lemma. I f there are two distinct type VU such that Proof: r pure subspaces . M A N f 0 and T(M) = T(N) , then r of k Let M = S l ( r M r T(VU) = T(M) . ,N = S and T(M) = T(N) = S. . where ^( ) ^ ) and N S ,S , S r are two dimensional subspaces of U . By hypothesis, M f\ N = S 1 1 Hence (sns) = ^(r) 1 2 ± 2 = <Y , Yo> • Then 3 9 2 3 T(S J and s 1 o ( r ) S . =f= 0 . Let S = <y , y > , S = <y , y > . Consider Hence S, (r) n 3 2 S f\ S (r) (r) = 2 2 2 3 3 ,S n S (r) (r) = <y > 2 X T i s a rank one preserver is' a two dimensional pure subspace, i t follows that <T(y*) ,'T(y^)> i s two dimensional. two distinct type Let g T s (r) 1 <y > ) f | / \ 2 <T(y^) , T(y,)> . Since <y , y^> scalars. 1 T(S ) = S, . because any r r subspaces of VU have at most one dimension in common. z = ay^ + 3y + y y Consider Hence 2 3 where a , 3 ,y S^ = <y^ , z> = <y^ , 3y + Y y 2 > 3 are a l l non-zero • Since 87. A 0 V ) we have Hence ^ <yh r 3 T(S^ T(S^ o 3<(6y+Yy ) > , S n S (r) (r) V ) S S 2 ) (~\ S ^ ) = 3 4 3 <T(y ) , T((3y +yy ) )> 1 . 2 Consequently, > 1 which i s two dimensional. 3 T( ,y >) = since 3 r r V<y^ , y , y^> 2 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 Consider the type one subspace w £ <y^ , y y^ , y^> • Let 2 y "U . Let P = y^ n W = <y^ , w> y^'U • Since r dim (P fl V<y , y 1 , y > ) = 3 , we have dim (T(P)O. ( ) ) L S 2 3 3 r • Since the maximal dimension of the intersections of two distinct maximal pure T(P) C s ^ subspaces i s 2 , we conclude that T ( W (r) ^ (r) - ) S < T ( y l ) ' T ( y l y 1 ' w ) • > . We observe that two dimensional pure subspace, <T(y^) , T(y^ dimensional. Hence .)= T(W / (rr r T(VTJ) = 4.9. S S, . (r) s y *w)> _ 1 - " w > a i s also two n By Theorem 4 . 7 , we conclude that ( j • This completes the proof. r Lemma. Suppose that for any two distinct type such that MON =f 0 , we have T(M) =f T(N) . Then a non-singular transformation on Proof: »yi S i n c e Let y Let T subspaces M ,N is induced by TJ . be any non-zero vector of independent vectors. r U . Let S^ = <y , y^> , S 2 y , y^ , y = <y , y > 2 . Let 2 be linearly 88. T(S ± . <> r of ) = S* 1, ' ( ? ) 9 (> (r) T s _ where ^(r) S x r U . Since S =f= S ,S Si , S' are two dimensional subspaces 1 2 f l S ' +' 0 , by hypothesis, (r) (r) Z Therefore Hence S' O (r) = T(S n S ) = <y > ( r ) ^(r) S' T(y ) = Ay' S* + S* (r) (r) 2 for some ,r \v) 1 y* e U . ± for some We claim that X in F . T(y y-U) = y' y'-U . Since T(y is a pure subspace, i t i s contained in a maximal pure subspace. i s contained in a type k . pure subspace 2 <_ k < r , then y' e g.^ This implies that g^ e W , a contradiction. in a type r subspace dim (S and n (r) dim (S 8 r-k' (k) g, 1 W a n d ^i < n e n c e T(S 1 (r) T(y < If T(y > yU) 1 y U ) = 2 implies that dim (T(S - ) 0 W, .) >_ 2 . {r) l K X ) r subspaces, i t follows that 2 , r Hence ; y-U) i s a type one pure subspace of VU y ) _> 2 ; ( r ) ) = W. v = T(S„ ) , a contradiction to our hypothesis. ^(r) r U i s contained dim (T(S )H W (r) y y - U ) = 2 implies that pure subspaces under the assumption that Since ~ y ' > y' e W . > r ) and T(S„ ) are both type (r) '\r) y-U) W ( ) , then f\ y T(S If T(y g , -W... . where r-k (k) (r) Since y-U) e T(y since there are no power type char F = 0 y-U) , i t follows that T(y or char F > r . y-U) = y 1 y'-U . 89. r-l r-l r-l By Theorem 4.7, l e t x^ , ••• » x be a basis of V U . Note that x^ 3 <_ dim U < r + 1 implies that and x_. are linearly independent. r >_ 3 . Clearly, i f i =(= j , then Consider any type one subspace r-l z , = Z X.x. r-l . , 1 1 i=l r - 1 * ^ "*" P fc z. 1 We z , -U . Let z, r-l 1 shall show that T(z^ to the contrary that v^ Z S a (i) T ( z z 1 t y e o n e - l ' ^ — (r) U r where S P u r e ° X. e F and i = 1 , ••• , I subspace. r T ^ Z Suppose 1 k"^(k) , 2 _< k < r , for some two dimensional subspace Z r - l " U ^ — S of U and some v , ••• , v . e U - S . 1 r-k n Let T(x. l T(x.) = n. x! i l l have j r r j f i , <x^ , x^> x!'U , i = 1 , ••• , t . Note that we l n. e F where l for some i = 1 , ••• , t . For r i s a two dimensional pure subspace of VU r r x. 'U) = x! i i implies that r _ r T(<x. , x.>) = <x! , x'. > Therefore i s a two dimensional pure subspace of VU . 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 > U S U ( U < x ! , x!>) . This i s possible since 1 r-k .i. I i < i+J r-l <xl^ , xj> are a l l proper subspaces of U . Let T(x^ -u) = x^ r r-l i > 2 , let T(x. «u) = x! — i l r For each We shall show that _ _ 1 -v . 1 *u. . l for i = 2 , ••• , t . Since i n r-l r-l r—1 r—1 <x, -u , x. 'U> i s a pure subspace for i > 2 <x! .v , x! .u.> l i — » l ' i i r <v.> , S l <u.> = <v> i v i s also 90. a pure subspace. By our choice of v , <xj^ , v , x_j> i s 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 l or <u.> = <w > , i > 2 . Either <u.> = <x'> l i — i 1 r-l r-l <u.> = <v> . If <u.> = <x'> , then <x' «v , x! *u.> i s a pure l l 1 I ' l l r-2 r-l <x| «v , x| > i s a pure subspace (Lemma 1.8). subspace implies that r-2 r-l Consequently, by Lemma 1.9, xj^ -v and x!^ contradiction. Therefore <u.> = <v> , i > 2 . I Now, we have have a common factor, a — u. = a. v i i for some a. e F , i > 2 and l — r-l i i 1=1 r-l t r-l = X, x' -v.+ E X. x! -a. v 1 1 . i i Z T( z Z]L r _l' u ) = T ( 1 X X U ) = 2 r-l = (X.x' + 1 Let T(z, 1 z -«u) = w. r-l 1 1 t E i=2 w r 0 In view of Lemma 1.8, <w^> = <v> w e v, r 1 v , "S,, , we have r-k (k) N 1 ( Z 1 Z r - l ' U ) ^ V l 1 4 = .0 1 for some <v> = <v.> l This contradicts our choice of v . Hence T 1 w . Then r r-l t r-l '(X, x! + E X. a. x! ")-v = w. 11 . i i i 1 i=2 w, 1 r-l X. a. x! )-v Vk' (k) S j , 1 £ j _< r . Since for some i or v e, s . 91. Similarly T(z z 1 -l" ^ £ ( ) ' U r S H E N C E r T ^ T " r one pure subspace of VU . In view of Theorem 3.13, * T Z R 1*^0 i s a type i s induced by a non- singular linear transformation on U . Combining the above two Lemmas, we have the following main result: r 4.10. Theorem. r 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 then either r or T is induced by a non-singular linear transformation on U T(VU) i s a type then T r . If 3 £ dim U < r + r subspace. In particular, i f T i s non-singular, i s 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 r subspace when 3 <_ dim U < r + 1 . such that i t s image i s a type 92. CHAPTER III RANK K PRESERVERS ON GRASSMANN SPACES Let closed f i e l d U be an n-dimensional vector space over an algebraically F of characteristic not equal to two. Let AU r r Grassmann product space of U . If T : /\U -*• AU then the structure of T is known: T be the r*"* 1 i s a rank one preserver, i s an r ^ compound of a non-singular linear transformation of U , except possibly when n = 2r , i n 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 i s also a rank one preserver on AU > provided that T i s nonsingular 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 i s called a rank 2k preserver n i f every rank 2k matrix in H i s mapped into another rank 2k matrix. c n Let 2 be a basis of U . Let ty : AU •+ H n u, , • • • , u I n ,x be a mapping defined by <J>(u.Au.) -= E and extend linearly to a l l - E -- » 1 < i < j < n 2 AU • E denotes the n-square matrix with 1 i , j and 0 elsewhere. It i s shown i n [15], that cf) i s 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 i n i s mapped under ty onto the set in position of a l l rank on 2 AU 2k matrices in H . Moreover, T i s a rank k preserver n -1 i f and only i f ty T ty i s a rank 2k preserver on H . n 93. 1. Definition. 2. Theorem. Let x^ , y^ , ••• , k I 2 z e R^(AU) , we write For each , R(z) = k . be 2k vectors of U . Then 2 x j[A i y 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 i n AU Proof: Clearly we are able to choose a vector w 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)) i s of rank T(z+Xw) = T(z) + XT(w) 2m . Since i s of rank k <j>(T(z) + XT(w)) = <j>(T(z)) + X <f>(T(w)) matrix for each non-zero T i s a rank k preserver, for a l l X =f= 0 . I t follows that is a rank 2k skew-symmetric X in F . On the other hand, since infinite f i e l d , we can choose a X o F i s an in F , X 4= 0 , such that o 1 <j>(T(z)) + A <}>(T(w)) i s of rank greater than or equal to 2m . Hence q 2m <^ 2k .» This proves the Lemma. r 4. Lemma. a rank Proof: If T : AU r AU is a rank one preserver, then k preserver for a l l possible From [18], we know that T i s also k. 2 T i s a compound of a non-singular linear 2 transformation of U . In view of Theorem 1.2.15, T i s a rank k preserver, k Now, let x = E i=l x i ^(M^) e where x^ i s of rank one for each i . Let 94. t E T(x) = i = 1 t E y i=l r y. E R (AU) i t k £ T(x.) , we have i=l = t 1 1 T(y ) e R^(AU) , i = 1 , , t . Since i t >_k . Consequently t = k . Therefore 2 Theorem. Let non-singular, then Proof: Case 1. T 2 1 i s a rank T is a rank k preserver. k k preserver, preserver. 2 T : AU + AU T Since t T (x) = T( E y.) = E T(y.) i=l i=l t <_ k . Now r and 5. r * y. e R-(AU) , 1 < i < t . i l where be a rank If T is i s a rank one preserver. dim U _> 2k + 2 . By the non-singularity of T , we 2 shall show that i f there i s an A e R (AU) such that t R(T(A)) = I <_ k , 2 then there i s a B e AU such that R(B) <_ t + 1 and R(T(B)) _> I + 1 . Indeed, suppose T(A) = u l A u + ... + u 2 for some independent vectors Extend u^ , • • • , u ^ u _ u £ 2 - £ l A i > *"* » £ u 2 to a basis 2 °^ ^ where u^ , • • • , 1 <_ I <_ k » "'* • u n °^ 2 Since T i s onto, there exists a vector z e AU such that 2 T(z) = 2 £ + i 2£+2 * u A u L e t z = z l + *** 2 + z e R, (AU) , m = 1 , ••• , s . Let m l z s e R S ^U) where T(z ) = E ot u. A u. where m . . mil l 1 Kj J J a .. £ F . Then mij T(z) = E (z ) = £ ( E a m=l i<j m=l u u )= « T A J J £ Au £ + . ^ " 95. This implies that for some m , we have a =j= 0 . ° o,2£+l,2-e+2 2 Hence there i s a w i n ^(AU) such that m T(w) = ^ Hence the 2l+l each a.j u.Au. , a , 21+2 2£+l,2-e+2 * F A i s i n f i n i t e and some non-zero A + b & l+l 2 l * 3+ 2l+2 i n F . Hence o Therefore l t 2 l + + 0 2 is a ^ 2 . For + 1 2t + 2 i n the upper l e f t corner <j>(T(A+Aw)) i s of the form a Since + entry of <b(T(w)) A i n F , the minor of order of the matrix u T b 2 ^ ^ 0 i s n o n - z e r o 1 f ) o r 4(T(A+A w)) has rank > 2t + 2 . o — R(A+A w) < R(A) + 1 . — Now, assume that rank one element ( ' ^ R(T(A+A w)) > I + 1 . However o + o — T i s not a rank one preserver. A^ is mapped to a rank j element, Then some j > 1 , since T is non-singular. By Lemma 3, j _< k . In view of the above argument, there i s an 2 A e AU such that R(A > <_ 2 and R ( T ( A ) ) ^ j + l > 2 . I f j + l < _ k , 2 2 2 2 continue the process as above, we see that eventually some element of AU 2 of rank less than or equal to k i s mapped to an element of AU greater than preserver. k . This contradicts Lemma 3. Hence of rank T i s a rank one 96. Case 2. dim U <_ 2k + 1 . We f i r s t note that the maximal 2 rank of a l l elements in AU is k is not a rank one preserver. T ^ of to a rank T Z T -1 Then some rank one element i s mapped under element, Z > 1 . If Z =|= k , by the non-singularity , there exists a - 1 (Theorem 2). Assume that B e AU such that R(B) £ 2 and R(T (B)) _1 >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 Therefore Hence 6. T ^ T i s a rank one preserver. T i s a rank By Lemma 4, k T ^ preserver. preserves a l l ranks. i s a rank one preserver. Remark. The proof of Theorem 6 i s analogous to that of Theorem 3.1 [1] and Theorem 1 [6]. 7. Remark. It i s well-known that the maximal dimension of a rank subspace (a subspace with every non-zero element of rank the vector space of a l l field 2 AU F , i s one. Hence the maximal dimension of a rank 2 A : AU H A. 2 <j)(R(AU)) k Theorem. then Proof: T (F) , k subspace in i s an onto isomorphism such 1 1 i s the set of a l l rank 2 8. n n-square matrices over an algebraically closed i s one i f dim U = 2k , since that n) in n Let 2k matrices in H n . 2 T : AU -* AU be a rank k preserver. If dim U = 2k , i s a rank one preserver. In view of Theorem 5, i t suffices to show that T i s non-singular. 9 7 . Assume that T i s singular. Then some 2 z e i s such that = 0 , z + 0 . Clearly t ={= k , since T i s 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) . T(z) Assume that Extend Let w w = u- A u_ + ••• + u_ . , /\ u„. where 1 2 2 j - l 2j u^ , ••• , u ^ * = A to a basis u^ , ••• , u ^ , ••• , + ••• + u k A u^ . Then for a , b a w + bw* = u A (bu -au ) + . . . + « J 3 2 1 A < b u 2J+r 2j-l a u of U . in F , ' 2 2 i= k - t. + ) bu 2j+2 2j+3 Au 2kA i ' * 2 T(aw+bw ) E R (AU) . b u * Clearly i f b f 0 , R(aw+bw ) = k If b = 0 and and hence + u fc * 2 T(aw+bw ) = T(aw) E R (AU) . Therefore a =j= 0 , then fc * if a and b are not both zero, then T(aw+bw )E R^(AU) . It follows * 2 that <T(w) , T(w )> i s a rank k subspace of AU with dimension 2 equal to two. k This contradicts the fact that the maximal dimension of a rank 2 subspace of AU i s one. Therefore T i s non-singular and the proof is complete. 9• Lemma. If dim <x^ , y^ , • • • , for all i = l , ••• , k + l , k > l , k+l Z i=l , J^.-^ then 2 x i A y i e \(AU) = 2k + 1 and x^ A y^ =J= 0 98. Proof: x i It i s clear » ^ i '» """ » such that A y ,y 1 ' a r li e N n e a r , • • •; , *x 1 has rank less than k + 1 TL J l dependent. v because Also there i s one j , 9j . • • • » V l , y k+1 > = 2k . Hence 2 e Example. ' v,V i » W k+1 Z x . A y . i s of rank at least •- , I ' i i=l k+1 Z x.Ay. . , i i=l proves that when .. ^(AU) . Since x. or y. £ ^ . y ^ - : by Corollary 1.2.6, 10. . 1=1 » k+l dim k+1 t x. y. k+1 Z x. A y. that i s of rank i w . k . This k. 2 k subspace of AU The following is a 3-dimensional rank dim U = 2 k + l , k > 2 . Let u^ , ••• , 2 k i + f, = u, A 1 f f be a basis of U . Let u 2 = 3 = lA \ u + + \+l A \ + 2 U u 2 k+2 A u u 2 iA \ \ + 2 AV 3+ + + + A + + 3 A" A 2 k u 2k+l - V l A 2k+l + + u 2 Clearly in f^ } > e ^(AU) • Let a^ , a^ , a^ be non-zero elements F . We have ^ V i = 3 3\+l A \ 2 + U + l ^ V k + 1 2\+2 ( + 3 + a 3 k+3 + •'••+ V l ^ l 2 k - l 2 2 k 3 2 k + l ( a u + a U + a U U ) + ) \ * l 2k 2 2k+l ( a U + a U ) 99. Since dim < 3 3 J ^ a . u ^ . , •• • ,^._ J^V2k-241'V u^, V Ufcfl> 3 b y Lemma 9, E a.f. is of rank i-1 1 2 ^2 + a 3 ^3 a r e °^ r a n k Hence a rank k + a " k , a^ f^ + a^ i , n Theorem 2, a^ f^ + a~ £~ B y , -*- °^ s are linearly independent and m subspace of We feel that every rank **2k+l <_ 3 . likely of dimension S m o s t Theorem. Proof: 2 2 T : AU ->• AU Let dim U = 2k + 1 is 3 . Then T T Choose a vector Then by u f o ^ A \ i s singular. where t q k ^i»^2'^3 > ^ S k preserver. Then k >_ 2 subspace in Suppose subspaces of 2 in R, (AU) Tc-t such that z + f o v + u 2 A \ + + 1 * ** + u j A j+k-l u c and for some 2 2 T(Rj^(AU)) C R^AU) t <_ k . Since 2 T(z + f ) = T(z) + T(f ) = T(f ) e Rj (AU) . Q k 2k is a rank one preserver. 2 z e R (AU) , T(z) = 0 f be a rank and the maximal dimension of a l l rank Assume that Denote k M ,. ,..(F) i s m+l,m+l r at most 4-dimensional (see [21]). 2 AU < r a n 2 subspace of AU . J 11. > k . It i s known that every rank * u 1 Similarly, using Lemma 9, we check that both a V2k 2 2W-l v i s of rank w h e r e , t =(= k . k . 3 = k - t . 100. Extend of a i f i lA V l f l = U f 2 = U 3 = l A + U U A u t o , " j ^ . ^ basis a u 1 , *" , u 2 k + 1 + 2 A \+2 2 A\ + U l + + + 3 \+3 A V * 2k + + u V A 2k+l u V-l + 2k+1 A U a , a, , a„ , a„ be four non-zero scalars in F . Then o ± I 5 3 Vfl = \+2 + \+l k+2 Let \ , • • • ., U . Consider f 1=0 , a \+2 l A + « A(a j +U . + l a o V j o k l\+l 2 k+2 3 k+3 A ( a U u 1 + a + a k + j + a U + a U 2 k + j + l 3 k+j+2 U + a U ) * * ' + )+ V + A ^ l 2k 2 2k+l^ a U +a U Since 3 d i m 3 V'^'V/AVlV. °iVj-l+l' j+l . I u n • 1=0 1=0 \' l 2k 2 2k+l a U + a by Lemma 9, U = 2 k + , J E 0 1 i^J+l ' "• ' 1=1 ' 1 Z a.f. i s of rank i=0 1 that > 3 k . Similarly, using Lemma 9, we check 1 a f + a, f. + a, f, , a 'f + a f„ + a, f_ are of rank o o 1 1 3 3 o o 2 2 3 3 0 Also we have 2 a f + a. f e P^CAU) Q ± ± and a Q f + a Q 1 k. 2 f + a~ f„ e R^CAU) 1 101. i = 1 , 2 , 3 , by Theorem 2. Since T i s a rank k ^(^ ) preserver and e Q 2 R^CAU) , from the above argument, we see that 3 3 I c. T ( f ) = Z T ( f . ) e ^(AU) i=0 i=0 2 ± i f one of c. + 0 . Hence Ci 3 £ d. T(f.) = 0 i-0 • 1 Therefore implies that a l l d. = 0 . 1 T ( f ) , ••• , T(f^) generate a Q 4-dimensional rank k subspace 2 of AU which i s a contradiction to our hypothesis. This proves that T Hence T i s non-singular. is a rank one preserver by Theorem 5. 2 12. Corollary. then T Let T be a rank 2 preserver on AU . If dim U = 5 , i s 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 13. Theorem. Let 2 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 2 of A^ n a s dim U _> 6 . It has been shown in [9], that every rank dimension _< n - 3 i f dim U = n >. 6 . 2 subspace 102. Assume that T i s not a rank one preserver. By Lemma 3, some rank one vector i s mapped to the zero vector or a rank 2 vector. If some rank one vector is mapped to the zero vector, i t i s clear that some rank one vector i s mapped to a rank 2 2 preserver. U l Hence 2 vector since T(u^ A v^) e R^AU) T i s a rank for some independent vectors • l * V Extend u, , v_ 1 1 to a basis u, , u„ , v, , • • • , v „ of TJ 1 / 1 n—z Consider the following vectors: f l = U f 2 = U n-2 f Let a. i 1 1 = U A v i A V 2 + l A V U n-2 A V + U 1 2 n-3 ' AV a „ be n - 2 elements in F . . Then n-z n-2 n-2 I a.f i=l 2 = u n-3 (} V i i=l A ) + ' 2 U A i+lV • i=l a n-2 n-2 n-3 E a.f. £ R,(AU) implies that E a.v. and E i=l i=l i-l2 1 1 1 1 dependent and thus when some a. 4 0 a = n 2 '*" = a = 2 1 1 a.,,v. 1 ^ ' l ' ® ' Hence a for 2 < i < n - 2 . Since T are linearly 1 ^ a n-2 i ^ i ^2^ ^ i=l i s a rank e 2 A 2 preserver 103. 2 T(u A v ) e R„(AU) , i t follows that and n-2 R(T( I a.f.)) = 2 whenever i=l 1 some a =|=0,l_<i_<n-2. This shows that i are linearly independent and T(f^) <T(f^) , ••• , ^ - 2 ^ T 1 > , ••• i s a r a n n , ( _) T f n ^ 2 subspace 2 2 of /\U of dimension equal to maximal dimension of a rank n - 2 . This contradicts the fact that the 2 subspace i s n - 3 . Hence T is a rank one preserver. 14. Remark. Theorem 13 i s also, obtained independently by M.J.S. Lim [10]. Combining Theorem 5, Theorem 8 and Theorem 13, we have 2 15. Theorem. or dim U = 2k Let or transformation of 2 T : /\U ~* AU k = 2 , then be a rank T k preserver. If T i s non-singular is a compound of a non-singular 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. singular or matrix P Let n = 2k such that in which case S S :H or n ->- H. n be a rank 2k preserver. If S k = 2 , then there exists a non-singular S(A) = PAP' for any A may possibly be of the form: in H n i s nonn-square , except when n = 4 , 104. *34 " 34 ° a S(A) = P 24 23 14 l a P' 12 " 24 " 14 a a Corollary. that S a 0 *12 A = [a..] , a.. = -a.. ij 17. l a " 23 " 13 where 13 ij Ji Let S : H n H' be a linear transformation such n - det A = det S(A) for a l l A in H n where has the form as described in Theorem 16 where Proof: By hypothesis, S i s a rank n n i s even. Then det P = + 1 . preserver on H n . Therefore this Corollary follows from Theorem 16 and the fact that 0 -a M2 12 0 det "13 14 *23 °24 34 -a 34 = det 13 " 23 ~ 14 ~ 24 a a *34 a a -a 34 -a 24 23 a 24 23 *14 "13 14 13 '12 12 105. BIBLIOGRAPHY [I] Leroy B. Beasley, Linear Transformations on Matrices: of Rank k The Invariance Matrices, Lin. Alg. and App. 3 (1970), 407-427. [2] H. Boerner, [3] Wei-Liang Chow, On the Geometry of Algebraic Homogeneous Spaces, Ann. Math. 50 (1949), 32-67. Larry J. Cummings, Linear Transformations of Symmetric Tensor Spaces which Preserve Rank 1 , Ph.D. Thesis, U.B.C. 1967. [4] [5] [6] ' Representations of Groups, Amesterdam, 1963. , Decomposable Symmetric Tensors, Pac. J. Math. 35 (1970), 65-77. D. Z. Djokovic, Linear T r a n s f i a Fixed Rank, Par Math. 30 (1969), 411-414. [7] W.H.Greub, Multilinear [8] M.J.S. Lim, Rank k Grassmann Fcoducts, Pac. J. Math. 29 (1969), 367-374. , L - 2 Subspaces of Grassmann Spaces, Pac. J. Math. 33(1970), 167-182. [9] [10] lgebrt ations of Tensor Products Preserving ringer-Verlag, N.Y., 1967. , 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] C F . 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 Can. J. Math. 22 (1970), 363-371. (G , a) Space, [18] R. Westwick, Linear Transformations on Grassmann Spaces, Pac. J. Math. 1 4 (1964), 1123-1127. [19] , 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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Rank preservers on certain symmetry classes of tensors Lim, Ming-Huat 1971
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Title | Rank preservers on certain symmetry classes of tensors |
Creator |
Lim, Ming-Huat |
Publisher | University of British Columbia |
Date Issued | 1971 |
Description | Let U denote a finite dimensional vector space over an algebraically closed field F . In this thesis, we are concerned with rank one preservers on the r(th) symmetric product spaces r/VU and rank k preservers on the 2nd Grassmann product spaces 2/AU. The main results are as follows: (i) Let T : [formula omitted] 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 linear transformation on U or [formula omitted] for some two dimensional sub-space W of U. (ii) Let [formula omitted] be a rank one preserver where r < s. If dim U ≥ s + 1 and the characteristic of F is zero or greater than s/r, then T is induced by s - r non-zero vectors of U and a non-singular linear transformation on U. (iii) Let T : [formula omitted] 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. |
Subject |
Calculus of tensors |
Genre |
Thesis/Dissertation |
Type |
Text |
Language | eng |
Date Available | 2011-05-03 |
Provider | Vancouver : University of British Columbia Library |
Rights | For non-commercial purposes only, such as research, private study and education. Additional conditions apply, see Terms of Use https://open.library.ubc.ca/terms_of_use. |
DOI | 10.14288/1.0302216 |
URI | http://hdl.handle.net/2429/34231 |
Degree |
Doctor of Philosophy - PhD |
Program |
Mathematics |
Affiliation |
Science, Faculty of Mathematics, Department of |
Degree Grantor | University of British Columbia |
Campus |
UBCV |
Scholarly Level | Graduate |
Aggregated Source Repository | DSpace |
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