SUBSPACES OF MATRICES WITH SPECIAL RANK PROPERTIES

Transcription

1 SUBSPACES OF MATRICES WITH SPECIAL RANK PROPERTIES JEAN-GUILLAUME DUMAS, ROD GOW, GARY MCGUIRE, AND JOHN SHEEKEY Absrac. Le K be a field and le V be a vecor space of finie dimension n over K. We invesigae properies of a subspace M of End K (V ) of dimension n(n r + 1) in which each non-zero elemen of M has rank a leas r and show ha such subspaces exis if K has a cyclic Galois exension of degree n. We also invesigae he maximum dimension of a consan rank r subspace of End K (V ) when K is finie. 1. Inroducion Le K be a field and le m and n be posiive inegers wih m n. Le M m n (K) denoe he vecor space of m n marices wih enries in K. When m = n, we wrie M n (K) in place of M n n (K). For any non-zero subspace U of M m n (K), we le U denoe he subse of non-zero elemens in U. Given a posiive ineger s, we le K s denoe he s-dimensional vecor space of row vecors of size s over K. We consider he elemens of M m n (K) as linear ransformaions from K m ino K n, he acion being defined by righ muliplicaion on elemens of K m. Le r be an ineger saisfying 1 r m. Much research has been devoed o he sudy of subspaces M, say, of M m n (K) which saisfy one of he hree following condiions: (1) Each elemen of M has rank a mos r. (2) Each elemen of M has rank r. We say ha M is a consan rank r subspace in his case. (3) Each elemen of M has rank a leas r. Paricular aenion has been focused on finding he maximum value of dim M in case M saisfies one of condiions (1), (2) or (3). Concerning condiion (1), noe ha if N denoes he subspace of M m n (K) consising of all marices whose boom m r rows are zero rows, each elemen of N has rank a mos r and dim N = nr. This simple observaion enables us o obain an upper bound for he dimension of a subspace saisfying condiion (3). Lemma 1. Suppose ha M is a subspace of M m n (K) wih rank T r for all T M. Then dim M n(m r + 1) Mahemaics Subjec Classificaion. 15A03, 15A33. Key words and phrases. rank, consan rank subspace, cyclic exension, marix, skewsymmeric marix. The auhors were suppored by Claude Shannon Insiue, Science Foundaion Ireland Gran 06/MI/006. 1

2 2 J.-G. DUMAS, R. GOW, G. MCGUIRE, AND J. SHEEKEY Proof. Replacing r by r 1, we know ha here exiss a subspace R, say, of M m n (K) of dimension n(r 1) whose elemens all have rank a mos r 1. Since M R = 0 and dim M m n (K) = mn, he desired inequaliy is immediae. While Lemma 1 is obained by elemenary means, i is ineresing o noe ha for cerain fields K, he bound we have obained is opimal. For suppose ha K admis a cyclic Galois exension of degree n. Then provided ha 1 r m n, Guralnick has shown in [8], Lemma 3.2, ha M m n (K) conains a subspace M wih rank T r for all T M and dim M = n(m r + 1). When K is finie, he exisence of such a subspace had been shown by Delsare in [4], Secion 6 (in paricular, Theorem 6.3). In general, however, here do no exis subspaces of he ype described in Lemma 1 which mee he upper bound for he dimension over arbiary fields K. Indeed, in unpublished work, referenced in [9], p.333, Meshulam has shown ha if K is algebraically closed and if m = n, he bound in Lemma 1 can be improved o dim M (n r + 1) 2. Furhermore, Roh has shown in [9], p.333, ha if K is infinie, here exiss a subspace M of M n (K) wih rank T r for all T M and dim M = (n r + 1) 2. Thus, Meshulam s bound is opimal for square marices in he algebraically closed case. We will invesigae he srucure of a subspace of m n marices of dimension n(m r + 1) whose non-zero elemens all have rank a leas r, and show some ineresing uniformiy in properies of hese subspaces. When K is finie, we will deermine combinaorially he number of elemens of any given rank in such a subspace, in erms of Gaussian coefficiens. This enumeraion has been performed previously by Delsare, [4], Theorem 5.6, who used complex characers defined on a finie abelian group o obain his conclusion. Using Delsare s characer formula, we show in Secion 4 ha if K is finie, a consan rank r subspace of M m n (K) has dimension a mos m + n r. This bound is generally poor for r less han m/2, and a bound much closer o n should be expeced. See, for example, Theorem 2 (and erraum) of [2]. Examples of Beasley, [1], and Boson, [3], show ha here are (n + 1)-dimensional consan rank n 1 subspaces of M n (F 2 ) for 3 n 5. The exisence of hese subspaces implies ha he bound above is opimal when m = n, r = n 1, K = 2 and 3 n 5. We show in his paper ha here is a 6-dimensional consan rank 3 subspace of M 4 5 (K) when K = 2, so ha he upper bound is again aained for r = 3, m = 4, n = 5. In Secion 6, we briefly describe some aspecs of rank quesions as hey relae o subspaces of skew-symmeric marices. Our main resul, Theorem 10, employs a characer formula of Delsare-Goehals, [5], o obain a dimension bound when n is odd for a consan rank subspace of n n skew-symmeric marices over a finie field. Again, his bound is unlikely o be opimal in mos cases. 2. Subspaces of marices of rank bounded below Le M be a subspace of M m n (K) each of whose non-zero elemens has rank a leas r. Suppose furhermore ha dim M equals n(m r + 1), he upper bound provided by Lemma 1. We will derive some properies of M in his secion. Lemma 2. Le M be a subspace of M m n (K) wih rank T r for all T M and suppose ha dim M = n(m r + 1). Le v 1,..., v m r+1 be linearly independen

3 SUBSPACES OF MATRICES 3 elemens of K m. Then given any m r +1 elemens w 1,..., w m r+1 of K m, here exiss a unique elemen T, say, of M wih for 1 i m r + 1. v i T = w i Proof. We may idenify K n(m r+1) wih he direc sum of m r + 1 copies of K n. Given he v i as above, we define a linear ransformaion θ : M K n(m r+1) by θ(t ) = (v 1 T,..., v m r+1 T ) for all T M. We claim ha ker θ = 0. For le T be an elemen of ker θ. Then ker T has dimension a leas m r + 1, since i conains he linearly independen vecors v 1,..., v m r+1. This conradics he supposiion ha rank T r and we deduce ha θ is injecive. Since dim M = dim K n(m r+1) = n(m r + 1), i follows ha θ is an isomorphism, and his esablishes he lemma. Definiion 1. Le M be a subspace of M m n (K) and le U be a subspace of K m. We se M U = {T M : ut = 0 for all u U}. M U is clearly a subspace of M. The following resul shows ha he subspaces M U exhibi uniform properies when M saisfies he hypoheses of Lemma 2. Theorem 1. Le M be a subspace of M m n (K) ha saisfies he hypoheses of Lemma 2. Then we have M U = 0 if dim U > m r and if dim U m r. dim M U = n(m + 1 r dim U) Proof. Suppose ha T M U. Then U ker T and hence r rank T m dim U. I follows ha M U = 0 if dim U > m r. Suppose now ha s = dim U m r. Le u 1,..., u s be basis vecors of U. We define a linear ransformaion φ : M K ns by φ(t ) = (u 1 T,..., u s T ) for all T M. Lemma 2 implies ha φ is surjecive. Furhermore, i is clear ha ker φ = M U. The formula for dim M U now follows from he rank-nulliy heorem. We noe wihou proof he following propery of he subspaces M U. Lemma 3. Given subspaces U and W of K m, we have M U M W = M U+W. Corollary 1. Le M be a subspace of M m n (K) ha saisfies he hypoheses of Lemma 2. Le U and W be subspaces of K m such ha dim(u + W ) > m r. Then M U M W = 0. Corollary 2. Le M be a subspace of M m n (K) ha saisfies he hypoheses of Lemma 2. Le U and W be differen subspaces of K m such ha dim U = dim W = m r. Then each elemen of M U and of M W has rank r and M U M W = 0.

4 4 J.-G. DUMAS, R. GOW, G. MCGUIRE, AND J. SHEEKEY Proof. Each elemen T of M U saisfies rank T r, since U ker T. However, rank T r, since T M U, and we deduce ha rank T = r. Clearly, he same conclusion holds for each elemen of M W. Now since U W, i follows ha dim(u + W ) > m r and herefore we obain by Corollary 1. M U M W = 0. We can now obain our main conclusion abou he srucure of he special subspaces of endomorphisms sudied in his secion. Theorem 2. Le M be a subspace of M m n (K) ha saisfies he hypoheses of Lemma 2. Then he subse of all elemens of rank r in M is he disjoin union of all he subses M U, as U ranges over all subspaces U of dimension m r in Km. Proof. We know ha he union of he subses M U is disjoin by Corollary 1. Now any elemen T of M wih rank T = r lies in he subse M W, where W = ker T. Since dim W = m r, we see ha he union described above comprises all elemens of rank r. We noe ha Theorem 2 applies o M = M m n (K), wih r = Enumeraion in he finie field case Le q be a power of a prime and le K = F q. Le M be a subspace of M m n (F q ), wih rank T r for all T M and dim M = n(m r + 1). We will enumerae he number of elemens in M of nulliy (equivalenly, of rank m ), where r m m. The answer depends only on q, m, n, r and, and no on how M is consruced. As we menioned earlier, he enumeraion was firs performed by Delsare in [4], Theorem 5.6, bu our mehod is differen, as we use only elemenary linear algebra and couning echniques, raher han complex characers. We noe ha he formulae obained generalize known resuls for he number of elemens of M m n (F q ) having a given nulliy, since M m n (F q ) is he unique subspace M when r = 1. We sar by inroducing some noaion. As we will be using conceps of nulliy, raher han rank, we se s = m r. We now define subses M and M of M by M = {T M : nulliy T = }, M = {T M : nulliy T } for 0 s. Using he noaion of Definiion 1, we have he following resul. Lemma 4. M = dim U= (M U ), he union being aken over all subspaces U of dimension. Proof. Suppose ha T M U. Then U ker T and hence nulliy T. I follows ha T M. Conversely, suppose ha T M. Then dim ker T and hence for any -dimensional subspace U of ker T, T M U. This implies he desired equaliy. The nex resul is he key o he evaluaion of M. We use he he familiar noaion [ n k] o denoe he q-gaussian coefficien which measures he number of subspaces of dimension k in an n-dimensional vecor space over F q.

5 SUBSPACES OF MATRICES 5 Lemma 5. We have M = Proof. We have seen ha [ ] m (q n(s +1) 1) M = dim U= s j=+1 (M U ). [ ] j M j. Le T M be an elemen saisfying nulliy T = i. Such a T occurs in exacly [ ] i subspaces M U, namely hose ha correspond o U being a -dimensional subspace of ker T. Thus, he elemens of rank in M are hose ha occur in exacly one subspace in he union. Couning he number of elemens in he union above, and aking ino accoun he mulipliciies due o any given elemen belonging o more ha one subspace, we obain [ ] [ ] m m (M U ) = (q n(s +1) 1). If we subrac from his sum he muliple conribuions due o elemens of nulliy greaer han, we obain [ ] m s [ ] j (q n(s +1) 1) M j j=+1 and his expression measures he number of elemens of nulliy, as required. We now use inducion o find M. Theorem 3. Le q be a power of a prime. Le M be a subspace of M m n (F q ), each of whose non-zero elemens has rank a leas r, and suppose ha dim M = n(m r + 1). Le s = m r. For 0 s, le M denoe he subse of elemens of nulliy in M. Then we have M = s +1 i=1 [ m s + 1 i ][ s + 1 i ] ( 1) s i+1 q (s i+1 2 ) (q in 1). Proof. We proceed by inducion on s. When s = 0, which occurs when = m r, he formula given in Lemma 5 implies ha [ ] m M s = (q n 1), s which is he value of M s prediced by he formula we wish o prove. Suppose now ha we have obained he formula for M +1,..., M s. We proceed o esablish he corresponding formula for M by inducion. We are hus assuming ha M j = s j+1 i=1 [ m s + 1 i ][ s + 1 i j ] ( 1) s j i+1 q (s j i+1 2 ) (q in 1) for + 1 j s. The coefficien of q in 1 in his expression is [ ][ ] ( 1) s j i+1 m s + 1 i q (s j i+1 2 ). s + 1 i j

6 6 J.-G. DUMAS, R. GOW, G. MCGUIRE, AND J. SHEEKEY Thus, applying he formula in Lemma 5, he coefficien of q in 1 in M is s+1 i [ ][ ][ ] j ( 1) s j i m s + 1 i q (s j i+1 2 ) s + 1 i j j=+1 and we wan o show ha his equals ( 1) s i+1 [ m s + 1 i ][ s + 1 i ] q (s i+1 2 ). Cancelling he common ( 1) s i erms, we herefore have o prove ha s+1 i [ ][ ][ ] j ( 1) j m s + 1 i q (s j i+1 2 ) = 0. s + 1 i j j= This is equivalen o showing ha he sum s+1 i [ ][ ] j s + 1 i ( 1) j q (s j i+1 2 ) j j= equals 0. We now se T = s i + 1. The sum above is hen T [ ][ ] j T ( 1) j j (T q 2 ). j j= By a well known propery of Gaussian coefficiens, [6], Exercise 2.6.2, we have [ ][ ] [ ][ ] j T T T = j T j and hus i suffices o prove ha T [ ] T ( 1) j j (T q 2 ) = 0. T j j= We now se l = T j and see ha he sum above is ransformed o T [ ] T ( 1) T ( 1) l q 2) (l. l l=0 This expression equals 0, by anoher well known propery of Gaussian coefficiens. See, for example, Formula 2, , in [6]. 4. Consan rank subspaces of marices We begin by skeching a known consrucion of n-dimensional consan rank r subspaces. Suppose ha L is a field exension of K of degree n. The regular represenaion of L over K provides us wih a subspace M n, say, of M n (K) in which each non-zero marix is inverible. Le T be an elemen of rank r in M m n (K). Then he subse T M n, consising of all lef muliples of elemens of M n by T, is a consan rank r subspace. We summarize his argumen as follows.

7 SUBSPACES OF MATRICES 7 Theorem 4. Suppose ha he field K has an exension field of degree n. Then here exiss a consan rank r subspace of M m n (K) of dimension n for each ineger r wih 1 r m. We are ineresed here in obaining a reasonable upper bound for he dimension of a consan rank r subspace of M m n (K). As far as we know, a he ime of his wriing, he bes general resul of his naure is he following, proved by Beasley and Laffey, [2], Theorem 2 (see also he erraum o his paper). Theorem 5. (Beasley and Laffey) Suppose ha 1 r m, K r + 1 and n 2r 1. Then he dimension of a consan rank r subspace of M m n (K) is a mos n. I is hus small finie fields which migh cause problems in aemping o obain a uniform resul abou he maximum dimension of a consan rank subspace, in line wih wha Theorems 4 and 5 sugges. We presen in Theorem 6 an upper bound for he dimension of a consan rank subspace over a finie field. In view of he resul of Beasley and Laffey, our resul is ineresing in ha involves no hypohesis abou he size of he field (oher han being finie), bu i is clearly generally weak compared wih Theorem 5 when m > 2r 1. Le p be a prime and le q be a power of p. Suppose for he remainder of his secion ha K = F q, he finie field of order q. Under addiion, M m n (F q ) is an elemenary abelian p-group of order q mn, and hus has q mn irreducible complex characers, which are homomorphisms from he addiive group ino he muliplicaive group of he complex numbers. We may describe hese characers in he following way. Le r : M m (F q ) F q denoe he usual (marix-heoreic) race funcion. Le τ : F q F p denoe he field-heoreic race funcion. Le ω be a primiive p-h roo of uniy in he complex numbers. Then for each S M m n (K), we define an irreducible characer λ S of by M m n (F q ) λ S (T ) = ω τ(r(st )) for all T M m n (F q ). (Here, T denoes he ranspose of T.) I is no difficul o show, using properies of he wo race funcions, ha we obain all irreducible characers of M m n (F q ) in his way. For 0 k m, le Ω k denoe he subse of all elemens of rank k in M m n (F q ). The funcion P k : M m n (F q ) C defined by P k = S Ω k λ S is he characer of a complex represenaion of M m n (F q ) (usually a reducible represenaion). Alhough expressed in slighly differen language, Delsare proves he following resul, [4], Theorem 3.1. Lemma 6. The characers P k ake equal (raional inegral) values on elemens of he same rank. Delsare uses P k (r) o denoe he value of P k on an elemen of rank r. We will only need he values P 1 (r) in our work here. See [4], Theorem A2 (noe ha he summaion index m in he prined formula mus be replaced by k).

8 8 J.-G. DUMAS, R. GOW, G. MCGUIRE, AND J. SHEEKEY Lemma 7. Wih he noaion inroduced above, we have P 1 (r) = (qm 1) q 1 + qn (q m r 1). q 1 We can now proceed o he proof of our main resul of his secion. Theorem 6. Le q be a power of he prime p and le M be a consan rank r subspace of M m n (F q ). Then dim M m + n r. Proof. Le = dim M. The resricion of he characer P 1 of M m n (F q ) o he subgroup M is also a characer of M and herefore elemenary characer heory implies ha he inner produc of his characer wih he rivial characer of M is an ineger. Now his inner produc is P 1 (0) + (q 1)P 1 (r) q, since all non-zero elemens of M have rank r by hypohesis. I follows ha P 1 (0) P 1 (r) (mod q ). Using Delsare s characer formula, Lemma 7, we obain I follows ha = dim M m + n r. q m+n r (q r 1) 0 (mod q ). The reader may compare our bound above wih he upper bound m + n 2r + 1 for he dimension of a consan rank r subspace of M m n (C), obained by Weswick in [10]. This upper bound may be improved o n r + 1 when n r + 1 does no divide (m 1)!/((r 1)!. The bound obained in Theorem 6 is paricularly poor when r = 1, since he correc upper bound for he dimension of a consan rank 1 subspace is n, a resul which holds for all fields. Noneheless, as we menioned in he inroducion, here are examples of (n + 1)-dimensional consan rank n 1 subspaces of M n (F 2 ) for 3 n 5. The exisence of hese subspaces implies ha he bound can be opimal in non-rivial ways. Example 1. Consider he 6-dimensional subspace U of M 4 5 (F 2 ) consising of he linear span of he marices , , , , , A consideraion of all cases shows ha U is a consan rank 3 space. By Theorem 5, a consan 3 rank subspace of M 4 5 (K) has dimension a mos 5 when K 4. Thus we see ha he resricion in he field size in Theorem 5 is appropriae when m = 4, n = 5 and r = 3..

9 SUBSPACES OF MATRICES 9 Example 2. If we adjoin an addiional zero row a he op of each marix of Example 3, he resuling marices span a 6-dimensional consan rank 3 subspace of M 5 (F 2 ). 5. A maximaliy resul for cerain consan rank subspaces Suppose ha M n (K) conains an n-dimensional consan rank n subspace, N, say. This occurs, for example, if K admis a field exension of degree n. Le T be any elemen of rank m 1 in M m n (K). The subspace M = T N is hen an n-dimensional consan rank m 1 subspace of M m n (K). We inend o show in his secion ha such an n-dimensional consan rank m 1 subspace is maximal when K is finie, ha is, he subspace is no conained in any (n + 1)-dimensional consan rank m 1 subspace. Definiion 2. Le R be a subspace of M m n (K). Given any non-zero vecor u in K m, we se R u = {T R : ut = 0}. This is in accordance wih he noaion used in Definiion 1. Lemma 8. Le R be a subspace of M m n (K) and le u be a non-zero elemen of K m. Suuppose ha dim R > n. Then dim R u > 0. Proof. We define a K-linear ransformaion ɛ : R K n by ɛ(s) = us for all S R. Since dim R > dim K n, ɛ is no injecive, and hence dim ker ɛ > 0. The desired resul now follows, since ker ɛ = R u. We omi he proof of he following simple fac. Lemma 9. Le M = T N be he n-dimensional consan rank m 1 subspace of M m n (K) described above. Le u be a basis vecor for he kernel of T. Then each nonzero elemen of M has he same kernel u. Lemma 10. Le R be a consan rank n 1 subspace of M m n (K) and le u, v be differen one-dimensional subspaces of V. Then R u R v = 0 Proof. This follows since a non-zero elemen in he inersecion would annihilae he linearly independen vecors u and v and hence have rank a mos m 2. Lemma 11. Le M = T N be he n-dimensional consan rank m 1 subspace of M m n (K) described in Lemma 9 and le u be a basis vecor for ker T. Suppose, if possible, ha here exiss an (n + 1)-dimensional consan rank m 1 subspace R of M m n (K) conaining M. Then if v K m is no a scalar muliple of u, we have while dim R v = 1, R u = M u = M.

10 10 J.-G. DUMAS, R. GOW, G. MCGUIRE, AND J. SHEEKEY Proof. Suppose ha v u. Then we know ha M v = 0. We hus have R v M = M v = 0. I follows ha dim R v 1, since M has codimension 1 in R. However, Lemma 8 implies ha dim R v 1 and hence dim R v = 1. Finally, we know ha R u conains M. If R u = R, hen R u has non-rivial inersecion wih any R v. This conradics Lemma 10 when u and v are linearly independen. We can now show ha here is no subspace R fulfilling he requiremens of Lemma 11 when K is finie. Theorem 7. Le M = T N be he n-dimensional consan rank m 1 subspace of M m n (K) described in Lemma 9. Then if K is finie, M is no conained in any larger consan rank m 1 subspace of M m n (K). Proof. Suppose by way of conradicion ha M is conained in he consan rank m 1 subspace R of dimension n + 1. Then we have R = R v, where v ranges over he one-dimensional subspaces of K m, and he subspaces R v have rivial inersecion wih each oher. Moreover, one subspace in he union above is M, he res are one-dimensional. Thus if K = q, couning non-zero vecors in R, we obain q n+1 1 = q n 1 + (qm q) (q 1). (q 1) This equaliy is clearly impossible when m n. The examples of Beasley, [1], and Boson, [3], imply ha he spaces M n (F 2 ) for 3 n 5 conain wo ypes of maximal consan rank n 1 subspaces. One ype has dimension n, he oher dimension n + 1. Similarly, M 4 5 (F 2 ) conains wo ypes of maximal consan rank 3 subspaces, one ype of dimension 5, he oher of dimension Subspaces of skew-symmeric marices wih special rank properies Specializing he rank heme described in he inroducion o his paper, i is an ineresing research problem o invesigae subspaces S, say, of skew-symmeric marices which possess one of he following hree properies. (1) Each elemen of S has rank a mos 2r. (2) Each elemen of S has rank 2r. (3) Each elemen of S has rank a leas 2r. We noe, as is well known, ha a skew-symmeric marix has even rank. Concerning condiion 3, we have he following resul, [7], Theorem 8. Theorem 8. Suppose ha he field K has a cyclic Galois exension field of odd degree n and le r be an ineger saisfying 2 2r n 1. Then here exiss a subspace S of skew-symmeric marices in M n (K) of dimension n(n 2r + 1)/2 in which each elemen T of S saisfies rank T 2r.

11 SUBSPACES OF MATRICES 11 Concerning Theorem 8, Delsare and Goehals show in [5], Theorem 4, ha if n is odd and K is finie, n(n 2r + 1)/2 is he maximum dimension of a subspace of skew-symmeric marices in M n (K) in which each non-zero elemen has rank a leas 2r. Their proof use characers defined on associaion schemes and we have no found a proof of his resul working in he conex of linear algebra, more especially, one which applies o infinie fields. (We do no know if he Delsare-Goehals upper bound even holds for infinie fields.) As far as consan rank subspaces of skew-symmeric marices are concerned, he following exisence heorem was esablished in [7]. Theorem 9. Suppose ha he field K has a cyclic Galois exension field of odd degree n and le s > 1 be a divisor of n. Then here exiss an n-dimensional consan rank n(s 1)/s subspace of skew-symmeric marices in M n (K). We do no know if here are values of r differen from n(s 1)/s for which here exiss an n-dimensional consan rank r subspace of skew-symmeric marices in M n (K). We conclude his secion by obaining a version of Theorem 6 for consan rank subspaces of skew-symmeric marices over finie fields. Our proof again uses characers of finie groups. Theorem 10. Le S be a consan rank 2r subspace of skew-symmeric marices in M n (F q ). Then dim S 2n 2r 1. Proof. Le A denoe he vecor space of n n skew-symmeric marices wih enries in F q. We may consider A o be an elemenary abelian p-group of order q nk, and S o be a subgroup of order q m, where m = dim S. For each ineger k wih 1 k [n/2], Delsare and Goehals consruc a (reducible) complex characer P k of A which akes he same inegral value on elemens of he same rank in A, [5], p.29. Following he noaion of [5], we le P 1 (r) denoe he value of P 1 on an elemen of rank 2r. By he formula (15) in [5], we have P 1 (r) = (q2 1) q qn (q 2 2r 1) q 2, 1 where n = is odd. In he case ha n is even, we have P 1 (r) = (qn 1) q qn 1 (q n 2r 1) q 2. 1 The res of he proof is idenical wih ha of Theorem 6. I may be of ineres o commen on he sharpness of he dimension bound obained above. When n is odd, here is an n-dimensional consan rank n 1 subspace of skew-symmeric marices in M n (F q ) and hus he bound given by he heorem is sharp in his case. When n is even, one can show ha he maximum dimension of a consan rank n subspace of skew-symmeric marices in M n (F q ) is n/2, whereas he heorem gives an upper bound of n 1 for he dimension. Thus he bound given by he heorem is weak in his case. On he oher hand, when m is an odd posiive ineger and n = 2m, we can consruc an m-dimensional consan rank m 1 subspace of skew-symmeric marices in M m (F q 2) and hen use he race map from F q 2 o F q o obain an n-dimensional consan rank n 2 subspace of skew-symmeric marices in M n (F q ). The heorem above gives n + 1 as he maximum dimension for such a consan rank subspace, so ha we are quie close

12 12 J.-G. DUMAS, R. GOW, G. MCGUIRE, AND J. SHEEKEY o a sharp bound here. A he ime of his wriing, we have no found any consan rank subspace of skew-symmeric marices in M n (F q ) of dimension greaer han n. 7. Consrucion of examples by field exension Le L be a field exension of K of finie degree m. Recall ha we may consider any vecor space over L o be a vecor space over K. In paricular, le V be a vecor space of finie dimension n over L and le V K denoe V considered as a vecor space over K. We have hen dim K V K = mn. Clearly, an L-linear endomorphism, σ, say, of V defines a K-linear endomorphism of V K, which we shall denoe by σ K. I is elemenary o see ha he mapping σ σ K is K-linear and injecive. Consequenly, we have a K-linear monomorphism End L (V ) End K (V K ) whose image has dimension mn 2 (and hence he monomorphism is an isomorphism only when L = K). The following fac abou his monomorphism is well known, bu we include a brief proof. Lemma 12. Wih he noaion inroduced above, rank σ K = m rank σ. Proof. Le N denoe he kernel of σ. Clearly, N K (N considered as a vecor space over K) is he kernel of σ K. We have hen rank σ = dim L V dim L N and rank σ K = dim K V K dim K N K = m dim L V m dim L N. The resul is immediae from hese equaliies. Suppose now ha M is a subspace of End L (V ) and le M K denoe is image under he mapping jus considered. Suppose ha he differen ranks of he non-zero elemens of M are r 1, r 2,..., r k. Then M K has dimension m dim L M and is non-zero elemens have rank mr 1, mr 2,..., mr k. This simple observaion provides a mehod for generaing subspaces of marices wih special rank properies. Suppose in paricular ha we have found an (n + 1)-dimensional consan rank n 1 subspace in M n (F q m), where m > 1. Then he consrucion described above creaes an m(n + 1)-dimensional consan rank m(n 1) subspace in M mn (F q ). Theorem 6 shows ha m(n + 1) is he maximum dimension for such a consan rank subspace. However, a his ime of wriing, he known examples are resriced o F 2, and we do no know wheher his small field is excepional in his heory. Le us assume for he res of his secion ha L is a separable exension of K. Le T denoe he race funcion from L o K. The separabiliy assumpion hen implies ha T maps L ono K. Le f : V V L be an L-valued bilinear form. We define a bilinear form f K : V K V K K by seing f K (u, v) = T (f(u, v)) for all u and v in V. We omi he formal proof of he following resul, which follows ha of Lemma 12.

13 SUBSPACES OF MATRICES 13 Lemma 13. Wih he noaion inroduced above, rank f K = m rank f. Furhermore, he mapping f f K is a K-linear monomorphism from he vecor space of all L-valued bilinear forms on V V (considered as a space over K) ino he vecor space of all K-valued bilinear forms on V K V K, whose image has dimension mn 2 over K. This monomorphism maps alernaing bilinear forms ino alernaing bilinear forms and symmeric bilinear forms ino symmeric bilinear forms. Corollary 3. Suppose ha K has a separable field exension L of degree m. Then here exiss a 3m-dimensional consan rank 2m subspace of skew-symmeric marices in M 3m (K) Proof. Le V be a 3-dimensional vecor space over L. The space of L-valued alernaing bilinear forms on V V is 3-dimensional and each non-zero elemen in his space has rank 2. The resul follows from Lemma 13. Thus, for example, aking K = R and L = C, we see ha here is a 6-dimensional consan rank 4 subspace of skew-symmeric marices in M 6 (R). Similarly, as we observed in he previous secion, when m is an odd posiive ineger and n = 2m, we can consruc an m-dimensional consan rank m 1 subspace of skew-symmeric marices in M m (F q 2) and hen obain an n-dimensional consan rank n 2 subspace of skew-symmeric marices in M n (F q ) for each prime power q. References [1] L. B. Beasley, Spaces of rank-2 marices over GF(2), Elecron. J. Linear Algebra 5 (1999), [2] L. B. Beasley and T. J. Laffey, Linear operaors on marices: he invariance of rank-k marices, Linear Algebra Appl. 133 (1990), Erraum, Linear Algebra Appl. 180 (1993), 2. [3] N. Boson, Spaces of consan rank marices over GF (2), Elecron. J. Linear Algebra 20 (2010), 1-5. [4] P. Delsare, Bilinear forms over a finie field, wih applicaions o coding heory, J. Combinaorial Theory Ser. A 25 (1978), [5] P. Delsare and J. M. Goehals, Alernaing bilinear forms over GF (q), J. Combinaorial Theory Ser. A 19 (1975), [6] I. P. Goulden and D. M. Jackson, Combinaorial Enumeraion, Wiley-Inerscience, New York, [7] R. Gow and R. Quinlan, Galois exensions and subspaces of alernaing bilinear forms wih special rank properies, Linear Algebra Appl. 430 (2009), [8] R. Guralnick, Inverible preservers and algebraic groups, Linear Algebra Appl. 212/213 (1994), [9] R. M. Roh, Maximum-rank array codes and heir applicaion o crisscross error correcion, IEEE Trans. Inform. Theory 37 (1991), [10] R. Weswick, Spaces of marices of fixed rank, Linear and Mulilinear Algebra 20 (1987),

14 14 J.-G. DUMAS, R. GOW, G. MCGUIRE, AND J. SHEEKEY Universié de Grenoble, France address: School of Mahemaical Sciences, Universiy College Dublin, Ireland address: School of Mahemaical Sciences, Universiy College Dublin, Ireland address: School of Mahemaical Sciences, Universiy College Dublin, Ireland address: