aa r X i v : . [ m a t h . C O ] M a r EXTREMAL COLLECTIONS OF k -UNIFORM VECTORS JOSEPH BRIGGS AND WESLEY PEGDEN
Abstract.
We show any matrix of rank r over F q can have ≤ (cid:0) rk (cid:1) ( q − k distinct columnsof weight k if k ≤ O q ( √ log r ) (up to divisibility issues), and ≤ (cid:0) rk (cid:1) ( q − r − k distinct columnsof co-weight k if k ≤ O q ( r / ). This shows the natural examples consisting of only r rowsare optimal for both, and the proofs will recover some form of uniqueness of these examplesin all cases. Introduction
The field of extremal combinatorics deals with the asymptotic study of how parameters growover increasing classes of discrete structures. Recently (see for example [4]), there has beengrowing interest in the study of an extremal theory for matroids. This includes an extremaltheory for representable matroids, whose ground set is the set of columns of some matrix(and independence is given by linear independence).One standard method for generating random representable matroids, see e.g. [7], is as follows.Construct a matrix representation M by generating m randomly chosen columns of somefixed weight k and length n . Indeed, when k = 2 and the base field is F , this gives thegraphic matroid of the Erd˝os-R´enyi random graph G n,m (of which M acts as the vertex-edgeincidence matrix).Our desire is to settle perhaps the most natural extremal question in this setting: how largecan m be, upon fixing the “size” of such a representable matroid? It makes little sense tofix the number n of rows, as then one can take all m = (cid:0) nk (cid:1) ( q − k weight- k column vectors,and every possible matrix will just consist of a subset of these columns. So instead, we fixthe rank.Let us take a step back. For a matrix M over the finite field F q , we are considering thefollowing question: Question 1.1.
What is the maximum number of distinct columns M can have, if eachcolumn has weight k , that is k nonzero entries, and M has rank ≤ r ? We denote this value by ex q ( r, k ). We can answer this question if r is large enough: Research supported in part by NSF grant DMS 1700365.
Theorem 1.2.
For all k and q , there is an R k,q such that for all r ≥ R k,q , ex q ( r, k ) = ((cid:0) r +1 k (cid:1) q = 2 and k even, (cid:0) rk (cid:1) ( q − k otherwise. When k = q = 2, this tells us that graphs of graphic matroid rank ≤ r have ≤ (cid:0) r +12 (cid:1) edges:this was previously noted in Theorem 2.8 of [6], where it was shown for every r (not justthose sufficiently large). Furthermore, the case q = 2 was a question asked by Ahlswede,Aydinian and Khachatrian [1]. Khachatrian (according to [2]) and Kramer [9] conjecturedthe above structure, and the latter proved it when the number of rows of the matrix is r + 1.Our result confirms their conjecture, but only once r is large enough.The nature of this question does not change much after replacing “nonzero” with “non- β ”for an arbitrary β ∈ F q (see Section 3, and more specifically Theorem 3.4, an affine variantwe will use to prove this result). This effectively answers both questions of this type over F . However, for other fields F q , “weight k ” and “ k Question 1.3.
What is the maximum number of distinct columns M can have, if eachcolumn has k zeros, and M has rank ≤ r ? Denoting this by ex q ( r, k ), we have a corresponding result: Theorem 1.4.
Suppose F q = F . For all k , there is an ¯ R k,q such that for all r ≥ ¯ R k,q , ex q ( r, k ) = (cid:18) rk (cid:19) ( q − r − k . Furthermore, in the context of both Theorem 1.2 and Theorem 1.4, we will show that the onlyexamples attaining the equality have exactly r nonzero rows (unless k = 0, see Corollary 2.3).This corresponds to the “uniqueness of the cliques” in Theorem 2.8 of [6], where additionalisolated vertices correspond to additional rows of all 0’s here.In fact, a result akin to Theorem 1.2 holds in a far more general setting. Suppose F is anarbitrary field (not necessarily finite). Let L = ( L , . . . , L s ) be a collection of disjoint finitesets L i ⊂ F ∗ of nonzero labels. Then, for each s -tuple k = ( k , . . . , k s ) of positive integers,an “( L , k ) -vector ” is defined to be one with exactly k i entries in L i for each i , and the restequal to 0. Thus, a binary vector of weight w is an ( L , k )-vector for L = ( { } ) and k = ( w ).The corresponding question in this setting is thus: Question 1.5.
What is the maximum number of distinct columns M can have, if eachcolumn is an ( L , k ) -vector, and M has rank ≤ r ? We denote this value by ex F , L ( r, k ).We will prove the following theorem in Section 3: XTREMAL COLLECTIONS OF k -UNIFORM VECTORS 3 Theorem 1.6.
Suppose s = 1 , so that k = ( k ) and L = ( L ) . Then there is an R k such thatfor all r ≥ R k : ex F ,L ( r, k ) = ((cid:0) r +1 k (cid:1) L = { ℓ } and ℓ · in F , | L | k (cid:0) rk (cid:1) otherwise.Alternatively, suppose each L i is a single element { ℓ i } for every i . Then, once r ≥ R k , ex F , L ( r, k ) = ((cid:0) r +1 k (cid:1) P ℓ i k i = 0 , L k (cid:0) r k (cid:1) otherwise,where by (cid:0) r k (cid:1) we mean the multinomial coefficient (cid:0) rk ,...,k s ,r − P i k i (cid:1) , and by L k we mean theproduct Q i ∈ [ s ] | L i | k i .Moreover, in both cases, any extremal matrix M has only r +1 or r nonzero rows respectively. Here, R k , R k respectively depend on | L | and L k , but not on the field F .We remark there is nonempty (albeit rather small) overlap between Theorem 1.6 and themain theorem of Ahlswede, Aydinian and Khachatrian [1]. They considered this questionin the case F = R , s = 1, and L = ( { } ), i.e. binary vectors over the reals of weight k , butmanaged to solve this for every r . In particular, the equalities given in (1.6) were shown tobreak down precisely once r < k . As with their question for q = 2, this leads us to ask howsmall R k,q can be made in general—we will discuss this a little more in Section 5. But, forbinary vectors over any field, the bound on R k,q we obtain remains the same.We close the introduction by noting Theorem 1.2 follows from the first part of Theorem 1.6.Indeed, let L := F × q = F q \{ } , a single list consisting of all nonzero elements of F q . Here, L = { ℓ } if and only if q = 2 and ℓ = 1, so the clause “ ℓ · k is even.2. Preliminaries, Notation
We first obtain nontrivial bounds for all 3 questions, by generalizing the setup further still.For an arbitrary set S ⊂ Z s ≥ of possible weight vectors, we say a column vector is an “( L , S ) -vector ” whenever it is an ( L , k )-vector for some k ∈ S , and denote by ex F , L ( r, S ) the maxi-mum size of a collection of ( L , S )-vectors whose rank is ≤ r . We can define ex q ( r, T ) , ex q ( r, T )correspondingly when T is just a subset of nonnegative integers, and specifically writeex q ( r, ≤ k ) , ex q ( r, ≤ k ) as shorthand for ex q ( r, { , , . . . , k } ) , ex q ( r, { , , . . . , k } ) respectively.We can form a poset structure (cid:22) on the set of weight vectors Z s ≥ of length s by saying k ′ (cid:22) k if and only if k ′ i ≤ k i in every coordinate i . Say that S ⊂ Z s ≥ is a down-set if k ′ ∈ S whenever k ∈ S and k ′ (cid:22) k . JOSEPH BRIGGS AND WESLEY PEGDEN
Lemma 2.1.
For any rank r , field F q and weight k : ex q ( r, ≤ k ) = X i ≤ k (cid:18) ri (cid:19) ( q − r − i . Also, for any field F , down-set S ⊂ Z s ≥ , weight vector k and list vector L , ex F , L ( r, S ) = X k ′ ∈ S (cid:18) r k ′ (cid:19) L k ′ , and hence for any F q , ex q ( r, ≤ k ) = X i ≤ k (cid:18) ri (cid:19) ( q − i . Proof.
It suffices to show “ ≤ ”, since the corresponding lower bounds are all immediate fromconsidering matrices with precisely r rows (with all columns of weight ≤ k in the first case,or all ( L , S )-vectors of length r in the second).For the second bound, given a matrix M of rank r , let C be its columns and W = hCi beits column space. Since the row rank of M is also r , there exists a subset I of r of its rowssuch that the projection W → W | I is an isomorphism. In particular, it is injective, andrestricts to an injection on the original vectors π : C ֒ → C| I . For any x ∈ C and i ∈ [ s ], thenumber of L i -entries in π ( x ) is ≤ that of x . Hence, if x is an ( L , k )-vector, then π ( x ) is an( L , k ′ )-vector for some k ′ (cid:22) k , and hence an ( L , S )-vector as S is a down-set. The desiredbound is then obtained by counting all ( L , S )-vectors in C| I ≃ F r .The proof of the bound on ex is identical, with “zero-entries” and “vectors with k ′ zeros” inplace of “ L i -entries” and “( L , k ′ )-vectors” respectively. (cid:3) Corollary 2.2.
For any q, F , r, k, k and L , ex q ( r, k ) ≤ X i ≤ k (cid:18) ri (cid:19) ( q − r − i , ex F , L ( r, k ) ≤ X k ′ (cid:22) k (cid:18) r k ′ (cid:19) L k ′ , and ex q ( r, k ) ≤ X i ≤ k (cid:18) ri (cid:19) ( q − i . In particular, we obtain the k = 0 case of Theorem 1.4: Corollary 2.3. ex q ( r,
0) = ( q − r . Furthermore, any matrix M with no zeros, of rank r ,with ( q − r distinct columns, has r rows u , . . . , u r such that every row is a scalar multipleof some u i .Proof. Taking k = 0 in the 3rd equality of Corollary 2.2 establishes ex q ( r, ≤ ( q − r .For any matrix M attaining this equality, in the above proof, we see that C| I consists of all column vectors with no zeros, that is, C| I ≃ ( F × q ) r . Letting u , . . . , u r denote the rows of M given by I , this says that for every v ∈ ( F × q ) r there is some j such that v = u ,j ...u r,j . Now,suppose there is another row u of M . Since rank( M ) = r = dim( hC| I i ) , the { u i } form a XTREMAL COLLECTIONS OF k -UNIFORM VECTORS 5 basis for the row space of M , so u = P l i u i for some scalars l i ∈ F q . As M has no zeros,0 = P l i u i,j = h x , v i for every j , writing x := ( l , . . . , l r ).Now, since u = 0, x = 0 so some l j = 0. Now take any j ′ ∈ [ r ] \{ j } . Consider the q vectors of the form v ( α, β ) := (1 , . . . , , α, , . . . , , β, , . . . , ∈ F rq with α, β in positions j, j ′ respectively. For each α = 0, we know h x , v (1 , α ) i ∈ F × q , and they are distinct since l j = 0. It follows h x , v (1 , i = 0. As q ≥
3, we similarly find another β ∈ F × q \{ } , also with h x , v ( β, i = 0 by the same logic. Subtracting these gives 0 = h x , (1 − β ) e j ′ i = (1 − β ) l j ′ ,hence l j ′ = 0.Since j ′ = j was arbitrary, u = l j u j , which is what we were trying to prove. (cid:3) Remark 2.4.
The final part of the argument showed that any vector over F q ( q ≥
3) ofweight ≥ Weight- k Proofs
Our proofs will first establish an affine variant of Theorem 1.6 for technical reasons. To stateit, define the a -rank , or affine rank , of a set of vectors to be the smallest r so that any subsetof r + 1 vectors yield an a -dependence , where by an a -dependence we mean a nontrivial lineardependence whose coefficients sum to 0 in F . Notation 3.1.
We denote by aex L ( r, k ) the maximum size of a collection of ( L , k ) vectorsof a -rank ≤ r . (As F will always be fixed, we drop the dependence in this notation.) Notice that, in general, the a -rank of a collection is at least the rank of the collection. Onthe other hand, the a -rank of the columns of a matrix M is the same as the rank of thematrix M with an additional row of all 1’s added. Thus, we have(1) rank( M ) ≤ a -rank( M ) = rank (cid:18) M · · · (cid:19) ≤ rank( M ) + 1 . Moreover,
Remark 3.2.
Suppose every L i = { ℓ i } has only one element, and that P ℓ i k i = 0. Thenaex L ( r, k ) = ex L ( r, k ). Indeed, this time (1 , . . . , ∈ rowspan( M ) for any matrix M whosecolumns are ( L , k )-vectors, and so its a -rank and rank coincide.In a similar spirit, we will also make frequent use of the following standard lemma: Lemma 3.3.
Let l, µ ∈ F be distinct. Then a -rank (cid:18) l · · · l µY v (cid:19) = a -rank (cid:18) l · · · lY (cid:19) + 1 , and hence equals a -rank ( Y ) + 1 (provided l = 0 ), for any vector v and matrix Y over F withthe same number of rows. JOSEPH BRIGGS AND WESLEY PEGDEN
Proof.
It suffices to show “ ≥ ”, the other direction being trivial.Let r = a -rank (cid:18) l · · · l µY v (cid:19) and take any r column vectors v , . . . , v r of Y . By definitionof r , there is an a -dependence among (cid:0) l v (cid:1) , . . . , (cid:0) l v r (cid:1) , (cid:0) µ v (cid:1) . Since the coefficients sum to 0,and l = µ, it follows the coefficient of (cid:0) µ v (cid:1) is 0, so in fact we have an a -dependence among (cid:0) l v (cid:1) , . . . , (cid:0) l v r (cid:1) . Thus, a -rank (cid:0) l ··· lY (cid:1) ≤ r −
1, as desired. (cid:3)
Theorem 3.4.
Suppose that either every L i in the list L has size 1, or that s = 1 (so L = ( L ) and k = ( k ) ). Then there is a Q k such that for all r ≥ Q k , aex L ( r, k ) = ((cid:0) r − k (cid:1) | L | k s = 1 , | L | > (cid:0) r k (cid:1) every | L i | = 1 . Moreover, any extremal collection must consist of vectors which are zero except in r − common positions (respectively, r common positions).Proof. For the lower bound, we simply take “all vectors of the maximum possible length”-but we must be cautious whether the maximum possible length is r or r −
1. First suppose L i = { ℓ i } for each i . Let M be the matrix whose columns are all (cid:0) r k (cid:1) ( L , k )-vectors of length r . Then rank( M ) = r − P ℓ i k i = 0, and r otherwise, but in both instances a -rank( M ) = r (see (1) and Remark 3.2, respectively).Meanwhile, if L = ( L ) and | L | ≥
2, then the matrix of all | L | k (cid:0) r − k (cid:1) ( L , k )-vectors oflength r − r −
1, and again by (1) has a -rank ≤ r .Write aex ∗ L ( r, k ) for aex L ( r + {∃ i : | L i | > } , k ). For the upper bound, we will prove aex ∗ L ( r, k ) ≤ L k (cid:0) r k (cid:1) for r ≥ Q k .To begin, we will show that for any nonzero k and for all r ≥ k k k + 2,(2) aex ∗ L ( r, k ) ≤ aex ∗ L ( r − , k ) + X i ∈ [ s ] | L i | · aex ∗ L ( r − , k − e i ) . where e i denotes the i th unit vector, so that k − e i = ( k , . . . , k i − , . . . , k s ) . To see this,we first show (2) without the stars. Consider any matrix M of a -rank ≤ r whose columnsare all ( L , k )-vectors. If all nonzero rows of M had all entries in S L i , then M has only k k k nonzero rows and in particular ≤ L k columns, independently of r . Plus, having alreadyestablished the lower bound in the theorem, we know L k ≤ (cid:0) r − k (cid:1) L k ≤ aex L ( r − , k ), onlyneeding r ≥ k k k . So WLOG, assume that the first row of M contains both a 0 and an ℓ ∈ S L i .Now let A ℓ be the set of vectors with ℓ in row 1, for each ℓ ∈ { } ∪ S L i . Both S ℓ =0 A ℓ and A are nonempty by assumption. Define A ′ ℓ to be the collection of vectors producedby removing the first coordinate from each vector in A ℓ . By Lemma 3.3, A ′ ℓ has a -rank ≤ r − ℓ ∈ { } ∪ S L i . Hence | A ′ ℓ | ≤ aex L ( r − , k − e i ) whenever ℓ ∈ L i while | A ′ | ≤ aex L ( r − , k ) . This establishes (2) for aex L ( r, k ). To add the stars, note that theshift in r of 1 in aex ∗ L occurs for all terms in the case s = 1 , | L | > no terms if | L i | = 1 for every i . XTREMAL COLLECTIONS OF k -UNIFORM VECTORS 7 The inequality (2) would suffice to prove Theorem 3.4 if we could establish a family of basecases for each nonzero k for the induction, since clearly aex ∗ L ( r, ) = 1 for every r and L .We do not know how to do this directly, however. Instead we define α k r = aex ∗ L ( r, k ) − L k (cid:18) r k (cid:19) and consider the sequence { α k r } r ∈ N . Now, (2) gives that for r ≥ k k k +2, α k r ≤ α k r − + P i ∈ [ s ] | L i | · α k − e i r − . By induction on k k k , we then have for Q ′ k := max {k k k + 2 } ∪ { Q k − e i : i ∈ [ s ] } that r − ≥ Q ′ k = ⇒ α k r ≤ α k r − . Observe that to prove Theorem 3.4, it suffices to show that for r − ≥ Q ′ k , Claim 3.5. α k r = α k r − = ⇒ ( α k r = 0 and any collection realizing aex ∗ L ( r, k ) must have support r ) . To prove Claim (3.5), let us suppose that r, k are such that α k r = α k r − , and r − ≥ Q ′ k , sothat(3) aex ∗ L ( r, k ) = aex ∗ L ( r − , k ) + X i ∈ [ s ] | L i | k i (cid:18) r − k − e i (cid:19) . Recall that in the decomposition above, A has size at most aex ∗ L ( r − , k ). Thus for anextremal collection for r, k where (3) holds, we have that(4) X ℓ =0 | A ℓ | ≥ X | L i | k i (cid:18) r − k − e i (cid:19) . Moreover, by induction on k k k , we have that the unique candidate for A ′ ℓ of size L k − e i (cid:0) r − k − e i (cid:1) is a collection of vectors whose support has size r −
1, for every ℓ ∈ L i . In fact, they are all L k − e i (cid:0) r − k − e i (cid:1) such possible vectors. The above inequality (4) is therefore an equality, alreadyestablishing α k r = 0.We next see these supports must be identical for every ℓ ∈ ∪ L i . For otherwise, for somedistinct ℓ, j ∈ ∪ L i we have supp( A ′ ℓ ) \ supp( A ′ j ) = ∅ , WLOG containing 2. That is, some u ′ ∈ A ′ j has u ′ ∈ ∪ L i while 2 is outside the support of A ′ j . Write A ′′ j for A ′ j upon deletion ofthis topmost row of all zeros. Then take some other v ′ ∈ A ′ ℓ with v ′ = 0 (using the structureof A ′ ℓ and that r > k k k ). Now expand along rows 1 and 2 in Lemma 3.3 in turn, writing v ′′ , u ′′ for the vectors obtained from v ′ , u ′ respectively by removing the first entry: a -rank( A ′ j ) = a -rank( A ′′ j ) ≤ a -rank (cid:18) j · · · j ℓA ′′ j v ′′ (cid:19) − ≤ a -rank j · · · j ℓ ℓ · · · u A ′′ j v ′′ u ′′ − ≤ r − , a contradiction.Finally, we check the support of A must be contained in the support of A ′ j (now equivalentfor any j ∈ ∪ L i ), establishing Claim (3.5), and thus also Theorem 3.4. Indeed, suppose A contains a vector u such that u ( t ) ∈ ∪ L i , where t is outside the support of A ′ j . This time we JOSEPH BRIGGS AND WESLEY PEGDEN consider three cases:
Case 1: All vectors v ∈ A satisfy v ( t ) ∈ ∪ L i , but A is nonconstant on row t . Decomposing A according to t -th entries, and applying Lemma 3.3 to each part we see | A | ≤ P | L i | · aex ∗ L ( r − , k − e i ). So by induction on k , | A | ≤ P | L i | · L k − e i (cid:0) r − k − e i (cid:1) = O ( r k k k− ) < aex ∗ L ( r − , k ) for r large enough, but this contradicts (3). Case 2: All vectors v ∈ A satisfy v ( t ) = ℓ ∈ L i . Deleting row t from A then does notaffect the a -rank, so in fact | A | ≤ aex L ( r − , k − e i ) ≤ O ( r k k k− ) by induction, again acontradiction. Case 3: There is a vector v ∈ A with v ( t ) = 0 . In this case, two applications of Lemma3.3, using rows t and 1 in turn, show a -rank( M ) ≥ a -rank( u | v | A ′ j ) = a -rank( v | A ′ j ) + 1 = a -rank( A ′ j ) + 2 = r + 1 , a contradiction. (cid:3) Remark 3.6.
We can obtain an explicit bound on Q k as follows. Claim (3.5) was sufficientto prove the theorem since { α k r } is bounded below by 0. But in fact, by recalling rank( M ) ≤ a -rank( M ) and applying Corollary 2.2, α k Q ′ k ≤ aex ∗ L ( Q ′ k , k ) ≤ ex L ( Q ′ k + 1 , k ) ≤ X k ′ (cid:22) k (cid:18) Q ′ k + 1 k ′ (cid:19) L k ′ ≤ (cid:18) Q ′ k + 1 k (cid:19) L k X k ′ (cid:22) k (cid:18) k k ′ k r (cid:19) k k − k ′ k ≤ (cid:18) Q ′ k + 1 k (cid:19) L k k k k X i =0 (cid:18) k k || i (cid:19) (cid:18) k k k r (cid:19) i = (cid:18) Q ′ k + 1 k (cid:19) L k (cid:18) k k k r (cid:19) k k k ≤ O (cid:0) ( Q ′ k | L | ) k k k (cid:1) , e.g. taking | L | := max i {| L i |} . So the decreasing sequence { α k r } stabilizes after ≤ O (cid:0) ( Q ′ k | L | ) k k k (cid:1) additional steps. Thus we can take Q k := Q ′ k + O (cid:0) ( Q ′ k | L | ) k k k (cid:1) in the theorem. This way, Q k is bounded by ( | L | + 1) O ( k k k ) as k k k → ∞ .We are now ready to prove Theorem 1.6, and make the transition from affine rank to usualrank. Proof of Theorem 1.6.
Note that if every L i = { ℓ i } and P ℓ i k i = 0, the a -rank and rankcoincide, and we are immediately done by Theorem 3.4.So next suppose some | L i | >
1. If we take the L k (cid:0) r k (cid:1) ( L , k )-vectors with some fixed supportof size r , then the rank is exactly r . This gives the lower bound.Now consider any collection of ( L , k )-vectors of rank at most r . By (1), the a -rank is ≤ r + 1 , and Theorem 3.4 shows the size is at most L k (cid:0) r k (cid:1) , along with the uniqueness of the equalitycase.Lastly we consider the case where ∀ iL i = { ℓ i } but P ℓ i k i = 0. XTREMAL COLLECTIONS OF k -UNIFORM VECTORS 9 For the lower bound, if we now take the L k (cid:0) r +1 k (cid:1) ( L , k )-vectors with some fixed support ofsize r + 1, then the rank is at most r , since they all lie in the subspace x · (1 , . . . ,
1) = 0.For the upper bound, any collection of ( L , k )-vectors of rank at most r has a -rank at most r + 1, and we finish by Theorem 3.4 again, this time concluding they number ≤ L k (cid:0) r +1 k (cid:1) . (cid:3) We may generalise Theorem 1.6 to an arbitrary set S ⊂ Z s ≥ of possible weight vectors asfollows. Recall that a vector is an ( L , S )-vector whenever it is an ( L , k )-vector for some k ∈ S , and that ex F , L ( r, S ) is the maximum size of a collection of ( L , S )-vectors whose rankis ≤ r . We say the pair ( L , k ) is shifted if every | L i | = 1, say L i = { ℓ i } , and P ℓ i k i = 0 in F . Thus, every ( L , k )-vector automatically lies in the hyperplane P x j = 0. Corollary 3.7.
Suppose that either | L i | = 1 for every i , or that s = 1 .Then there is an R S such that for all r ≥ R S , ex F , L ( r, S ) = P k ∈ S (cid:0) r +1 k (cid:1) ∀ k ∈ S ( L , k ) is shifted , P k ∈ S L k (cid:0) r k (cid:1) ∀ k ∈ S, ( L , k ) is not shifted.Moreover, any extremal matrix M has only r + 1 or r nonzero rows respectively.Proof. The equalities follow directly from Theorem 1.6, simply by decomposing a givencollection of ( L , S )-vectors into respective ( L , k )-vectors.Now, let M be any extremal matrix, and for each k ∈ S , write A k for its submatrix of( L , k )-vector columns. By the equality case of Theorem 1.6, each A k has only r + 1 or r nonzero rows (in the respective cases). These supports are then identical, for otherwiserank( M ) ≥ r + 1 is witnessed by adding any ( L , k ′ )-vector v with a nonzero entry outside ofthe support of A k to A k . (cid:3) One may hope to combine the shifted and not-shifted vectors together in S , but havingexactly r rows appears not to always be optimal in this case. Indeed, consider vectors ofweight 1 or 4 over F (so L = ( { } ), S = { , } ). Then one may vainly hope all (cid:0) r (cid:1) + (cid:0) r (cid:1) possible ( L , S )-vectors in a matrix with exactly r rows is optimal. But this can be improvedto (cid:0) r +14 (cid:1) columns in a matrix with r + 1 rows, by restricting to just the vectors of weight4: without any weight-1 vectors, all columns will lie in the hyperplane P x j = 0, reducingtheir rank from r + 1 to r . 4. k Zeros Proofs
We now proceed with the proof of Theorem 1.4. First, we establish a standard countingfunction:
Lemma 4.1.
For any sequence u , u , . . . of nonzero elements of F q , and any number n , thenumber of vectors x ∈ ( F × q ) n orthogonal to ( u , . . . , u n ) is a (0) n = q (cid:0) ( q − n + ( − n ( q − (cid:1) . Proof.
More generally, let a ( β ) n := | S βn | , where S βn := { x ∈ ( F × q ) n : x u + · · · + x n u n = β } .Since x βx is a bijection S n → S βn for every β ∈ F × q , it follows a ( β ) n = a (1) n . Furthermore, | S αn +1 | = P β = α | S βn | for any α ∈ F q , since any vector in F β = α S βn can be uniquely extendedto a vector in S αn , since v n = 0 . Thus, we have the recursive relations for each n ≥ a (0) n +1 = ( q − a (1) n ,a (1) n +1 = ( q − a (1) n + a (0) n . Since a ( β )0 = β =0 , the results a (0) n = q (cid:0) ( q − n +( − n ( q − (cid:1) and a (1) n = q (cid:0) ( q − n +( − n +1 (cid:1) follow by a trivial induction (or may be derived directly using generating functions). (cid:3) For a fixed r , we use X as shorthand for F rq . Furthermore, for each n ≤ r , we denote by X ≥ n and X = n , the sets of vectors of weight ≥ n and exactly n respectively. Immediatelynote that | X = n | = (cid:0) rn (cid:1) ( q − n for every n . Lemma 4.2.
Suppose v ∈ X ≥ has i ≥ non-zero entries, and W is its orthogonal comple-ment W := v ⊥ = { x ∈ X : x · v = 0 } . Then | X = r − k ∩ W | = 1 q (cid:18) rk (cid:19) ( q − r − k + ( − i ( q − r − i − k +1 k X s =0 (1 − q ) s (cid:18) is (cid:19)(cid:18) r − ik − s (cid:19)! ≥ q (cid:18) rk (cid:19) ( q − r − k (cid:18) − q − (cid:19) for r sufficiently large.Proof. WLOG, v = ( v , . . . , v i , , . . . ,
0) where v , . . . , v i are all non-zero.For each S ∈ (cid:0) [ r ] k (cid:1) , let W S := { x ∈ X = r − k ∩ W : { j ∈ [ r ] : x j = 0 } = S } , so we maydecompose W as S ks =0 S | S ∩ [ i ] | = s W S .We claim that, if | S ∩ [ i ] | = s , then | W S | = q (cid:0) ( q − r − k + ( − i + s ( q − r − i − k + s +1 (cid:1) . Sincethere are (cid:0) is (cid:1)(cid:0) r − ik − s (cid:1) such S ∈ (cid:0) [ r ] k (cid:1) with | S ∩ [ i ] | = s , the above decomposition gives the result.To see the claim, note x ∈ W S ⇔ x j = 0 ∀ j ∈ Sx j = 0 ∀ j ∈ [ r ] \ S P j ∈ [ r ] \ S x j v j = 0 ⇔ x j = 0 ∀ j ∈ Sx j = 0 ∀ j ∈ [ r ] \ ( S ∪ [ i ]) x j = 0 ∀ j ∈ [ i ] \ S ∧ P j ∈ [ i ] \ S x j v j = 0 . Applying the lemma to ( u , . . . , u n ) := proj [ i ] \ S ( v ) and noting n = i − s , we see there are q (cid:0) ( q − i − s + ( − i − s ( q − (cid:1) ways to choose the entries of x in S ∩ [ i ]. Furthermore, thereare ( q − | [ r ] \ ( S ∪ [ i ]) | = ( q − r − k − i + s ways to choose the entries of x in [ r ] \ ( S ∪ [ i ]) , and sothere are q (cid:0) ( q − r − k + ( − i + s ( q − r − i − k + s +1 (cid:1) such x in total. XTREMAL COLLECTIONS OF k -UNIFORM VECTORS 11 We now proceed to prove the claimed inequality. Let a s := (cid:0) is (cid:1)(cid:0) r − ik − s (cid:1) ( q − s , for each0 ≤ s ≤ k .Note that, for s < i, a s a s +1 = ( s +1)( r − k − i + s +1)( i − s )( k − s )( q − is an increasing function in s (and for s > ia s = 0 anyway). Hence the sequence { a s } is unimodal , i.e. consists of a (possibly empty)monotonically increasing subsequence followed by a decreasing subsequence. In particular,the alternating sum P ks =0 ( − s a s is bounded above by max s { a s } . Let us fix the s attainingthis maximum.Now, (cid:0) rk (cid:1) ( q − i − a s = ( q − i − − s (cid:0) rk (cid:1)(cid:0) is (cid:1)(cid:0) r − ik − s (cid:1) ≥ q − i ≥ s + 2, since the denominator is a single term in the identity P s ′ (cid:0) is ′ (cid:1)(cid:0) r − ik − s ′ (cid:1) = (cid:0) rk (cid:1) .Else, i ∈ { s, s + 1 } . We check the lower bound still holds here:When i = s + 1, the above is ( rk ) i ( r − ik − i +1 ) ≥ ( rk ) i ( r − ik − i ) = ( ri ) i ( ki ) ≥ ( r − i ( k − (using i ≥ r ≥ k ).Similarly, if i = s , the above is ( rk ) ( q − ( r − ik − i ) = ( ri ) ( q − ( ki ) ≥ ( r − ( q − k − .So these are both still ≥ q −
1, assuming r ≥ max { q / k / , qk } . In summary,1 q ( q − r − i − k +1 (cid:18) rk (cid:19) ( q − i − + k X s ′ =0 ( − s ′ + i a s ′ ! ≥ q ( q − r − i − k +1 (cid:18)(cid:18) rk (cid:19) ( q − i − − a s (cid:19) ≥ q (cid:18) rk (cid:19) ( q − r − k (cid:18) − q − (cid:19) . (cid:3) Our remaining tool is a standard observation in abstract linear algebra. For a vector space V over F q , recall that the dual space V ∗ consists of all linear functions V → F q , and has thesame dimension as V (when finite). Lemma 4.3.
Suppose f , . . . , f r ∈ V ∗ and T rj =1 V j = { } . Then f , . . . , f r are linearlyindependent. We are now in a position to prove the nonzero case of Theorem 1.4. This time, the extremalmatrices cannot have any duplicate rows, nor scalings thereof.
Theorem 4.4.
Let k ≥ . Then ex q ( r, k ) = (cid:0) rk (cid:1) · ( q − r − k , provided r ≥ max { q k, q / k / } .Furthermore, the unique extremal example is a matrix M consisting of only r rows and allpossible columns.Proof. We may assume rank( M ) = r , and that all rows are distinct. Let Y denote the setof columns of M , with span h Y i = V . If r ′ denotes the number of rows of M , then V ≤ F r ′ q is a subspace of dimension r . For each j ∈ [ r ′ ], we have V ′ j := { y ∈ F r ′ q : y j = 0 } is codimension-1 in F r ′ q , and hence V j := V ′ j ∩ V is codimension- ≤ V . Whenever dim( V j ) = r −
1, we say that row j is nontrivial , and observe V j = { y ∈ V : f j ( y ) = 0 } for some f j ∈ V ∗ , the dual space of V (namely, linear functions V → F q ). Otherwise, V j = V , and we say row j is trivial . (In fact,every trivial row of M is necessarily all zeros, but we will not need this for the argument.)In fact, since the row rank of M is r , we see there are some r rows I which are linearlyindependent. Thus, the projection onto just these coordinates is a linear isomorphism Π I : V → F Iq ( ≃ F rq ). In this way, for every nontrivial j , Π I ( V j ) ≤ F Iq is also a codimension-1subspace.We have by assumption that every y ∈ Y is in exactly k of the { V j } counting multiplicities,and hence in κ of the { V j : j nontrivial } , where κ := k − |{ j trivial }| .As such, for every y ∈ Y , Π I ( y ) is in exactly k of the mapped subspaces F := { Π I ( V j ) : j nontrivial } (viewed as a multiset). In particular, F contains every coordinate subspace e ⊥ ℓ at least once (as e ⊥ ℓ = Π I ( V ℓ ) for each ℓ ∈ I ). Every Π I ( y ) is in ≤ κ of these coordinatesubspaces, and hence has ≤ κ zeros. Deduce Π I ( Y ) ⊂ X ≥ r − κ , so we immediately obtain | Y | = | Π I ( Y ) | ≤ (cid:0) rκ (cid:1) ( q − r − κ + (cid:0) rκ − (cid:1) ( q − r − κ +1 + · · · + ( q − r . Of course, that wassomething we already established in Corollary 2.2, but we will need this setup to help removethe trailing terms.Suppose first that there is some W ∈ F which is not a coordinate hyperplane. We will showthat | Y | is too small in this case. Now, by dimension counting, W ⊥ = h v i for some v ∈ X .Plus, as W is not a coordinate hyperplane, v has ≥ r − κ in W .In fact, Π I ( Y ) ⊂ ( X = r − κ \ W ) ∪ ( X ≥ r − κ +1 ), since all vectors in X ≤ r − κ − ∪ ( X = r − κ ∩ W ) are in ≥ κ + 1 spaces in F . Also note that | X ≥ r − κ +1 | ≤ | X = r − κ +1 | = 2 (cid:0) rκ − (cid:1) ( q − r − κ +1 provided r ≥ qκ . Putting these together with Lemma 4.2, | Y | = | Π I ( Y ) | ≤ | X = r − κ | − | X = r − κ ∩ W | + | X ≥ r − κ +1 |≤ (cid:18) rκ (cid:19) ( q − r − κ − q (cid:18) rκ (cid:19) ( q − r − κ (cid:18) − q − (cid:19) + 2 (cid:18) rκ − (cid:19) ( q − r − κ +1 < (cid:18) rκ (cid:19) ( q − r − κ , if r ≥ q κ. As such, we may assume every Π I ( V j ) ∈ F is some coordinate hyperplane e ⊥ ℓ . However, westill are not yet sure that the original subspaces { V j } were distinct (in the way that the { V ′ j } are): there may be collisions upon intersection with V .For each x ∈ F rq , we denote by Z x its zero-set { ℓ : x ℓ = 0 } . Also, letting w ( x ) := |{ W ∈F : x ∈ W }| (counting multiplicities), we see that every x ∈ Π I ( Y ) has w ( x ) = κ . Forma poset structure on X = F rq by x ≺ y ⇔ Z x ) Z y : thus, ( X, (cid:22) ) looks like a blowup ofthe Boolean lattice where each vector of weight n has been blown up ( q − n times. Also,since F contains each coordinate subspace at least once, w is a strictly increasing functionon ( X, (cid:22) ), and hence Π I ( Y ) forms an antichain. XTREMAL COLLECTIONS OF k -UNIFORM VECTORS 13 This satisfies a LYM-type inequality (see e.g. [3] for an exposition we will mimic here): forany arbitrary A ⊂ X , write A = i := A ∩ X = i for each i ≤ r . Then a random maximal chain C in X satisfies E [ | C ∩ A | ] = P i ≤ r | A = i || X = i | by symmetry. For the antichain A := Π I ( Y ), deducethis is ≤
1. Furthermore, with r ≥ qk ≥ qκ , we have | X = r − κ | > | X = r − κ +1 | > · · · > | X = r | ,and hence 1 ≥ X i ≤ r | A = i || X = i | = X r − κ ≤ i ≤ r | A = i || X = i | ≥ X r − κ ≤ i ≤ r | A = i || X = r − κ | = | A || X = r − κ | , so | Y | = | A | ≤ | X = r − κ | = (cid:0) rκ (cid:1) ( q − r − κ ≤ (cid:0) rk (cid:1) ( q − r − k is immediate. In the equality case, k = κ , and all rows were nontrivial. Also, every | A = i || X = i | = | A = i || X = r − k | for i > r − k , hence they areall 0, from which it follows Π I ( Y ) = A = X = r − k .Deduce F = { e ⊥ , . . . , e ⊥ r } with no repeated subspaces, so r ′ = r and M only had r rowsoriginally. (cid:3) Concluding Remarks and Further Questions
In light of Theorems 1.2 and 1.4, one may naively hope that ex q ( r, k ) = ex q ( r, r − k ) for somereasonable values of r, k and q , but this is very far from being true, so there is a limit to howsmall we can make R k,q and ¯ R k,q . Indeed, ex q ( r, k ) is an increasing function of k (for fixed r and q ), since adding a row of zeros does not increase the rank of a matrix. Plus, whileadding rows of all 1’s might increase the rank, it does not increase the a -rank, so aex L ( r, k )is also an increasing function of k .Even more strikingly, ex q ( r, k ) = ex q ( r, k ) = 0 for negative k , whereas ex q ( r, k ) , ex q ( r, k ) canbe defined for k > r and are clearly positive: in fact, ex q ( r, ( q − q r − ) = ex q ( r, q r − ) = q r − any k , since an F q -vector space of dimension r only has q r distinct elements in total, including 0.The matrix M attaining the above is simply the dual Hamming code [8] of length q r , asnoted in the concluding section of Ahlswede, Aydinian and Khachatrian [1] (and in fact, wasshown to be essentially the unique such matrix up to repetition by Bonisoli [5]). Explicitly,we list all q r vectors F rq = { v , . . . , v q r } as the rows of a matrix A , then let the columns of M consist of all nonzero vectors in the column space of A = ( u | . . . | u r ), so rank( M ) = r .Now, the i -th entry of a column P l j u j of M is zero if and only if v i is in the hyperplane { P j l j x j = 0 } . This is true for exactly q r − such vectors v i ∈ F rq , and hence every columnof M has weight ( q − q r − .So, we know these theorems cannot be extended arbitrarily. But we can still ask about thethreshold functions: Question 5.1.
How small can R k,q and ¯ R k,q be made in Theorems 1.2 and 1.4? Theorem 1.4 was established directly, obtaining the result for ¯ R k,q = O q ( k / ). In sharpcontrast, the proof of Theorem 1.2 used an induction for which we were unable to directlyestablish a base case, and is only known for R k,q = 2 O q ( k ) . Now, ¯ R k,q can’t be made independent of q . Once r < qk , we note that (cid:0) rk (cid:1) ( q − r − k < (cid:0) rk − (cid:1) ( q − r − k +1 , so (for example) our usual example is beaten by a matrix consisting of allco-weight k − r , and then appending a row of all 0’s. Perhaps it is stilltrue that ex q ( r, k ) = (cid:0) rk ′ (cid:1) ( q − r − k ′ for some k ′ < k , using matrices with lots of empty rows.Similar logic shows that R k,q can’t be made smaller than qq − k . Yet, it is still plausible thate.g. ex (2 k, k ) = (cid:0) kk (cid:1) for every odd k , and even that R k,q can be made independent of q .Furthermore, we wonder whether Theorem 1.4 can be generalized in a similar fashion toTheorem 1.6. This leads to questions that lie strictly between the original two, the simplestinstance of which is the following: Question 5.2.
Does every rank- r matrix over F have ≤ (cid:0) rk (cid:1) · r columns with exactly k entries either 0 or 1 (for all r sufficiently large)? Acknowledgements
We would like to thank Boris Bukh for helpful discussions and for suggestions significantlyimproving a previous draft of the paper, and Imre Leader for informing us of the relevance of[1]. We are also indebted to an anonymous referee for their very careful reading and findingof a number of mistakes in an earlier version of the paper.
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