welsh’s problem on the number of bases of matroids
TRANSCRIPT
Welsh’s problem on the number of bases of matroids
Edward S. T. Fan∗1 and Tony W. H. Wong†2
1Department of Mathematics, California Institute of Technology2Department of Mathematics, Kutztown University of Pennsylvania
February 24, 2014
Abstract
In this paper, we study a problem raised by Dominic Welsh on the existence ofmatroids with prescribed size, rank, and number of bases. In case of corank being atmost 3, by an explicit construction of matrices, we are able to show the existence ofsuch matroids, except the counterexample found by Mayhew and Royle. This idea alsoextends to give asymptotic results on the existence of matroids with any fixed corank.
1 Background
Let (n, r, b) be a triple of integers such that 0 < r ≤ n and 1 ≤ b ≤(nr
). Given a triple (n, r, b),
Welsh [?] asked if there exists a matroid with n elements, rank r, and exactly b bases. Sucha matroid would be called an (n, r, b)-matroid. It was conjectured that an (n, r, b)-matroidexists for every such triple, until Mayhew and Royle [?] found the lone counterexample todate, namely (n, r, b) = (6, 3, 11). However, they proposed the following conjecture.
Conjecture 1.1. An (n, r, b)-matroid exists for all 0 < r ≤ n and 1 ≤ b ≤(nr
)except
(n, r, b) = (6, 3, 11).
In this paper, we verify this conjecture for a large family of triples (n, r, b).
2 Introduction
A matroid M = (X, I) is a combinatorial structure defined on a finite ground set X of nelements, together with a family I of subsets of X called independent sets, satisfying thefollowing three properties:
∗[email protected]†[email protected]
1
1. ∅ ∈ I, or I is nonempty.
2. If I ∈ I and J ⊆ I, then J ∈ I. This is sometimes known as hereditary property.
3. If I, J ∈ I and |J | < |I|, then there exists x ∈ I\J such that J ∪ {x} ∈ I. This isknown as augmentation property or exchange property.
A maximal independent set is called a basis of the matroid. A direct consequence of theaugmentation property is that all bases have the same cardinality. This cardinality is definedas the rank ofM, and the difference between the size of X and the rank ofM is defined asthe corank of M.
One of the most important examples of matroids is a linear matroid. A linear matroidis defined from a matrix A over a field F, where X is the set of columns of A, and anindependent set I ∈ I is a collection of columns which is linearly independent over F. For alinear matroid, the rank is exactly the rank of the matrix A, and the corank is the differencebetween the number of columns and the rank of A. Since the elements of a linear matroidare the columns of a matrix, a linear matroid is also called a column matroid.
In this paper, we are going to show that a linear (n, r, b)-matroid exists for a large familyof triples (n, r, b). Throughout this paper, A is an r × n matrix over Q with rank r andcorank k = n − r, and the number of r × r invertible submatrices of A is b. Note that bis an invariant if we perform row operations on A or permute the columns of A, so we canalways assume that A = (Ir|M), where M is an r × k matrix. We start with the followingobservation that there is a natural duality between the rank and corank of a linear matroid.
Proposition 2.1. The number of invertible square submatrices of M is b, and the numberof invertible k× k submatrices of (Ik|M>) is also b. (Here, we adopt the convention that anempty matrix is invertible.)
Proof. Let S be the set of all invertible r × r submatrices of A, where r ≥ 0, and let T bethe set of all invertible square submatrices of M .
Let S be a matrix in S with columns i1, i2, . . . , ir, where i1, . . . , ij ≤ r and ij+1, . . . , ir > r.Then there is a bijection between S and T which sends S to the square submatrix of Mwith rows {1, 2, . . . , r}\{i1, i2, . . . , ij} and columns ij+1 − r, ij+2 − r, . . . , ir − r. Hence, thenumber of invertible square submatrices of M is b.
It is then obvious that the number of invertible square submatrices of M> is b, and theabove bijective argument in turn implies that the number of invertible k × k submatrices of(Ik|M>) is b.
In view of this proposition, to determine the existence of a linear (n, r, b)-matroid, wewould try to construct an r × k matrix M with exactly b invertible square submatrices,where k = n − r. Such a matrix M will be called an (r, k, b)-matrix. Note that we onlyneed to consider the case when k ≤ r for our purpose of studying this Welsh’s problem,since if k > r, we can instead consider the existence of an (k, r, b)-matrix. Also note thatthe existence of a linear (n, r, b)-matroid is equivalent to the existence of an (r, k, b)-matrix.Next, we observe that the naive “extension by zero” construction exhibits a useful relationbetween the existences of various (r, k, b)-matrices.
2
Lemma 2.2. If there exists an (r0, k0, b)-matrix, then for all r ≥ r0, k ≥ k0, (r, k, b)-matrixexists.
Proof. If M0 is an (r0, k0, b)-matrix, by building an r × k matrix M such that M0 is asubmatrix and all the extra entries are 0’s, M is an (r, k, b)-matrix.
Combining the results from Sections ?? and ??, we show that Conjecture ?? holds forall 1 ≤ b ≤
(r+3r
), except that (n, r, b) = (6, 3, 11). In Section ??, we prove Conjecture ??
for all large r and large b. We strengthen this result in Section ?? such that Conjecture?? holds for all large enough r, regardless the values of n and b. This gives the closest tocomplete answer to the Welsh’s problem and provides a plausible reason of the existence ofcounterexample for small values of r.
3 Existence of linear matroids with corank at most 2
In this section, we will study the existence of (r, k, b)-matrices for k ≤ 2. The cases k = 0and k = 1 are very straight forward. For the first nontrivial case k = 2, the following lemmain number theory will help us to show the existence of matrices satisfying the parameters(r, 2, b).
Lemma 3.1. Let s ≥ 5 be a positive integer, and let k be a nonnegative integer such thatk ≤ s2−5s
4. Then there exist nonnegative integers a1, a2, . . . , as such that a1 +a2 + · · ·+as = s
and a21 + a22 + · · ·+ a2s = s+ 2k.
Proof. For 5 ≤ s ≤ 32, we verified the lemma by Mathematica; for s ≥ 33, we will use stronginduction on s.
Suppose the statement is true for all integers u such that 5 ≤ u < s for some s ≥ 33, i.e.,for all nonnegative integers k′ ≤ u2−5u
4, there exist nonnegative integers a1, . . . , au such that
a1 + · · ·+ au = u and a21 + · · ·+ a2u = u+ 2k′.
Let t and k be integers such that 0 < t ≤ s − 5 and 0 ≤ k − t2−t2≤ (s−t)2−5(s−t)
4. Then
u := s − t falls in the range 5 ≤ u < s, and k′ := k − t2−t2≤ u2−5u
4. By the induction
hypothesis, there are nonnegative integers a1, . . . , as−t such that a1 + · · ·+ as−t = s− t anda21+· · ·+a2s−t = s−t+2
(k− t2−t
2
)= s+2k−t2. If we set as−t+1 = t and as−t+2 = · · · = as = 0,
then a1 + · · ·+ as = s and a21 + · · ·+ a2s = s+ 2k, implying that the statement holds true for
k satisfying 0 ≤ k − t2−t2≤ (s−t)2−5(s−t)
4, or equivalently, t2−t
2≤ k ≤ 3t2−2st+3t+s2−5s
4.
It now suffices to show that the union of the intervals I(t) :=[t2−t2, 3t
2−2st+3t+s2−5s4
]for
0 < t ≤ s− 5 covers[0, s
2−5s4
]when s ≥ 33. Let α(t) = t2−t
2and β(t) = 3t2−2st+3t+s2−5s
4.
Claim 1. s2−5s4≤ β(t) if and only if t ≥ 2
3s − 1, which is attainable for some t in the range
0 < t ≤ s− 5 if s ≥ 12.
Proof of claim 1. This inequality holds if and only if 3t2 − 2st+ 3t ≥ 0, which is equivalentto t ≥ 2
3s− 1 since t is positive. We finish by noticing that when s ≥ 12, s− 5 ≥ 2
3s− 1.
3
Claim 2. α(t− 1) ≤ α(t) ≤ β(t).
Proof of claim 2. The first inequality holds since α(t) is an increasing function for t ≥ 1,since α′(t) = 2t−1
2> 0 when t ≥ 1, considering α as a continuous function on R. The second
inequality holds if and only if 5(s− t) ≤ (s− t)2, which is always true since t ≤ s− 5.
Claim 3. α(t) ≤ β(t− 1) if and only if t ≤ 2s+1−√16s+1
2.
Proof of claim 3. This inequality holds if and only if t2 − (2s + 1)t + s2 − 3s ≥ 0, which
occurs if and only if t ≤ 2s+1−√16s+1
2or t ≥ 2s+1+
√16s+1
2. However, t ≥ 2s+1+
√16s+1
2is rejected
since t < s.
By claims 2 and 3, if t ≤ 2s+1−√16s+1
2, then I(1) ∪ · · · ∪ I(t − 1) ∪ I(t) forms one closed
interval. If⌈23s− 1
⌉≤ 2s+1−
√16s+1
2, then claim 1 implies that
[0, s
2−5s4
]⊆d 23 s−1e⋃t=1
I(t).
To obtain⌈23s− 1
⌉≤ 2s+1−
√16s+1
2, we look for integers s satisfying 2
3s ≤ 2s+1−
√16s+1
2, or
equivalently, 3√
16s+ 1 ≤ 2s+ 3. This inequality holds if 33s ≤ s2, or s ≥ 33.
Theorem 3.2. If k ≤ 2, then for all integers b such that 1 ≤ b ≤(r+kr
), there exists an
(r, k, b)-matrix M .
Proof. It is trivial for k = 0. If k = 1, then M is a column vector with the first b− 1 entries1’s and the rest 0’s.
If k = 2, let the first column of M have the first s entries 1’s, the second column havethe first s entries nonzero, and the rest be all 0’s. Furthermore, assume that there are ai i’sin the second column, 1 ≤ i ≤ s, where a1 +a2 + · · ·+as = s. Then the number of invertiblesquare submatrices of M is
1 + 2s+∑
i<j aiaj = 1 + 2s+ 12(∑
i ai)2 − 1
2(∑
i a2i )
= 1 + 2s+ 12s2 − 1
2(∑
i a2i ),
and we would like to set it to be b, which gives s+ 2((s+22
)− b) =
∑i a
2i .
By Lemma ??, if 0 ≤(s+22
)− b ≤ s2−5s
4, or equivalently s2+11s+4
4≤ b ≤ s2+3s+2
2, there is
a solution for ai’s. It is easy to check that the intervals [ s2+11s+4
4, s
2+3s+22
] cover all integersb ≥ 39, and the only missing integers are in [1, 20] ∪ [22, 26] ∪ [29, 32] ∪ [37, 38]. Here, wefinish the proof by constructing M explicitly for each of these b’s.
In the following table, 0 represents a column vector of all 0’s (possibly of length 0), whichfills up the column so that M has r rows.
b = 1 2 3 4 5 6 7 8
M = 0 01 00 0
1 10 0
1 00 10 0
1 11 00 0
1 11 20 0
1 11 01 00 0
1 11 11 00 0
4
b = 9 10 11 12 13 14 15 16
M =
1 11 21 00 0
1 11 21 30 0
1 11 11 01 00 0
1 11 21 01 00 0
1 11 11 21 00 0
1 11 21 31 00 0
1 11 21 31 40 0
1 11 11 01 00 10 0
b = 17 18 19 20 22 23 24 25
M =
1 11 11 21 01 00 0
1 11 21 31 01 00 0
1 11 11 21 31 00 0
1 11 21 31 41 00 0
1 11 11 11 21 01 00 0
1 11 11 21 21 01 00 0
1 11 11 21 31 01 00 0
1 11 21 31 41 01 00 0
b = 26 29 30 31 32 37 38
M =
1 11 11 21 31 41 00 0
1 11 11 21 31 01 01 00 0
1 11 21 31 41 01 01 00 0
1 11 11 21 21 31 01 00 0
1 11 11 21 31 41 01 00 0
1 11 11 21 21 31 01 01 00 0
1 11 11 21 31 41 01 01 00 0
Theorem 3.3. Conjecture ?? holds for all 1 ≤ b ≤(r+2r
).
Proof. By Theorem ?? and Lemma ??, for all 1 ≤ b ≤(r+2r
), for all r and k such that(
r+kr
)≥ b, (r, k, b)-matrices and hence linear (n, r, b)-matroids exist.
4 Existence of linear matroids with ample amount of
bases
In Section ??, we used an induction argument together with computational exhaustion toshow the existence of linear (n, r, b)-matroids with 1 ≤ b ≤
(r+2r
). In this section, we will
establish an asymptotic existence result by studying a special class of matrices obtained froma generic hyperplane arrangement. A generic hyperplane arrangement is a finite collectionH = {Γ1,Γ2, . . . ,Γ`} of (k − 1)-dimensional linear subspaces of Qk satisfying
5
1. The hyperplanes Γj’s do not contain any coordinate directions, i.e. ei /∈ Γj for alli = 1, 2, . . . , k and j = 1, 2, . . . , `, where ei = (0, . . . , 1, . . . , 0) is the i-th coordinatevector.
2. No hyperplanes are of the form {(x1, x2, . . . , xk) : xi = xj} for some i 6= j.
3. For all j = 1, 2, . . . ,min(k, `), the intersection of any j hyperplanes from H is a (k−j)-dimensional subspace.
Given a generic hyperplane arrangement H together with integers n0, n1, . . . , n` ≥ k, aset of generic (n0, n1, . . . , n`)-vectors from H is a collection of vectors, where nj of themcome from Γj and n0 of them from the complement of the union of the hyperplanes, suchthat
1. No vector lies in more than one hyperplane.
2. Any k − 1 vectors are linearly independent.
3. Any k vectors are linearly dependent only when they are all chosen from the samehyperplane.
It is well known that such generic hyperplane arrangements and sets of generic vectorsof any size always exist. The former can be shown by choosing normal vectors in generalpositions and the latter follows easily from an inductive construction.
Now, given nonnegative integers ak, ak+1, . . . , ar withr∑s=k
ass ≤ r, we first pick a generic
hyperplane arrangement H of size ` =r∑s=k
as. Then we choose a set of generic (n0, n1, . . . , n`)-
vectors from H, where n0 = r−r∑s=k
ass, and among n1, n2, . . . , n`, there are exactly as copies
of s for each s = k, k + 1, . . . , r. If we put these vectors together as row vectors of a matrix,we get an r × k matrix M .
This construction builds an (r, k, b)-matrix M with b =(nr
)− b, where b =
r∑s=k
as(sk
)denotes the number of singular r × r submatrices in A, or the number of singular squaresubmatrices in M . In the following proposition, we determine the range of b that is achievableby our construction.
Proposition 4.1. For each k ≥ 3, there exists r0 ∈ N such that for all integers r ≥ r0, andall 0 ≤ b ≤
(r+k−1k−1
), there exist nonnegative integers ak, ak+1, . . . , ar such that
(i)r∑s=k
as(sk
)= b, and
(ii)r∑s=k
ass ≤ r.
6
Proof. Let κ be a real number greater than (k1k + 1)k. Then κ > (k
1k )k + k · k 1
k > 2k.Let b0 = b. For each integer i ≥ 0, if bi ≥
(κk
), choose si ∈ N such that
(sik
)≤ bi <
(si+1k
).
Note that 1 ≤ bi/(sik
)<(si+1k
)/(sik
)which is less than 2 since
(si+1k
)≥ 2
(sik
)if and only if
si ≤ 2k− 1. Let asi = 1, and bi+1 = bi−(sik
). If bi <
(κk
)for some i ≥ 0, then let ak = bi and
all other undetermined as be 0.
From this definition, condition (i) is clearly satisfied. For condition (ii),r∑s=k
ass = s0 +
s1 + · · · + sm + bm+1 for some m ≥ 0. To give an upper bound to this sum, let fk(x) =(xk
)and fk−1(x) =
(xk−1
). Note that they are both strictly increasing functions when x ≥ k, so
f−1k fk−1 is also a strictly increasing function.For all i ≥ 0, bi+1 = bi −
(sik
)<(si+1k
)−(sik
)=(sik−1
). Hence, s0 ≤ (f−1k fk−1)(r + k − 1)
and si+1 ≤ f−1k (bi+1) < (f−1k fk−1)(si). When x ≥ κ > 2k,
f−1k fk−1(x) < f−1k(xk−1
(k−1)!
)= f−1k
((k1/kx(k−1)/k)
k
k!
)< k
1kx
k−1k + k − 1,
which is less than (k1k +1)x
k−1k when x > (k−1)
kk−1 . Since 2k−1 = (1+1)k−1 > 1+k−1 > k−1,
we have (k − 1)1
k−1 < 2, implying 2k > (k − 1)k
k−1 . Therefore,
f−1k fk−1(x) <(k
1k + 1
)x
k−1k
for all x ≥ 2k. In particular,
s0 <(k
1k + 1
)(r + k − 1)
k−1k , (1)
and for i ≥ 1,
si ≤ (f−1k fk−1)i+1(r + k − 1) < (k
1k + 1)1+
k−1k
+( k−1k
)2+···+( k−1k
)i(r + k − 1)(k−1k
)i+1
≤ (k1k + 1)k(r + k − 1)(
k−1k
)i+1
.
As (f−1k fk−1)(sm) < κ, we would like to find the minimum integer m′ such that
(k1k + 1)k(r + k − 1)(
k−1k
)m′+2 ≤ κ.
This is equivalent to
(m′ + 2) log(k−1k
)≤ log log
(κ
(k1/k+1)k
)− log log(r + k − 1),
orm′ ≥
(log log(r + k − 1)− log log
(κ
(k1/k+1)k
))/log(
kk−1
)− 2. (2)
Combining (??) and (??), we have s0 + s1 + · · · sm + bm+1 ≤ ms0 +(κk
)≤ m′s0 +
(κk
)=
O(rk−1k log log r) for fixed k, which is less than r when r is large.
This gives the following immediate asymptotic result on the existence of linear (n, r, b)-matroids.
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Theorem 4.2. For each fixed k ≥ 3, let r0 ∈ N be as defined in Proposition ??. Then forall integers r ≥ r0, for all integers b such that b ≥
(r0+k−1
k
), linear (n, r, b)-matroids always
exist, where n = r + k.
Proof. By Proposition ??, for every integer r ≥ r0, for all integers b such that(r+k−1k
)=(
r+kk
)−(r+k−1k−1
)≤ b ≤
(r+kk
), we can construct an (r, k, b)-matrix following the procedures
introduced at the beginning of Section ??. Finally, we are done by noticing that the intervals[(r0+k−1+i
k
),(r0+k+i
k
)], i = 0, 1, . . . , r − r0, cover all integers
(r0+k−1
k
)≤ b ≤
(nr
).
In fact, we can remove the lower bound on the values of b in Theorem ??.
Theorem 4.3. For each fixed k ≥ 3, there exists R ∈ N such that for all r ≥ R, linear(n, r, b)-matroids always exist, where n = r + k.
The proof of this theorem will be shown at the end of Section ??.
5 Existence of linear matroids with corank at most 3
In this section, we will show that when k = 3, there always exists a linear (r+3, r, b)-matroidexcept (n, r, b) = (6, 3, 11). For 4 ≤ r ≤ 10, the existence of an (r, 3, b)-matrix will be verifiedthrough explicit constructions, which is described in Appendix ??. Then we will base on themethod employed in Proposition ?? and modify it into an inductive argument.
Proposition 5.1. For each r ≥ 49 and for all 0 ≤ b ≤(r+22
), there exist nonnegative integers
a3, a4, . . . , ar such that
(i)r∑s=3
as(s3
)= b, and
(ii)r∑s=3
ass ≤ r.
Proof. Let the statement of the proposition be denoted by P(r) for r ≥ 49. In Appendix??, we show that P(r) holds for 49 ≤ r ≤ 203. Suppose that P(r) is true for all 49 ≤ r ≤ r0for some r0 ≥ 203. We would like to check the validity of P(r0 + 1).
Let b be an integer such that 0 ≤ b ≤(r0+32
). If b ≤
(r0+22
), we can apply P(r0) to b to
obtain nonnegative integers a3, a4, . . . , ar0 and ar0+1 = 0 such that
(i)r0+1∑s=3
as(s3
)= b, and
(ii)r0+1∑s=3
ass ≤ r0 ≤ r0 + 1,
so it suffices to consider(r0+22
)< b ≤
(r0+32
). Note that there exists a unique integer s0 such
that(s03
)≤ b <
(s0+13
).
Set b′:= b−
(s03
). If b
′= 0, we are done. If b
′ ≥ 1, we have 1 ≤ b′<(s0+13
)−(s03
)=(s02
).
Clearly, s0 < r0 + 2, so we can apply the induction hypothesis P(s0 − 2) to b′
as long ass0 ≥ 51. From this, we can find nonnegative integers a3, a4, . . . , as0−2 with
8
(i)s0−2∑s=3
as(s3
)= b
′, and
(ii)s0−2∑s=3
ass ≤ s0 − 2.
By setting as0−1 = 0, as0 = 1, and as0+1 = · · · = ar0+1 = 0, we have
(i)r0+1∑s=3
as(s3
)= b, and
(ii)r0+1∑s=3
ass ≤ 2s0 − 2.
It suffices to show that s0 ≥ 51 and 2s0 − 2 ≤ r0 + 1. First, observe that for r0 ≥ 203, wehave (
s0+13
)> b >
(r0+22
)≥(2052
)= 20910,
or equivalently, (s0 + 1)s0(s0 − 1) > 125460. A straight forward calculation shows that itholds if and only if s0 ≥ 51.
Next, for any integer x ≥ 1,(x+323
)−(x+32
)= (x+3)(x+1)(x−1)
48− (x+3)(x+2)
2
= 148
(x+ 3)(x2 − 24x− 49) ≥ 0
if and only if x2 − 24x − 49 ≥ 0. By solving the quadratic inequality, it is easy to see thatit holds for all integers x ≥ 26. Since r0 ≥ 203 and
(x3
)is a strictly increasing function for
x ≥ 3, we have (s03
)≤ b ≤
(r0 + 3
2
)≤( r0+3
2
3
),
which implies r0+32≥ s0 or 2s0 − 2 ≤ r0 + 1.
To complete the case for k = 3, we still need to consider r = 3 and 11 ≤ r ≤ 48.When r = 3, Theorem ?? implies that linear (6, 3, b)-matroid exists for 1 ≤ b ≤ 10, and thefollowing constructions produce the (3, 3, b)-matrices M , where 12 ≤ b ≤ 20.
b = 12 13 14 15
M =0 0 11 1 11 1 1
0 0 11 0 11 1 0
0 0 11 0 21 1 1
0 1 10 1 21 1 2
b = 16 17 18 19 20
M =0 1 10 1 21 1 3
0 1 11 0 11 1 0
0 1 11 0 11 1 3
0 1 11 1 21 2 5
1 1 11 2 31 3 6
9
As for 11 ≤ r ≤ 48, we have to slightly modify the generic hyperplane arrangementconstruction as described in Section ?? to reduce the number of rows r needed to produce amatrix M with a prescribed number of singular submatrices b. For simplicity, we will onlyshow the construction for k = 3, but the corresponding one for general k is similar. Thereare three types of modification:
1. Allow some of the 2-planes to contain exactly one coordinate direction, i.e. 2-planeswith equations ax + by = 0, by + cz = 0, or ax + cz = 0, where a, b, c 6= 0, and pickvectors from these 2-planes on the same line parallel but distinct to the axis. Each ofthese modified 2-planes increases b by changing a summand of
(s3
)to(s+13
). In other
words, we save one row every time we modify a 2-plane.
2. Take one to three 2-planes to be those orthogonal to a coordinate axis, i.e. x = 0,y = 0, or z = 0. The resulting b will increase by changing a summand of
(s3
)to(s+23
).
In other words, we can save two rows for up to three times.
3. For each vector chosen from the complement of the union of hyperplanes, we duplicatethem t times in the matrix M . Each such duplication will increase the resulting b by(t3
)+(t2
)(r − t+ 3).
Proposition 5.2. For each r such that 11 ≤ r ≤ 48, and for each 0 ≤ b ≤(r+22
), there
exists an(r, 3,
(r+3r
)− b)-matrix M .
Proof. For each r and b such that 11 ≤ r ≤ 48 and 0 ≤ b ≤(r+22
), let b0 = b. For
each integer i ≥ 0, let ti be largest integer such that(ti3
)+(ti2
)(r − ti + 3) ≤ b, let ci =⌊
bi/((
ti3
)+(ti2
)(r − ti + 3)
)⌋, and let bi+1 = bi − ci
((ti3
)+(ti2
)(r − ti + 3)
). We repeat the
process until bi+1 <(23
)+(22
)(r − 2 + 3) = r + 1. In this process, we are building an r × 3
matrix M using only rows from the Type 3 modification.After using only rows from Type 3 modification, we still need to pick appropriate rows
to give an additional bi+1 singular submatrices if bi+1 > 0. Note that bi+1 ≤ r ≤ 48. Thefollowing is a list of partitions of integers from 1 to 48. Each summand
(s+23
)corresponds
to s rows from Type 2 modification, and summand(s+13
)corresponds to s rows from Type
1 modification.
10
1 =(1+23
)25 =
(4+23
)+(2+23
)+(1+23
)2 =
(1+23
)+(1+23
)26 =
(4+23
)+(2+23
)+(1+23
)+(2+13
)3 =
(1+23
)+(1+23
)+(1+23
)27 =
(4+23
)+(2+23
)+(1+23
)+(2+13
)+(2+13
)4 =
(2+23
)28 =
(4+23
)+(2+23
)+(2+23
)5 =
(2+23
)+(1+23
)29 =
(4+23
)+(2+23
)+(2+23
)+(2+13
)6 =
(2+23
)+(1+23
)+(1+23
)30 =
(4+23
)+(3+23
)7 =
(2+23
)+(1+23
)+(1+23
)+(2+13
)31 =
(4+23
)+(3+23
)+(1+23
)8 =
(2+23
)+(2+23
)32 =
(4+23
)+(3+23
)+(1+23
)+(2+13
)9 =
(2+23
)+(2+23
)+(1+23
)33 =
(4+23
)+(3+23
)+(1+23
)+(2+13
)+(2+13
)10 =
(3+23
)34 =
(4+23
)+(3+23
)+(2+23
)11 =
(3+23
)+(1+23
)35 =
(5+23
)12 =
(3+23
)+(1+23
)+(1+23
)36 =
(5+23
)+(1+23
)13 =
(3+23
)+(1+23
)+(1+23
)+(2+13
)37 =
(5+23
)+(1+23
)+(1+23
)14 =
(3+23
)+(2+23
)38 =
(5+23
)+(1+23
)+(1+23
)+(2+13
)15 =
(3+23
)+(2+23
)+(1+23
)39 =
(5+23
)+(2+23
)16 =
(3+23
)+(2+23
)+(1+23
)+(2+13
)40 =
(5+23
)+(2+23
)+(1+23
)17 =
(3+23
)+(2+23
)+(1+23
)+(2+13
)+(2+13
)41 =
(5+23
)+(2+23
)+(1+23
)+(2+13
)18 =
(3+23
)+(2+23
)+(2+23
)42 =
(5+23
)+(2+23
)+(1+23
)+(2+13
)+(2+13
)19 =
(3+23
)+(2+23
)+(2+23
)+(2+13
)43 =
(5+23
)+(2+23
)+(2+23
)20 =
(4+23
)44 =
(5+23
)+(2+23
)+(2+23
)+(2+13
)21 =
(4+23
)+(1+23
)45 =
(5+23
)+(3+23
)22 =
(4+23
)+(1+23
)+(1+23
)46 =
(5+23
)+(3+23
)+(1+23
)23 =
(4+23
)+(1+23
)+(1+23
)+(2+13
)47 =
(5+23
)+(3+23
)+(1+23
)+(2+13
)24 =
(4+23
)+(2+23
)48 =
(5+23
)+(3+23
)+(1+23
)+(2+13
)+(2+13
)From this table, we can deduce directly the number of additional rows from Type 1 and
Type 2 modifications we need to obtain b singular square submatrices in M . Together withthe rows from Type 3 modification, it remains to verify that the total number of rows wehave used is at most r. Once again, we employ Mathematica to finish the verification, andthe program codes are provided in Appendix ?? for reference.
Theorem 5.3. Linear (n, r, b)-matroids exist for all 1 ≤ b ≤(r+3r
), as long as
(nr
)≥ b,
except (n, r, b) = (6, 3, 11).
Proof. Propositions ?? and ?? imply that for all r ≥ 11, linear (n, r, b)-matroids exist forall(r+23
)=(r+33
)−(r+22
)≤ b ≤
(r+33
). Hence, we are done by connecting with the explicit
constructions for 3 ≤ r ≤ 10.
Now, we have all the tools for proving Theorem ??.
Proof of Theorem ??. For each i = 0, 1, 2, . . . , k − 3, let r0(k − i) ∈ N be the constantsobtained by applying Proposition ?? on k − i. Note that r0(k) ≥ r0(k − 1) ≥ · · · ≥ r0(3).Let R0 = r0(k). Then there exist integers R1, R2, . . . , Rk−3 such that
1. R0 ≤ R1 ≤ R2 ≤ · · · ≤ Rk−3, and
11
2.(Ri+k−i−1
k−i
)≤(Ri+1+k−(i+1)
k−(i+1)
)for all i = 0, 1, 2, . . . , k − 4.
Let R = Rk−3, and fix r ≥ R. By Theorem ??, an (r, k, b)-matrix exists for all integersb such that
(R0+k−1
k
)≤ b ≤
(r+kr
). By Proposition ??, for each i = 1, 2, . . . , k − 3, since
Ri ≥ r0(k − i), an (Ri, k − i, b)-matrix exists for all integers b such that(Ri+k−i−1
k−i
)≤ b ≤(
Ri+k−ik−i
). By Lemma ??, for each i = 1, 2, . . . , k − 3, an (r, k, b)-matrix exists for all b
such that(Ri+k−i−1
k−i
)≤ b ≤
(Ri+k−ik−i
). Our definition of Ri’s implies that
[(R0+k−1
k
),(r+kr
)]∪
k−3⋃i=1
[(Ri+k−i−1
k−i
),(Ri+k−ik−i
)]covers all integers
(R+23
)≤ b ≤
(r+kr
). Therefore, an (r, k, b)-matrix
exists for all b ≥(R+23
).
Finally, we are done by Theorem ??, which says an (r, k, b)-matrix exists for all b ≤(R+33
).
6 Conclusion and remarks
In our treatment, we always assume the base field is Q. In fact, the same argument worksfor any infinite field. In particular, if we consider the algebraic closure Fp for some prime p,the constructed (r, k, b)-matrix naturally descends to some finite extension of Fp. However,we cannot ensure there is a fixed finite extension which captures all of them. In any case,the matroid structure arising from matrices will not be affected.
All our results focus on the situation when n and r are very close together. Recall fromLemma ??, we only need to consider r ≤ n ≤ 2r. Hence, our next goal is to investigate thecase when n is close to 2r. In view of the counterexample of the non-existence of (6, 3, 11)-matroids, this latter goal should be much harder. But the general direction towards acomplete solution to Conjecture ?? will be another asymptotic result for the existence of(r, k, b)-matrices for large enough k which should isolate a finite number of cases for directchecking.
A Construction of (r + 3, 3, b)-matroids for 4 ≤ r ≤ 10
If we can construct a matrix A′ = (Ir′|M ′), where M ′ is an r′ × 3 matrix, such that A′ hasexactly b invertible r′ × r′ submatrices, then for all r ≥ r′, the matrix A = (Ir|M), whereM is obtained by appending r − r′ rows of all zeros to M ′, has exactly b invertible r × rsubmatrices. Hence, for r ≥ 5, we only need to construct matrices with b invertible r × rsubmatrices, where
(r+23
)< b ≤
(r+33
).
Theorem ?? implies that when r = 4 and 1 ≤ b ≤(r+22
)= 15, there exists a matrix A of
size r × (r + 3) such that the number of invertible r × r submatrices is b. When r = 4 and15 < b ≤
(r+33
)= 35, we give the construction of A = (Ir|M) as follows.
12
b = 16 17 18 19 20 21
M =
1 0 01 0 01 1 01 0 1
1 0 01 1 11 1 11 1 1
1 0 01 0 00 1 11 0 1
1 0 00 1 01 1 01 0 1
1 0 00 1 11 1 11 1 1
1 0 00 1 00 1 11 0 1
b = 22 23 24 25 26 27 28
M =
1 0 01 0 11 1 11 0 2
1 0 01 1 11 1 12 0 1
1 1 01 0 11 0 11 1 1
1 0 00 1 11 1 01 0 1
1 1 01 1 00 1 11 0 1
1 0 01 1 12 1 11 2 1
0 1 11 0 11 1 11 2 1
b = 29 30 31 32 33 34 35
M =
1 1 00 1 11 0 11 1 1
1 0 11 2 00 1 22 1 2
0 1 11 1 11 1 22 0 1
1 2 01 0 20 2 12 1 2
0 1 11 0 22 1 22 2 1
1 2 31 3 22 0 23 1 1
1 2 31 4 42 1 23 3 2
When r = 5 and 35 < b ≤(r+33
)= 56, we give the construction of A = (Ir|M) as follows.
b = 36 37 38 39 40 41 42
M =
1 0 03 0 01 1 12 1 33 1 3
2 0 00 3 00 1 22 3 02 3 1
1 1 31 1 32 1 32 2 24 4 4
3 0 00 1 40 2 21 2 03 1 0
1 0 00 3 30 3 31 1 32 1 3
2 0 00 3 21 1 22 2 03 3 2
1 0 01 0 10 2 22 3 02 3 3
b = 43 44 45 46 47 48 49
M =
1 0 01 3 21 4 02 1 02 2 2
0 2 30 2 31 2 21 2 34 0 2
1 0 32 0 12 0 22 4 33 2 3
3 0 00 1 21 2 11 2 33 2 4
0 1 30 2 21 1 32 2 12 2 3
0 1 31 2 32 1 32 3 12 3 1
2 0 01 1 31 2 21 3 43 3 1
b = 50 51 52 53 54 55 56
M =
0 1 30 2 21 0 31 1 03 1 4
1 2 21 3 32 3 33 1 24 4 0
1 1 01 3 31 3 42 3 33 0 1
0 2 21 0 31 1 21 4 22 3 1
0 2 11 1 21 1 41 2 31 3 1
1 2 21 2 32 1 33 1 24 3 1
1 2 42 3 43 2 23 4 34 2 3
13
When r = 6 and 56 < b ≤(r+33
)= 84, we give the construction of A = (Ir|M) as follows.
b = 57 58 59 60 61 62 63
M =
3 0 03 0 01 1 22 1 02 4 43 2 2
1 0 02 0 00 2 21 3 03 1 24 0 2
1 0 00 1 12 2 03 2 03 3 24 3 0
1 0 03 0 00 2 43 1 24 1 34 2 2
1 0 00 3 20 3 21 1 41 2 03 2 0
1 0 00 1 20 2 20 3 41 0 11 1 1
2 0 00 1 10 3 31 2 41 4 43 1 3
b = 64 65 66 67 68 69 70
M =
1 0 01 1 11 3 12 1 32 2 22 3 1
3 0 00 2 31 1 42 1 02 2 04 4 4
1 0 00 1 31 1 12 0 13 2 04 1 0
1 0 01 2 22 1 03 0 23 2 34 0 4
1 0 01 0 31 2 33 0 13 2 23 3 1
2 0 00 2 21 1 11 3 02 2 43 1 2
1 0 31 1 00 3 33 3 04 0 34 3 4
b = 71 72 73 74 75 76 77
M =
1 1 01 3 01 3 13 0 13 3 23 4 0
1 2 02 0 12 1 02 3 23 3 04 4 3
0 1 21 1 22 1 22 2 32 3 23 3 2
1 1 11 3 12 4 33 0 34 4 24 4 3
1 1 01 2 22 4 33 1 04 3 34 4 3
0 2 31 2 31 3 21 4 32 2 34 4 3
0 2 31 2 11 3 23 0 24 0 44 1 0
b = 78 79 80 81 82 83 84
M =
0 1 31 2 32 1 12 1 22 2 23 1 0
0 2 31 0 31 4 32 1 43 1 03 1 3
1 0 32 2 32 3 23 1 34 1 24 1 3
1 1 21 2 31 4 42 1 33 0 23 2 4
0 3 21 4 42 3 43 1 23 2 04 4 2
1 4 42 3 43 3 23 4 14 2 44 4 3
1 5 42 3 13 2 14 1 44 3 34 4 1
When r = 7 and 84 < b ≤(r+33
)= 120, we give the construction of A = (Ir|M) as
follows.
14
b = 85 86 87 88 89 90
M =
3 0 00 4 01 2 01 4 13 0 14 0 14 2 0
4 0 00 5 00 4 31 4 12 3 03 0 34 5 0
2 0 02 0 01 3 52 4 43 4 04 4 05 2 3
0 0 40 4 00 1 50 2 41 0 41 3 21 3 3
5 0 00 1 11 0 11 4 41 4 42 4 43 2 1
0 0 20 0 40 4 21 3 02 4 03 4 14 4 1
b = 91 92 93 94 95 96
M =
3 0 00 1 20 3 11 1 01 2 05 3 05 3 2
2 0 00 4 02 0 23 2 23 3 34 4 05 1 4
2 0 00 2 21 0 12 2 02 2 04 3 15 4 4
0 3 00 1 20 1 40 4 31 5 53 0 44 1 1
0 3 00 0 41 1 42 0 32 4 33 0 14 3 4
1 1 11 1 11 3 11 3 33 4 44 5 45 4 0
b = 97 98 99 100 101 102
M =
3 0 00 1 00 3 51 1 21 2 13 0 13 2 3
0 3 00 0 30 3 53 1 13 3 44 2 25 1 4
3 0 00 0 21 3 21 3 43 3 34 2 45 5 1
2 0 00 3 31 1 21 1 32 2 33 1 25 1 2
0 1 00 3 20 4 11 4 14 1 24 4 44 5 1
4 0 02 3 32 5 43 3 04 2 05 2 25 4 2
b = 103 104 105 106 107 108
M =
2 0 01 0 11 1 22 3 12 4 03 1 03 2 1
0 2 40 3 22 5 13 3 34 4 04 4 45 0 3
0 2 40 2 41 1 41 4 21 5 02 4 04 0 2
1 0 00 1 43 3 14 1 04 3 44 4 35 0 2
0 3 11 0 21 1 01 1 41 1 41 2 34 3 3
1 1 31 2 01 2 41 2 43 2 34 0 25 3 2
15
b = 109 110 111 112 113 114
M =
0 2 40 3 41 2 31 4 12 3 34 0 44 4 4
0 2 30 2 52 0 32 1 33 4 34 3 45 0 2
0 1 10 3 41 2 32 4 33 2 25 2 25 5 3
1 1 11 4 33 0 53 1 13 3 04 1 44 3 0
0 1 10 3 11 0 21 1 42 3 24 4 15 2 0
0 1 20 2 21 0 31 3 23 1 43 3 15 4 0
b = 115 116 117 118 119 120
M =
1 1 41 3 31 4 52 1 12 3 22 4 43 4 2
0 1 51 3 31 4 43 3 23 5 24 1 45 3 0
0 1 31 0 31 4 12 1 14 3 04 4 54 5 1
1 2 52 3 44 0 24 3 04 4 15 2 15 4 3
0 3 51 1 51 3 41 4 23 2 34 1 35 3 2
2 3 43 6 14 2 54 5 15 1 25 4 35 5 4
When r = 8 and 120 < b ≤(r+33
)= 165, we give the construction of A = (Ir|M) as
follows.
b = 121 122 123 124 125 126
M =
5 0 00 1 00 5 00 5 21 1 01 1 51 2 13 5 6
1 0 01 0 00 2 34 0 45 0 15 2 06 4 66 6 5
0 1 00 4 00 5 41 3 02 1 42 5 33 3 65 5 0
0 3 50 4 42 1 12 2 22 4 43 1 15 4 46 2 3
0 2 00 2 00 1 50 3 22 5 44 3 45 0 45 5 4
3 0 04 0 01 3 02 0 43 3 63 5 55 1 05 4 5
b = 127 128 129 130 131 132
M =
0 1 00 0 30 2 10 5 11 4 22 2 64 0 55 0 5
0 0 20 0 20 1 40 4 41 1 32 3 64 1 55 1 1
4 0 00 0 40 1 40 5 21 0 43 2 05 2 16 5 0
0 2 00 1 20 1 20 5 43 0 43 1 43 4 05 4 2
0 2 00 0 40 2 30 5 21 1 34 0 25 3 45 5 3
0 1 01 2 01 4 01 5 01 1 13 2 24 1 45 2 5
16
b = 133 134 135 136 137 138
M =
0 3 01 1 02 5 02 5 34 1 44 2 35 4 65 5 0
0 1 30 6 52 0 22 1 02 2 23 3 35 4 36 0 6
0 0 21 1 01 3 52 0 52 3 13 4 04 0 35 0 2
0 1 00 4 41 2 51 4 33 5 04 2 04 4 06 4 3
2 0 00 5 01 1 42 2 44 0 34 5 25 4 05 5 1
2 0 00 2 02 5 53 0 43 5 43 6 24 0 14 4 5
b = 139 140 141 142 143 144
M =
5 0 01 5 42 0 62 5 33 0 43 4 44 0 25 2 3
0 1 20 2 41 4 33 4 03 4 43 4 44 3 35 3 1
3 0 01 4 33 0 53 1 34 2 05 0 55 3 56 1 3
0 1 01 2 11 4 34 1 54 3 54 3 55 6 66 0 2
1 0 01 3 12 2 42 5 13 2 44 3 54 3 55 2 0
4 0 00 1 31 1 51 2 21 5 52 3 32 4 04 3 4
b = 145 146 147 148 149 150 151
M =
2 2 02 4 03 1 43 2 23 3 03 6 54 1 24 3 6
0 3 00 2 51 1 41 2 44 0 34 0 44 1 55 1 0
1 5 11 5 42 0 52 1 23 1 43 1 44 0 54 4 0
0 5 22 0 42 1 12 2 22 4 23 0 34 0 55 2 0
0 0 10 5 22 5 13 0 33 4 14 1 44 3 55 5 1
0 2 21 2 51 3 52 1 12 1 32 2 04 4 05 4 6
1 5 34 1 14 1 14 1 54 3 54 6 35 1 05 1 3
b = 152 153 154 155 156 157 158
M =
0 5 01 4 43 2 53 4 24 3 54 5 04 5 26 4 1
0 2 11 0 31 1 21 4 12 6 05 3 35 5 05 5 4
0 1 40 5 51 2 52 2 22 2 32 4 23 1 23 4 2
0 1 30 3 13 0 43 1 54 3 14 5 35 2 26 0 2
0 2 10 5 21 0 62 1 22 5 43 4 25 1 35 5 5
0 4 11 4 22 1 13 3 24 1 05 0 15 2 25 5 2
1 4 41 6 32 0 42 0 62 2 32 4 13 5 14 4 5
17
b = 159 160 161 162 163 164 165
M =
0 4 12 1 53 2 53 3 33 4 23 4 44 2 45 5 2
0 1 31 2 31 4 01 4 34 2 15 1 15 2 65 3 5
1 5 01 6 24 4 25 2 35 4 45 4 55 5 26 5 5
1 4 52 2 13 2 13 5 24 0 34 6 25 1 55 4 3
0 1 41 2 31 4 52 1 12 3 44 4 35 0 45 3 2
0 3 21 3 41 5 33 5 23 6 54 2 65 1 25 6 3
1 4 53 5 43 7 24 5 35 1 46 6 76 7 67 2 4
When r = 9 and 165 < b ≤(r+33
)= 220, we give the construction of A = (Ir|M) as
follows.
b = 166 167 168 169 170 171
M =
2 0 05 0 02 0 52 6 25 2 05 4 05 6 06 4 36 5 3
1 0 00 0 10 0 50 4 11 0 61 1 01 6 11 6 42 6 2
0 4 00 5 02 2 33 5 24 5 24 5 25 0 36 2 36 2 4
0 1 00 5 00 1 41 0 21 0 52 0 52 0 65 3 45 5 5
0 1 00 4 00 3 50 3 61 0 31 0 42 3 64 6 26 1 2
1 0 00 3 00 6 02 2 42 3 33 2 13 2 66 3 16 6 0
b = 172 173 174 175 176 177 178
M =
1 0 00 1 00 4 01 4 11 6 45 0 55 0 65 4 36 2 1
0 0 50 0 61 3 62 0 42 6 24 2 55 0 35 2 45 2 5
0 0 50 0 60 5 12 1 52 2 53 0 54 4 25 2 06 0 6
5 0 06 0 00 1 30 5 51 0 63 1 03 6 24 1 54 6 0
0 0 21 0 61 3 32 0 52 2 02 3 03 0 64 0 25 4 3
0 5 00 6 01 1 21 6 23 1 64 0 64 1 05 5 66 3 5
0 3 00 3 00 2 30 4 34 2 54 5 65 5 46 0 16 4 1
b = 179 180 181 182 183 184 185
M =
0 1 00 0 10 6 52 0 42 4 02 6 03 4 26 0 16 1 4
0 0 30 0 40 4 61 2 31 5 12 3 43 6 64 2 35 6 6
6 0 00 0 21 2 02 4 44 2 04 5 45 4 46 2 26 3 1
2 0 00 1 10 2 22 5 42 6 44 2 24 5 25 6 66 6 4
1 0 00 0 60 4 10 4 51 6 02 3 23 6 64 4 05 5 1
0 0 60 5 50 6 51 0 21 1 03 4 16 0 36 1 66 1 6
1 0 00 6 00 3 40 5 33 0 33 2 64 2 05 5 56 1 5
18
b = 186 187 188 189 190 191 192
M =
0 6 61 2 12 4 22 4 45 1 45 2 45 5 05 5 06 2 4
4 0 01 4 03 1 43 2 54 4 25 0 55 1 06 4 36 6 3
0 3 00 2 50 3 10 4 31 2 42 2 32 2 63 0 45 0 1
0 0 50 3 20 3 21 1 01 5 22 3 63 2 44 4 65 2 0
0 2 00 4 41 1 61 3 11 4 12 1 25 3 55 4 36 4 0
0 2 00 1 10 5 10 5 41 5 03 1 24 5 65 2 46 4 2
0 0 20 5 31 1 42 0 43 0 54 3 15 2 25 3 66 0 1
b = 193 194 195 196 197 198 199
M =
1 5 62 1 52 4 03 5 44 1 04 1 65 3 25 4 05 4 0
3 0 01 5 02 2 13 4 03 4 13 6 34 6 55 2 16 2 6
0 0 50 5 11 5 61 6 12 1 13 1 33 4 03 4 05 1 4
0 0 60 2 22 1 54 1 55 0 65 2 06 0 26 4 36 5 0
1 6 23 4 04 0 24 2 54 2 66 0 36 2 16 3 06 4 2
0 0 10 1 30 5 61 2 61 6 43 6 24 1 14 2 16 2 2
0 3 20 6 41 2 41 4 11 6 42 0 13 6 14 5 46 1 6
b = 200 201 202 203 204 205 206
M =
1 0 00 2 11 4 11 6 63 4 04 1 64 3 04 6 26 2 1
1 2 63 3 33 4 23 6 05 0 45 2 05 5 46 1 56 6 0
0 1 00 4 11 3 12 2 22 2 44 2 54 5 64 6 26 2 3
1 6 42 0 42 3 23 1 13 6 14 2 25 4 46 0 56 3 2
0 1 21 3 31 4 42 0 22 5 33 2 25 5 45 5 56 2 4
1 5 51 6 22 1 44 1 34 4 45 0 35 3 65 3 66 1 4
0 2 20 5 60 6 31 2 31 4 11 6 42 6 44 5 34 5 5
b = 207 208 209 210 211 212 213
M =
0 2 30 3 11 4 12 1 13 4 04 0 45 4 56 2 36 2 5
1 4 61 5 52 3 34 4 15 1 35 1 35 5 25 6 56 2 4
0 6 44 0 45 3 36 2 66 3 36 4 46 5 36 5 46 6 1
0 2 30 3 11 0 22 5 23 3 63 5 35 0 45 2 66 4 3
1 2 32 5 65 0 65 1 35 3 56 0 66 4 16 4 56 6 0
0 4 61 0 51 2 21 3 13 1 55 4 26 0 46 1 16 6 4
0 6 11 4 12 1 52 4 63 1 54 0 64 2 35 2 25 4 2
19
b = 214 215 216 217 218 219 220
M =
1 0 14 3 54 5 35 1 35 2 25 2 46 1 46 3 66 4 5
0 4 61 3 22 1 63 6 04 1 44 2 14 3 15 5 35 6 6
0 1 61 2 01 6 13 1 43 1 53 2 13 5 24 4 35 2 5
1 0 61 4 11 6 32 5 14 1 04 5 34 6 65 1 15 5 6
1 0 61 4 41 5 12 2 33 5 44 6 14 6 55 3 65 4 2
1 3 61 5 34 1 34 6 65 2 76 1 66 4 36 5 17 2 5
2 1 62 4 12 6 53 1 34 5 15 2 35 6 16 1 16 5 2
When r = 10 and 220 < b ≤(r+33
)= 286, we give the construction of A = (Ir|M) as
follows.
b = 221 222 223 224 225 226
M =
1 0 01 0 00 0 30 6 51 0 61 4 42 1 23 1 15 2 26 3 0
0 3 00 5 00 1 30 6 22 1 03 1 63 4 04 2 25 4 06 1 4
1 0 02 0 04 0 02 2 63 3 54 3 15 3 66 0 56 1 46 5 1
0 1 00 2 00 0 20 1 52 4 22 4 32 6 43 4 03 4 44 1 0
2 0 00 2 00 1 10 6 33 4 23 4 63 4 64 5 54 5 55 0 3
3 0 00 0 20 1 42 1 22 1 62 2 23 4 05 6 06 3 06 6 6
b = 227 228 229 230 231 232
M =
6 0 06 0 00 6 01 0 11 1 51 5 24 6 15 5 26 5 26 6 5
0 0 40 4 20 5 50 6 30 6 61 1 01 4 21 6 12 2 62 4 5
0 4 00 6 00 0 51 3 62 0 62 1 64 0 15 1 06 1 36 4 6
6 0 00 0 50 5 30 6 22 0 34 0 35 2 25 3 66 0 36 5 0
0 0 10 1 30 2 10 3 30 6 64 5 45 0 45 1 25 2 46 3 6
2 0 03 0 00 6 02 1 52 5 33 5 33 6 15 3 45 3 56 6 0
20
b = 233 234 235 236 237 238
M =
5 0 00 1 50 5 61 3 44 0 14 0 14 2 04 4 34 6 06 3 0
3 0 00 5 01 5 52 3 02 5 23 6 04 2 04 3 15 6 25 6 5
6 0 00 1 10 3 10 4 10 5 52 0 12 3 02 4 23 2 06 4 1
3 0 00 0 30 4 10 5 51 3 02 1 62 4 43 6 65 0 46 1 5
0 2 00 0 20 3 60 6 41 0 42 1 04 4 14 4 15 1 46 2 6
0 0 40 0 50 3 21 1 11 6 22 5 42 5 63 0 64 6 56 6 1
b = 239 240 241 242 243 244
M =
1 0 11 0 31 0 41 5 02 0 62 2 63 5 44 0 34 3 55 1 0
0 3 00 3 00 4 11 4 42 0 52 1 32 1 52 2 42 4 42 5 0
0 0 50 1 40 2 20 4 41 5 53 1 54 1 04 4 26 2 26 5 4
0 1 00 6 02 4 63 2 33 6 44 0 14 0 54 2 24 3 35 0 3
0 0 40 5 22 1 02 2 03 2 04 6 24 6 46 2 36 4 06 5 5
0 6 00 0 40 1 30 3 42 6 13 5 35 0 55 1 65 4 06 3 3
b = 245 246 247 248 249 250 251
M =
2 0 00 2 00 1 31 1 11 6 63 3 03 5 55 0 25 5 66 4 1
1 0 06 0 00 6 41 4 01 5 11 6 52 6 22 6 53 4 55 6 6
0 3 00 6 00 4 61 3 43 3 14 1 54 2 35 0 35 6 66 4 0
0 1 50 3 30 6 61 5 32 0 32 1 32 2 04 2 66 0 46 3 1
0 1 30 1 60 1 61 5 13 0 33 0 53 2 04 3 04 4 05 2 5
0 3 00 0 51 4 62 1 22 5 03 0 34 1 15 5 16 0 16 5 1
1 0 00 1 41 1 31 1 31 5 61 6 12 5 62 6 32 6 45 5 6
21
b = 252 253 254 255 256 257 258
M =
2 0 00 2 10 6 61 3 41 5 52 0 42 1 04 0 24 3 14 6 0
1 0 00 4 31 1 01 2 01 3 21 4 22 5 43 0 33 6 36 6 4
0 1 30 3 20 6 31 0 11 1 22 1 32 6 15 0 25 0 55 2 2
0 2 00 2 30 2 40 2 53 6 44 4 04 5 34 5 55 0 15 4 5
0 3 01 0 52 1 62 4 42 6 44 0 25 1 45 1 45 3 66 6 3
6 0 00 6 41 2 61 5 01 5 33 4 03 5 65 1 25 6 06 6 3
0 0 50 5 52 1 43 3 53 6 04 0 14 0 54 2 25 6 36 6 3
b = 259 260 261 262 263 264 265
M =
0 5 51 4 01 4 53 3 23 4 43 6 43 6 64 6 65 2 26 1 4
0 6 01 0 21 0 61 2 51 4 11 6 12 6 13 5 15 0 26 4 2
0 1 30 1 30 5 62 2 62 6 24 3 54 4 66 1 56 2 26 4 6
0 2 52 5 03 1 54 1 34 4 25 1 35 1 36 2 16 2 06 5 6
0 0 42 1 62 2 22 5 53 0 34 0 34 3 54 6 15 1 16 4 5
1 0 53 0 13 1 23 3 13 6 05 0 55 4 45 6 06 2 06 5 3
6 0 00 2 61 0 53 3 24 1 34 3 56 1 66 2 06 2 26 6 3
b = 266 267 268 269 270 271 272
M =
0 1 60 5 52 3 02 3 22 5 02 5 22 5 34 1 24 1 66 5 2
4 0 00 4 32 2 42 3 13 6 14 2 45 3 25 6 26 0 16 1 4
0 2 30 5 23 3 44 4 65 0 45 2 55 3 66 0 56 0 66 1 1
1 2 01 6 52 2 02 5 63 5 44 0 14 0 44 3 34 4 25 0 2
0 3 50 4 30 5 41 4 51 6 32 3 63 4 43 4 64 1 06 0 2
1 5 21 5 62 0 22 0 43 1 63 2 43 6 34 5 34 6 45 4 1
0 4 41 2 21 4 11 5 01 6 32 5 03 3 23 5 33 6 35 6 4
22
b = 273 274 275 276 277 278 279
M =
0 5 40 5 51 2 41 3 12 6 53 1 63 5 04 4 45 2 25 3 1
1 5 52 1 53 6 14 5 65 0 55 2 05 5 26 2 26 3 36 5 2
0 1 61 0 41 5 52 4 13 1 13 2 34 2 04 4 65 1 66 2 0
1 1 21 1 32 5 53 3 14 6 55 2 06 0 16 2 26 5 16 5 6
2 5 42 6 42 6 53 2 13 2 33 4 54 6 26 1 56 4 36 6 5
0 4 10 4 61 5 12 4 32 6 63 1 13 1 43 5 64 2 05 6 3
1 5 42 3 32 4 12 5 53 1 04 0 35 1 56 3 56 4 36 5 0
b = 280 281 282 283 284 285 286
M =
0 1 41 3 41 6 32 3 62 5 32 5 54 5 65 1 65 2 15 5 3
0 1 31 4 21 6 42 6 23 5 63 6 55 0 25 2 36 3 56 5 4
0 5 61 4 01 5 31 6 52 1 12 5 43 3 53 4 34 6 26 4 5
0 5 21 2 21 4 11 6 42 5 63 4 15 1 65 6 16 2 36 5 0
1 2 51 4 11 6 32 4 33 1 64 1 34 6 55 1 25 3 36 1 3
3 6 44 5 74 7 65 4 15 5 66 2 56 6 16 7 77 1 77 6 3
1 6 72 5 32 7 44 1 54 5 75 3 25 6 15 7 36 6 57 3 6
B Proof of Proposition ?? for 49 ≤ r ≤ 203
We use Mathematica to check the validity of Proposition ?? for 49 ≤ r ≤ 203.
IntDiv[n , d ] := Block[{}, (n - Mod[n, d])/d];
Do[bin = Table[Binomial[s, 3], {s, r}]; a = Table[0, {s, r}];Do[btemp = bbar;
Do[a[[s]] = IntDiv[btemp, bin[[s]]];
btemp = Mod[btemp, bin[[s]]], {s, r, 3, -1}];If[(Sum[a[[s]]*s, {s, 3, r}]) > r,
Print[r, " ", bbar, " ", Sum[a[[s]]*s, {s, 3, r}]]],{bbar, 0, Binomial[r + 2, 2]}],
{r, 49, 203}]
As there is no output after these lines finish running, we finish our verification.
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C Proof of Proposition ?? for 11 ≤ r ≤ 48
From the list in Proposition ??, we use leng below to record the number of rows from Type1 and Type 2 modifications to obtain additional singular submatrices.
leng = {0, 1, 2, 3, 2, 3, 4, 6, 4, 5, 3, 4, 5, 7, 5, 6, 8, 10, 7, 9,
4, 5, 6, 8, 6, 7, 9, 11, 8, 10, 7, 8, 10, 12, 9, 5, 6, 7, 9, 7, 8,
10, 12, 9, 11, 8, 9, 11, 13};Do[tbin = Table[Binomial[t, 3] + Binomial[t, 2] (r - t + 3), {t, r}];
a = Table[0, {s, r}]; max = 0;
Do[btemp = bbar;
Do[a[[s]] = IntDiv[btemp, tbin[[s]]];
btemp = Mod[btemp, tbin[[s]]], {s, r, 2, -1}];If[Sum[a[[s]]*s, {s, 2, r}] + leng[[btemp + 1]] > max,
max = Sum[a[[s]]*s, {s, 2, r}] + leng[[btemp + 1]]],
{bbar, Binomial[r + 2, 2]}];If[r < max, Print[r, " ", max]], {r, 11, 48}]
As there is no output after these lines finish running, we finish our verification.
References
[1] D. Mayhew and G. F. Royle, Matroids with nine elements, J. Combinatorial TheorySer. B 98 (2008), 415–431.
[2] V. Sivaraman, On a conjecture of Welsh, preprint.
[3] D. Welsh, Combinatorial problems in matroid theory, Combinatorial Math. and itsApplications (Proc. Conf. Oxford 1969), Academic Press, London, 1971.
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