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Volume 14, Issue 1 (Jan 2016)

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The Bruhat rank of a binary symmetric staircase pattern

Zhibin Du
  • School of Mathematics and Statistics, Zhaoqing University, Zhaoqing 526061, Guangdong, China
  • Email:
/ Carlos M. da Fonseca
  • Corresponding author
  • Department of Mathematics, Kuwait University, Safat 13060, Kuwait
  • Email:
Published Online: 2016-11-03 | DOI: https://doi.org/10.1515/math-2016-0079

Abstract

In this work we show that the Bruhat rank of a symmetric (0,1)-matrix of order n with a staircase pattern, total support, and containing In, is at most 2. Several other related questions are also discussed. Some illustrative examples are presented.

KeyWords: Permutation matrix; Bruhat order; Bruhat shadow; Bruhat rank; Inversion

MSC 2010: 05B20; 06A07; 15A36

1 The Bruhat shadow

Setting L2=(0110)andI2=(1001), the standard inversion-reducing interchange process applied to a permutation matrix P, replaces a 2 × 2 submatrix equals to L2 by I2, for short, L2I2.

Given two permutation matrices P and Q of the same order, Q is below P in the Bruhat order and written as GB P, if Q can be obtained from P by a sequence of L2I2 interchanges. The Bruhat order in terms of permutation matrices has attracted considerable attentions recently [26].

If Sn denotes the set of all permutation matrices of order n, a nonempty subset 𝓙 of Sn is called a Bruhat ideal if P ∈ 𝓘 and QB P imply that Q ∈ 𝓘. A principal Bruhat idealP〉 is an ideal generated by a single permutation matrix P. Denoting the Boolean sum of two (0, 1)-matrices A and B by A +b B, the Bruhat shadow of 𝓘 is the matrix 𝒮(𝓘) = +b{Q ∈ 𝓘}.

As an example, setting P=(001000000100100000000001010000000010)andQ=(100000010000000010001000000100000001),(1) we have S(P,Q)=(111000111100111110011111011111000011).

If 𝓘 = 〈P〉, then we simply write 𝒮(P): the Bruhat shadow of P.

Observe that 𝒮(〈P, Q〉) has a staircase pattern with I6, P, Q ⩽ 𝒮(〈P, Q〉) (see [2, Theorem 2.4]). We recall that a (0, 1)-matrix A = (aij) has a staircase pattern provided the 1’s in each row and each column occur consecutively and the leftmost (resp., rightmost) 1 in a row occurs above or to the left of the leftmost (resp., rightmost) 1 in the next row. Here, without loss of generality, we confine our attention to indecomposable staircase patterns, i.e., our matrices cannot be represented as the direct sum of two matrices.

Suppose now that we have a set of indices 1 = i1 < i2 < ... < ipn, such that ri1=ri1+1==ri21<ri2=ri2+1==ri31<<rip=rip+1==rn(2) is a sequence with integers in the set {1, . . .,n}, and rii. This sequence r = r1, ...,rn is called a right-sequence or, for short, r-sequence. Analogously, for a given set of indices 1 ⩽ j1 < j2 < ... < jq = n, the sequence 𝓵 = 𝓵1, ...,𝓵n with integers in the set {1, ...,n}, and 𝓵ii, such that l1=l2==lj1<lj1+1==lj2<<ljq1+1==ljq(3) is called a left-sequence or, briefly, 𝓵-sequence.

We interchange the roles of r and 𝓵 in [2] since we believe this is a more natural convention.

When aij = 1, for 1 ⩽ jri, where r1 ⩽ ... ⩽ rn = n, and aij = 0, otherwise, we call A = (aij) a full staircase pattern.

For any n × n permutation matrix P, corresponding to a permutation σ of the set {1, ...,n}, we may associate the permutation σ = (σ(1) ... σ(k) ... σ(n)) with an 𝓵- and r-sequences: 𝓵k is the smallest integer in the set {σ(k), ...,σ(n)} and rk is the largest integer in the set {σ(1), ...,σ(k)}. For the permutation matrix P in (1), we have 𝓵 = 1, 1, 1, 2, 2, 5 and r = 3, 4, 4, 6, 6, 6.

If one defines the (0, 1)-matrix A(P) = (auv) of order n having a staircase pattern such that auv = 1 if 𝓵uvru, and 0 otherwise, for 1 ⩽ u, vn, then 𝒮(P) = A(P) (see [2, Theorem 2.2]). Thus, the Bruhat shadow of a permutation matrix is determined by the 𝓵- and r-sequences. With P as in (1), we have S(P)=A(P)=(111000111100111100011111011111000011).

The remainder of this paper is organized as follows. In Section 2 we give an explicit characterization for the 𝓁-and r-sequences of the Bruhat shadow of a permutation matrix. The main section will be devoted to the symmetric permutation matrices, especially for their Bruhat shadows and Bruhat ranks. In the last section we extend the results on the numbers of inversions established in [2]. Several examples illustrate the results. To our knowledge, this is the first attempt to partially answer the questions left open by Brualdi and Dahl in [2].

2 Characterizing the Bruhat shadow

The characterization of the Bruhat shadow of a permutation matrix was first established by Brualdi and Dahl in [2, Theorem 3.5].

[2] Let A be a (0, 1) -matrix. Then A is the Bruhat shadow of a permutation matrix if and only if A has a staircase pattern and the matrix A' of order m obtained from A by striking out the rows and columns of its extreme positions satisfies ImA'.

Taking into account Theorem 2.1, given an r-sequence (2) and an 𝓵-sequence (3), we may characterize explicitly these sequences such that both define the Bruhat shadow of an indecomposable permutation matrix.

The two sequences (2) and (3) are, respectively, the r- and 𝓁-sequences of an indecomposable permutation matrix if and only if

  • (i)

    {i1, ...,ip} ∩ {j1, ...,jq} = ∅,

  • (ii)

    {ri1, ...,rip} ∩ {𝓁j1, ...,𝓁jq} = ∅ , and

  • (iii)

    if κx is the x-th lowest integer in {1, ...,n} not in {ri1, ...,rip, 𝓁j1, ...,𝓁jq} , and τx is the x-th lowest integer in {1, ..., n} not in {i1, ..., ip, j1, ..., jq} , then 𝓁τx < κx < rτx.

Let us start by assuming that (2) and (3) are, respectively, the r- and 𝓁-sequences of a permutation matrix P. Since P is indecomposable and contains exactly one entry 1 in each row and each column and 0’s elsewhere, the assertion (i) and (ii) are immediate. According to Theorem 2.1, m = npq, the diagonal entries of A' are equal to 1. Note that the diagonal entries of A' are the (τt, κt)-entry in A, for t = 1, ..., m, thus the inequalities in (iii) follow clearly.

Conversely, condition (i) guarantees that there exists exactly one 1 in each row, while condition (ii) says that there is exactly one 1 in each column. If {ri1, ..., rip, 𝓁j1, ..., 𝓁jq} = {1, ..., n}, then the result comes immediately. Otherwise, let κ1 be the lowest positive integer not in {ri1, ..., rip, 𝓁j1, ..., 𝓁jq} and let τ1 be the lowest positive integer not in {i1, ..., ip, j1, ..., jq}. From condition (iii), we have 𝓁τ1 < κ1 < rτ1, which means that the (τ1, κ1)-entry of A is 1. If {ri1, ..., rip, 𝓁j1, ..., 𝓁jq, κ1} = {1, ..., n}, then in Theorem 2.1, m = 1 and A' = A[τ1, κ1], the submatrix of A resulting from the retention of the row and column indexed by τ1 and κ1, respectively. Otherwise, we proceed analogously to choose κ2 as the lowest positive integer not in {ri1, ..., rip, 𝓁j1, ..., 𝓁jq, κ1}, and let τ2 be the lowest positive integer not in {i1, ..., ip, j1, ..., jq, τ1}. From condition (iii), we have 𝓁τ2 < κ2 < rτ2, i.e., the (τ2, κ2)-entry of A is 1. In the end we would get the submatrix A' = A[τ1, ..., τm, κ1, ..., κm] with ImA'. □

We remark that in the condition (iii) of the previous theorem, we are considering the integers in {1, ..., n} ordered increasingly. □

Let us consider the r-sequence r = 5, 7, 7, 7, 7, 7, 7 and the 𝓁 -sequence 𝓁 = 1, 1, 1, 2, 2, 4, 4 . Then p = 2, q = 3, i1 = 1, i2 = 2, j1 = 3, j2 = 5, and j3 = 7. Now, we have {1, 2} ∩ {3, 5; 7} = ∅ and {5, 7} ∩ {1, 2, 4} = ∅. Moreover, κ1 = 3 and κ2 = 6. On the other hand, τ1 = 4 and τ2 = 6. It is straightforward to see now that 𝓁4 < 3 < r4 and 𝓁6 < 6 < r6. In another words, following the notation of [2], interchanging the roles of r and 𝓁 , as we pointed out previously, we have

and σ gives rise to the permutation matrix whose Bruhat shadow is (1111100111111111111110111111011111100011110001111).

Let us consider now the r-sequence r = 6, 6, 7, 7, 7, 7, 7 and the 𝓁-sequence 𝓁 = 1, 1, 2, 2, 4, 5, 5. Here p = 2, q = 4, i1 = 1, i2 = 3, j1 = 2, j2 = 4, j3 = 5, and j4 = 7 ( We have {1, 3} ∩ {2, 4, 5, 7} = ∅ and {6, 7} ∩ {1, 2, 4, 5} = ∅. Moreover, κ1 = 3 and τ1 = 6( However, 𝓁6 < 3 < r6 is obviously false.

Example 2.4 shows that the matrix defined in [2, pp.31] (1111110111111001111110111111000111100001110000111) is not the Bruhat shadow of any permutation matrix.

When a permutation matrix is symmetric, its Bruhat shadow will clearly be symmetric. In this case, any characterization depends on one of the sequences defined previously. We will consider, for example, the integral sequence r introduced in (2). Then, the 𝓁-sequence of the symmetric permutation matrix is i1,,i1ri1,i2,,i2ri2ri1,...,ip,,ipriprip1

Note that, in this case, jt = rit and 𝓁jt = it, for t = 1, ..., p.

As a consequence of Theorem 2.2, we have the following:

The integral sequence (2) is the r-sequence of a symmetric permutation matrix if and only if {i1, ..., ip} ∩ {ri1, ..., rip} = ∅.

We only need to observe that 𝓁κ < κ < rκ, in particular for each κ ∉ {i1, ..., ip, ri1, ..., rip}. □

This corollary considerably simplifies [2, Theorem 3.6].

3 Bruhat rank: the symmetric case

The term rank ρ(A) of a (0, 1)-matrix A is defined to be the maximum number of 1’s of A with no two of the 1’s in the same row or column of A. By the Kτonig-Egérvàry Theorem [8, pp.55-56], ρ(A) is equal to the minimum number of rows and columns of A which together contain all the 1’s of A. The term rank is a well-studied quantity associated with any (0, 1)-matrix. The matrix A is said to have total support provided ρ(A(i, j)) = n − 1, for each pair of i, j with aij = 1, where A(i, j) denotes the (n − 1) × (n − 1) submatrix obtained from A by deleting row I and column j [1]. Mirsky and Perfect [7] showed that a matrix has total support if and only if there exists a doubly stochastic matrix having the same zero pattern. Equivalently, A has total support if every nonzero element of A occurs in the positive main diagonal under a column permutation.

Brualdi and Dahl [2] defined the Bruhat rank of an n × n(0, 1)-matrix M, denoted by rB(M), with a staircase pattern and having total support with InM, as the smallest number of permutation matrices P, such that if 𝓘 is the Bruhat ideal generated by the P’s, then 𝒮(𝓘) = M. For example, a full staircase pattern has Bruhat rank 1. Another more elaborated example is the following staircase pattern M=(11111111111111111111).

It is not hard to check that the Bruhat rank for this matrix is 3, since a possible minimal decomposition of M is M=(1111111111111)+b(11111111111111)+b(11111111111), a decomposition as a Boolean sum of the Bruhat shadows of three permutation matrices.

In [2], the authors proposed the following general problem:

questionQuestion 3.1

Determine either a formula for the Bruhat rank of M or an algorithm for evaluating it.

In this work, we will answer Question 3.1, in the symmetric case. First, we observe that Corollary 2.5 provides a characterization of Bruhat rank 1 matrices. Interestingly, for the remaining cases the Bruhat rank is always 2.

The Bruhat rank of any symmetric (0, 1) -matrix of order n, with a staircase pattern and InM, is at most 2.

Let us denote by M such a matrix. Let A be the largest upper left principal submatrix of M, such that A is the Bruhat shadow of some permutation matrix. Assume that n1 is the order of A. If M = A, then rB(M) = 1. Otherwise, we have the following block decomposition for M :

where C is the matrix of the form

with ñ1 ⩾ 1 and F has an (n11) × m1 full staircase pattern. Consequently, F has no zero columns.

If m2 = 0, then B = Jn−n1 and

where Jk denotes the k × k all 1’s matrix. Therefore rB(M) = 2. Actually, in this case, one of the matrices is a block decomposition of two principal submatrices with Bruhat rank 1 and the other is a off-diagonal blocks decomposition, with 1’s in the extreme positions not included in the diagonal blocks.

Let us assume now that m2 > 0. Note that B has a symmetric staircase pattern. Suppose that (k, rk) is the entry of the first 1 in the block B verifying k > n1 and rk > n1 + m1. Then we partition B into 2 × 2 blocks:

where J* is of order kn1 − 1. If B1 has Bruhat rank 1, then we get the decomposition

Otherwise, the process now follows on B1 in the same manner until the order n is exhausted. □

We provide next three illustrative examples of Theorem 3.2.

Let us consider the matrix M=(1111111111111111111111111111111111111111111111111111111111).

Then

Let us consider now the matrix M=(11111111111111111111111111111111111111111111111111111111111111111111111111111111111).

Then

We analyze a final example, also mentioned in [2, pp.32], considering now the symmetric tridiagonal (0, 1) -matrix: M=(1111111111111111111).

In this simple case, we have

4 Inversions

The number of inversions of a permutation matrix P, inv(P), can be interpreted as the number of submatrices of P equal to L2. In [2], for a given (0, 1)-matrix with a full staircase pattern, it was shown that the maximum number of inversions of a permutation matrix PA is equal to βA=k=1pnk[(nk1)/2+rikik+1+1],(4) where the r-sequence is defined as in (2) and nk = ik+1ik, for k = 1, ..., p, assuming that ip+1 = n + 1. Moreover, it was considered a simple algorithm for constructing a permutation matrix PA. It turned out that such algorithm gave rise to a permutation matrix with the largest possible number of inversions. Following the same technique used in Lemma 4.1 and Theorem 4.2 of [2], we may establish a more general result for Bruhat rank one matrices.

Let A be a (0, 1) -matrix with rB(A) = 1 defined by the r - and 𝓁 -sequences in (2) and (3) , respectively. The permutation matrix P corresponding to the permutation σ with maximum number of inversions, with PA, is given by the following algorithm:

  1. Set P = 0.

  2. For k = 1, ..., p, replace the (ik, rik ) -entry by 1 , i.e., make σik = rik.

  3. For k = 1, ..., q, replace the (jk, 𝓁jk ) -entry by 1 , i.e., make σjk = 𝓁jk.

  4. Set I = {i1, ..., ip, j1, ..., jq}.

  5. If I = {1, ..., n} , then the process stops.

  6. If I ≠ {1, ..., n} , let u be the lowest positive integer not in I.

  7. Let v be the largest integer such that 𝓁u < v < ru and v ≠ σk, for kI.

  8. Replace the (u, v)-entry by 1 , i.e., make σu = v.

  9. Replace I by I ∪{u} in Step 4. and proceed.

Note that in the above algorithm, the condition rB(A) = 1 is required, since the condition (iii) of Theorem 2.2 would guarantee the existence of integer v in Step 7.

We observe that for the case of a full staircase pattern given by A, i.e., when 𝓁 = 1, ..., 1, we have rB(A) = 1 and, as a consequence, we derive the next corollary.

If A has a full staircase pattern, then βA=s=p1is<k<is+1σisσk(5) with ip+1 = n + 1.

The full staircase pattern gave in [2, pp.37] is defined by the r-sequence 4,6,6,6,7,7,11,13,13,13,13,13,13.

Taking into account (4), we have βA=3+9+3+4+15=34.

From (5), we get βA=1+3+4+5+12+5+1+3=34.

In general we have:

If A has a staircase pattern with rB(A) = 1 , then βA=s=p1is<k<is+1σisσk#{t|t>kandσk<σt<σis}(6) with ip+1 = n + 1.

Setting A=(1110000111110011111001111100011111101111110000011), then the optimal permutation matrix is P=(0010000000010000010001000000000000101000000000010).

Furthermore, from (6) we have βA=(721)+(76)+(54)+(511)=9.

In the same fashion as the proof of [2, Corollary 4.3], by a sequence of L2I2 interchanges, it is possible to provide the range for the number of inversions of permutation matrices PA, where rB(A) = 1.

For each integer 0 ⩽ kβA, with rB(A) = 1 , there is a permutation matrix PA with inv(P) = k.

Acknowledgement

We would like to thank both anonymous referees for their comments.

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About the article

Received: 2016-02-29

Accepted: 2016-10-10

Published Online: 2016-11-03

Published in Print: 2016-01-01


Citation Information: Open Mathematics, ISSN (Online) 2391-5455, DOI: https://doi.org/10.1515/math-2016-0079.

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© Du and da Fonseca, published by De Gruyter Open. This work is licensed under the Creative Commons Attribution-NonCommercial-NoDerivatives 3.0 License. BY-NC-ND 3.0

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