On Matrices with Signed Null Spaces

We denote by Q(A) the set of all matrices with the same sign pattern as A. A matrix A has signed null-space provided there exists a set S of sign patterns such that the set of sign patterns of vectors in the null-space of Ã is S for each Ã ∈ Q(A). Some properties of matrices with signed null-spaces are investigated.


Introduction.
The sign of a real number a is defined by 0 if a = 0, and 1 if a > 0.
A sign pattern is a (0, 1, −1)-matrix.The sign pattern of a matrix A is the matrix obtained from A by replacing each entry with its sign.We denote by Q(A) the set of all matrices with the same sign pattern as A.
Let A be an m by n matrix and b an m by 1 vector.The linear system Ax = b has signed solutions provided there exists a collection S of n by 1 sign patterns such that the set of sign patterns of the solutions to Ax = b is S for each A ∈ Q(A) and b ∈ Q(b).This notion generalizes that of a sign-solvable linear system (see [1] and references therein).The linear system, Ax = b, is sign-solvable provided each linear system Ax = b ( A ∈ Q(A), b ∈ Q(b)) has a solution and all solutions have the same sign pattern.Thus Ax = b is sign-solvable if and only if Ax = b has signed solutions and the set S has cardinality 1.
The matrix A has signed null-space provided Ax = 0 has signed solutions.Thus A has signed null-space if and only if there exists a set S of sign patterns such that the set of sign patterns of vectors in the null-space of A is S for each A ∈ Q(A).An L-matrix is a matrix A, with the property that each matrix in Q(A) has linearly independent rows.A square L-matrix is a sign-nonsingular (SN S)-matrix.A totally L-matrix is an m by n matrix such that each m by m submatrix is an SN S-matrix.It is known that totally L-matrices have signed null-spaces [3].We also have the fact as a corollary of some results in this paper.Thus matrices with signed null-spaces generalize totally L-matrices.
A vector is mixed if it has a positive entry and a negative entry.A matrix is rowmixed if each of its rows is mixed.A signing is a nonzero diagonal (0, 1, −1)-matrix.

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A signing is strict if each of its diagonal entries is nonzero.A matrix B is strictly row-mixable provided there exists a strict signing D such that BD is row-mixed.
In this paper, some properties of matrices with signed null-spaces are investigated, and we show that there exists an m by n matrix A with signed null-space such that A contains a totally L-matrix with m rows as its submatrix and the columns of A are distinct up to multiplication by −1 for any n ∈ {m, m + 1, . . ., 2m}.
We use the following standard notation throughout the paper.If k is a positive integer, then k denotes the set {1, 2, . . ., k}.Let A be an m by n matrix.If α is a subset of {1, 2, . . ., m} and β is a subset of {1, 2, . . ., n}, then A[α|β] denotes the submatrix of A determined by the rows whose indices are in α and the columns whose indices are in β.We sometimes use In particular, A(−|β) denotes the submatrix obtained from A by deleting columns whose indices are in β.We write diag(d 1 , d 2 , . . ., d n ) for the n by n diagonal matrix whose (i, i)-entry is d i .Let J m,n denote the m by n matrix, all of whose entries are 1, and let e i denote the column vector, all of whose entries are 0 except for the ith entry, which is 1.

Matrices with signed null-space.
We say that an m by n matrix A = [a ij ] contains a mixed cycle provided there exist distinct i 1 , i 2 , . . ., i k and distinct j 1 , j 2 , . . ., j k such that An m by n (0, 1, −1)-matrix has signed mth compound provided each of its m by m submatrices having term rank m is an SN S-matrix.
We make use of the following results of matrices with signed null-spaces.Theorem 2.1 (see [3]).Let A be a strictly row-mixable m by n matrix.Then the following three conditions are equivalent.
(a) A has signed null-space.(b) A has term rank m, and its mth compound is signed.(c) AD has no mixed cycle for each strict signing such that AD is row-mixed.Theorem 2.2 (see [2], [3]).If a strictly row-mixable matrix A has signed nullspace, then there exist matrices B and C (possibly with no rows) and nonzero vectors b and c such that B and C are strictly row-mixable matrices with signed null-spaces, B b and c C have signed null-spaces, and, up to permutation of rows and columns, The converse also holds.
Let A be an m by n (0, 1, −1)-matrix.The matrix B is conformally contractible to A provided there exists an index k such that the rows and columns of B can be permuted so that B has the form where x = [x 1 , . . ., x m ] T and y = [y 1 , . . ., y m ] T are (0, 1, −1)-vectors such that x i y i ≥ 0 for i = 1, 2, . . ., m, and the sign pattern of x + y is the kth column of A.
Let B be conformally contractible to A. It is known that A has signed null-space if and only if B has signed null-space [3].All matrices we consider from now on are assumed to be (0, 1, −1)-matrices.
Theorem 2.3 (see [4]).Let an m by n matrix A have a k by k + 1 submatrix B whose complementary submatrix in A has term rank m − k.If there is a matrix B * obtained from B by replacing some nonzero entries with 0's if necessary such that J 2,3 is the zero pattern of a matrix obtained from B * by a sequence of conformal contractions, then A does not have signed null-space.
Let M be an m by n strictly row-mixable matrix of the form

M has signed null-space if and only if
has signed null-space.
Proof.Let M have signed null-space, and let C be any m is an SN S-matrix, or C(m + 1|m + 1) has identically zero determinant.Therefore, C is an SN S-matrix, or C has identically zero determinant.Let C contain both the (n − 1)th column and the nth column of A. Then C(m + 1|m + 1) is an SN S-matrix, or C(m + 1|m + 1) has identically zero determinant.If C(m + 1|m + 1) has identically zero determinant, then there exists an s by t zero submatrix of C(m + 1|m + 1) such that s + t = m + 1.From this, it is easy to show that C has a p by q zero submatrix such that p + q = m + 2; i.e., C has identically zero determinant.Let C(m + 1|m + 1) be an SN S-matrix.Since C is conformally contractible to C(m + 1|m + 1), C is also an SN S-matrix.Thus the (m + 1)th compound of A is signed.Since M has signed null-space, the term rank of M is m, and hence the term rank of A is m + 1.Thus A has signed null-space by Theorem 2.1.The converse is trivial.
We say that A is a single extension of M in Proposition 2.4.Proposition 2.4 means that a strictly row-mixable matrix has signed null-space if and only if its single extension has signed null-space. Let be an m by n matrix, and let is also an SN S-matrix, C is an SN S-matrix.
The converse is trivial.
We say that H is a double extension of G in Proposition 2.5.That G should have a row with exactly three ones is necessary in Proposition 2.5.For example, let Then B is a double extension of A that has signed null-space.But B[1, 2, 3, 4|1, 2, 3, 4] is a mixed submatrix of A, and hence B does not have signed null-space.Corollary 2.6.Every totally L-matrix has signed null-space.Proof.From Propositions 2.4 and 2.5, we have the result.Proposition 2.7.Let A be a strictly row-mixable m by n matrix with no duplicate columns up to multiplication by −1.If A has signed null-space and is not conformally contractible to a matrix, then it has at least two rows with exactly three nonzero entries.
Proof.Without loss of generality, we may assume that each row of A has at least three nonzero entries and A has no zero column.Notice that m ≥ 2 comes from the condition.We prove the result by induction on m.Trivially, we have the result for m = 2. Let m ≥ 3.By Theorem 2.2, A can be rearranged as Thus M is a row-mixed matrix with signed null space, which is impossible by Theorem 2.1.Hence p = 1.Therefore, every duplicate column of B is of the form e i or −e i for some i.
Hence B has only one pair of duplicate columns, which are e i or −e i for some i(< k).
The matrix obtained from B by deleting one of the duplicate columns, which are e i or −e i , satisfies the conditions of the hypothesis if its ith row has at least three nonzero entries.This implies that B has at least one row with exactly three nonzero entries.
Similarly, C has a row i with exactly three nonzero entries for some i( = 1).Hence C has at least one row with exactly three nonzero entries.Therefore, A has at least two rows with exactly three nonzero entries.Similarly, in the case in which A[γ|δ] has one of the unit vectors ±e 1 as a column, we have the result.Assume that A[α|β] and A[γ|δ] do not have the unit vectors ±e k and ±e 1 , respectively, as columns.Since b is nonzero, the k by s + 1 matrix B * obtained from A[α|β] by adding e k as a column is a strictly row-mixable matrix with no duplicate columns up to multiplication by −1.Since B has signed null-space, B * also has signed null-space.Applying the similar method above to B * , we can derive that B has at least one row with exactly three nonzero entries.Similarly, C also has at least one row with exactly three nonzero entries.Hence we have the result when k > 1 and l > 1.
Let k = 1.Then s = 1 since the columns of A are distinct up to multiplication by −1.Hence we may assume that A = [a ij ] is of the form 1 c  O C .
If C has no duplicate columns up to multiplication by −1, then we have the result for C by induction, and hence we have the result for A. Let C have duplicate columns up to multiplication by −1.Then the duplicate columns of C are of the form e i or −e i for some i, as we have shown before.This implies that the number of identical columns of C up to multiplication by −1 is at most 3. Therefore, we may assume that the zero pattern of A is of the form , where u = (1, 1, 0), v = (1, 1), w = (1, 0), and x = (1, 1, 1), and the unspecified entries are zero.Let be the set of indices of columns in A corresponding to S T .Then we may also assume that A[γ \ {1}| ] has no duplicate columns up to multiplication by −1, and the columns are also different from the ones of A(1| ) up to multiplication by −1.If S T is vacuous, we are done since l ≥ 3 and every row but the first row of A has at least three nonzero entries.Let only T be vacuous.Notice that each column of S has at least two nonzero entries.Hence each row of S has at most one nonzero entry.For, suppose that a row of S has two nonzero entries.Since the columns of A[γ \ {1}| ] are distinct up to multiplication by −1, we may assume that there exists a submatrix of A whose zero pattern is where * is 0 or 1.By Theorem 2.3, A does not have signed null-space.This is a contradiction.Next, suppose that a row r of A[γ \ {1}| n ] has four nonzero entries.
Since each row of S has at most one nonzero entry and each column of S has at least two nonzero entries, we have a submatrx of A whose zero pattern is which is also impossible by Theorem 2.3.Hence each row of A[γ \ {1}| n ] has exactly three nonzero entries.Thus we have the result when T is vacuous.Let T be nonvacuous.Notice that the submatrix of A corresponding to T is a strictly row-mixable matrix with signed null-space.Let T be the matrix obtained from T by deleting zero columns.Then we may assume that T is of the form [O T ].If the submatrix A of A corresponding to T has no duplicate columns up to multiplication by −1, then A has at least two rows with exactly three nonzero entries by induction.Hence we have the result.Suppose that A has duplicate columns up to multiplication by −1.
It is easy to show that such columns of A have exactly one nonzero entry as we have shown above.We want to show that the number of identical columns of A is at most three.Suppose that there are four identical columns in A up to multiplication by −1.
We may assume that the zero pattern of the submatrix consisting of such duplicate columns of A is of the form Since A[γ \{1}| ] has no duplicate columns up to multiplication by −1, we may assume that A[γ \ {1}| ] has a submatrix whose zero pattern is where * is 0 or 1. Hence we can have a submatrix N of A whose zero pattern is where * is 0 or 1.By Theorem 2.3, A does not have signed null-space.This is a contradiction.Thus we can assume that T is of the form where T 1 is a block diagonal matrix whose diagonal blocks are either and the submatrix of A corresponding to has no duplicate columns up to multiplication by −1.Continuing this process, we can assume that T is of the form where T i = [O T i ] for i = 1, 2, . . ., q and T i are block diagonal matrices whose diagonal blocks are either [1 1] or [1 1 1] for i = 1, 2, . . ., q − 1.Let λ i be the set of indices of rows in A corresponding to T i .Let i and δ i be the set of indices of nonzero columns in A and zero columns in A corresponding columns are nonzero and distinct up to multiplication by −1.If A contains an m by m + 2 totally L-matrix with k double-extensions, then n ≤ 2m − 2k.
Proof.We will prove the result by induction on k.Let T k be an m by m + 2 totally L-matrix with k double-extensions contained in A. Notice that each column of A which does not correspond to T k has exactly one zero entry by Lemma 3.2.If k = 0, then it is known [1] that T k has a signed rth compound for each r = 1, 2, . . ., m. Hence we can have the identity matrix I m as a submatrix of A. Since T 0 has exactly two columns with exactly one nonzero entry, n ≤ m + 2 + (m − 2) = 2m = 2m − 2k.Let k = 0.By Proposition 2.4 and Lemma 3.2, we may assume that A is of the form Then A(m − 1, m|n − 1, n) has signed null-space, and it contains an m − 2 by m totally L-matrix with k − 1 double-extensions.The columns of A(m − 1, m|n − 1, n) are distinct up to multiplication by −1 because, if not, then A 3 has a column of the forms (0, 1) T or (0, −1) T , say, (0, 1) T .Then A has a submatrix Corollary 3.4.Let T be an m by m + 2 totally L-matrix which contains no single-extensions.Then there is no m by n matrix A with signed null-space such that A contains T properly, and the columns of A are nonzero and distinct up to multiplication by −1.
Proof.Let A be an m by n matrix with signed null-space, and let A contain T .Since T contains no single-extensions, l = 0. Hence n ≤ m + l + 2 = m + 2. Hence A = T .
Let M be an m by n matrix of the form in (2.1) with signed null-space, and let A be the m + 1 by n + 2 matrix such that is an m by 2m − 2k matrix whose columns are distinct up to multiplication by −1, and it has signed null-space.
Let j be the index of a row of T k used when a double-extension is done, and suppose that T k does not have e j as a column.[T k e j ] has a submatrix of the form in (3.1), and hence it does not have signed null-space, as we have shown in the proof of Proposition 3.
has identically zero determinant by Theorem 2.1.Thus C is an SN S-matrix, or C has identically zero determinant.Hence we may assume that C does not contain the last column of A. If C contains neither the n − 1th column nor the nth column, then clearly C has identically zero determinant.If C contains only one of the (n−1)th column or the nth column, then C(m + 1|m + 1) is an m by m submatrix of M .Hence C(m + 1|m + 1)

1 . 1 .
where matrices B and C (possibly with no rows) are strictly row-mixable matrices which have signed null-spaces, and vectors b and c are nonzero.B b and c C also have signed null-spaces.Let A[α|β] = B b and A[γ|δ] = c C such that |α| = k, |β| = s, |γ| = l, and |δ| = t.Then k + l − 1 = m and s + t = n.Let k > 1 and l > 1.If A[α|β] has one of the unit vectors ±e k as a column, then we can assume that A[α|β] is of the form B O b Let B have no duplicate columns up to multiplication by −1.By induction, B and hence A have at least two rows with exactly three nonzero entries.Thus we are done.Therefore, we assume that B has duplicate columns up to multiplication by −Then b = 0.If b has at least two nonzero entries, then A[α|β] is a strictly rowmixable matrix with no duplicate columns up to multiplication by −1.Since A[α|β] is not conformally contractible to a matrix, B has at least one row with exactly three nonzero entries.Let b have exactly one nonzero entry.Let the columns 1, 2 of B be a pair of duplicate columns up to multiplication by −1, and let p be the number of nonzero entries in the column 1 of B .Let D be a strict signing such that M = B D is row-mixed.Since B has signed null-space, M has no mixed cycle, and hence the columns 1 and 2 of M must be identical or p = 1.If p ≥ 2, then the matrix M obtained from M by multiplying the column 2 by −1 has a mixed cycle.

. 1 )
which is not an SN S-matrix.Since A contains an m by m + 2 totally L-matrix, the complementary submatrix to B in A has term rank m − 4. Hence A does not have signed null-space by Theorem 2.1.This is a contradiction.Therefore, n − 2 ≤ 2(m − 2) − 2(k − 1) = 2m − 2k − 2 by induction.Thus we have n ≤ 2m − 2k.Let l be the number of single-extensions contained in A. Then we have l = m − 2k − 2. Hence we can restate the result of Proposition 3.3 in terms of l: n ≤ m + l + 2.
3. Let T k be the set of all matrices of the form in (3.2).Notice that columns of A ∈ T k are nonzero and distinct up to multiplication by −1.We can express the m by n matrices A with n = 2m − 2k in Proposition 3.3 in terms of elements of T k .Proposition 3.5.In Proposition 3.3, n = 2m − 2k if and only if there exists a permutation matrix Q such that A is equal to T Q up to multiplication of rows and columns by −1 for some T ∈ T k .Proof.Let A be an m by n matrix such that A = T Q for some permutation matrix Q and T ∈ T k .Then m = 2k + l + 2, and hence n = m +2+l = m +2+(m − 2 − 2k) = 2m − 2k.Conversely, let A be an m by 2m − 2k matrix satisfying the conditions in Proposition 3.3.Let T k be an m by m + 2 totally L-matrix with k double-extensions contained in A. Then there exists a permutation matrix Q and strict signings D, E such that DAQE is a submatrix of matrix T of the form in (3.2) by Lemma 3.2 and the remark above.Since T is an m by 2m − 2k matrix, A = DT Q −1 E. Since T ∈ T k , we have the result.Corollary 3.6.Let m be a positive integer with m ≥ 2, and let n be any integer in {m, m + 1, . . ., 2m}.Then there exists an m by n matrix A with signed null-space such that A contains a totally L-matrix with m rows as its submatrix and the columns of A are nonzero and distinct up to multiplication by −1.Proof.Let n be any integer in {m, m + 1, . . ., 2m}.If n ≤ m + 2, then we can take an m by n totally L-matrix as such a matrix A. If n > m + 2, there exists an m by m + 2 totally L-matrix T n−m−2 with n − m − 2 single-extensions.Hence there exists an m by n matrix A ∈ T n−m−2 which contains T n−m−2 by the remark above.
5. The m by n strictly row-mixable matrix G has signed nullspace if and only if H has signed null-space.Proof.Let G have signed null-space, and let C = [c ij ] be an m + 2 by m + 2 submatrix of H.That is, C = H[ * |β] for some β ⊂ n + 2 .If n + 2 ∈ β, then H[ m + 1 |β \ {n + 2}] is an SN S-matrix, or it has identically zero determinant since H(m + 2|n + 2) is a single extension of G. Hence C is an SN S-matrix, or C has identically zero determinant.Similarly, we can show that C is an SN S-matrix or C has identically zero determinant if n + 1 ∈ β.Hence we may assume that β contains neither n+1 nor n+2.Then it is easy to show that C has identically zero determinant if β contains at most two among n − 2, n− 1, and n.Let {n − 2, n− 1, n} ⊂ β.Then H[ m |β \ {n − 1, n}] is an SN S-matrix or it has identically zero determinant since G has signed null-space.If H[ m |β \ {n − 1, n}] has identically zero determinant, then clearly C has identically zero determinant.Let H[ m |β \ {n − 1, n}] be an SN S-matrix.Then H is conformally contractible to M , A(−|n + 2) has signed null-space.< i 2 < • • • < i l be the set of indices of rows used when single-extensions are constructed in T k .Notice that T k does not contain any e ij , j = 1, 2, . . ., l.The remark above and Proposition 2.5 imply that T = [T k e i1 e i2 . . .e i l ] (3.2) Since M has signed null-space, A has signed null-space by Theorem 2.1.Let T k be an m by m + 2 totally L-matrix with k double-extensions.Let {i 1 , i 2 , . . ., i l } with i 1