rank inequalities over semirings

J. Korean Math. Soc. 42 (2005), No. 2, pp. 223–241

RANK INEQUALITIES OVER SEMIRINGS

LeRoy B. Beasley and Alexander E. Guterman

Abstract. Inequalities on the rank of the sum and the product oftwo matrices over semirings are surveyed. Preferences are given tothe factor rank, row and column ranks, term rank, and zero-termrank of matrices over antinegative semirings.

1. Introduction

During the past century a lot of literature has been devoted to in-vestigations of semirings. Briefly, a semiring is essentially a ring whereonly the zero element is required to have an additive inverse. Therefore,all rings are also semirings. Moreover, among semirings there are suchcombinatorially interesting systems as the Boolean algebra of subsets ofa finite set(with addition being union and multiplication being intersec-tion), nonnegative integers and reals(with the usual arithmetic), fuzzyscalars(with fuzzy arithmetic), etc. Matrix theory over semirings is anobject of much study in the last decades, see for example [9]. In particu-lar, many authors have investigated various rank functions for matricesover semirings and their properties, see [1, 3, 6, 7, 8, 12] and referencesthere in.

There are classical inequalities for the rank function ρ of sums andproducts of matrices over fields, see, for example [10, 11]:The rank-sum inequalities:

| ρ(A)− ρ(B) |≤ ρ(A + B) ≤ ρ(A) + ρ(B);

Sylvester’s laws:

ρ(A) + ρ(B)− n ≤ ρ(AB) ≤ min{ρ(A), ρ(B)}

Received September 2, 2003.2000 Mathematics Subject Classification: 15A.Key words and phrases: rank functions, semirings, inequalities.The research of the second author was partially supported by RFBR.

224 LeRoy B. Beasley and Alexander E. Guterman

and the Frobenius inequality:

ρ(AB) + ρ(BC) ≤ ρ(ABC) + ρ(B),

where A,B, C are real or complex conformal matrices.These inequalities may or may not hold when S is not a field.In the present paper we compare different rank functions for matrices

over semirings. For these rank functions we investigate the semiringversions of the above mentioned classical inequalities for the sum andproduct of matrices. Numerous examples are given to illustrate thebehaviour of rank functions under consideration. In particular, we showthat our bounds are exact and the best possible. Namely not only pairsof matrices satisfying the case of equality are presented but for any givenr and s it is proved that there exist matrices A and B of ranks r and srespectively such that the equality holds.

Our paper is organized as follows. In section 2 we collect all necessarydefinitions and notations. In section 3 we compare different semiringrank functions for a given matrix. In subsequent sections, we investigateupper and lower bounds on these introduced rank functions for sumsand products of matrices: section 4 is devoted to factor rank, section 5is devoted to term rank, section 6 is devoted to zero-term rank, andsection 7 is devoted to row and column rank functions.

2. Definitions and notations

Definition 2.1. A semiring, S, consists of a nonempty set S andtwo binary operations, addition and multiplication, such that:

• S is an Abelian monoid under addition(identity denoted by 0);• S is a semigroup under multiplication(identity, if any, denoted by

1);• multiplication is distributive over addition on both sides;• s0 = 0s = 0 for all s ∈ S.

In this paper we will always assume that there is a multiplicative identity1 in S.

Definition 2.2. A semiring is called antinegative if no nonzero ele-ment has an additive inverse.

Definition 2.3. Let S be a set of subsets of a given set M contain-ing the empty set and M , the sum of two subsets is their union, andthe product is their intersection. If S is closed under addition(unions)and multiplication(intersections), M is a semiring. The zero element is

Rank inequalities over semirings 225

the empty set and the identity element is the whole set M . We callthese types of semirings, or any semiring which is isomorphic to these,Boolean semirings.

It is straightforward to see that a Boolean semiring is commutativeand antinegative. If S consists of only the empty subset and M , M 6= ∅,then it is called a binary Boolean semiring(or {0, 1}-semiring) and isdenoted by B.

Definition 2.4. A semiring is called a chain semiring if the set S istotally ordered with universal lower and upper bounds and the opera-tions are defined by a + b = max{a, b} and a · b = min{a, b}.

Let Mm,n(S) denote the set of m× n matrices with entries from thesemiring S. Throughout we assume that m ≤ n. The matrix In is then× n identity matrix, Jm,n is the m× n matrix of all ones, Om,n is them×n zero matrix. We omit the subscripts when the sizes of matrices isobvious from the context, and we write I, J , and O, respectively. Thematrix Ei,j , called a cell, denotes the matrix with exactly one nonzeroentry, that being a one in the (i, j) entry. Let Ri denote the matrix whoseith row is all ones and is zero elsewhere, and Cj denote the matrix whosejth column is all ones and is zero elsewhere. Let Uk denote the k × kmatrix of all ones above and on the main diagonal, Lk denote k × kstrictly lower triangular matrix of ones. A line of matrix A is a row orcolumn of the matrix A. We denote by A⊕B the block-diagonal matrix

of the form[

A OO B

]. Note that in this sense the operation ⊕ is not

commutative. We say that the matrix A dominates the matrix B if andonly if bi,j 6= 0 implies that ai,j 6= 0, and we write A ≥ B or B ≤ A. IfA and B are (0, 1)-matrices and A ≥ B we let A \B denote the matrixC where

ci,j ={

0 if bi,j = 1ai,j otherwise .

Below we recall well-known rank concepts for matrices over semirings.The detailed information on this subject can be found in [1, 2, 4, 8].

Definition 2.5. The matrix A ∈ Mm,n(S), A 6= O is said to be offactor rank k (rank(A) = k) if k is the smallest positive integer suchthat there exist matrices B ∈ Mm,k(S) and C ∈ Mk,n(S) such thatA = BC. The factor rank the zero matrix, O, shall be equal to 0.


Proposition 2.6. The factor rank of A is equal to the minimumnumber of factor rank-1 matrices whose sum is A.

Proof. For rank(A) = k and A = BC, B ∈ Mm,k(S) and C ∈Mk,n(S) we have A =

∑ki=1 bici where bi denotes the ith column of

B and ci denotes the ith row of C. Thus the factor rank of A is atleast the minimum number of factor rank-1 matrices whose sum is A. IfA =

∑li=1 bici where bi ∈ Mm,1(S) and ci ∈ M1,n(S) then A = BC

where B = [b1,b2, · · · ,bl] and C =

c1

c2...cl

. Thus, the factor rank of A

is at most the minimum number of factor rank-1 matrices whose sum isA. The proposition follows.

Definition 2.7. A matrix A ∈Mm,n(S) is said to be of term rank k(t(A) = k) if the minimum number of lines needed to include all nonzeroelements of A is equal to k.

Let us denote by tc(A) the least number of columns needed to includeall nonzero elements of A and by tr(A) the least number of rows neededto include all nonzero elements of A.

Definition 2.8. A generalized diagonal of a matrix A ∈ Mm,n(S)is a set of min{m,n} entries of A such that no row or column containstwo of these entries.

Proposition 2.9. [4, Theorem 1.2.1] The term rank of A is themaximum number of nonzero entries in some generalized diagonal of A.

Definition 2.10. The matrix A ∈Mm,n(S) is said to be of zero-termrank k(z(A) = k) if the minimum number of lines needed to include allzero elements of A is equal to k.

A vector space is usually only defined over fields or division rings,and modules are generalizations of vector spaces defined over rings. Wegeneralize the concept of vector spaces to semiring vector spaces definedover arbitrary semirings.

Definition 2.11. Given a semiring S, we define a semiring vector spa-ce, V (S), to be a nonempty set with two operations, addition and scalarmultiplication such that V (S) is closed under addition and scalar mul-tiplication, addition is associative and commutative, and such that forall u and v in V (S) and r, s ∈ S


1. There exists a 0 such that 0 + v = v,2. 1v = v = v1,3. rsv = r(sv),4. (r + s)v = rv + sv, and5. r(u + v) = ru + rv.

Definition 2.12. A set of vectors, S, from a semiring vector space,V (S) is called linearly independent if there is no vector in S that canbe expressed as a nontrivial linear combination of the others. The set islinearly dependent if it is not independent.

Note that, unlike vectors over fields, there are several ways to defineindependence, we will use the definition above.

Definition 2.13. A collection, B, of linearly independent vectors issaid to be a basis of the semiring vector space V (S) if its linear spanis V (S). The dimension of V (S) is a minimal number of vectors in anybasis of V (S).

Definition 2.14. The matrix A ∈Mm,n(S) is said to be of row(resp.column) rank k(r(A) = k) if the dimension of the linear span of therows(resp. columns) of A is equal to k.

Definition 2.15. The matrix A ∈Mm,n(S) is said to be of spanningrow(resp. column) rank k(sr(A) = k) if the minimal number of rows(respcolumns) that span all rows (resp. columns) of A is k.

Definition 2.16. The matrix A ∈Mm,n(S) is said to be of maximalrow(resp. column) rank k(mr(A) = k) if it has k linearly independentrows(resp. columns) and any (k + 1) rows(resp. columns) are linearlydependent.

3. Rank relations

If the semiring S is a subsemiring of a field F , let ρ(A) denote therank of the matrix A as an element of Mm,n(F). If the semiring Scoincides with the field F , then, ρ(A) = rank(A) = r(A) = c(A) =sr(A) = sc(A) = mr(A) = mc(A). Note that ρ(A) is invariant of thefield chosen to contain S. Over more general semirings, the situationis more complicated. Namely, the following inequalities are true formatrices with entries from arbitrary semirings:

Proposition 3.1. Let A ∈Mm,n(S) then


1. rank(A) ≤ min{r(A), c(A)};2. r(A) ≤ sr(A) ≤ mr(A); c(A) ≤ sc(A) ≤ mc(A);3. rank(A) ≤ t(A);4. If S is a subsemiring of a field F then ρ(A) ≤ rank(A).

Proof. 1. See [3, Lemma 2.3].2. Follows directly from the definitions.3. Let A ∈ Mm,n(S). Suppose that the term rank of A is k. By

Konig’s Theorem, see [4, Chapter 1.2], there exist permutation matricesP ∈Mm,m(S) and Q ∈Mn,n(S) such that

PAQ =[

A1 A2

A3 O(m−r)×(n−s)

],

where A1 ∈Mr,s(S), A2 ∈Mr,n−s(S), A3 ∈Mm−r,s(S), and r + s = k.Now, define

X =[

Ir Or×s

O(m−r)×r A3

]and Y =

[A1 A2

Is Os×(n−s)

].

Then PAQ = XY and for B = P tX and C = Y Qt we have A = BC,and B ∈Mm,k(S), C ∈Mk,n(S). Thus rank(A) ≤ k = t(A).

4. It is well-known that the factorization of A into the product ofm×ρ(A) and ρ(A)×n matrices exists over F and there is no factorizationof A into the product of m× k and k × n matrices for k < ρ(A). Thusrank(A) ≥ ρ(A) since S is a subsemiring of F .

Example 3.2.

1. Note that the inequality max{r(A), c(A)} ≤ t(A) does not holdover any antinegative semiring S, since for

A =

1 0 1 11 1 0 00 1 1 0

∈M3,4(S)

one has that r(A) = t(A) = 3, c(A) = 4.2. The inequality min{r(A), c(A)} ≤ t(A) does not hold over Z+,

since for

A =

3 5 75 0 07 0 0

∈M3,3(Z+)

one has that r(A) = c(A) = 3, t(A) = 2.


3. The spanning column rank may actually exceed the column rankover some semirings. For example, we consider A = (3−√7,

√7−

2) ∈ M1,2(Z[√

7]+). Thus sc(A) = 2 since 3 − √7 6= α(√

7 − 2)and α(3 − √

7) 6= √7 − 2 in Z[

√7]+. However, c(A) = 1 since

1 = (3−√7) + (√

7− 2) generates the column space of A, see [8]for the details.

4. The maximal column rank may actually exceed the spanning col-umn rank over some semirings. For example, we consider A =(4 − √

7,√

7 − 2, 1) ∈ M1,3(Z[√

7]+). Thus sc(A) = 1, since 1spans all columns of A. However similar to the previous exampleone can see that mc(A) = 2.

Remark 3.3. If S = B is a binary Boolean semiring then z(A) =t(J \A) for all A ∈Mm,n(S).

4. The factor rank

The inequalities for factor rank differ according to the type of arith-metic within the semiring. Let us show that the standard lower boundfor the rank of sum of two matrices is not valid in general.

Example 4.1. Let S be a Boolean semiring. Then the inequalityrank(A + B) ≥ |rank(A)− rank(B)| need not hold.

Let us consider A = K7 =

0 1 1 1 1 1 11 0 1 1 1 1 11 1 0 1 1 1 11 1 1 0 1 1 11 1 1 1 0 1 11 1 1 1 1 0 11 1 1 1 1 1 0

and B = I7

then, 1 = rank(J7) = rank(A + B) but over any Boolean semiringrank(K7) ≤ 5 due to the factorization

K7 =

1 0 1 0 11 1 0 0 00 1 0 1 10 1 1 1 00 1 1 0 10 0 1 1 11 0 1 1 0

0 0 1 1 1 1 01 0 0 0 0 1 10 1 1 0 0 0 01 1 0 0 1 0 00 0 0 1 0 0 1


(see [5] for more details). Thus |rank(A) − rank(B)| = rank(B) −rank(A) ≥ 7− 5 = 2 > 1 = rank(A + B).

However, the following bounds are true.

Proposition 4.2. Let S be an antinegative semiring, A,B ∈Mm,n(S).Then

1. rank(A + B) ≤ min{rank(A) + rank(B), m, n};

2. rank(A + B) ≥

rank(A) if B = Orank(B) if A = O

1 if A 6= O and B 6= O.

These bounds are exact, the upper bound is the best possible and thelower bound is the best possible over Boolean semirings.

Proof. 1. By Proposition 2.6 we have the first inequality. To provethat this bound is exact and the best possible, for each pair (r, s), 0 ≤r, s ≤ n consider the matrices Ar = Ir ⊕ On−r and Bs = On−s ⊕ Is inthe case m = n. It is routine to generalize the above example to thecase m 6= n.

2. Since A + B = O if and only if both A = O and B = O, wehave rank(A + B) ≥ 1 unless A = B = O, and clearly if A = O,rank(A + B) = rank(B) and the second inequality is established. Forthe exactness in the second inequality, let A = B = E1,1. To show thatthis bound is the best possible, consider the following family of matrices:for each pair (r, s), 0 ≤ r, s ≤ m consider the matrices

Ar =[

Ur Jr,n−r

Jm−r,r Jm−r,n−r

]

and

Bs =[

Js,n−s Ls + Is

Jm−s,n−s Jm−s,s

].

Then rank(Ar) = r, rank(Bs) = s and over Boolean semirings rank(Ar+Bs) = 1 since over Boolean semirings Ar + Bs = Jm,n.

Let us see that the lower bound for the factor rank of a product oftwo matrices is not parallel to Sylvester’s lower inequality.

Example 4.3. Let S be a Boolean semiring,

A =

1 0 0 · · · 01 1 0 · · · 01 0 1 · · ·...

......

. . ....

1 0 0 · · · 1


and B = At, then rank(A) = rank(B) = n and rank(AB) = 1 6≥rank(A) + rank(B)− n = n since AB = J .

The parallel proposition for the Sylvester inequalities is:

Proposition 4.4. Let S be an antinegative semiring, A ∈Mm,n(S)and B ∈Mn,r(S). Then

1. rank(AB) ≤ min{rank(A), rank(B)},2. Assume S has no zero divisors, then

rank(AB) ≥{

0 if rank(A) + rank(B) ≤ n1 if rank(A) + rank(B) > n

.

These bounds are exact, the upper bound is the best possible and thelower bound is the best possible over Boolean semirings that have nosero divisors.

Proof. 1. The first inequality is verified as if the semiring were a field.2. To prove the second part, let us note that if rank(A)+rank(B) ≤ n

it is possible that AB = O, for example if A = E1,1 and B = O1 ⊕ Xfor any matrix X ∈ Mn−1,r−1(S). However, if rank(A) + rank(B) > n,AB 6= O. To see this, suppose that AB = O, then for some permutation

matrix Q, AQ = [A1|O] where A1 has k columns and QtB =[

OB1

]

where B1 has at most n − k rows, since there are no zero divisors inS. But then, rank(A) + rank(B) ≤ k + (n − k) ≤ n, a contradiction.For exactness one can take matrices from Example 4.3. In the casem = n = k in order to show that this bound is the best possible weconsider the family of matrices

Ar =[

Lr Or,n−r

Jn−r,r On−r,n−r

]

and

Bs =[

Us Js,n−s

On−s,s On−s,n−s

]

for each pair (r, s), 0 ≤ r, s ≤ n. Then rank(Ar) = r, rank(Bs) = s andrank(ArBs) = 1 if r, s 6= 0. It is routine to generalize the above exampleto the case m 6= n 6= k.

Example 4.5. The triple (A, I, At), where A is as in Example 4.3,provides a counterexample to Frobenius inequality in the Boolean case.

If S is a subsemiring of <+, the positive real numbers, better boundscan be found. However the standard lower bound for the rank of sumof two matrices is not valid in this case either.


Example 4.6. Let S be a subsemiring of <+. Then the inequalityrank(A + B) ≥ |rank(A)− rank(B)| need not hold.

Let r, s ≥ 4 and s < n− 4. Let us consider

A′ =

1 2 3 41 1 1 11 0 1 00 0 2 2

and

B′ =

0 0 0 00 0 0 00 1 0 10 1 0 1

.

Note that rank(A′) = 4, ρ(A′) = 3, rank(B′) = ρ(B′) = 1 and rank(A′+B′) = ρ(A′ + B′) = 2.

Let

A =

A′ O4,r−4 O4,n−r

Or−4,4 Lr−4 + Ir−4 Or−4,n−r

Om−r,4 Om−r,r−4 Om−r,n−r

and

B =

B′ O4,1 O4,s−1 O4,n−s−4

Os−1,4 Os−1,1 Us−1 Os−1,n−s−4

Om−s−3,4 Om−s−3,1 Om−s−3,s−1 Om−s−3,n−s−4

.

Here we have that rank(A) = r, ρ(A) = r − 1, rank(B) = ρ(B) = sand rank(A + B) =| r − s | −1 < | rank(A) − rank(B) | if S is asubsemiring of <+. Note if r = s + 3, reversing the roles of A′ and B′in A and B also gives the corresponding example.

Proposition 4.7. Let S ⊆ <+, A,B ∈Mm,n(S). Then

1. rank(A + B) ≤ min{rank(A) + rank(B), m, n},2. rank(A + B) ≥ |ρ(A)− ρ(B)|.

These bounds are exact and the best possible.

Proof. 1. By Proposition 4.2 we have the first inequality.2. By Proposition 3.1.4 the second inequality follows. For the exactnessone can take A = E1,1+E1,2+E2,1, B = E2,2. In order to prove that thisbound is the best possible for each pair (r, s), 0 ≤ r, s ≤ m we considerthe family of matrices

Ar =[

Lr + Ir Or,n−r

Om−r,r Om−r,n−r

]


and

Bs =[

0 Us Os,n−s

0 Om−s,s Om−s,n−s

].

Then rank(Ar) = r = ρ(Ar), rank(Bs) = s = ρ(Bs) and rank(Ar+Bs) =| s− r |.

In the following example we see that the standard lower bounds forthe Sylvester and Frobenius inequalities also are not valid in the caseS ⊆ <+.

Example 4.8. Let S = <+ and let us consider A =

0 1 1 11 0 1 11 1 0 41 1 4 0

and B =

1 1 0 01 1 0 00 0 1 10 0 1 1

, then ρ(A) = 3, rank(A) = 4, ρ(B) =

rank(B) = 2 and AB =

1 1 2 21 1 2 22 2 4 42 2 4 4

, so rank(AB) = 1. Thus,

1 = rank(AB) 6≥ rank(A)+rank(B)−n = 4+2−4 = 2 and 6 = 4+2 =rank(AI) + rank(IB) 6≤ rank(AIB) + rank(I) = 1 + 4 = 5.

However the multiplicative upper bounds and the following general-izations of lower bounds are true.

Proposition 4.9. Let S ⊆ <+, A ∈Mm,n(S), B ∈Mn,k(S). Then

1. rank(AB) ≤ min{rank(A), rank(B)},2. rank(AB) ≥

{0 if ρ(A) + ρ(B) ≤ n,

ρ(A) + ρ(B)− n if ρ(A) + ρ(B) > n.


Proof. 1. The justification for the upper bounds is the same as if thesemiring were a field.

2. The lower bounds are the bounds for any real matrices by Proposi-tion 3.1.4. For the exactness in both cases one can take A = E11, B = I.For the proof that lower bound is the best possible we consider

Ar =[

Ir Or,n−r

On−r,r On−r,n−r

]


and

Bs =[

On−s,n−s On−s,s

Os,n−s Is

]

for each pair (r, s), 0 ≤ r, s ≤ n in the case m = n = k. Then rank(Ar) =r, rank(Bs) = s, rank(ArBs) = 0 if r+s ≤ n, and rank(ArBs) = r+s−nif r + s > n. It is routine to generalize the above example to the casem 6= n 6= k.

Proposition 4.10. Let S ⊆ <+, A ∈ Mm,n(S), B ∈ Mn,k(S), andC ∈Mk,l(S). Then

ρ(AB) + ρ(BC) ≤ rank(ABC) + rank(B).

This bound is exact and the best possible.

Proof. We have ρ(AB) + ρ(BC) ≤ ρ(ABC) + ρ(B) by Frobenius’inequality, and ρ(ABC)+ρ(B) ≤ rank(ABC)+rank(B) since rank(X) ≥ρ(X) for all nonnegative real matrices A by Proposition 3.1.4. For theexactness take A = B = C = E1,1. To prove that this bound is thebest possible we consider the following family of matrices: in the casem = n = k = l for given r, s let us take

Ar =[

Ir Or,n−r

On−r,r On−r,n−r

], Cs =

[Is Os,n−s

On−s,s On−s,n−s

], and

Br,s =[

It Ot,n−t

On−t,t On−t,n−t

]

where t is the greatest integer less than r+s2 . Then, if r ≤ s, ρ(ArBr,s)+

ρ(Br,sCs) = r + t = rank(ArBr,sCs) + rank(Br,s) and if s ≤ r thenρ(ArBr,s)+ρ(Br,sCs) = s+t = rank(ArBr,sCs)+rank(Br,s). It is routineto generalize the above example to the case m 6= n 6= k, n 6= l.

5. The term rank

The following inequalities are true for the term rank:

Proposition 5.1. Let S be an arbitrary antinegative semiring. Forany matrices A,B ∈Mm,n(S) we have:

t(A + B) ≤ min{t(A) + t(B),m, n}.This bound is exact and the best possible.


Proof. This inequality follows directly from the definition of termrank. The substitution Ar = Ir ⊕ On−r, Bs = On−s ⊕ Is for each pair(r, s), 0 ≤ r, s ≤ n shows that this bound is exact and the best possiblein the case m = n. It is routine to generalize this example to the casem 6= n.

Example 5.2. A nontrivial additive lower bound for the term rankof a sum does not hold over an arbitrary semiring. It is enough to takeA = B = Jm,n over a field whose characteristics is equal to 2. Thent(A + B) = t(0) = 0.

However for antinegative semirings there is a lower bound for theterm rank of a sum which is better than the one for fields or arbitrarysemirings. Namely, the following is true.

Proposition 5.3. Let S be an antinegative semiring. For any ma-trices A,B ∈Mm,n(S) the following inequality holds:

t(A + B) ≥ max{t(A), t(B)}.This bound is exact and the best possible.

Proof. This inequality follows from the antinegativity of S, i.e., a+b 6=0 for any a, b ∈ S, a 6= 0, and the definition of the term rank. To provethat this bound is exact and the best possible we consider the matricesAr = Ir ⊕ On−r, Bs = Is ⊕ On−s for each pair (r, s), 0 ≤ r, s ≤ n inthe case m = n. It is routine to generalize this example to the casem 6= n.

Example 5.4. A nontrivial multiplicative lower bound does not holdover an arbitrary semiring. It is enough to take A = B = Jn over a fieldwhose characteristic is a divisor of n. Then t(AB) = t(nJn) = 0.

Over an antinegative semiring the Sylvester lower bound holds:

Proposition 5.5. Let S be an antinegative semiring without zerodivisors. Then for any A ∈ Mm,n(S), B ∈ Mn,k(S) the followinginequality holds:

t(AB) ≥{

0 if t(A) + t(B) ≤ n,t(A) + t(B)− n if t(A) + t(B) > n.

This bound is exact and the best possible.

Proof. Let A ∈ Mm,n(S), B ∈ Mn,k(S) be arbitrary matrices,t(A) = tA, t(B) = tB. Then A and B have generalized diagonals withtA and tB nonzero elements, respectively. Denote them by DA and DB,


respectively. Then AB ≥ DADB since S is antinegative. Since theproduct of two generalized diagonal matrices, which have tA and tBnonzero entries, respectively, has at least tA + tB − n nonzero entries,the inequality follows.

In order to show that this bound is exact and the best possible foreach pair (r, s), 0 ≤ r, s ≤ n let us take Ar = Ir⊕On−r, Bs = On−s⊕ Is

in the case m = n. It is routine to generalize this example to the casem 6= n.

Example 5.6. The inequality t(AB) ≤ min(t(A), t(B)) does nothold. It is enough to take A = C1, B = R1. Then t(AB) = t(Jn) = n >1.

However the following inequality is true

Proposition 5.7. Let S be an antinegative semiring. Then for anyA ∈ Mm,n(S), B ∈ Mn,k(S) the inequality t(AB) ≤ min(tr(A), tc(B))holds. This is exact and the best possible bound.

Proof. This inequality is a direct consequence of the definition of theterm rank and antinegativity. The exactness follows from Example 5.6.In order to prove that this bound is the best possible, for each pair(r, s), 0 ≤ r ≤ m, 0 ≤ s ≤ k, consider the family of matrices Ar =E1,1 + . . . + Er,1 and Bs = E1,1 + . . . + E1,s.

Example 5.8. For an arbitrary semiring, the triple (C1, R1, 0) is acounterexample to the term rank version of the Frobenius inequality,since t(C1R1) + t(R10) = n > t(C1R10) + t(R1) = 1. However if S is asubsemiring of <+ the following obvious version is true:

ρ(AB) + ρ(BC) ≤ t(ABC) + t(B)

6. Zero-term rank

Proposition 6.1. Let S be an antinegative semiring. For A,B ∈Mm,n(S) one has that

0 ≤ z(A + B) ≤ min{z(A), z(B)}These bounds are exact and the best possible.

Proof. The lower bound follows from the definition of the zero-termrank function.

In order to check that this is exact and the best possible for eachpair (r, s), 0 ≤ r, s ≤ min{m,n} let us consider the family of matrices


Ar = J \ (r∑

i=1Ei,i), Bs = J \ (

s∑i=1

Ei,i+1) if s < min{m,n} and Bs =

J \ (∑s−1

i=1 Ei,i+1 + Es,1) if s = min{m, n}. Then z(Ar) = r, z(Bs) = sby definition and z(Ar + Bs) = 0 by antinegativity.

The upper bound follows directly from the definition of zero-termrank and from the antinegativity of S. For the proof of its exactness letus take A = J and B = 0. In order to check that this bound is the bestpossible we consider the following family of matrices: for each pair (r, s),0 ≤ r, s ≤ min{m,n} let us consider the matrices Ar = J \ (

∑ri=1 Ei,i)

and Bs = J \ (∑s

i=1 Ei,i).

Proposition 6.2. Let S be an antinegative semiring withour zerodivisors. For A ∈Mm,n(S), B ∈Mn,k(S) one has that

0 ≤ z(AB) ≤ min{z(A) + z(B), k,m}These bounds are exact and the best possible for n > 2.

Proof. The lower bound follows from the definition of the zero-termrank function. In order to show that this bound is exact and the bestpossible let us consider the family of matrices: for each pair (r, s), 0 ≤r ≤ min{m,n}, 0 ≤ s ≤ min{k, n}, we take Ar = J \ (

∑ri=1 Ei,i),

Bs = J \ (∑s

i=1 Ei,i+1) if s < min{k, n} and Bs = J \ (∑s−1

i=1 Ei,i+1 +Es,1) if s = min{k, n}. Then z(Ar) = r, z(Bs) = s by definition and ifn > 2 then ArBs does not have zero elements by antinegativity. Thusz(ArBs) = 0.

The upper bound follows directly from the definition of zero-termrank and from the antinegativity of S.

In order to show that this bound is exact and the best possible let usconsider the family of matrices: for each pair (r, s), 0 ≤ r ≤ min{m, n},0 ≤ s ≤ min{k, n}, we take Ar = J \ (

r∑i=1

Ri) and Bs = J \ (s∑

i=1Ci).

Example 6.3. The triple (C1, I, R1) is a counterexample to the zero-term rank version of the Frobenius inequality, since z(C1) + z(R1) =2n− 2 > z(C1R1) + z(I) = n for n > 2.

7. Row and column ranks

The standard upper bound in the additive inequalities is not validfor row and column ranks.


Example 7.1. Consider

A =

1 0 00 1 01 1 01 0 00 1 0

, B =

0 2 20 2 00 0 20 6 20 4 6

.

Then it is easy to see that r(A) = r(B) = sr(A) = sr(B) = mr(A) =mr(B) = 2. However,

r(A + B) = r

1 2 20 3 01 1 21 6 20 5 6

= 5 = sr(A + B) = mr(A + B)

over Z+.

Proposition 7.2. Let S be an antinegative semiring. Then for 0 6=A,B ∈Mm,n(S) one has that

1 ≤ c(A + B), r(A + B), sr(A + B), sc(A + B), mr(A + B),mc(A + B)

These bounds are exact over any antinegative semiring and the bestpossible over Boolean semirings.

Proof. These inequalities follow directly from the definition of theserank functions and the condition that A, B 6= 0. For the exactnessone can take A = B = E1,1. Let S be a Boolean semiring. For each

pair (r, s), 0 ≤ r, s ≤ m we consider the matrices Ar = J \ (r∑

i=1Ei,i),

Bs = J \(s∑

i=1Ei,i+1) if s < m and Bs = J \(

∑s−1i=1 Ei,i+1)+Es,1 if s = m.

Then

c(Ar) = r(Ar) = sr(Ar) = sc(Ar) = mr(Ar) = mc(Ar) = r,

c(Bs) = r(Bs) = sr(Bs) = sc(Bs) = mr(Bs) = mc(Bs) = s

by definition and Ar + Bs = J has row and column ranks equal to 1.Thus, these bounds are the best possible over Boolean semirings.

Proposition 7.3. Let S be a subsemiring in <+. Then for A,B ∈Mm,n(S) one has that

c(A + B), r(A + B), sr(A + B), sc(A + B),

mr(A + B), mc(A + B) ≥ |ρ(A)− ρ(B)|



Proof. These inequalities follow directly from the fact that ρ(X) ≤r(X), c(X) for all X ∈Mm,n(<+), Proposition 3.1.2, and correspondinginequalities for matrices with coefficients from the field <. For the proofof exactness consider matrices A = E1,1 + . . . + En−1,n−1, B = J \A. In order to show that these bounds are the best possible one cantake the family of matrices Ar, Bs that show that the lower bound inProposition 4.7 is the best possible.

Proposition 7.4. Let S be an antinegative semiring without zerodivisors. For 0 6= A ∈Mm,n(S), 0 6= B ∈Mn,k(S) and c(A)+r(B) > n,one has that

1 ≤ c(AB), r(AB), sr(AB), sc(AB), mr(AB),mc(AB)

These bounds are exact over any antinegative semiring without zerodivisors and the best possible over Boolean semirings without zero divi-sors.

Proof. For an arbitrary antinegative semiring, if c(A) = i and r(B) =j then A has at least i nonzero columns while B has at least j nonzerorows. Thus, if i+j > n, AB 6= O and hence these bounds are established.For the proof of exactness let us take A = B = E1,1.

Let S be a semiring with Boolean arithmetic and without zero divi-sors. In the case m = n = k for each pair (r, s), 1 ≤ r, s ≤ n let us

consider the matrices Ar =r∑

i=1Ei,i +

m∑i=1

Ei,1, Bs =s∑

i=1Ei,i +

n∑i=1

E1,i.

Then

c(Ar) = r(Ar) = sr(Ar) = sc(Ar) = mr(Ar) = mc(Ar) = r,

c(Bs) = r(Bs) = sr(Bs) = sc(Bs) = mr(Bs) = mc(Bs) = s

by definition and ArBs = J . Thus c(ArBs) = r(ArBs) = sr(ArBs) =sc(ArBs) = mr(ArBs) = mc(ArBs) = 1. It is routine to generalize thisexample to the case m 6= n 6= k.

Note that the condition that c(A) + r(B) > n is necessary becausefor A = Ik ⊕O and B = O ⊕ Ij we have AB = O whenever k + j ≤ n.

Proposition 7.5. Let S be a subsemiring in <+. Then for A ∈Mm,n(S) B ∈Mn,k(S) one has that

c(AB), sc(AB), mc(AB),r(AB), sr(AB),mr(AB) ≥

{0 if ρ(A) + ρ(B) ≤ n,

ρ(A) + ρ(B)− n if ρ(A) + ρ(B) > n



Proof. These inequalities follow directly from the fact that ρ(X) ≤r(X), c(X) for all X ∈ Mm,n(<+), Proposition 3.1.2, and correspond-ing inequalities for matrices with coefficients from the field <. For theexactness one can take A = B = I. In order to show that these boundsare the best possible one can take the family of matrices Ar, Bs thatshow that the lower bound in Proposition 4.9 is the best possible.

The following example, given in [8] for the spanning column rank,shows that standard analogs for upper bound of the rank of product oftwo matrices do not work for row and column ranks.

Example 7.6. Let A = (3, 7, 7) ∈ M1,3(Z+), B =

1 1 10 1 10 0 1

.

Then c(A) = sc(A) = mc(A) = 2, c(B) = sc(B) = mc(B) = 3, andc(AB) = c(3, 10, 17) = sc(AB) = mc(AB) = 3.

However, the following upper bounds are true due to [8].

Proposition 7.7. Let S be an antinegative semiring. For A ∈Mm,n(S), B ∈Mn,k(S), one has that

c(AB) ≤ c(B), sc(AB) ≤ sc(B),mc(AB) ≤ mc(B),

r(AB) ≤ r(A), sr(AB) ≤ sr(A),mr(AB) ≤ mr(A)These bounds are exact and the best possible.

Proof. The inequality sc(AB) ≤ sc(B) is proved in [8, Formula 2.4].The other inequalities can be proved in essentially the same way. Forthe proof that these bounds are exact and the best possible, considerAr = Ir ⊕ On−r and Bs = Is ⊕ On−s for each pair r, s, 1 ≤ r, s ≤ nin the case m = n. It is routine to generalize this example to the casem 6= n.

Example 7.8. The triple (A, I,B), where A and B are from Proposi-tion 7.4 is a counterexample to the corresponding Frobenius inequalitiesif r and s are chosen such that r + s > n + 1.

The authors would like to thank Professor Seok-Zun Song for pointingour several errors in an earlier version.

References

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[8] S. -G. Huang and S. -Z. Song, Spanning column ranks and there preservers ofnonnegative matrices, Linear Algebra Appl. 254 (1997), 485–495.

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[10] M. Marcus and H. Minc, A Survey of Matrix Theory and Matrix Inequalities,Allyn and Bacon, Inc. Boston, 1964.

[11] G. Marsaglia and G. P. H. Styan, Equalities and inequalities for ranks of matricesLinear and Multilinear Algebra 2 (1974), 269–292.

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LeRoy B. BeasleyDepartment of Mathematics and StatisticsUtah State UniversityLogan, Utah 84322-4125, USAE-mail : [email protected]

Alexander E. GutermanDepartment of Mathematics and MechanicsMoscow State UniversityMoscow 119992, RussiaE-mail : [email protected]

rank inequalities over semirings

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