The Boolean rank of a nonzero m × n Boolean matrix A is the least integer k such that there are an m × k Boolean matrix B and a k × n Boolean matrix C with A = BC . In 1984, Beasley and Pullman showed that a linear operator preserves the Boolean rank of any Boolean matrix if and only if it preserves Boolean ranks 1 and 2. In this paper, we extend this characterization of linear operators that preserve the Boolean ranks of Boolean matrices. We show that a linear operator preserves all Boolean ranks if and only if it preserves two consecutive Boolean ranks if and only if it strongly preserves a Boolean rank k with 1 ≤ k ≤ min{ m,n }. The binary Boolean algebra consists of the set = {0, 1} equipped with two binary operations, addition and multiplication. The operations are defined as usual except that 1 + 1 = 1. Let denote the set of all m × n Boolean matrices with entries in . The usual definitions for adding and multiplying matrices apply to Boolean matrices as well. Throughout this paper, we shall adopt the convention that 3 ≤ m ≤ n unless otherwise specified. The ( Boolean ) rank, b(A) , of nonzero A ∈ is the least integer k such that there are Boolean matrices B ∈ and C ∈ with A = BC . It follows that 1 ≤ b(A) ≤ m for all nonzero A ∈ . The Boolean rank of the zero Boolean matrix is 0. A mapping T : → is called a linear operator if T ( αA + βB ) = αT(A) + βT(B) for all A,B ∈ and for all α,β ∈ . A linear operator T on is called a ( P,Q )- operator if there are permutation matrices P and Q of orders m and n , respectively, such that T(X) = PXQ for all X , or m = n and T(X) = PX^tQ for all X , where X^t is the transpose of X . Let 1 ≤ k ≤ m . For a linear operator T on , we say that Beasley and Pullman ( [1] ) have characterized linear operators on that preserve Boolean rank as follows: Theorem 1.1. For a linear operator T on , the following are equivalent : The characterization of linear operators on vector space of matrices which leave functions, sets or relations invariant began over a century ago when in 1897 Fröbenius [7] characterized the linear operators that leave the determinant function invariant. Since then, several researchers have investigated the preservers of nearly every func- tion, set and relation on matrices over fields. See [6 , 7] for an excellent survey of Linear Preserver Problems through 2001. For Boolean matrix and Boolean rank are important research topics on matrix theory. See [4 , 5] for detailed contents and applications on Boolean matrix theory. Recently Beasley and Song ( [3] ) have obtained a new characterization of Theorem 1.1: For a linear operator T on , T preserves Boolean rank if and only if T preserves Boolean ranks 1 and k , where 1 ＜ k ≤ m . They also have obtained characterizations of the linear transformations that preserve term rank between different matrix spaces over semirings containing the binary Boolean algebra in [2] . In this paper, we extend Theorem 1.1 to any two consecutive Boolean rank preservers. Furthermore we obtain other characterizations. The matrix O is an arbitrary zero matrix and J_m,n is the m × n matrix all of whose entries are 1. A matrix in is called a cell if it has exactly one 1 entry. We denote the cell whose one 1 entry is in the (i, j)^th position by E_i,j . Further we let ε_m,n be the set of all cells in . That is, ε_m,n = { E_i,j | 1 ≤ i ≤ m , 1 ≤ j ≤ n }. If A and B are Boolean matrices in , we say that A dominates B (written B ⊑ A or A ⊒ B ) if a_i,j = 0 implies b_i,j = 0 for all i and j . This provides a reflexive and transitive relation on . For Boolean matrices A and B in with B ⊑ A , we define A ＼ B to be the Boolean matrix C such that c_i,j = 1 if and only if a_i,j = 1 and b_i,j = 0 for all i and j . Lemma 2.1 ( [1] ). If T is a linear operator on , then T is invertible if and only if T permutes ε_m,n . A Boolean matrix L ∈ is called a line matrix if L = or L = for some i ∈ {1, …, m } or for some j ∈ {1, …, n }: R_i = is the i th row matrix and C_j = is the j th column matrix . For a linear operator T on , we say that T preserves line matrices if T(L) is a line matrix for every line matrix L . Lemma 2.2. Let T be an invertible linear operator on . Then T preserves line matrices if and only if T is a (P,Q)-operator . Proof . By Lemma 2.1, T permutes ε_m,n and hence T(J_m,n) = J_m,n . Let T preserve all line matrices. Now we will claim that either If m ≠ n , (1) is satisfied since T is invertible and preserves all line matrices. Thus we assume that m = n . Suppose that the claim is not true. Then there are distinct row matrices R_i and R_j (or column matrices C_i and C_j ) such that T(R_i) is a row matrix and T(R_j) is a column matrix. But then T(J_m,n) = T(R₁) +⋯ T(R_i) +⋯+ T(R_j) +⋯+ T(R_n) cannot dominate J_m,n . This contradicts T(J_m,n) = J_m,n . Hence the claim is true. Case (1): We note that T(R_i) = R_α(i) for all i and T(C_j) = C_β(j) for all j , where α and β are permutations of {1, …, m } and {1, …, n }, respectively. Then for any cell E_i,j , we have T(E_i,j) = E_α(i),β(j) . Let P and Q be the permutation matrices corresponding to α and β , respectively. Then for any Boolean matrix we have Hence T is a ( P,Q )-operator. Case (2): We note that m = n , T(R_i) = C_α(i) for all i and T(C_j) = R_β(j) for all j , where α and β are permutations of {1, ⋯, n }. By a parallel argument similar to Case (1), we obtain that T(X) is of the form T(X) = PX^tQ , and thus T is a ( P,Q )-operator. The converse is obvious. □ For nonzero A ∈ , it is well known ( [1] ) that b(A) is the least integer k such that A is the sum of k Boolean matrices of Boolean rank 1. This establishes the following: Lemma 2.3. For Boolean matrices A and B in , we have b(A + B) ≤ b(A) + b(B). Theorem 2.4. Let T be an invertible linear operator on and 1 ≤ k ≤ m . Then T preserves Boolean rank k if and only if T is a (P;Q)-operator . Proof . By Lemma 2.1, T permutes ε_m,n . Assume that T preserves Boolean rank k . Now, we will show that T preserves line matrices, and then T is a ( P , Q )-operator by Lemma 2.2. For the case of k = 1, it is clear that T preserves line matrices since the Boolean rank of every line matrix is 1. Thus we assume that k ≥ 2. Suppose that T does not preserve a line matrix. Then there are two distinct cells E_i,j and E_s,t that are not dominated by the same line matrix such that T(E_i,j) and T(E_s,t) are dominated by the same line matrix. Without loss of generality, we assume that T ( E _1,1 + E _2,2 ) = E _1,1 + E _1,2 . So, we have a contradiction for the case of k = 2. Hence we assume that k ≥ 3. Then for the Boolean matrix A = E _3,3 + ⋯ + E_k,k , we have b ( E _1,1 + E _2,2 + A ) = k . But by Lemma 2.3, b(T(E_1,1 + E_2,2 + A)) ≤ b(T(E_1,1 + E_2,2)) + b(T(A)) ≤ 1 + (k − 2) = k − 1, a contradiction to the fact that T preserves Boolean rank k . Hence T preserves line matrices. The converse is obvious. □ An operator T on is singular if T(X) = O for some nonzero X ∈ ; otherwise T is nonsingular . In fact, if T is a singular linear operator on , then we can easily check that T(E) = O for some cell E . Further, if T is a ( P,Q )-operator on , then T is nonsingular. Example 3.1. For 1 ≤ k ≤ m , let A = E _1,1 + E _2,2 + ⋯ + E_k,k 2 . Define an operator T on by T(O) = O and T(X) = A for all nonzero X 2 . Clearly, T is linear, nonsingular and preserves Boolean rank k , while T does not preserve Boolean rank. The number of nonzero entries of a Boolean matrix A ∈ is denoted by #( A ). Lemma 3.2. Let 1 ≤ k ＜ m and 1 ≤ l ≤ m . Assume that T is a linear operator on . If then T is nonsingular. Proof . If T is singular, then T(E) = O for some cell E . Hence we have a contradiction for the case of k = l = 1. Thus we assume that k, l ≥ 2. Now, choose Boolean matrices A and B with E ⊑ A and E ⊑ B such that b(A) = #( A ) = k + 1 and b(B) = #( B ) = l . It follows that b ( A ＼ E ) = k and b ( B ＼ E ) = l − 1. But then T(A) = T ( A ＼ E ) + T(E) = T ( A ＼ E ) contradicts the condition (i) and T(B) = T ( B ＼ E ) + T(E) = T ( B ＼ E ) contradicts the condition (ii). Hence T is nonsingular. □ Lemma 3.3. Let T be a linear operator on . If then T maps cells to cells. Proof . If T preserves Boolean rank k and k + 1 with 1 ≤ k ≤ m − 1, or T strongly preserves Boolean rank k with 1 ≤ k ≤ m , then T is nonsingular by Lemma 3.2. Suppose that T does not map cells to cells, in particular suppose that T(E) dominates two cells for some cell E . By permuting we may assume that T(E) ⊒ E + F for some cell F ≠ E . If E and F are in the same row, we may assume by permuting that E = E _1,k+1 and F = E _1,k . If E and F are in the same column, we may assume by permuting that E = E _k+1,1 and F = E _k,1 . If E and F are in different rows and different columns, we may assume by permuting that E = E _1,k+1 and F = E _2,k−1 . For 1 ≤ r ≤ m , let W_r = [ ] where = 0 if and only if i + j ≤ r . Then b(W_r) = r . Since E ⊑ W _k+1 and F ⋢ W _k+1 , we have that b ( W _k+1 + E ) = k +1 and b ( W _k+1 + F ) = k . Let L = T^d where d is chosen so that L is idempotent ( L ² = L ). Then, L preserves Boolean ranks k and k + 1 for case (i), or L strongly preserves Boolean rank k for case (ii) and L(E) ⊒ E + F . Now, since L(E) = F + X for some Boolean matrix X , L(E) + F = (X + F) + F = X + F = L(E) and since L is idempotent, L(E) = L²(E) = L(L(E)) = L(L(E) + F) = L²(E) + L(F) = L(E) + L(F) = L(E + F). That is, L ( E + F ) = L(E) . Thus if Y is any Boolean matrix which dominates E , we have that L ( Y + F ) = L(Y) since L ( Y + F ) = L ( Y + E + F ) = L(Y) + L ( E + F ) = L(Y) + L(E) = L ( Y + E ) = L(Y) . Thus, L(W_k+1) = L(W_k+1 + F). However, b ( W _k+1 ) = k + 1, b ( W _k+1 + F ) = k and L preserves both Boolean rank k and k +1 for case (i) or L strongly preserves Boolean rank k for case (ii). Thus, we have k + 1 = b(L(W_k+1)) = b(L(W_k+1 + F)) = k, which is a contradiction for the both cases (i) and (ii). Therefore T maps cells to cells. □ Theorem 3.4. Let T be a linear operator on . Then T preserves Boolean rank if and only if Proof . Let T preserve Boolean ranks k and k +1 or T strongly preserves Boolean rank k . Then T maps cells to cells by Lemma 3.3. Now, suppose that T is not invertible. Then T(E) = T(F) for some distinct cells E and F by Lemma 2.1. If b ( E + F ) = 2, choose a Boolean matrix A ∈ with b(A) = #( A ) = k −1 such that b ( E + A ) = k and b ( E + F + A ) = k + 1. But then k + 1 = b ( T ( E + F + A )) = b ( T ( E + A )) = k , a contradiction for both cases (i) and (ii). For the case of b ( E + F ) = 1, we may assume, without loss of generality, that E = E _1,1 and F = E _1,2 . Let B = E _2,1 + E _2,2 + E _3,3 + ⋯ + E _k+1,k+1 . Then b ( E + F + B ) = k and b ( E + B ) = k + 1. But then k = b ( T ( E + F + B )) = b ( T ( E + B )) = k + 1, a contradiction for both cases (i) and (ii). Thus T is invertible. By Theorem 2.4, T is a ( P, Q )-operator and hence T preserves Boolean rank by Theorem 1.1. The converse is obvious. □ Recently Beasley and Song ( [3] ) showed that for a linear operator T on , T preserves Boolean rank if and only if T preserves Boolean ranks 1 and k , where 2 ≤ k ≤ m . Now we summarize our results by: Theorem 3.5. Let T be a linear operator on . Then the following are equivalent : As a concluding remark, we suggest to prove the following conjecture: Conjecture 3.6. Let T be a linear operator on . Then T preserves Boolean rank if and only if T preserves any two Boolean ranks h and k with 1 ≤ h ＜ k ≤ m ≤ n .