Possible numbers of ones in 0 1 matrices with a given rank

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1 Linear and Multilinear Algebra, Vol, No, 00, Possible numbers of ones in 0 1 matrices with a given rank QI HU, YAQIN LI and XINGZHI ZHAN* Department of Mathematics, East China Normal University, Shanghai 000, China Communicated by RB Bapat (Received 1 October 00) We determine the possible numbers of ones in a 0 1 matrix with given rank in the generic case and in the symmetric case There are some unexpected phenomena The rank symmetric case is subtle Keywords 0 1 Matrix; Rank; Number of ones; Symmetric matrix AMS Classifications 1A0; 1A; 0B0 1 Introduction A 0 1 matrix is a matrix whose entries are either 0 or 1 Such matrices arise frequently in combinatorics and graph theory It is known [1, p ] that the largest number of ones in an n n nonsingular 0 1 matrix is n n þ 1 Interpreting nonsingularity as full rank, we may ask further the question What are the possible numbers of ones in a 0 1 matrix with given rank? We will answer this question in the generic case and in the symmetric case The rank symmetric case is subtle Valiant [] defined the rigidity R A (k) of a matrix A to be the minimal number of entries in the matrix that have to be changed in order to reduce the rank of A to less than or equal to k So our work is along lower bounds on rigidity of explicit matrices See [] In section we prove the main results In section we give some examples *Corresponding author zhan@mathecnueducn Linear and Multilinear Algebra ISSN print ISSN 1-19 online ß 00 Taylor & Francis Group Ltd http//wwwtandfcouk/journals DOI /

2 Q Hu et al Main results THEOREM 1 Let k, n be positive integers with k n There exists an n n 0 1 matrix of rank k with exactly d ones if and only if (i) d ¼ xy for some integers x and y with 1 x n, 1 y n when k ¼ 1; (ii) k d n k þ 1 when k Proof First note that any n n 0 1 matrix of rank k has at least k ones and has at most n k þ 1 ones Thus the condition k d n k þ 1 is always necessary Throughout, we denote by f (A) the number of ones in a 0 1 matrix A (i) Let A be an n n 0 1 matrix of rank 1 Let be any nonzero row of A Then each row is either equal to or equal to 0 Suppose contains y ones and A has x nonzero rows Then f ðaþ ¼xy The if part is obvious (ii) It suffices to prove the if part The case k is covered by Theorem (iii), which follows So we need to prove only the case k ¼ here Let d n 1 For every such d, we will exhibit an n n 0 1 matrix of rank with d ones Let B ¼ Then f ðbþ ¼ Next starting with B we increase the number of ones by one in each step, up to n 1 At the same time all these matrices are of rank, which can be seen by looking at the rows Keeping the entry Bð, 1Þ ¼0 fixed and successively changing the 0 s in the first two rows to 1 s, we obtain the matrix B 1 ¼ Then in B 1 successively setting B 1 ði,1þ¼1, i ¼,,, n we obtain the matrix C 1 ¼ In general we denote by B t the n n 0 1 matrix whose first t þ 1 rows consist of ones and other rows consist of zeros except that the first column has only one nonzero entry

3 at (1, 1) position, and denote by C t the matrix obtained from B t by making the first column all ones except that the (, 1) entry is 0, t ¼ 1,, n 1 Thus B ¼ Note that f ðb Þ¼f ðc 1 Þþ1 Set B ði,1þ¼1 and retain the other entries for i ¼,,, n successively We obtain Observe that Possible numbers of ones in 0 1 matrices with a given rank C ¼ B ¼ and f ðb Þ¼f ðc Þþ1 Repeating the above process by using the last n entries of the first column we finally obtain the matrix C n 1 whose only zero entry is C n 1 ð, 1Þ and f ðc n 1 Þ¼n 1 This completes the proof g Now we turn to the study of symmetric 0 1 matrices To establish the second main result, we first prove two lemmas Denote by A½1,,, kš the principal submatrix of A lying in the first k rows and first k columns LEMMA Let A be a symmetric complex matrix with rankðaþ ¼k If the first k rows of A are linearly independent, then A½1,,, kš is nonsingular Proof In fact it is easy to show the following more general result Let A be an m n complex matrix with rankðaþ ¼k If the first k rows of A are linearly independent and the first k columns of A are linearly independent, then A½1,,, kš is nonsingular This follows since the rank remains unchanged after deleting the last m k rows of A, and then deleting the last n k columns of the resulting matrix g LEMMA Let A be a symmetric 0 1 matrix with rankðaþ ¼ Let and be two linearly independent rows of A and be any row of A Then ¼, or ¼, or ¼ 0

4 8 Q Hu et al Proof ¼ u þ v for some real numbers u, v Since and are linearly independent, it is not hard to show that u, v f0, 1, 1g By simultaneous row and column permutations if necessary, without loss of generality we may suppose that, and are respectively, the first, second, and i th rows By Lemma, A½1, Š must be one of the following four matrices ,,,, which are the only nonsingular symmetric 0 1 matrices Let A ¼ða ij Þ We have a i1 ¼ ua 11 þ va 1, a i ¼ ua 1 þ va, ð1þ a ii ¼ ua i1 þ va i ¼ u a 11 þ uva 1 þ v a ðþ We need to show ðu, vþ fð1, 0Þ, ð0, 1Þ, ð0, 0Þg Suppose this is not the case Then ðu, vþ fð1, 1Þ, ð1, 1Þ, ð 1, 1Þ, ð 1, 1Þ, ð0, 1Þ, ð 1, 0Þg (i) (ii) (iii) (iv) a 11 a 1 a 1 a ¼ 1 1 By (1), we must have u ¼ 1, v ¼ 1 But then by (), a ii ¼ 1, contradicting the fact that A is a 0 1 matrix a 11 a 1 a 1 a ¼ 0 1 By (1), u ¼ v ¼ 1 But then by (), a ii ¼, a contradiction a 11 a 1 a 1 a ¼ 0 1 In the same way as in Case (ii) we have a contradiction a 11 a 1 a 1 a ¼ By (1), u ¼ 1, v ¼ 1 But then by (), a ii ¼ 1, contradictions complete the proof a contradiction The above g

5 Possible numbers of ones in 0 1 matrices with a given rank 9 We remark that the conclusion of Lemma does not hold when rank Consider E ¼ The first three rows of E are linearly independent, and the fourth row is the difference of the first two rows So E is a symmetric 0 1 matrix of rank But the fourth row is not equal to any of the first three rows and it is not a zero row This example can be extended in an obvious way to a matrix of arbitrary rank k and of order k þ 1 THEOREM Let k, n be positive integers with k n There exists an n n symmetric 0 1 matrix of rank k with exactly d ones if and only if (i) d ¼ x for some integer x with 1 x n when k ¼ 1; (ii) 8 >< s t d ¼ s þ t > st when k ¼ ; (iii) k d n k þ 1 when k for some integers s and t with 1 t < s n, or for some integers s and t with s 1, t 1 and s þ t n, or for some integers s and t with s 1, t 1 and s þ t n Proof (i) Let A ¼ða ij Þ be an n n symmetric 0 1 matrix of rank 1 If all the diagonal entries of A are 0, then since A has at least one 1, A has a submatrix of the form ½ 0 1 Š which is nonsingular Thus rankðaþ, contradicts the rank 1 assumption So A has a diagonal entry, say, a ii ¼ 1 Suppose the i th row contains x 1 s Any other row is either a zero row or equal to the i th row If a ji ¼ 0, then the jth row is a zero row; if a ji ¼ 1, then the jth row is equal to the i th row But by the symmetry the i th column contains x 1 s Hence A has x rows equal to the i th row and has the remaining rows equal to 0 Therefore, A has x 1 s Conversely, for 1 x n let J x denote the all-one matrix of order x Then the n n matrix ½ J x Š is a symmetric 0 1 matrix of rank 1 and has x 1 s (ii) Let A ¼ða ij Þ be an n n symmetric 0 1 matrix of rank By simultaneous row and column permutations if necessary, we may suppose that the first two rows are linearly independent By Lemma, A½1, Š is nonsingular Thus A½1, Š is one of the following four matrices 1 1,, , If A½1, Š ¼½ Š, then we may interchange the first two rows and interchange the first two columns so that A½1, Š ¼½ 1 1 Š Therefore we need to consider only the first three possibilities It is easy to check that the conclusions are true for n ¼ Next we assume n

6 0 Q Hu et al Case 1 A½1, Š ¼½ 1 1 Š By Lemma, each row of A is equal to the first row, or the second row, or 0 For every i, ða i1, a i Þ is equal to ð1, 1Þ, or ð1, 0Þ, or ð0, 0Þ Correspondingly the i th row is equal to the first row, or the second row, or 0 Suppose that the first row contains s 1 s and that there are t rows equal to the second row Using the symmetry of A and considering the numbers of 1 s in the first and second columns, we see that 1 t < s n, there are s t rows equal to the first row, and the second row contains s t 1 s The number of 1 s in A is ðs tþs þ tðs tþ ¼s t Denote by J p,q the p q matrix all of whose entries are equal to 1, by 0 p, q the p q zero matrix We write J p and 0 p for J p,p and 0 p, p, respectively Then for any 1 t < s n, the matrix G 1 ¼ 1 1 J 1, s t 1 J 1, t 1, n s J 1, s t 1, t 1, n s J s t 1, 1 J s t 1, 1 J s t 1 J s t 1, t s t 1, n s J t 1, t 1, 1 J t 1, s t t t 1, n s 0 n s, n s, n s, s t n s, t n s is an n n symmetric 0 1 matrix of rank with s t 1 s Case A½1, Š ¼½ 0 1Š Suppose the first row of A contains s 1 s and the second row contains t 1 s Obviously s 1, t 1, and s þ t n by applying Lemma to the columns of A Using the same analysis as in Case 1, we deduce that the number of 1 s in A is s þ t Conversely, for any s 1, t 1withsþtn, the matrix J 1, s 1, t 1, n s t 0 1, s 1 J 1, t 1, n s t G ¼ J s 1, s 1, 1 J s s 1, t s 1, n s t 0 t 1, 1 J t 1, t 1, s 1 J t t 1, n s t 0 n s t, n s t, n s t, s n s t, t n s t is an n n symmetric 0 1 matrix of rank with s þ t 1 s Case A½1, Š ¼½ 0 1 Š Suppose the first row of A contains s 1 s and the second row contains t 1 s Then s 1, t 1, s þ t n Using the same analysis once more as in Case 1, we deduce that the number of 1 s in A is st Conversely, for any s 1, t 1withsþtn, the matrix G ¼ 0 1 J 1, s 1, t 1, n s t 0 1, s 1 J 1, t 1, n s t J s 1, s 1, s 1 J s 1, t s 1, n s t 0 t 1, 1 J t 1, 1 J t 1, s t t 1, n s t 0 n s t, n s t, n s t, s n s t, t n s t is an n n symmetric 0 1 matrix of rank with st 1 s (iii) For any rank k 1, the number d of ones obviously satisfies k d n k þ 1 Let I k be the identity matrix of order k Let H k ¼ðh ij Þ be the k k matrix with h ii ¼ 0 for i ¼,,, n and with all other entries equal to 1 Then H k is nonsingular In fact,

7 Possible numbers of ones in 0 1 matrices with a given rank 1 Hk 1 ¼½ k J 1, k 1 J k 1, 1 I k 1 Š So both ½ I k Š and Z n, k ½ J n k J n k, k J k, n k H k Š are of rank k and they have k and n k þ 1 1 s, respectively This shows that the lower bound k and the upper bound n k þ 1 are attained Now assume k Let P(n, k) denote the proposition that for every positive integer d with k d n k þ 1 there exists an n n symmetric 0 1 matrix of rank k with exactly d ones It remains to prove Pðn, kþ We divide the proof into three steps Step 1 Pðn,Þ is true for all n We use induction on n The following matrices 0 0, , , 0 1 1, are of rank and have,,,, ones, respectively So Pð, Þ holds Denote by k n the set of n n symmetric 0 1 matrices of rank k Suppose Pðn,Þ holds, ie, there are matrices in n with d ones for d n We will show that Pðn þ 1, Þ holds, ie, there are matrices in nþ1 with d ones for d ðnþ1þ The range d n is covered by P(n,) Just add one zero row and one zero column to the attaining matrices of order n Consider J n J n, 1 J n, 1 Z n, ¼ J 1, n 0 1 n J 1, n Our strategy is to change the entries of the submatrix in the right-bottom corner of Z n, and add one row and one column to Z n, so that the resulting matrix is in nþ1 and has the required number of 1 s In the following matrices A j, for each j ¼ 1,,,, A j ½n 1, n, n þ 1Š is nonsingular and hence the last three rows of A j are linearly independent, while every other row is a linear combination of these three rows Thus A j nþ1 Let J n J n, 1 J n, n, 1 J A 1 ¼ 1, n 0 J 1, n , n 0 The number of 1 s in A 1 is f ða 1 Þ¼n 1 Let J n J n, 1 J n, n, 1 J A ¼ 1, n 1 J 1, n 0 0 1, n 0 0 1

8 Q Hu et al Then f ða Þ¼n Let J n J n, 1 J n, n, 1 J A ¼ 1, n 1 J 1, n 1 0 1, n 0 Then f ða Þ¼n þ 1 Let A ¼ J n T a n 1 where ¼ða 1, a,, a n,1,0þ and a i ¼ 0 or 1 which will be specified For any d with n þ d ðnþ1þ, if d ¼ n þ þ p for some nonnegative integer p, set a 1 ¼ a ¼¼a p ¼ 1, a pþ1 ¼¼a n 1 ¼ 0; if d ¼ n þ þ p þ 1 for some nonnegative integer p, set a 1 ¼ a ¼¼a p ¼ a n 1 ¼ 1, a pþ1 ¼¼a n ¼ 0 Then f ða Þ¼d Therefore Pðn þ 1, Þ holds Step P(n, k) implies Pðn þ 1, k þ 1Þ Suppose P(n, k) holds Then for every d with k d n k þ 1 there exists an A d k n with f ða dþ¼d We have B d ½ A d Škþ1 nþ1 and f ðb dþ¼d þ 1 So the range k þ 1 d n k þ is attained by these B d Denote by Z n,k the n n 0 1 matrix whose only zero entries are the last k 1 diagonal entries Our strategy is to construct matrices W j kþ1 nþ1 for j ¼ 1,, based on Z n,k with desired numbers d of 1 s For each j, the principal submatrix of A j lying in the last k þ 1 rows is nonsingular and every other row is a linear combination of the last k þ 1 rows Thus W j kþ1 nþ1 We will omit the verifications of these easily seen facts Let W 1 ¼½ Z n, k T 0 Š where ¼ð0, 0,,0,1Þ Then f ðw 1Þ¼n k þ Let W ¼½ Z n, k T 1 Š where is as mentioned earlier Then f ðw Þ¼n k þ Let J n kþ1 J n kþ1, k 1 T W ¼ J k 1, n kþ1 H k 1! T! a n 1 where H k 1 is the 0 1 matrix of order k 1 whose only zero entries are the last k diagonal entries, ¼ða 1,, a n k,1þ,! ¼ð0, a n kþ1,, a n Þ, and the a i are to be specified For any d with n k þ d ðnþ1þ ðkþ1þþ1, if d ¼ n k þ þ p, set a 1 ¼¼a p ¼ 1, a pþ1 ¼¼a n 1 ¼ 0; if d ¼ n k þ þ p þ 1, set a 1 ¼¼a p ¼ a n 1 ¼ 1, a pþ1 ¼¼a n ¼ 0 Then f ðw Þ¼d Thus we have proved Pðn þ 1, k þ 1Þ Step P(n, k) is true for all n k By Step 1, Pðn k þ, Þ is true Using Step we have the following implications Pðn k þ, Þ )Pðn k þ, Þ))Pðn, kþ This completes the proof g

9 Possible numbers of ones in 0 1 matrices with a given rank Examples Example 1 By Theorem 1 (i), f1,,,,, 8, 9, 1, 1g ¼fdjThere is a 0 1 matrix of rank 1 with d 1 sg Note that,, 10, 11, 1, 1, 1 are missing By Theorem (i), f1,, 9, 1g ¼fdjThere is a symmetric 0 1 matrix of rank 1 with d 1 sg Example By Theorem (ii), f,,,,,, 8, 10, 1, 1g ¼fdjThere is a symmetric 0 1 matrix of rank with d 1 sg Note that 9, 11, 1, 1 are missing Example By Theorem (iii), f,,,,, 8, 9, 10, 11, 1, 1, 1g ¼fdjThere is a symmetric 0 1 matrix of rank with d 1 sg Acknowledgments The authors are grateful to the referees for their careful reading of the manuscript and kind comments Supported by SRF for ROCS, SEM References [1] Halmos, PR, 199, Linear Algebra Problem Book, Mathematical Association of America, Washington, DC [] Valiant, LG, 199, Graph-theoretic arguments in low level complexity In Mathematical Foundations of Computer Science, Lecture Notes in Comput Sci, Vol (Berlin Springer), pp 1 1 [] Shokrollahi, MA, Spielman, DA and Stemann, V, 199, A remark on matrix rigidity Information Processing Letters, (), 8 8