Factor-set of binary matrices and Fibonacci numbers
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Author's personal copy Applied Mathematics and Computation 236 (2014) 235–238
Contents lists available at ScienceDirect
Applied Mathematics and Computation journal homepage: www.elsevier.com/locate/amc
Factor-set of binary matrices and Fibonacci numbers Krasimir Yordzhev Faculty of Mathematics and Natural Sciences, South–West University, 2700 Blagoevgrad, Bulgaria
a r t i c l e
i n f o
a b s t r a c t The article discusses the set of square n n binary matrices with the same number of 1’s in each row and each column. An equivalence relation on this set is introduced. Each binary matrix is represented using ordered n-tuples of natural numbers. We are looking for a formula which calculates the number of elements of each factor-set by the introduced equivalence relation. We show a relationship between some particular values of the parameters and the Fibonacci sequence. Ó 2014 Elsevier Inc. All rights reserved.
Keywords: Fibonacci number Binary matrix Equivalence relation Factor-set
1. Introduction A binary (or boolean, or (0,1)-matrix) is a matrix whose all elements belong to the set B ¼ f0; 1g. With Bn we will denote the set of all n n binary matrices. Let n and k be integers such that 0 6 k 6 n; n P 2. We let Kkn denote the set of all n n binary matrices in each row and each column of which there are exactly k in number 1’s. Let us denote with kðn; kÞ ¼ jKkn j the number of all elements of Kkn . There is not any known formula to calculate the kðn; kÞ for all n and k. There are formulas for the calculation of the function kðn; kÞ for each n for relatively small values of k, more specifically, for k ¼ 1; k ¼ 2 and k ¼ 3. We do not know any formula to calculate the function kðn; kÞ for k > 3 and for all positive integer n. It is easy to prove the following well-known formula
kðn; 1Þ ¼ n!: The following formula
kðn; 2Þ ¼
X 2x2 þ3x3 þþnxn ¼n
ðn!Þ2 xr r¼2 xr !ð2rÞ
Qn
is well known [8]. One of the first recursive formulas for the calculation of kðn; 2Þ appeared in [1] (see also [4, p. 763]).
h i kðn; 2Þ ¼ 1 nðn 1Þ2 ð2n 3Þkðn 2; 2Þ þ ðn 2Þ2 kðn 3; 2Þ for n P 4; 2 kð1; 2Þ ¼ 0; kð2; 2Þ ¼ 1; kð3; 2Þ ¼ 6: Another recursive formula for the calculation of kðn; 2Þ occurs in [3].
2 n kðn; 2Þ ¼ ðn 1Þnkðn 1; 2Þ þ ðn1Þ kðn 2; 2Þ for n P 3; 2 kð1; 2Þ ¼ 0; kð2; 2Þ ¼ 1:
E-mail address: [email protected] http://dx.doi.org/10.1016/j.amc.2014.03.073 0096-3003/Ó 2014 Elsevier Inc. All rights reserved.
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K. Yordzhev / Applied Mathematics and Computation 236 (2014) 235–238
The next recursive system is to calculate kðn; 2Þ.
kðn þ 1; 2Þ ¼ nð2n 1Þkðn; 2Þ þ n2 kðn 1; 2Þ pðn þ 1Þ; n P 2; 2 2 pðn þ 1Þ ¼ n ðn1Þ ½8ðn 2Þðn 3Þkðn 2; 2Þ þ ðn 2Þ2 kðn 3; 2Þ 4pðn 1Þ; n P 4; 4 kð1; 2Þ ¼ 0; kð2; 2Þ ¼ 1; pð1Þ ¼ pð2Þ ¼ pð3Þ ¼ 0; pð4Þ ¼ 9; where pðnÞ identifies the number of a special class of K2n -matrices [9]. The following formula is an explicit form for the calculation of kðn; 3Þ.
kðn; 3Þ ¼
n!2 X ð1Þb ðb þ 3cÞ!2a 3b ; 6n a!b!c!2 6c
where the sum is done as regards all ðnþ2Þðnþ1Þ solutions in nonnegative integers of the equation a þ b þ c ¼ n [7]. As it is noted 2 in [6], this formula does not give us good opportunities to study behavior of kðn; 3Þ. Let A; B 2 Kkn . We will say that A B, if A is obtained from B by moving some rows and/or columns. Obviously, the relation defined like that is an equivalence relation. We denote with
lðn; kÞ ¼ Kkn= ;
ð1Þ
the number of equivalence classes on the above defined relation. Problem 1. Find lðn; kÞ for given integers n and k; 1 6 k < n. The task of finding the cardinal number lðn; kÞ of the factor set Kkn= for all integers n and k; 1 6 k 6 n is an open scientific problem. This is the subject of discussion in this article. We will prove that there is a relationship between some particular values of the parameters of lðn; kÞ of and the Fibonacci sequence. Specifically, we will prove that if k ¼ n 2 then lðn; kÞ coincides with the Fibonacci numbers for all n 2 N; n P 2. 2. Canonical binary matrices Let N be the set of natural numbers and let
T n ¼ ha1 ; a2 ; . . . ; an i j ai 2 N; 0 6 ai 6 2n 1; i ¼ 1; 2; . . . ; n :
ð2Þ
With ‘‘ xv and let A0 be obtained from A by changing the locations of rows with the number u and v, while the remaining rows stay in their places. Then cðA0 Þ < cðAÞ. (b) Let yu > yv and let A0 be obtained from A by changing the locations of columns with the number u and v, while the remaining columns stay in their places. Then rðA0 Þ < rðAÞ.
Proof. (a) It is easy to see that rðA0 Þ < rðAÞ. Let A ¼ ½aij nn , where aij 2 f0; 1g; 1 6 i; j 6 n. Then the representation of xu and xv in binary notation be respectively as follows:
xu ¼ au1 au2 aun ; xv ¼ av 1 av 2 av n :
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K. Yordzhev / Applied Mathematics and Computation 236 (2014) 235–238
where auj ; av j 2 f0; 1g; j ¼ 1; 2; . . . ; n. If xu or xv can be written with less than n digits in binary notation then this number is completed from the left with the necessary number of zeros. Since xv < xu , then there is an integer m 2 f1; 2; . . . ; kg, such that auj ¼ av j for j < m; aum ¼ 1 and av m ¼ 0. Let cðAÞ ¼ hy1 ; y2 ; . . . ; yk i and let cðA0 Þ ¼ hz1 ; z2 ; . . . ; zk i. Then yj ¼ zj for j < m. Let the representation of ym and zm in binary notation be respectively as follows:
ym ¼ a1m a2m au1 m aum av m anm ; zm ¼ a1m a2m au1 m av m aum anm : Since aum ¼ 1 and av m ¼ 0, then zm < ym , whence it follows that cðA0 Þ < cðAÞ. (b) Analogously to (a). h
Corollary 1. Let A 2 Bn . Then rðAÞ is the minimal element in the set frðBÞ j B Ag if and only if cðAÞ is the minimal element in the set fcðBÞ j B Ag. h Corollary 1 gives us reason to formulate the following definition: Definition 2. The matrix A 2 Bn we will call a canonical matrix if rðAÞ is the minimal element in the set frðBÞ j B Ag and cðAÞ is the minimal element in the set fcðBÞ j B Ag.
Let M is arbitrary subset of Bn . Obviously, there is a unique canonical matrix in every equivalence class of factor-set M= and if rðAÞ ¼ hx1 ; x2 ; . . . ; xn i; cðAÞ ¼ hy1 ; y2 ; . . . ; yn i then x1 6 x2 6 6 xn and y1 6 y2 6 6 yn . Therefore the number of canonical matrices in M gives the cardinality of the factor-set M= . Lemma 2. If A ¼ ½aij 2 Kkn is a canonical matrix then
a1 1 ¼ a1 2 ¼ ¼ a1 nk ¼ 0; a1 nkþ1 ¼ a1 nkþ2 ¼ ¼ a1 n ¼ 1; a1 1 ¼ a2 1 ¼ ¼ ank 1 ¼ 0; ankþ1 1 ¼ ankþ2 1 ¼ ¼ an 1 ¼ 1; Proof. Immediately. h Corollary 2. If A ¼ ½aij 2 Kkn is a canonical matrix and rðAÞ ¼ hx1 ; x2 . . . xn i; cðAÞ ¼ hy1 ; y2 ; . . . yn i, then x1 ¼ y1 ¼ 2k 1.
h
3. The function lðn; kÞ and Fibonacci numbers Obviously, when k ¼ 0, the ‘‘zero’’ n n matrix ½0nn is the only matrix in the set K0n . When k ¼ n, there is only one n n binary matrix of Knn , and this is the matrix ½1nn all elements of which are equal to 1. Therefore
lðn; 0Þ ¼ lðn; nÞ ¼ 1:
ð4Þ
When k ¼ 1 if A 2 K1n is a canonical matrix, then rðAÞ ¼ h1; 2; 4; . . . ; 2n1 i, i.e. Ais a binary matrix with 1 in the second (not leading) diagonal and 0 elsewhere. For k ¼ n 1, if A 2 Kn1 is a canonical matrix, then A is a binary matrix with 0 in the n leading diagonal and 1 elsewhere. Hence
lðn; 1Þ ¼ lðn; n 1Þ ¼ 1:
ð5Þ 1
As is well known [2,5], the sequence ffn gn¼0 of Fibonacci numbers is defined by the recurrence relation
f0 ¼ f1 ¼ 1;
f n ¼ fn1 þ fn2
for n ¼ 2; 3; . . .
In this section, we will prove that the sequence flðk þ 1
ð6Þ 1 2; kÞgk¼0
coincides with the Fibonacci sequence (6). 1
Theorem 1. Let the sequence flðk þ 2; kÞgk¼0 is defined by (1) where n ¼ k þ 2, and let ffk gk¼0 be the Fibonacci sequence (6). Then for every integer k ¼ 0; 1; 2; 3; . . . the equality
lðk þ 2; kÞ ¼ fk is true.
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K. Yordzhev / Applied Mathematics and Computation 236 (2014) 235–238
Proof. When k ¼ 0 the assertion follows from (6) and (4). When k ¼ 1 the assertion follows from (6) and (5). When k ¼ 2 there are two canonical matrices A1 ; A2 2 K24 such that rðA1 Þ ¼ h3; 3; 12; 12i and rðA2 Þ ¼ h3; 5; 10; 12i. This fact is verified immediately. Therefore
lð2; 0Þ ¼ f0 ; lð3; 1Þ ¼ f1 and lð4; 2Þ ¼ f2 : Let k be an arbitrary positive integer such that k P 3 and let A ¼ ½ai j 2 Kkkþ2 ; 1 6 i; j 6 k þ 2 be a canonical matrix. Then, according to Lemma 2 a1 1 ¼ a1 2 ¼ a2 1 ¼ 0 and a1 3 ¼ a1 4 ¼ ¼ a1 n ¼ a3 1 ¼ a4 1 ¼ ¼ an 1 ¼ 1. Therefore, the following two cases are possible: (i) a2 2 ¼ 0, i.e., A is of the form
2
0
0 1 1
3
60 0 1 17 7 6 7 6 7 1 1 A¼6 7 6 7 6 .. .. 5 4. . B 1 1 We denote by M1 the set of all canonical matrices of this kind. Let A be an arbitrary matrix of M1 . In A, we remove the first and second rows and the first and second columns. We obtain the matrix B 2 Kk2 k . It is easy to see that Bis the canonical matrix. Conversely, let B ¼ ½bi j 2 Kkk2 (k P 3) and let B be a canonical matrix. From B we obtain the matrix A ¼ ½ai j 2 Kkkþ2 as follows: a1 1 ¼ a1 2 ¼ a2 1 ¼ a2 2 ¼ 0; a1 j ¼ a2 j ¼ 1; 3 6 j 6 k þ 2 and ai 1 ¼ ai 2 ¼ 1; 3 6 i 6 k þ 2. For each i; j 2 f3; 4; . . . ; k þ 2g we assume ai j ¼ bi2 j2 . It is easy to see that the so obtained matrix A is a canonical matrix of the set Kkkþ2 . Therefore, jM1 j ¼ lðk; k 2Þ for any integer k P 3. (ii) a2 2 ¼ 1, i.e., A is of the form
2
0
0 1 1 1
3
60 1 0 1 17 7 6 7 6 7 61 0 7 6 A¼6 7 7 61 1 7 6. . 7 6. . 5 4. . 1 1 Let M2 be the set of all canonical matrices of this kind and let A ¼ ½ai j ; a2 2 ¼ 1 be an arbitrary matrix of M2 . We change a2 2 from 1 to 0 and remove the first row and the first column of A. In this way we obtain a matrix of Kk1 kþ1 , which is easy to see that it is canonical. k1 Conversely, let B ¼ ½bi j 2 Kkþ1 and let B be a canonical matrix. According to Lemma 2 b1 1 ¼ b1 2 ¼ b2 1 ¼ 0. We change b1 1 from 0 to 1. In B, we add a first row and a first column and get the matrix A ¼ ½ai j 2 Kkkþ2 , such that a1 1 ¼ a1 2 ¼ a2 1 ¼ 0; a1 j ¼ 1 for j ¼ 3; 4; . . . ; k þ 2; ai 1 ¼ 1 for i ¼ 3; 4; . . . ; k þ 2 and asþ1 tþ1 ¼ bs t for s; t 2 f1; 2; . . . ; k þ 1g. It is easy to see that the resulting matrix A is canonical and A 2 M2 . Therefore, jM2 j ¼ lðk þ 1; k 1Þ for every integer k P 3. If M is the set of all canonical matrices, M # Kkkþ2 , then obviously
M1 \ M2 ¼ ; and M1 [ M2 ¼ M: Therefore
lðk þ 2; kÞ ¼ jMj ¼ jM1 j þ jM2 j ¼ lðk; k 2Þ þ lðk þ 1; k 1Þ; for all integers k P 3, which proves the theorem.
h
References [1] [2] [3] [4] [5] [6] [7] [8] [9]
H. Anand, V.C. Dumir, H. Gupta, A combinatorial distribution problem, Duke Math. J. (33) (1966) 757–769. K. Atanassov, V. Atanassova, A. Shannon, J. Tumer, New Visual Perspectives on Fibonacci Numbers, World Scientific, 2002. I. Good, J. Grook, The enumeration of arrays and generalization related to contingency tables, Discrete Math. (19) (1977) 23–45. H. Gupta, G.L. Nath, Enumeration of stochastic cubes, Not. Am. Math. Soc. (19) (1972) A-568. T. Koshy, Fibonacci and Lucas Numbers with Applications, John Wiley & Sons, 2011. R.P. Stanley, Enumerative Combinatorics, vol. 1, Wadword & Brooks, California, 1986. M.L. Stein, P.R. Stein, Enumeration of stochastic matrices with integer elements, Tech. Rep. LA-4434, Los Alamos Scientific Laboratory (1970). V.E. Tarakanov, Enumeration of stochastic cubes, Comb. Anal. (5) (1972) 4–15. in Russian. K. Yordzhev, Combinatorial problems on binary matrices, Math. Edu. Math. (24) (1995) 288–296.
Author's personal copy Applied Mathematics and Computation 236 (2014) 235–238
Contents lists available at ScienceDirect
Applied Mathematics and Computation journal homepage: www.elsevier.com/locate/amc
Factor-set of binary matrices and Fibonacci numbers Krasimir Yordzhev Faculty of Mathematics and Natural Sciences, South–West University, 2700 Blagoevgrad, Bulgaria
a r t i c l e
i n f o
a b s t r a c t The article discusses the set of square n n binary matrices with the same number of 1’s in each row and each column. An equivalence relation on this set is introduced. Each binary matrix is represented using ordered n-tuples of natural numbers. We are looking for a formula which calculates the number of elements of each factor-set by the introduced equivalence relation. We show a relationship between some particular values of the parameters and the Fibonacci sequence. Ó 2014 Elsevier Inc. All rights reserved.
Keywords: Fibonacci number Binary matrix Equivalence relation Factor-set
1. Introduction A binary (or boolean, or (0,1)-matrix) is a matrix whose all elements belong to the set B ¼ f0; 1g. With Bn we will denote the set of all n n binary matrices. Let n and k be integers such that 0 6 k 6 n; n P 2. We let Kkn denote the set of all n n binary matrices in each row and each column of which there are exactly k in number 1’s. Let us denote with kðn; kÞ ¼ jKkn j the number of all elements of Kkn . There is not any known formula to calculate the kðn; kÞ for all n and k. There are formulas for the calculation of the function kðn; kÞ for each n for relatively small values of k, more specifically, for k ¼ 1; k ¼ 2 and k ¼ 3. We do not know any formula to calculate the function kðn; kÞ for k > 3 and for all positive integer n. It is easy to prove the following well-known formula
kðn; 1Þ ¼ n!: The following formula
kðn; 2Þ ¼
X 2x2 þ3x3 þþnxn ¼n
ðn!Þ2 xr r¼2 xr !ð2rÞ
Qn
is well known [8]. One of the first recursive formulas for the calculation of kðn; 2Þ appeared in [1] (see also [4, p. 763]).
h i kðn; 2Þ ¼ 1 nðn 1Þ2 ð2n 3Þkðn 2; 2Þ þ ðn 2Þ2 kðn 3; 2Þ for n P 4; 2 kð1; 2Þ ¼ 0; kð2; 2Þ ¼ 1; kð3; 2Þ ¼ 6: Another recursive formula for the calculation of kðn; 2Þ occurs in [3].
2 n kðn; 2Þ ¼ ðn 1Þnkðn 1; 2Þ þ ðn1Þ kðn 2; 2Þ for n P 3; 2 kð1; 2Þ ¼ 0; kð2; 2Þ ¼ 1:
E-mail address: [email protected] http://dx.doi.org/10.1016/j.amc.2014.03.073 0096-3003/Ó 2014 Elsevier Inc. All rights reserved.
Author's personal copy 236
K. Yordzhev / Applied Mathematics and Computation 236 (2014) 235–238
The next recursive system is to calculate kðn; 2Þ.
kðn þ 1; 2Þ ¼ nð2n 1Þkðn; 2Þ þ n2 kðn 1; 2Þ pðn þ 1Þ; n P 2; 2 2 pðn þ 1Þ ¼ n ðn1Þ ½8ðn 2Þðn 3Þkðn 2; 2Þ þ ðn 2Þ2 kðn 3; 2Þ 4pðn 1Þ; n P 4; 4 kð1; 2Þ ¼ 0; kð2; 2Þ ¼ 1; pð1Þ ¼ pð2Þ ¼ pð3Þ ¼ 0; pð4Þ ¼ 9; where pðnÞ identifies the number of a special class of K2n -matrices [9]. The following formula is an explicit form for the calculation of kðn; 3Þ.
kðn; 3Þ ¼
n!2 X ð1Þb ðb þ 3cÞ!2a 3b ; 6n a!b!c!2 6c
where the sum is done as regards all ðnþ2Þðnþ1Þ solutions in nonnegative integers of the equation a þ b þ c ¼ n [7]. As it is noted 2 in [6], this formula does not give us good opportunities to study behavior of kðn; 3Þ. Let A; B 2 Kkn . We will say that A B, if A is obtained from B by moving some rows and/or columns. Obviously, the relation defined like that is an equivalence relation. We denote with
lðn; kÞ ¼ Kkn= ;
ð1Þ
the number of equivalence classes on the above defined relation. Problem 1. Find lðn; kÞ for given integers n and k; 1 6 k < n. The task of finding the cardinal number lðn; kÞ of the factor set Kkn= for all integers n and k; 1 6 k 6 n is an open scientific problem. This is the subject of discussion in this article. We will prove that there is a relationship between some particular values of the parameters of lðn; kÞ of and the Fibonacci sequence. Specifically, we will prove that if k ¼ n 2 then lðn; kÞ coincides with the Fibonacci numbers for all n 2 N; n P 2. 2. Canonical binary matrices Let N be the set of natural numbers and let
T n ¼ ha1 ; a2 ; . . . ; an i j ai 2 N; 0 6 ai 6 2n 1; i ¼ 1; 2; . . . ; n :
ð2Þ
With ‘‘ xv and let A0 be obtained from A by changing the locations of rows with the number u and v, while the remaining rows stay in their places. Then cðA0 Þ < cðAÞ. (b) Let yu > yv and let A0 be obtained from A by changing the locations of columns with the number u and v, while the remaining columns stay in their places. Then rðA0 Þ < rðAÞ.
Proof. (a) It is easy to see that rðA0 Þ < rðAÞ. Let A ¼ ½aij nn , where aij 2 f0; 1g; 1 6 i; j 6 n. Then the representation of xu and xv in binary notation be respectively as follows:
xu ¼ au1 au2 aun ; xv ¼ av 1 av 2 av n :
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K. Yordzhev / Applied Mathematics and Computation 236 (2014) 235–238
where auj ; av j 2 f0; 1g; j ¼ 1; 2; . . . ; n. If xu or xv can be written with less than n digits in binary notation then this number is completed from the left with the necessary number of zeros. Since xv < xu , then there is an integer m 2 f1; 2; . . . ; kg, such that auj ¼ av j for j < m; aum ¼ 1 and av m ¼ 0. Let cðAÞ ¼ hy1 ; y2 ; . . . ; yk i and let cðA0 Þ ¼ hz1 ; z2 ; . . . ; zk i. Then yj ¼ zj for j < m. Let the representation of ym and zm in binary notation be respectively as follows:
ym ¼ a1m a2m au1 m aum av m anm ; zm ¼ a1m a2m au1 m av m aum anm : Since aum ¼ 1 and av m ¼ 0, then zm < ym , whence it follows that cðA0 Þ < cðAÞ. (b) Analogously to (a). h
Corollary 1. Let A 2 Bn . Then rðAÞ is the minimal element in the set frðBÞ j B Ag if and only if cðAÞ is the minimal element in the set fcðBÞ j B Ag. h Corollary 1 gives us reason to formulate the following definition: Definition 2. The matrix A 2 Bn we will call a canonical matrix if rðAÞ is the minimal element in the set frðBÞ j B Ag and cðAÞ is the minimal element in the set fcðBÞ j B Ag.
Let M is arbitrary subset of Bn . Obviously, there is a unique canonical matrix in every equivalence class of factor-set M= and if rðAÞ ¼ hx1 ; x2 ; . . . ; xn i; cðAÞ ¼ hy1 ; y2 ; . . . ; yn i then x1 6 x2 6 6 xn and y1 6 y2 6 6 yn . Therefore the number of canonical matrices in M gives the cardinality of the factor-set M= . Lemma 2. If A ¼ ½aij 2 Kkn is a canonical matrix then
a1 1 ¼ a1 2 ¼ ¼ a1 nk ¼ 0; a1 nkþ1 ¼ a1 nkþ2 ¼ ¼ a1 n ¼ 1; a1 1 ¼ a2 1 ¼ ¼ ank 1 ¼ 0; ankþ1 1 ¼ ankþ2 1 ¼ ¼ an 1 ¼ 1; Proof. Immediately. h Corollary 2. If A ¼ ½aij 2 Kkn is a canonical matrix and rðAÞ ¼ hx1 ; x2 . . . xn i; cðAÞ ¼ hy1 ; y2 ; . . . yn i, then x1 ¼ y1 ¼ 2k 1.
h
3. The function lðn; kÞ and Fibonacci numbers Obviously, when k ¼ 0, the ‘‘zero’’ n n matrix ½0nn is the only matrix in the set K0n . When k ¼ n, there is only one n n binary matrix of Knn , and this is the matrix ½1nn all elements of which are equal to 1. Therefore
lðn; 0Þ ¼ lðn; nÞ ¼ 1:
ð4Þ
When k ¼ 1 if A 2 K1n is a canonical matrix, then rðAÞ ¼ h1; 2; 4; . . . ; 2n1 i, i.e. Ais a binary matrix with 1 in the second (not leading) diagonal and 0 elsewhere. For k ¼ n 1, if A 2 Kn1 is a canonical matrix, then A is a binary matrix with 0 in the n leading diagonal and 1 elsewhere. Hence
lðn; 1Þ ¼ lðn; n 1Þ ¼ 1:
ð5Þ 1
As is well known [2,5], the sequence ffn gn¼0 of Fibonacci numbers is defined by the recurrence relation
f0 ¼ f1 ¼ 1;
f n ¼ fn1 þ fn2
for n ¼ 2; 3; . . .
In this section, we will prove that the sequence flðk þ 1
ð6Þ 1 2; kÞgk¼0
coincides with the Fibonacci sequence (6). 1
Theorem 1. Let the sequence flðk þ 2; kÞgk¼0 is defined by (1) where n ¼ k þ 2, and let ffk gk¼0 be the Fibonacci sequence (6). Then for every integer k ¼ 0; 1; 2; 3; . . . the equality
lðk þ 2; kÞ ¼ fk is true.
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K. Yordzhev / Applied Mathematics and Computation 236 (2014) 235–238
Proof. When k ¼ 0 the assertion follows from (6) and (4). When k ¼ 1 the assertion follows from (6) and (5). When k ¼ 2 there are two canonical matrices A1 ; A2 2 K24 such that rðA1 Þ ¼ h3; 3; 12; 12i and rðA2 Þ ¼ h3; 5; 10; 12i. This fact is verified immediately. Therefore
lð2; 0Þ ¼ f0 ; lð3; 1Þ ¼ f1 and lð4; 2Þ ¼ f2 : Let k be an arbitrary positive integer such that k P 3 and let A ¼ ½ai j 2 Kkkþ2 ; 1 6 i; j 6 k þ 2 be a canonical matrix. Then, according to Lemma 2 a1 1 ¼ a1 2 ¼ a2 1 ¼ 0 and a1 3 ¼ a1 4 ¼ ¼ a1 n ¼ a3 1 ¼ a4 1 ¼ ¼ an 1 ¼ 1. Therefore, the following two cases are possible: (i) a2 2 ¼ 0, i.e., A is of the form
2
0
0 1 1
3
60 0 1 17 7 6 7 6 7 1 1 A¼6 7 6 7 6 .. .. 5 4. . B 1 1 We denote by M1 the set of all canonical matrices of this kind. Let A be an arbitrary matrix of M1 . In A, we remove the first and second rows and the first and second columns. We obtain the matrix B 2 Kk2 k . It is easy to see that Bis the canonical matrix. Conversely, let B ¼ ½bi j 2 Kkk2 (k P 3) and let B be a canonical matrix. From B we obtain the matrix A ¼ ½ai j 2 Kkkþ2 as follows: a1 1 ¼ a1 2 ¼ a2 1 ¼ a2 2 ¼ 0; a1 j ¼ a2 j ¼ 1; 3 6 j 6 k þ 2 and ai 1 ¼ ai 2 ¼ 1; 3 6 i 6 k þ 2. For each i; j 2 f3; 4; . . . ; k þ 2g we assume ai j ¼ bi2 j2 . It is easy to see that the so obtained matrix A is a canonical matrix of the set Kkkþ2 . Therefore, jM1 j ¼ lðk; k 2Þ for any integer k P 3. (ii) a2 2 ¼ 1, i.e., A is of the form
2
0
0 1 1 1
3
60 1 0 1 17 7 6 7 6 7 61 0 7 6 A¼6 7 7 61 1 7 6. . 7 6. . 5 4. . 1 1 Let M2 be the set of all canonical matrices of this kind and let A ¼ ½ai j ; a2 2 ¼ 1 be an arbitrary matrix of M2 . We change a2 2 from 1 to 0 and remove the first row and the first column of A. In this way we obtain a matrix of Kk1 kþ1 , which is easy to see that it is canonical. k1 Conversely, let B ¼ ½bi j 2 Kkþ1 and let B be a canonical matrix. According to Lemma 2 b1 1 ¼ b1 2 ¼ b2 1 ¼ 0. We change b1 1 from 0 to 1. In B, we add a first row and a first column and get the matrix A ¼ ½ai j 2 Kkkþ2 , such that a1 1 ¼ a1 2 ¼ a2 1 ¼ 0; a1 j ¼ 1 for j ¼ 3; 4; . . . ; k þ 2; ai 1 ¼ 1 for i ¼ 3; 4; . . . ; k þ 2 and asþ1 tþ1 ¼ bs t for s; t 2 f1; 2; . . . ; k þ 1g. It is easy to see that the resulting matrix A is canonical and A 2 M2 . Therefore, jM2 j ¼ lðk þ 1; k 1Þ for every integer k P 3. If M is the set of all canonical matrices, M # Kkkþ2 , then obviously
M1 \ M2 ¼ ; and M1 [ M2 ¼ M: Therefore
lðk þ 2; kÞ ¼ jMj ¼ jM1 j þ jM2 j ¼ lðk; k 2Þ þ lðk þ 1; k 1Þ; for all integers k P 3, which proves the theorem.
h
References [1] [2] [3] [4] [5] [6] [7] [8] [9]
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