Zero-Sum Triangles for Involutory, Idempotent, Nilpotent and Unipotent Matrices
Pengwei Hao1, Chao Zhang2, Huahan Hao3
Abstract: In some matrix formations, factorizations and transformations, we need special matrices with some properties and we wish that such matrices should be easily and simply generated and of integers. In this paper, we propose a zero-sum rule for the recurrence relations to construct integer triangles as triangular matrices with involutory, idempotent, nilpotent and unipotent properties, especially nilpotent and unipotent matrices of index 2. With the zero-sum rule we also give the conditions for the special matrices and the generic methods for the generation of those special matrices. The generated integer triangles are mostly newly discovered, and more combinatorial identities can be found with them.
Keywords: zero-sum rule, triangles of numbers, generation of special matrices, involutory, idempotent, nilpotent and unipotent matrices
1. Introduction
Zero-sum was coined in early 1940s in the field of game theory and economic theory, and the term “zero-sum game” is used to describe a situation where the sum of gains made by one person or group is lost in equal amounts by another person or group, so that the net outcome is neutral, or the sum of losses and gains is zero [1]. It was later also frequently used in psychology and political science. In this work, we apply the zero-sum rule to three adjacent cells to construct integer triangles. Pascal's triangle is named after the French mathematician Blaise Pascal, but the history can be traced back in almost 2,000 years and is referred to as the Staircase of Mount Meru in India, the Khayyam triangle in Persia (Iran), Yang Hui's triangle in China, Apianus's Triangle in Germany, and Tartaglia's triangle in Italy [2]. In mathematics, Pascal's triangle is a triangular array of the binomial coefficients, symmetrically with both boundary edges being all 1s and every interior number being the sum of the two numbers directly above it. More generic (푎푛, 푏푛)-Pascal's triangle [3] keeps the recurrence rule of the original Pascal's triangle unchanged for the interior cells of the triangle but uses integer sequences 푎푛 and 푏푛 for the input of the two boundary edges of the triangle. With Pascal's triangle, there have been quite a few topics in mathematics related, a staggering number of combinatorial identities found [4], and a growing number of applications developed. Placing a Pascal's triangle into a matrix, we can use a unit lower triangular matrix to represent the triangle. In [5] for reversible resampling, we have found the matrix relations between Pascal's triangle matrix, Vandermonde matrix, matrices of the falling and rising factorial polynomials, and matrices of Stirling numbers of the first kind and the second kind. Idempotent matrices, also known as projection matrices and closely related to involutory matrices, are used extensively in linear regression analysis [6] and econometrics [7], and involutory matrices, are used in quantum mechanics for Pauli spin matrices [8], and Hill cipher to avoid inversion, where the decryption needs the inverse of the key-matrix [9]. Idempotent matrices are diagonalizable and the eigenvalues are either 0 or 1 [10]. Involutory matrices are also diagonalizable and the eigenvalues are either −1 or 1 [11]. The matrices have many interesting properties [11]: If 푻 is idempotent, 푻2 = 푻, then 2푻 − 푰 and 푰 − 2푻 are involutory. If 푻 is involutory, 푻2 = 푰, then we have: 푻푇푻 and 푻푻푇 are symmetric and orthogonal; (푰 + 푻)(푰 − 푻) = 0; and (푰 + 푻)/2 and (푰 − 푻)/2 are idempotent. A matrix is similar to its inverse if and only if the matrix is the product of two involutory matrices, and a matrix of determinant 1 is the product of up to 4 involutory matrices. Unipotent matrices, closely related to nilpotent matrices, are popularly used in linear transformations and matrix factorizations, e.g. unipotent lower triangular matrices for Gauss elimination and triangular factorizations [12]. Unipotent matrices and nilpotent matrices are not diagonalizable, and the eigenvalues of unipotent and nilpotent matrices are all 1s and all 0s, respectively [11]. A unipotent matrix 푺 can be represented with a nilpotent matrix 푵, 푺 = 푰 + 푵, and we have (푺 − 푰)푘 = 푵푘 = 0, where 푘 is the index of the nilpotent matrix and the index of the unipotent matrix. For a unipotent matrix of index 2 and a nilpotent matrix of index 2, 푘 = 2, we have (푺 − 푰)2 = 푵2 = 0 and 푺−1 = (푰 + 푵)−1 = 푰 − 푵.
1 Queen Mary University of London, E1 4NS, UK, [email protected] 2 Peking University, Beijing, 100871, China, [email protected] 3 University of Cambridge, Cambridge, CB2 1TL, UK, [email protected] Fong and Sourour showed that a matrix of determinant 1 is the product of 3 unipotent matrices [13], and Wang and Wu proved that any unipotent matrix is the product of 2 unipotent matrices of index 2 and any matrix of determinant 1 is the product of 4 unipotent matrices of index 2 [14] or 4 unit triangular matrices (PLUS factorization) [15]. In this paper, we propose the zero-sum recurrence rule for triangle construction, take the triangles as triangular matrices to show the relations and properties, and investigate how the triangles as triangular matrices can be idempotent, involutory, nilpotent and unipotent. We first introduce the zero-sum rule and the general term formula in Section 2, give two lemmas as the recursive formula for the squared triangle in Section 3 and the product of the triangle with the boundary in Section 4, in Section 5-8, we present four theorems for the methods to construct the triangles with various properties, and finally in Section 9 we wish to express some of our thoughts about the zero-sum triangles beyond this paper.
2. The Zero-Sum Rule and the Triangles
The triangular zero-sum rule, we propose, is to make zero the sum of three adjacent numbers for all the interior cells, every interior number and the two numbers directly above it, as shown in Figure 1, 푡(푖, 푘 − 1) + 푡(푖, 푘) + 푡(푖 + 1, 푘) = 0, where 푖 ≥ 푘 ≥ 1.
Figure 1. Three cells involved in zero-sum rule See the two following triangles (1&2) for example of the zero-sum triangles: 1 1 1 0 2 -1 1 -1 1 1 -1 1 1 0 0 0 2 0 0 -1 1 -1 0 0 1 1 -2 0 1 1 1 0 1 0 -1 0 2 1 2 -1 -2 -1 1 -1 -1 -1 1 1 1 1 -3 -3 -1 3 3 1 1 0 2 2 0 -2 -2 0 2 2 6 4 -2 -6 -4 -1 1 -1 -2 -4 -2 2 4 2 1 1 -4 -8 -10 -2 8 10 5 1 Triangle 1 Triangle 2
If we use 푡(푖, 푘) to denote the 푘-th cell of the 푖-th row in the triangle, 푖 ≥ 푘 ≥ 0, the integer sequences of the two boundary edges can be given as 푎푛 and 푏푛, or: 푡(0,0) = 푎0 = 푏0, 푡(1,0) = 푎1, 푡(2,0) = 푎2, … , 푡(푖, 0) = 푎푖, …, (1) 푡(1,1) = 푏1, 푡(2,2) = 푏2, … , 푡(푖, 푖) = 푏푖, … To avoid involving exterior cells, the zero-sum rule can be equivalently expressed for three (푖, 푘) adjacent cells of the triangle as : 푡(푖 − 1, 푘 − 1) + 푡(푖 − 1, 푘) + 푡(푖, 푘) = 0 (푖 − 1 ≥ 푘 ≥ 1) 푡(푖, 푘 − 1) + 푡(푖, 푘) + 푡(푖 + 1, 푘) = 0 (푖 ≥ 푘 ≥ 1) (2) 푡(푖, 푘) + 푡(푖, 푘 + 1) + 푡(푖 + 1, 푘 + 1) = 0 (푖 − 1 ≥ 푘 ≥ 0) To find the (푖 + 1)-th row with the 푖-th row, the recurrence rule for the interior cells of the triangle can also be written as: 푡(푖 + 1, 푘) = −푡(푖, 푘 − 1) − 푡(푖, 푘) (푖 ≥ 푘 ≥ 1) (3) Placing the 푛-row triangle into a lower triangular matrix, we have the 푛-row matrix as:
푡(0,0) 0 푎0 0 푻푛 = [ ⋮ ⋯ ⋯ ] = [ ⋮ ⋯ ⋯ ] (4) 푡(푛 − 1,0) ⋯ 푡(푛 − 1, 푘 − 1) ⋯ 푡(푛 − 1, 푛 − 1) 푎푛−1 ⋯ 푡(푛 − 1, 푘 − 1) ⋯ 푏푛−1 where the (푖, 푘) entries of the matrix and the cells of the triangle are related as: 0 (푖 < 푘) 푻 (푖, 푘) = { (5) 푛 푡(푖 − 1, 푘 − 1) (푖 ≥ 푘 ≥ 1) With two boundary sequences and the zero-sum recurrence rule, the explicit formula of the general term to directly compute the interior cells (푖 > 푗 > 0) is : 푖−푗 푗 푖 − 푘 − 1 푖 − 푘 − 1 푡(푖, 푗) = ∑ ( ) (−1)푖−푘푎 + ∑ ( ) (−1)푖−푘푏 (6) 푗 − 1 푘 푖 − 푗 − 1 푘 푘=1 푘=1 which can be easily proved similar to the general term formula for the generalized Pascal's triangle [3], in respect that every row downwards negates the sum of the two cells in the row above. It is easy to see that the zero-sum triangles are symmetric, 푡(푖, 푗) = ±푡(푖, 푖 − 푗), if 푎푘 = ±푏푘 for all 푘, and the middle cell in an even indexed row is 0 if 푎푘 + 푏푘 = 0 and is the largest if 푎푘 = 푏푘 for all 푘. With the general term formula, a variety of sums of the triangle cells can be found. The sums for the first 2 rows are trivial, and the sum of all the cells in the 푖-th row 푆푖 (푖 ≥ 2) and the sum of all the cells in the whole 푛 + 1 row triangle (푛 ≥ 2) S are: 푖 푖−1 푖−푗 푗 푖 − 푘 − 1 푖 − 푘 − 1 푆 = ∑ 푡(푖, 푗) = 푎 + 푏 + ∑ {∑ ( ) (−1)푖−푘푎 + ∑ ( ) (−1)푖−푘푏 } 푖 푖 푖 푗 − 1 푘 푖 − 푗 − 1 푘 푗=0 푗=1 푘=1 푘=1 푖−1 푖−푘 푖−1 푖 − 푘 − 1 푖 − 푘 − 1 = 푎 + 푏 + (−1)푖 ∑ {(−1)푘푎 ∑ ( ) + (−1)푘푏 ∑ ( )} 푖 푖 푘 푗 − 1 푘 푖 − 푗 − 1 푘=1 푗=1 푗=푘 푖−1 푖 푘 푖−푘−1 푘 푖−푘−1 = 푎푖 + 푏푖 + (−1) ∑{(−1) 푎푘2 + (−1) 푏푘2 } 푘=1 푖−1 푖−푘−1 = 푎푖 + 푏푖 − ∑(−2) {푎푘 + 푏푘} 푘=1 푛 푛 푖 푛 푛 푖−1 푖−푘−1 푆 = ∑ 푆푖 = ∑ ∑ 푡(푖, 푗) = 푎0 + ∑{푎푖 + 푏푖} − ∑ ∑(−2) {푎푘 + 푏푘} 푖=0 푖=0 푗=0 푖=1 푖=2 푘=1 푛 푛−1 푛 푖−푘−1 = 푎0 + ∑{푎푖 + 푏푖} − ∑{푎푘 + 푏푘} ∑ (−2) 푖=1 푘=1 푖=푘+1 푛 푛−1 1 = 푎 + ∑{푎 + 푏 } + ∑ ((−2)푛−푘 − 1){푎 + 푏 } 0 푖 푖 3 푘 푘 푖=1 푘=1 푛−1 1 = 푎 + 푎 + 푏 + ∑(2 + (−2)푛−푘){푎 + 푏 } 0 푛 푛 3 푘 푘 푘=1 The two sums are 0 if 푎푘 + 푏푘 = 0 for all 푘, and we also have a recursive relation of the sums between two adjacent rows: 푖 푖−1 푖−2 푖−푘−1 푖−푘−2 푆푖 = ∑ 푡(푖, 푗) = {푎푖 + 푏푖} − ∑(−2) {푎푘 + 푏푘} = {푎푖 + 푏푖} − {푎푖−1 + 푏푖−1} + 2 ∑(−2) {푎푘 + 푏푘} 푗=0 푘=1 푘=1 푖−2 푖−푘−2 = {푎푖 + 푏푖} + {푎푖−1 + 푏푖−1} − 2 {{푎푖−1 + 푏푖−1} − ∑(−2) {푎푘 + 푏푘}} 푘=1
= {푎푖 + 푏푖} + {푎푖−1 + 푏푖−1} − 2푆푖−1
2 3. The Squared Matrices of the Triangles, 푻푛
We have our first lemma first: 2 Lemma 1: For a zero-sum triangle 푻푛, the elements of the matrix squared, 푻푛, has recursive relations as: 0 (푖 < 푗) 2 2 푏푖 (푖 = 푗) 푻푛(푖 + 1, 푗 + 1) = (7) 푠0(푖) (푖 > 푗 = 0) 2 {푻푛(푖, 푗) + 푡(푖, 푗){푏푖−1 + 푏푖} − 푡(푖 − 1, 푗 − 1){푏푗−1 + 푏푗} (푖 > 푗 > 0) where we use the following notation for simplification: 푖 2 푠0(푖) = 푻푛(푖 + 1,1) = ∑ 푡(푖, 푘)푎푘 (8) 푘=0 The details of 푠0(푖) and more are given in the next section.
Proof: 2 For the first two cases, it is simply true. Since 푻푛 is a lower triangular matrix, 푻푛 is also lower triangular, 2 2 2 2 푻푛(푖 + 1, 푗 + 1) = 0 if 푖 < 푗, and the diagonal elements are just squared, 푻푛(푖 + 1, 푖 + 1) = 푡 (푖, 푖) = 푏푖 . 2 For all the other elements of 푻푛 at (푖 + 1, 푗 + 1), 푖 > 푗 ≥ 0, 푖, 푗 = 0,1,2,3, … , 푛 − 1, 푛 푖 2 푻푛(푖 + 1, 푗 + 1) = ∑ 푻푛(푖 + 1, 푘)푻푛(푘, 푗 + 1) = ∑ 푡(푖, 푘)푡(푘, 푗) 푘=1 푘=푗 For 푖 > 푗 = 0, more relations and the proof are given in the next section by using the zero-sum rule. For 푖 > 푗 > 0, we use the zero-sum rule (Eq.2) to substitute the interior elements with their two adjacent ones above or reversely to combine two adjacent ones into one below for simplification: 푖 푖−1 2 푻푛(푖 + 1, 푗 + 1) = ∑ 푡(푖, 푘)푡(푘, 푗) = 푡(푖, 푗)푡(푗, 푗) + 푡(푖, 푖)푡(푖, 푗) + ∑ 푡(푖, 푘)푡(푘, 푗) 푘=푗 푘=푗+1 푖−1
= 푡(푖, 푗){푏푖 + 푏푗} + ∑ {푡(푖 − 1, 푘 − 1) + 푡(푖 − 1, 푘)}{푡(푘 − 1, 푗 − 1) + 푡(푘 − 1, 푗)} 푘=푗+1
= 푡(푖, 푗){푏푖 + 푏푗} 푖−1 + ∑ {푡(푖 − 1, 푘 − 1)푡(푘 − 1, 푗 − 1) + 푡(푖 − 1, 푘 − 1)푡(푘 − 1, 푗) − 푡(푖 − 1, 푘)푡(푘, 푗)} 푘=푗+1 푖−1
= 푡(푖, 푗){푏푖 + 푏푗} + ∑ 푡(푖 − 1, 푘 − 1)푡(푘 − 1, 푗 − 1) + 푡(푖 − 1, 푗)푡(푗, 푗) 푘=푗+1 − 푡(푖 − 1, 푖 − 1)푡(푖 − 1, 푗) 푖−2
= 푡(푖, 푗)푏푖 + ∑ 푡(푖 − 1, 푘)푡(푘, 푗 − 1) + {푡(푖, 푗) + 푡(푖 − 1, 푗)}푏푗 − 푡(푖 − 1, 푗)푏푖−1 푘=푗 푖−1
= 푡(푖, 푗)푏푖 + ∑ 푡(푖 − 1, 푘)푡(푘, 푗 − 1) − 푡(푖 − 1, 푗 − 1)푡(푗 − 1, 푗 − 1) 푘=푗−1 − 푡(푖 − 1, 푖 − 1)푡(푖 − 1, 푗 − 1) + {푡(푖, 푗) + 푡(푖 − 1, 푗)}푏푗 − 푡(푖 − 1, 푗)푏푖−1 푖−1
= 푡(푖, 푗)푏푖 + ∑ 푡(푖 − 1, 푘)푡(푘, 푗 − 1) − 푡(푖 − 1, 푗 − 1)푏푗−1 푘=푗−1 − {푡(푖 − 1, 푗 − 1) + 푡(푖 − 1, 푗)}푏푖−1 − 푡(푖 − 1, 푗 − 1)푏푗 푖−1
= 푡(푖, 푗)푏푖 + ∑ 푡(푖 − 1, 푘)푡(푘, 푗 − 1) − 푡(푖 − 1, 푗 − 1)푏푗−1 + 푡(푖, 푗)푏푖−1 푘=푗−1 2 − 푡(푖 − 1, 푗 − 1)푏푗 = 푻푛(푖, 푗) + 푡(푖, 푗){푏푖−1 + 푏푖} − 푡(푖 − 1, 푗 − 1){푏푗−1 + 푏푗} where we use the following identities for the recursive formula: 푖−1 2 ∑ 푡(푖 − 1, 푘)푡(푘, 푗 − 1) = 푻푛(푖, 푗) 푘=푗−1 ∎
4. The Triangles Multiplied with the Boundary
To make the proof simpler, this section gives another lemma for the triangle matrix multiplied with the left 푇 boundary column in three cases: with the boundary itself, 풔0 = 푻푛풂0, where 풂0 = [푎0, 푎1, … , 푎푛−1] , with the 푇 boundary shifted up one cell, 풔1 = 푻푛풂1, where 풂1 = [푎1, 푎2, … , 푎푛] , and with the boundary shifted down one 푇 cell, 풔−1 = 푻푛풂−1, where 풂−1 = [0, 푎0, 푎1, … , 푎푛−2] . The elements of the triangle-boundary product vectors can be expressed with the triangle cells as: 푖 푖 푖 2 푠0(푖) = 푻푛(푖 + 1,1) = ∑ 푡(푖, 푘)푎푘 , 푠1(푖) = ∑ 푡(푖, 푘)푎푘+1 , 푠−1(푖) = ∑ 푡(푖, 푘)푎푘−1 (9) 푘=0 푘=0 푘=1 2 with 푠0(0) = 푎0푏0 = 푎0, 푠1(0) = 푎1푏0 = 푎0푎1, 푠−1(1) = 푎0푏1, 푖 ≥ 1. We can find a few related recursive formulas as in Lemma 2 for the triangle-boundary multiplications, which are more accessible in the proofs in the next sections: Lemma 2: 푠 (0) = 푎2, 푠 (0) = 푎 푎 , 푠 (1) = 푎 푏 (base case) 0 0 1 0 1 −1 0 1 푠−1(푖) = 푎푖−1{푏푖−1 + 푏푖} − 푠−1(푖 − 1) − 푠0(푖 − 1) (푖 ≥ 2) 푠0(푖) = 푎0{푎푖−1 + 푎푖} + 푎푖{푏푖−1 + 푏푖} − 푠0(푖 − 1) − 푠1(푖 − 1) (푖 ≥ 1) (10)
푠0(푖) = 푎푖{푏푖 + 푏푖+1} − 푠−1(푖) − 푠−1(푖 + 1) (푖 ≥ 1) {푠1(푖) = 푎0{푎푖 + 푎푖+1} + 푎푖+1{푏푖 + 푏푖+1} − 푠0(푖) − 푠0(푖 + 1) (푖 ≥ 1) Proof: We prove them by using the zero-sum rule (Eq.2) to substitute the interior elements with their two adjacent ones if there exist. 푖 푖−1
푠−1(푖) = ∑ 푡(푖, 푘)푎푘−1 = 푡(푖, 푖)푎푖−1 − ∑{푡(푖 − 1, 푘 − 1) + 푡(푖 − 1, 푘)}푎푘−1 푘=1 푘=1 푖−1 푖−1
= 푎푖−1푏푖 − ∑ 푡(푖 − 1, 푘)푎푘 + 푡(푖 − 1, 푖 − 1)푎푖−1 − ∑ 푡(푖 − 1, 푘)푎푘−1 푘=0 푘=1 = 푎푖−1{푏푖−1 + 푏푖} − 푠0(푖 − 1) − 푠−1(푖 − 1) 푖 푖−1
푠0(푖) = ∑ 푡(푖, 푘)푎푘 = 푡(푖, 0)푎0 + 푡(푖, 푖)푎푖 − ∑{푡(푖 − 1, 푘 − 1) + 푡(푖 − 1, 푘)}푎푘 푘=0 푘=1 푖−1 푖−1
= 푎푖푎0 + 푎푖푏푖 − ∑ 푡(푖 − 1, 푘)푎푘+1 + 푡(푖 − 1, 푖 − 1)푎푖 − ∑ 푡(푖 − 1, 푘)푎푘 + 푡(푖 − 1,0)푎0 푘=0 푘=0 = 푎0{푎푖−1 + 푎푖} + 푎푖{푏푖−1 + 푏푖} − 푠1(푖 − 1) − 푠0(푖 − 1) 푖 푖−1
푠0(푖) = ∑ 푡(푖, 푘)푎푘 = 푡(푖, 푖)푎푖 − ∑{푡(푖, 푘 + 1) + 푡(푖 + 1, 푘 + 1)}푎푘 푘=0 푘=0 푖 푖+1
= 푎푖푏푖 − ∑ 푡(푖, 푘)푎푘−1 − ∑ 푡(푖 + 1, 푘)푎푘−1 + 푡(푖 + 1, 푖 + 1)푎푖 푘=1 푘=1 = 푎푖{푏푖 + 푏푖+1} − 푠−1(푖) − 푠−1(푖 + 1) 푖 푖−1
푠1(푖) = ∑ 푡(푖, 푘)푎푘+1 = 푡(푖, 푖)푎푖+1 − ∑{푡(푖, 푘 + 1) + 푡(푖 + 1, 푘 + 1)}푎푘+1 푘=0 푘=0 푖 푖+1
= 푎푖+1푏푖 − ∑ 푡(푖, 푘)푎푘 + 푡(푖, 0)푎0 − ∑ 푡(푖 + 1, 푘)푎푘 + 푡(푖 + 1,0)푎0 + 푡(푖 + 1, 푖 + 1)푎푖+1 푘=0 푘=0 = 푎0{푎푖 + 푎푖+1} + 푎푖+1{푏푖 + 푏푖+1} − 푠0(푖) − 푠0(푖 + 1) ∎
5. The Triangles for Idempotent Matrices
A matrix 푻 is idempotent if 푻2 = 푻. In the squared triangle matrix as in Eq.(7), for the diagonal entries, to make 2 푏푖 = 푏푖, we have 푏푖 = 0 or 푏푖 = 1. 2 2 For 푖 > 푗 > 0, if we have 푏푖−1 + 푏푖 = 푏푗−1 + 푏푗 = 1, then we have 푻푛(푖 + 1, 푗 + 1) = 푻푛(푖, 푗) + 푡(푖, 푗) − 푡(푖 − 1, 푗 − 1). To make it, we can simply set the integer sequences 푏푖 as alternatively 0 and 1 : 푡(푖, 푖) = 푏푖 = 1 − 푏푖−1 = mod(푖 + 푎0, 2) (11) with which we also have 푏푖푏푖−1 = 0. With the right boundary edge numbers given in Eq.(11), the elements of the squared triangle matrix, Eq.(7), are: 0 (푖 < 푗)
2 푏푖 (푖 = 푗) 푻푛(푖 + 1, 푗 + 1) = 푠0(푖) (푖 > 푗 = 0) 2 {푻푛(푖, 푗) + 푡(푖, 푗) − 푡(푖 − 1, 푗 − 1) (푖 > 푗 > 0) 2 For 푖 > 푗 ≥ 0, it is easy to verify with Eq.(7&10) if 푠0(푖) = 푎푖 and 푻푛(푖, 푗) = 푡(푖 − 1, 푗 − 1), and we then have 2 푻푛 = 푻푛, which is an idempotent matrix. To meet the conditions, we use 푠1(푖) for the solution and then we prove 2 푠−1(푖) = 0, 푠0(푖) = 푎푖, 푠1(푖) = 푎0{푎푖 + 푎푖+1} − 푎푖, and 푻푛 = 푻푛 as follows. If 푎0 = 0, 푡(2푚, 2푚) = 푏2푚 = mod(2푚 + 푎0, 2) = 푎0 = 0, then we use equation 푠0(2푚) = 푎2푚 for the solution of the even indexed boundary numbers 푎2푚 : 2푚−1
푎2푚 = 푠0(2푚) = ∑ 푡(2푚, 푘)푎푘 (12) 푘=1 In such a case, 푏2푚+1 = mod(2푚 + 1 + 푎0, 2) = 1, we need to prove the following equalities hold: 2 푠−1(2푚 + 1) = 0, 푠0(2푚 + 1) = 푎2푚+1, 푠1(2푚) = −푎2푚, 푻푛(2푚 + 2, 푗 + 1) = 푡(2푚 + 1, 푗) (13) If 푎0 = 1 , 푡(2푚 − 1,2푚 − 1) = 푏2푚−1 = mod(2푚 − 1 + 푎0, 2) = 1 − 푎0 = 0 , then we use equation 푠1(2푚 − 1) = 푎2푚 for the solution of 푎2푚 : 2푚−2
푎2푚 = 푠1(2푚 − 1) = ∑ 푡(2푚 − 1, 푘)푎푘+1 (14) 푘=0 In such a case, 푏2푚 = mod(2푚 + 푎0, 2) = 1, we need the following equalities hold: 2 푠−1(2푚) = 0, 푠0(2푚) = 푎2푚, 푠1(2푚) = 푎2푚+1, 푻푛(2푚 + 1, 푗 + 1) = 푡(2푚, 푗) (15) By induction we give the proof below that with Eq.(12&14) no matter what 푎2푚+1 are, or 푎2푚+1 can be any independent input, the equalities in Eq.(13) or Eq.(15) hold. In summary, a sufficient condition for the triangle matrix to be idempotent is the boundary integer sequences to be given by Eq.(11&12), or by Eq.(11&14), as descripted in the following theorem with Eq.(12&14) combined into one Eq.(16) and the boundary relations Eq.(13&15) combined into Eq.(17).
Theorem 1: The matrices of the zero-sum triangles are idempotent if the triangles are generated with the boundary sequences: 2푚−1 푠 (2푚) = ∑ 푡(2푚, 푘)푎 (푏 = 푎 = 0) 0 푘 2푚 0 푘=1 푎2푚 = (16) 2푚−2 푠1(2푚 − 1) = ∑ 푡(2푚 − 1, 푘)푎푘+1 (푏2푚 = 푎0 = 1) { 푘=0 where 푚 ≥ 1, 푎2푚+1 can be freely chosen, and 푏2푚+1 = 1 − 푏2푚. The multiplication of the triangle matrix with the left boundary gives: 푠 (0) = 푎 , 푠 (0) = 푎 푎 , 푠 (1) = 0 (base case) { 0 0 1 0 1 −1 (17) 푠0(푖) = 푎푖, 푠1(푖) = 푎0{푎푖 + 푎푖+1} − 푎푖, 푠−1(푖) = 0 (푖 ≥ 1) Proof: 2 2 2 For 푖 > 푗 > 0, staring with 푻푛(1,1) = 푎0 = 푎0 = 푡(0,0) = 푻푛(1,1) and the recursive formula, 푻푛(푖 + 1, 푗 + 2 2 1) = 푻푛(푖, 푗) + 푡(푖, 푗) − 푡(푖 − 1, 푗 − 1) , by induction we can easily have 푻푛(푖 + 1, 푗 + 1) = 푡(푖, 푗) from 2 푻푛(푖, 푗) = 푡(푖 − 1, 푗 − 1). So we only need to prove for 푠0(푖) = 푎푖 such that the left boundary elements of the 2 squared triangle also meet the idempotent property: 푻푛(푖 + 1,1) = 푠0(푖) = 푎푖 = 푡(푖, 0) = 푻푛(푖 + 1,1). For 푛 = 1,2,3, the triangles are formed as: 푎0 0 0 푎0 0 푻1 = 푎0, 푻2 = [ ] , 푻3 = [푎1 푏1 0 ] 푎1 푏1 푎2 −푎1 − 푏1 푏2 2 2 2 Let 푎1 be free, it is easy to verify that 푻1 = 푻1 , 푻2 = 푻2 , 푠0(0) = 푎0 = 푎0, 푠0(1) = 푎1푎0 + 푏1푎1 = 푎1, 푠−1(1) = 푎0푏1 = 0, 푠1(0) = 푎0푎1, and with the recursive formulas, 푠−1(2) = 푎1 − 푠−1(1) − 푠0(1) = 0. 2 With 푚 = 1, 푛 = 2푚 + 1 = 3, we can have 푻3 = 푻3 if we set as in Eq.(16), 1 푠 (2) = ∑ 푡(2, 푘)푎 = 푡(2,1)푎 = −푎 (푎 + 푏 ) (푏 = 푎 = 0) 0 푘 1 1 1 1 2 0 푘=1 푎2 = 0 2 푠1(1) = ∑ 푡(1, 푘)푎푘+1 = 푡(1,0)푎1 = 푎1 (푏2 = 푎0 = 1) { 푘=0 Using definitions Eq.(9) and the recursive formulas Eq.(10), we have for 푎0 = 0, 푠1(1) = 푎0{푎1 + 푎2} + 푎2 − 푠0(1) − 푠0(2) = 푎0{푎1 + 푎2} − 푎1 , 푠−1(3) = 푎2 − 푠−1(2) − 푠0(2) = 푎2 − 0 − 푎2 = 0 , and for 푎0 = 1 , 푠0(2) = 푎0{푎1 + 푎2} + 푎2 − 푠0(1) − 푠1(1) = 푎2 = 푎0{푎1 + 푎2} − 푎1 , 푠−1(3) = 푎2 − 푠−1(2) − 푠0(2) = 0. Or by combing the above we have 푠1(1) = 푎0{푎1 + 푎2} − 푎1, 푠0(1) = 푎1, 푠−1(1) = 0. Eq.(17) holds for 푖 = 1. For 푚 = 1, 푛 = 2푚 + 2 = 4, with the zero-sum rule we give 푻4 as: 푎0 0 0 0 푎1 푏1 0 0 푻4 = [ ] 푎2 −푎1 − 푏1 푏2 0 푎3 푎1 + 푏1 − 푎2 푎1 + 푏1 − 푏2 푏3 We have 푠1(2) = 푎2푎1 + 푡(2,1)푎2 + 푏2푎3 = 푎2(−푏1) + 푎0푎3 = 푎0{푎2 + 푎3} − 푎2 , 푠0(3) = 푎0{푎2 + 푎3} + 푎3 − 푠0(2) − 푠1(2) = 푎3 and 푠−1(4) = 푎3 − 푠−1(3) − 푠0(3) = 0. Put equalities of index 2 together, and we see Eq.(17) holds for 푖 = 2 as 푠1(2) = 푎0{푎2 + 푎3} − 푎2, 푠0(2) = 푎2, 푠−1(2) = 0. With Eq.(16), we find the solution for the next even indexed boundary number: 푠0(2푚) (푎0 = 0) 푎2푚 = { 푠1(2푚 − 1) (푎0 = 1) and we assume two steps of the three equalities in Eq.(17) hold for integer 푖 = 2푚 − 1 and 푖 = 2푚 (푚 > 1): 푠1(2푚 − 1) = 푎0{푎2푚−1 + 푎2푚} − 푎2푚−1, 푠0(2푚 − 1) = 푎2푚−1, 푠−1(2푚 − 1) = 0 푠1(2푚) = 푎0{푎2푚 + 푎2푚+1} − 푎2푚, 푠0(2푚) = 푎2푚, 푠−1(2푚) = 0 Then with the recursive relations in Eq.(10), we can have 푠−1(푖) and 푠0(푖) in Eq.(17) hold for 푖 = 2푚 + 1 as : 푠−1(2푚 + 1) = 푎2푚{푏2푚 + 푏2푚+1} − 푠−1(2푚) − 푠0(2푚) = 푎2푚 − 0 − 푎2푚 = 0 푠0(2푚 + 1) = 푎0{푎2푚 + 푎2푚+1} + 푎2푚+1{푏2푚 + 푏2푚+1} − 푠0(2푚) − 푠1(2푚) = 푎0{푎2푚 + 푎2푚+1} + 푎2푚+1 − 푎2푚 − 푎0{푎2푚 + 푎2푚+1} + 푎2푚 = 푎2푚+1 Hence by induction, we have 푠−1(푖) = 0 and 푠0(푖) = 푎푖 hold. Then with the recursive relation for 푠1(푖) we have the last one in Eq.(17) : 푠1(2푚 + 1) = 푎0{푎2푚+1 + 푎2푚+2} + 푎2푚+2{푏2푚 + 푏2푚+1} − 푠0(2푚 + 1) − 푠0(2푚 + 2) = 푎0{푎2푚+1 + 푎2푚+2} + 푎2푚+2 − 푎2푚+1 − 푎2푚+2 = 푎0{푎2푚+1 + 푎2푚+2} − 푎2푚+1 which shows that 푠1(푖) = 푎0{푎푖 + 푎푖+1} − 푎푖 also holds. ∎
With Theorem 1, for 푎0 = 1, we can find the first few dependent even indexed boundary cells are: 푎2 = 2 3 4 5 6 2 2 3 푎1 , 푎4 = (1 − 푎1)푎1 + (2푎1 − 1)푎3, 푎6 = 4푎1 − 5푎1 + 2푎1 + 푎3 + 7푎1 푎3 + 푎3 − 푎1(4푎3 + 1) − 2푎5 + 6 7 8 5 4 3 2푎1(푎5 − 2푎3), 푎8 = −39푎1 + 22푎1 − 5푎1 + 푎1 (12푎3 + 35) − 푎1 (41푎3 + 16) + 푎1 (56푎3 − 4푎5 + 3) + 2 5푎5 + 푎3(2푎5 − 5푎3 − 3) + 푎1 (11푎5 − 2푎3(20 + 3푎3)) − 3푎7 + 푎1(푎3(11푎3 + 16) + 2(푎7 − 6푎5)). 3 For 푎0 = 0 , the first few even indexed boundary cells are: 푎2 = −푎1(1 + 푎1), 푎4 = 푎1(1 + 푎1) − 4 5 6 3 2 2(1 + 푎1)푎3, 푎6 = −24푎1 − 11푎1 − 2푎1 − (−5 + 푎3)푎3 + 푎1(−26 + 4푎3) + 푎1 (−14 + 13푎3) + 푎1(−3 + 6 7 8 4 5 2 14푎3 − 2푎5) − 3푎5, 푎8 = 151푎1 + 42푎1 + 5푎1 + 푎1(359 − 71푎3) + 푎1 (301 − 12푎3) + 12푎3 + 14푎5 − 3 2 2푎3(14 + 푎5) + 4푎1(64 − 42푎3 + 푎5) + 푎1 (101 + 푎3(−199 + 6푎3) + 17푎5) + 푎1(17 + 푎3(−118 + 17푎3) + 26푎5 − 2푎7) − 4푎7. Since no division is involved in Eq.(16), the triangle only has integers as long as the odd indexed boundary numbers are input with integers. An example of idempotent triangle is Triangle 1 shown on the left in Section 2, 푖 where we use 푎푖 = 1. More examples can be with 푎푖 = 0, 푎푖 = (−1) , or with 푎푖 using a repeating pattern of 4 integers {0, −1,0,1}. The first 9 rows of two more triangles (3&4) are given below: 1 0 -1 0 -1 1 1 1 1 0 0 0 -1 -2 -2 0 1 0 0 1 1 3 4 2 1 0 -1 0 -1 0 -1 -4 -7 -6 -3 0 -1 1 1 1 1 1 1 5 11 13 9 3 1 0 0 -2 -2 -2 -2 0 -1 -6 -16 -24 -22 -12 -4 0 1 0 2 4 4 4 2 1 1 7 22 40 46 34 16 4 1 0 -1 -2 -6 -8 -8 -6 -3 0 Triangle 3 Triangle 4
In fact, if we remove the leftmost boundary cells, the triangle with 푎0 = 1 becomes the triangle with 푎0 = 0, and vice versa. Based on Theorem 1, we can also see that the even indexed rows are linear combinations of the odd indexed rows above, and the even indexed columns are linear combinations of the odd indexed columns rightwards.
6. The Triangles for Involutory Matrices
For a given idempotent matrix 푻, 푻2 = 푻, we can easily find two involutory matrices, 2푻 − 푰 and 푰 − 2푻, but we can directly find more general involutory matrices with the triangles constructed with the zero-sum rule. 2 Based on the recursive formula of the squared triangle matrix, Eq.(7), for the diagonal entries, to make 푏푖 = 1, 2 we have 푏푖 = 1 or 푏푖 = −1. For 푖 > 푗 > 0, if we have 푏푖−1 + 푏푖 = 푏푗−1 + 푏푗 = 0, then we have 푻푛(푖 + 1, 푗 + 2 1) = 푻푛(푖, 푗). To make 푏푖−1 + 푏푖 = 푏푗−1 + 푏푗 = 0, we can simply set the boundary sequence 푏푖 as alternatively −1 and 1, 푖 푖 푡(푖, 푖) = 푏푖 = (−1) 푏0 = (−1) 푎0 (18) Then the elements of the squared triangle matrix are: 0 (푖 < 푗)
2 1 (푖 = 푗) 푻푛(푖 + 1, 푗 + 1) = 푠0(푖) (푖 > 푗 = 0) 2 {푻푛(푖, 푗) (푖 > 푗 > 0) 2 2 For 푖 > 푗, if 푠0(푖) = 0 and 푻푛(푖, 푗) = 0, we have 푻푛 = 푰, which is an involutory matrix. To meet the conditions, we use 푠0(2푚) = 0 for the solution and then we prove 푠0(2푚 + 1) = 0, 푠−1(푖) = 푖 2 (−1) , 푠1(푖) = 푎0{푎푖 + 푎푖+1}, and 푻푛 = 푰 as follows. 2푚 If 푖 = 2푚 is even, 푏푖 = 푏2푚 = 푡(2푚, 2푚) = (−1) 푏0 = 푎0, we can solve the equation 푠0(2푚) = 0 to find the even indexed boundary numbers 푎2푚: 2푚 2푚−1
푠0(2푚) = 0 = ∑ 푡(2푚, 푘)푎푘 = 푡(2푚, 0)푎0 + 푡(2푚, 2푚)푎2푚 + ∑ 푡(2푚, 푘)푎푘 푘=0 푘=1 2푚−1
= 2푎0푎2푚 + ∑ 푡(2푚, 푘)푎푘 푘=1 With which we easily give the solution : 2푚−1 푎 푎 = − 0 ∑ 푡(2푚, 푘)푎 (19) 2푚 2 푘 푘=1 In summary, a sufficient condition for the triangle matrix to be involutory is the boundary integer sequences to be given by Eq.(18&19), as descripted in the following theorem.
Theorem 2: The matrices of the triangles are involutory if the triangles are generated with the boundary sequences 2푚−1 푎 푎 = − 0 ∑ 푡(2푚, 푘)푎 , 푏 = 푎 (20) 2푚 2 푘 2푚 0 푘=1 where 푚 ≥ 1, 푎0 = ±1, 푎2푚+1 can be freely chosen, and 푏2푚+1 = 1 − 푏2푚. The multiplication of the triangle matrix with the left boundary gives: 푠0(0) = 1, 푠1(0) = 푎0푎1, 푠−1(1) = −1 (base case) { 푖 푠0(푖) = 0, 푠1(푖) = 푎0{푎푖 + 푎푖+1}, 푠−1(푖) = (−1) (푖 ≥ 1) (21)
Proof: 2 2 2 For 푖 > 푗 > 0, if we have 푠0(푖 − 1) = 푻푛(푖, 1) = 0, with the recursive formula, 푻푛(푖 + 1, 푗 + 1) = 푻푛(푖, 푗), 2 2 by induction we can easily have 푻푛(푖, 푗) = 0 and 푻푛(푖 + 1, 푗 + 1) = 0 for all 푖 > 푗 > 1. So we only need to prove with the left boundary elements 푠0(푖) = 0 for 푖 > 0. For even indexed boundary elements, 푎2푚, we use Eq.(20) for the solution, so its corresponding equation 푠0(2푚) = 0 always holds. Then the proof is needed for the odd indexed boundary elements, or 푠0(2푚 + 1) = 0, and we do it by induction. For 푛 = 1,2,3, with 푎0 = ±1, the triangles are formed as: 푎0 0 0 푎0 0 푻1 = 푎0, 푻2 = [ ] , 푻3 = [푎1 −푎0 0 ] 푎1 −푎0 푎2 푎0 − 푎1 푎0 2 2 2 Let 푎1 be free, it is easy to verify that 푻1 = 푰 and 푻2 = 푰, 푠0(0) = 푎0 = 1, 푠1(0) = 푎0푎1, 푠−1(1) = 푎0푏1 = −1, 푠0(1) = 푎1푎0 + 푏1푎1 = 0. 2 With 푚 = 1, 푛 = 2푚 + 1 = 3, we can have 푻3 = 푰 if we find 푎2 by using Eq.(20): 2푚−1 푎 푎 푎 푎 = 푎 = − 0 ∑ 푡(2푚, 푘)푎 = − 0 푡(2,1)푎 = 0 푎 (푎 − 푎 ) 2 2푚 2 푘 2 1 2 1 1 0 푘=1 2 Then we can verify 푠1(1) = 푎1 + 푏1푎2 = 푎0{푎1 + 푎2} , 푠0(2) = 푎2푎0 + 푡(2,1)푎1 + 푏2푎2 = 0 , 푠−1(2) = 2 푡(2,1)푎0 + 푏2푎1 = 1, and 푻3 = 푰, and give 푻4 as: 푎0 0 0 0 푎 −푎 0 0 1 0 푎0 푻4 = 푎 (푎 − 푎 ) 푎 − 푎 푎 0 2 1 1 0 0 1 0 푎0 푎 − (푎 − 푎 )(푎 − 2푎 ) 푎 − 2푎 −푎 [ 3 2 1 0 1 0 1 0 0] 2 It is also easy to verify 푠1(2) = 푎0{푎2 + 푎3}, 푠0(3) = 0, 푠−1(3) = −1, and 푻4 = 푰. For the boundary conditions, we can summarize the equalities above as : 푠0(1) = 0, 푠1(1) = 푎0{푎1 + 푎2}, 푠−1(1) = −1 푠0(2) = 0, 푠1(2) = 푎0{푎2 + 푎3}, 푠−1(2) = 1 If we use Eq.(20) to find 푎2푚, we have 푠0(2푚) = 0. Assume Eq.(21) holds for integer 푖 = 2푚, 2푚 푠0(2푚) = 0, 푠1(2푚) = 푎0{푎2푚 + 푎2푚+1}, 푠−1(2푚) = (−1) = 1 Then we can find those for 푖 = 2푚 + 1 with the recursive formulas in Eq.(10). For 푠−1(푖) we have: 푠−1(2푚 + 1) = 푎2푚{푏2푚 + 푏2푚+1} − 푠−1(2푚) − 푠0(2푚) = −1 푖 By induction, we have 푠−1(푖) = (−1) holds, and with Eq.(10) again for 푠0(푖) we have : 푠0(푖) = 푎푖{푏푖 + 푏푖+1} − 푠−1(푖) − 푠−1(푖 + 1) = 0 Finally, with the recursive formula Eq.(10) for 푠1(푖), we have 푠1(푖) = 푎0{푎푖 + 푎푖+1} + 푎푖+1{푏푖 + 푏푖+1} − 푠0(푖) − 푠0(푖 + 1) = 푎0{푎푖 + 푎푖+1} ∎
1 1 With 푎 = ±1 , the first few dependent boundary cells are: 푎 = 푎 (푎 푎 − 1), 푎 = (푎 (2 + 4푎2 − 0 2 2 1 0 1 4 8 1 1 1 푎 (5푎 + 푎3 − 8푎 )) − 12푎 ), 푎 = (−8푎 − 38푎3 − 8푎5 + 40푎 + 40푎2푎 − 40푎 + 푎 (28푎2 + 25푎4 + 0 1 1 3 3 6 16 1 1 1 3 1 3 5 0 1 1 6 3 2 푎1 − 8푎1 푎3 + 8푎3 + 푎1(−68푎3 + 16푎5))). Since a division is involved in Eq.(20), if all the odd indexed boundary numbers are integers, the triangle does not have to be all integers. If we expand the solution for 푎2푚, Eq.(20), to find all the terms involving 푎2푚−1, then we can choose some 푎2푚−1 to make 푎2푚 integers: 2푚−1 2푚−2 푎 푎 푎 = − 0 ∑ 푡(2푚, 푘)푎 = 0 {−푡(2푚, 1)푎 − 푡(2푚, 2푚 − 1)푎 − ∑ 푡(2푚, 푘)푎 } 2푚 2 푘 2 1 2푚−1 푘 푘=1 푘=2 2푚−2 푎 = 0 {{푡(2푚 − 1,0) + 푡(2푚 − 1,1)}푎 − 푡(2푚, 2푚 − 1)푎 − ∑ 푡(2푚, 푘)푎 } 2 1 2푚−1 푘 푘=2 2푚−2 푎 = 0 {{푎 − 푡(2푚, 2푚 − 1)}푎 + 푎 푡(2푚 − 1,1) − ∑ 푡(2푚, 푘)푎 } 2 1 2푚−1 1 푘 푘=2 With the general formula for the interior cells Eq.(6) we can find 푡(2푚, 2푚 − 1) as : 1 2푚−1 2푚 − 푘 − 1 2푚 − 푘 − 1 푡(2푚, 2푚 − 1) = ∑ ( ) (−1)2푚−푘푎 + ∑ ( ) (−1)2푚−푘푏 2푚 − 1 − 1 푘 2푚 − 2푚 + 1 − 1 푘 푘=1 푘=1 2푚−1 2푘 = −푎1 + ∑ (−1) 푎0 = 푎0(2푚 − 1) − 푎1 푘=1 Then all the terms involving 푎2푚−1 can be {푎1 − 푡(2푚, 2푚 − 1)}푎2푚−1 = {2푎1 − 푎0(2푚 − 1)}푎2푚−1 = 2(푎1 − 푎0푚)푎2푚−1 + 푎0푎2푚−1 which gives 2푚−2 푎 푎 = 0 {푎 푎 + 2(푎 − 푎 푚)푎 + 푎 푡(2푚 − 1,1) − ∑ 푡(2푚, 푘)푎 } 2푚 2 0 2푚−1 1 0 2푚−1 1 푘 푘=2 We can just choose an even or an odd number for 푎2푚−1 to make the number inside the curly brackets an even number. Therefore, to make all the triangle cells integers, the odd indexed left boundary cells are not completely free to choose, but just limited to odd or even numbers. An example of involutory triangle is Triangle 2 shown on the right in Section 2, where we use 푎푖 = 1 + 푖 mod(푖, 2). More examples can be with 푎푖 = 0, 푎푖 = 3, 푎푖 = (−1) , or with 푎푖 using a repeating pattern of 4 integers {1,0, −2,1}. The first 9 rows of two more triangles (5&6) are given below: 1 1 3 -1 1 -1 3 -2 1 0 0 1 3 -1 1 -1 -2 0 -1 -1 3 -2 0 0 1 1 2 1 2 1 3 -1 2 0 -1 -1 1 -3 -3 -3 -3 -1 3 -2 -1 -2 1 2 1 0 2 6 6 6 4 1 3 -1 3 3 1 -3 -3 -1 -2 -2 -8 -12 -12 -10 -5 -1 3 -2 -2 -6 -4 2 6 4 1 1 4 10 20 24 22 15 6 1 Triangle 5 Triangle 6
Similar to the idempotent triangles, if we remove the leftmost boundary cells, the triangle with 푎0 = 1 becomes the triangle with 푎0 = −1, and vice versa. In addition, if we negate all the numbers in an involutory triangle, it is 2 2 also involutory, or if 푻 = 푰, (−푻) = 푰. This actually switches between the triangles with 푎0 = 1 and the triangles with 푎0 = −1.
7. The Triangles for Nilpotent Matrices of Index 2
If a triangular matrix is nilpotent, it should be strictly triangular and all the main diagonal entries should be 0. The triangles we have in the above sections are not, but we can shift a triangular matrix downwards or leftwards such that they become strictly triangular with all the main diagonal entries being 0. For square matrices, to shift a matrix down by one row, we fill a row of zeros at the top and remove the bottom row. To shift left by one column, we fill a column of zeros on the right and remove the first column. The shift operator is the same as matrix multiplication with the shift-left matrix, 푩퐿, a lower subdiagonal matrix with 1s as all subdiagonal entries, left-multiply to shift-down and right-multiply to shift-left. With the idempotent triangle matrices, we can convert them into nilpotent matrices by shifting: 푵퐷 = 푩퐿푻푛 and 푵퐿 = 푻푛푩퐿. With the idempotent triangles given by Theorem 1 above, we have the following theorem.
Theorem 3: If an idempotent matrix of zero-sum triangle 푻푛 is shifted one cell downwards, 푵퐷 = 푩퐿푻푛, or one 2 2 cell leftwards, 푵퐿 = 푻푛푩퐿, the matrices are nilpotent matrices of index 2, 푵퐷 = 푵퐿 = 0.
Proof: Let 푻푛 be an idempotent matrix of the triangle generated with the zero-sum rule, and we still use the same notations as in Section 5, cells 푻푛(푖 + 1, 푗 + 1) = 푡(푖, 푗) and the boundary cells meet the conditions as in Theorem 1. Use 푵퐷 and 푵퐿 to represent the matrices by shifting 푻푛 down one row and shifting 푻푛 left one column, respectively, which are equivalent to 푵퐷 = 푩퐿푻푛 and 푵퐿 = 푻푛푩퐿. The elements of the matrices are: 0 (푖 = 0) 푵퐷(푖 + 1, 푗 + 1) = { 푻푛(푖, 푗 + 1) (푖 > 0) 0 (푗 = 푛 − 1) 푵퐿(푖 + 1, 푗 + 1) = { 푻푛(푖 + 1, 푗 + 2) (푗 < 푛 − 1) The elements of 푻푛푩퐿푻푛 for 푖 > 푗 are 푛−1
(푻푛푩퐿푻푛)(푖 + 1, 푗 + 1) = (푻푛푵퐷)(푖 + 1, 푗 + 1) = ∑ 푻푛(푖 + 1, 푘 + 1)푵퐷(푘 + 1, 푗 + 1) 푘=0 푛−1 푖
= ∑ 푻푛(푖 + 1, 푘 + 1)푻푛(푘, 푗 + 1) = ∑ 푡(푖, 푘)푡(푘 − 1, 푗) 푘=1 푘=푗+1 For 푗 = 0, 푡(푘 − 1,0) = 푎푖−1, with the boundary multiplication equalities in Theorem 1, we have the first column all 0s : 푖 푖
(푻푛푩퐿푻푛)(푖 + 1,1) = ∑ 푡(푖, 푘)푡(푘 − 1,0) = ∑ 푡(푖, 푘)푎푖−1 = 푠−1(푖) = 0 푘=1 푘=1 For 푗 > 0, we have all the elements in a column equal to the column on the left up to a sign: 푖 푖
(푻푛푩퐿푻푛)(푖 + 1, 푗 + 1) = ∑ 푡(푖, 푘)푡(푘 − 1, 푗) = − ∑ 푡(푖, 푘){푡(푘 − 1, 푗 − 1) + 푡(푘, 푗)} 푘=푗+1 푘=푗+1 푖 푖 = 푡(푖, 푗)푡(푗 − 1, 푗 − 1) − ∑ 푡(푖, 푘)푡(푘 − 1, 푗 − 1) + 푡(푖, 푗)푡(푗, 푗) − ∑ 푡(푖, 푘)푡(푘, 푗) 푘=푗 푘=푗 2 = 푡(푖, 푗){푏푗−1 + 푏푗} − (푻푛푩퐿푻푛)(푖 + 1, 푗) − 푻푛(푖 + 1, 푗 + 1)
= 푡(푖, 푗){푏푗−1 + 푏푗} − (푻푛푩퐿푻푛)(푖 + 1, 푗) − 푻푛(푖 + 1, 푗 + 1)
= 푡(푖, 푗){푏푗−1 + 푏푗} − (푻푛푩퐿푻푛)(푖 + 1, 푗) − 푡(푖, 푗) = −(푻푛푩퐿푻푛)(푖 + 1, 푗) With the first column all 0s, this gives 푻푛푩퐿푻푛 = 0. 2 2 Thus, we have 푵퐷 = 푩퐿푻푛푩퐿푻푛 = 0 and 푵퐿 = 푻푛푩퐿푻푛푩퐿 = 0, and both 푵퐷 and 푵퐿 are nilpotent matrices of index 2. ∎
8. The Triangles for Unipotent Matrices of Index 2
With the nilpotent matrices generated by shifting an idempotent triangle, we can have their corresponding unipotent matrices of index 2, 푺퐷 = 푰 + 푵퐷 = 푰 + 푩퐿푻푛 and 푺퐿 = 푰 + 푵퐿 = 푰 + 푻푛푩퐿 . Then we have the following theorem for a bidiagonal matrix with both diagonals all 1s, a full-sized Jordan block of eigenvalue 1.
Theorem 4: If 푵퐷 and 푵퐿 are nilpotent matrices generated by shifting an idempotent matrix of zero-sum triangle one cell downwards and one cell leftwards, respectively, the product of the corresponding unipotent matrices makes a bidiagonal matrix with both diagonals all 1s: 푺퐷푺퐿 = (푰 + 푩퐿푻푛)(푰 + 푻푛푩퐿) = 푰 + 푩퐿.
Proof: With the idempotent property, we can easily have: 푺퐷푺퐿 = (푰 + 푩퐿푻푛)(푰 + 푻푛푩퐿) = 푰 + 푩퐿푻푛 + 푻푛푩퐿 + 푩퐿푻푛푻푛푩퐿 = 푰 + 푩퐿푻푛 + 푻푛푩퐿 + 푩퐿푻푛푩퐿 They are all lower triangular matrices, and the main diagonal elements of the triangular unipotent matrices are all 1s, so we only consider the lower elements for the proof. For the elements below the lower subdiagonal, 푖 − 2 ≥ 푗 ≥ 0, we have : (푩퐿푻푛 + 푻푛푩퐿 + 푩퐿푻푛푩퐿)(푖 + 1, 푗 + 1) = 푻푛(푖, 푗 + 1) + 푻푛(푖 + 1, 푗 + 2) + 푻푛(푖, 푗 + 2) = 푡(푖 − 1, 푗) + 푡(푖, 푗 + 1) + 푡(푖 − 1, 푗 + 1) = 0 The sum is 0, for the three positions are just at the right place for the zero-sum rule Eq.(2), thus all the elements below the subdiagonal are 0s. For the elements on the lower subdiagonal, 푖 − 1 = 푗 ≥ 0, we have them all 1s : (푩퐿푻푛 + 푻푛푩퐿 + 푩퐿푻푛푩퐿)(푖 + 1, 푗 + 1) = 푡(푖 − 1, 푗) + 푡(푖, 푗 + 1) + 0 = 푏푗 + 푏푗+1 = 1 In summary, the matrix product 푺퐷푺퐿 is a bidiagonal matrix with both diagonals all 1s, or 푺퐷푺퐿 = 푰 + 푩퐿. ∎
Such unipotent matrices can be used in matrix construction or factorization for Jordan canonical form blocks with prescribed eigenvalues or Frobenius normal form blocks with prescribed companion matrices, and the unipotent matrices are easy to generate recursively with integers only. By this theorem, we know that any unit triangular matrix is the product of two triangular unipotent matrices of index 2, any unipotent matrix can be a product of two unipotent matrices of index 2, and the factorization is not unique. For example with Theorem 4, we use the idempotent matrix of 푻4 as we give in the proof of Theorem 1: 푎0 0 0 0 0 0 0 0 푎1 푏1 0 0 1 0 0 0 푻4 = [ ] , 푩퐿 = [ ] 푎2 −푎1 − 푏1 푏2 0 0 1 0 0 푎3 푎1 + 푏1 − 푎2 푎1 + 푏1 − 푏2 푏3 0 0 1 0 Then we have 0 0 0 0 0 0 0 0 푎0 0 0 0 푏1 0 0 0 푵퐷 = 푩퐿푻푛 = [ ] , 푵퐿 = 푻푛푩퐿 = [ ] 푎1 푏1 0 0 −푎1 − 푏1 푏2 0 0 푎2 −푎1 − 푏1 푏2 0 푎1 + 푏1 − 푎2 푎1 + 푏1 − 푏2 푏3 0 0 0 0 0 0 0 0 0
2 0 0 0 0 2 0 0 0 0 푵퐷 = [ ] = 0, 푵퐿 = [ ] = 0 푎0푏1 0 0 0 푏1푏2 0 0 0 −푎0(푎1 + 푏1) + 푎1푏2 푏1푏2 0 0 푏1(푎1 + 푏1 − 푏2) − (푎1 + 푏1)푏3 푏2푏3 0 0 1 0 0 0 1 0 0 0 푎0 1 0 0 푏1 1 0 0 푺퐷푺퐿 = (푰 + 푵퐷)(푰 + 푵퐿) = [ ] ∙ [ ] 푎1 푏1 1 0 −푎1 − 푏1 푏2 1 0 푎2 −푎1 − 푏1 푏2 1 푎1 + 푏1 − 푎2 푎1 + 푏1 − 푏2 푏3 1 1 0 0 0 1 0 0 0 푎0 + 푏1 1 0 0 1 1 0 0 = [ ] = [ ] = 푰 + 푩퐿 (−1 + 푏1)푏1 푏1 + 푏2 1 0 0 1 1 0 −(푎1 + 푏1)(−1 + 푏1 + 푏2) (−1 + 푏2)푏2 푏2 + 푏3 1 0 0 1 1 2 where we use 푏0 = 푎0, 푏푖+1 = 1 − 푏푖, (1 − 푏푖)푏푖 = 0, 푏푖 = 푏푖.
9. Further Thoughts beyond
In the zero-sum triangles, in view of the general term of the interior cells Eq.(6), if the both boundary inputs are not all zeros, as long as they are nonzeros, positive or negative, or mixed with zeros, the further from the boundary edges, the bigger the cell numbers. In the left Triangle 7 below, for the boundary input, only 0s on the left edge, −1s and 1s on the right edge, the total input is zero, but the further from the boundaries, in the middle of the 9th row, we have a cell with a number as big as 33. 0 1 0 -1 -1 0 0 1 1 -1 1 1 0 -1 -2 -1 -1 0 -2 0 0 1 3 3 -1 -1 1 2 2 1 0 -1 -4 -6 -2 1 -1 0 -3 -4 -3 0 0 1 5 10 8 1 1 -1 1 3 7 7 3 1 0 -1 -6 -15 -18 -9 -2 -1 -1 0 -4 -10 -14 -10 -4 0 0 1 7 21 33 27 11 3 1 -1 1 4 14 24 24 14 4 1 Triangle 7 Triangle 8
In the right-hand Triangle 8, for the boundary input, only −1s on the left edge, 0s and 1s on the right edge, the total input is negative, but the further from the boundaries, in the middle of the 9th row, we have 2 cells with numbers as big as 24. According to Stirling's approximation to the central and maximal binomial coefficient of the binomial distribution [16], the triangle numbers further from the boundaries can be increasing at a geometric rate. If we go back to zero-sum's origin in economic game theory and view a zero-sum triangle through the lens of game theory, we can draw some interesting parallels between our matrix construction and a trading machine simulation. The boundary edge numbers can be seen as initial balances of “low-level” virtual traders following the zero-sum trading rule, while the positive and negative integers can be seen as the trader's cash surpluses and deficits, or credits and debts. Integers further away from the boundary edge can be seen “higher-level” traders' balances, which were determined by the trading patterns of “lower-level” traders. As we can see in Triangles 7&8, when trading moves further away from the starting positions at triangle boundaries, in terms of absolute magnitude, “higher-level” traders can see increasingly large balance surpluses and deficits, or like market bubbles of huge cash waves and financial crises of disastrous cash losses. This is the case even if the starting boundary inputs are small. This phenomenon lends itself to be compared with the real-world financial crises in 2008, where inter- linked financial institutions traded on highly-aggregated financial derivatives, whose default led to a disastrous outcome on the whole economy [17]. The zero-sum triangles may be a simple tool to construct integer balance triangles and to analyse the economic agents' trading behaviours that abide certain rules, so the potential areas of applications could include agent-based financial modelling and behaviour simulation of the economic agents in a zero-sum trading environment. The properties of the zero-sum triangles as triangular matrices should be able to find their explanations and applications in mathematical economics, which merits further investigation. Every matrix property of the zero-sum triangles may lead to a number of combinatorial identities, compared to those related to Pascal's triangle [4], and even a specific triangle has some identity relations in combinatorics. With the zero-sum rule, the proofs of the identities may also be much easier to clinch. For the triangle construction itself, if three boundary edges are involved, a pyramid of cells can be built with the zero-sum rule, or more edges for higher dimensional pyramids. And the triangles can also be extended to rectangles by using negative indices and indices bigger than the boundary, and extending the triangle from the boundary edges outwards to exterior cells with the zero-sum rule, similar to that with Pascal's triangle [16].
Acknowledgements: This work was supported by the National Nature Science Foundation of China under Grant 62071013 and National Key R&D Program of China under Grant 2018AAA0100300.
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