JOURNAL OF THE CHUNGCHEONG MATHEMATICAL SOCIETY Volume 33, No. 4, November 2020 http://dx.doi.org/10.14403/jcms.2020.33.4.413 ZERO COMMUTING MATRIX WITH PASCAL MATRIX Eunmi Choi* Abstract. In this work we study 0-commuting matrix C(0) under relationships with the Pascal matrix C(1). Like kth power C(1)k is an arithmetic matrix of (kx + y)n with yx = xy (k ≥ 1), we express C(0)k as an arithmetic matrix of certain polynomial with yx = 0. 1. Introduction An arithmetic matrix of a polynomial f(x) is a matrix consisting of coefficients of f(x). The Pascal matrix is a famous arithmetic matrix n n P n k n−k of (x + y) = k x y , in which the commutativity yx = xy is k=0 assumed tacitly. As a generalization, with two q-commuting variables x n and y satisfying yx = qxy (q 2 Z), the arithmetic matrix of (x + y) is called a q-commuting matrix denoted by C(q) [1]. The C(q) is composed hii (1−qi)(1−qi−1)···(1−qi−j+1) of q-binomials = 2 j [7]. When q = ±1, C(1) j q (1−q)(1−q )···(1−q ) is the Pascal matrix and C(−1) is the Pauli Pascal matrix [4]. A block 2i 3 i i h1 i matrix form C(−1) = 4i2i i 5 with a 2 × 2 matrix i = 11 looks very i3i 3i i ··· "1 # 11 similar to C(1) [5]. In particular if q = 0 then C(0) = 111 . The 1111 n × n matrix C(0) was studied as an adjacency matrix of relation ≤ on 2 1 3 −1 −1 1 f1; 2; ··· ; ng and its inverse C(0) = 4 −1 1 5 is a sum of nilpotent −11 ··· matrix and identity matrix ([2], p.5) Received May 12, 2020; Accepted August 20, 2020. 2010 Mathematics Subject Classification: 11B37, 11B39. Key words and phrases: q-commuting matrix, Pauli Pascal matrix, arithmetic matrix. 414 Eunmi Choi In the work we simply denote C(0) by Z and the Pascal matrix C(1) by P , and study Z under relationships with P . Since P k is an arith- metic matrix of (kx + y)n with yx = xy for k ≥ 1 [6], we develop interconnections of Zk with P as well as an arithmetic matrix of certain polynomial. −1 For our notations, write P = [ei;j] and inverse P = [vi;j], so vi;j = i−j i−ji (−1) ei;j = (−1) j for i ≥ j ≥ 0. As usual, Mn denotes a n × n matrix M. 2. Zk (k = 2; 3) as arithmetic matrices "1 # 21 3 21 3 11 2 21 3 3 1 Note Z = 111 , Z = 4321 5 and Z = 46 31 5. For each 1111 4321 10631 k (k) (k) (k) (k) k ≥ 1, write Z = y0 j y1 j y2 j · · · where yj (j ≥ 0) are jth column vectors. 2y1;1 y1;2 y1;3 ···3 y y y ··· Lemma 2.1. Let Y = y(1)j y(2)j y(3)j · · · = 2;1 2;2 2;3 be a 0 0 0 4y3;1 y3;2 y3;3 ···5 ··· (k) k matrix composed of 0th column vector y0 of each Z (k ≥ 1). Then (k) yi;j (i; j ≥ 1) satisfies yi;j = yi−1;j + yi;j−1. In the kth column y0 , the ith entry is yi;k = ei+k−2;k−1. "1 1 1 1 1# 1 2 3 4 5 Proof. Clearly Y = 1 3 6 10 15 is a symmetric type Pascal matrix 1 4 10 20 35 2 1 3 2 1 3 (3) 3 (4) 4 satisfying yi;j = yi−1;j + yi;j−1 [3]. From y = 6 6 7, y = 6107 and 0 4105 0 4205 ··· ··· 2 1 3 5 (5) i(i+1) i+1 i(i+1)(i+2) i+2 y = 6157, we notice yi;3 = = , yi;4 = = 0 4355 2 2 6 3 ··· i+3 (k) (k) i+k−3 and yi;5 = 4 in y0 (k = 3; 4; 5). In y0 , assume yi−1;k = k−1 = ei+k−3;k−1. Then we have yi;k = yi−1;k + yi;k−1 = ei+k−3;k−1 + ei+k−3;k−2 = ei+k−2;k−1. Zero commuting matrix with Pascal matrix 415 "ek−1;k−1# Thus Zk is determined by y(k) = ek;k−1 by shifting one space 0 ek+1;k−1 ··· down at each column. For Pascal matrix P = C(1) of (x + y)n with yx = xy, the power P k is an arithmetic matrix of (kx + y)n. Indeed 21 3 21 3 2 1 3 1 P 2 = 64 4 1 7, P 3 = 69 6 1 7 yield expansions of (2x + y)n and 48 12 6 1 5 427 27 9 1 5 ··· ··· (3x + y)n. An analogy question is that how Zk = C(0)k is related to polynomials (kx + y)n with yx = 0. For k ≥ 1, let A(k) be an arithmetic matrix of (kx + y)n with yx = 0. Theorem 2.2. Let diag[ki] be a diagonal matrix having diagonal entries ki (i ≥ 0). Then A(k) = diag[ki] Z diag[ki]−1 and A(k)Z = ZA(k). 21 3 21 3 2 1 3 1 Proof. Observe A(2) = 422 2 1 5 and A(3) = 432 3 1 5. Now for 23 22 21 33 32 31 ··· ··· any k, assume (kx + y)n = knxn + kn−1xn−1y + ··· + kxyn−1 + yn. Then (kx+y)n+1 = (kx+y)n(kx+y) = kn+1xn+1+knxny+···+kxyn+yn+1, 21 3 k 1 so A(k) = 4k2 k 1 5 = diag[ki]Zdiag[ki]−1. Thus ZA(k) = A(k)Z since k3 k2 k 1 ··· (k) sum of entries of 0th column in An equals that of nth row, and all entries of Z are 1. 20 3 20) 3 . 10 0 k . Let J = 4 10 5 = I . Then J = 4 . 5 for k > 0 by taking 10 0 k ::: I k proper size matrices. Clearly Jn is a nilpotent matrix satisfying Jn = 0 for k ≥ n, and we may consider J 0 = I. (2) (2) T T Theorem 2.3. Let Ad = A +J 2 and Zc2 = −Z2 +J 2 . Then for " k k #k k k k I2 : 0 0 I2 : 0 0 any n ≥ 4, A(2) = Z2 and Z2 = A(2) . n n [(2) n−4 n n Z[2 J n−4 An−2 J(n−2)×2 n−2 (n−2)×2 20010···03 0001···0 2T ···0 Proof. The transpose matrix Jk = 4 ···15 is upper triangular, 0 ···0 0 ···0 (2) 2 d(2) so with lower triangular matrices Ak = [ai;j] and Zk = [zi;j], the Ak = 416 Eunmi Choi (2) 2T 2 2 2T [^ai;j] = Ak + Jk and Zck = [^zi;j] = −Zk + Jk satisfya ^i;j = ai;j, z^i;j = −zi;j if i ≥ j, anda ^i;j =z ^i;j = 1 or 0 according to i = j + 1 or otherwise. Indeed, 21 0 1 3 21 3 20 0 1 0 3 2 1 0 1 0 0 0 1 2 (2) (2) T 2 1 62 2 1 0 1 7 Ad = A +J 2 = 22 2 1 + 6 ::: 07 = 6··· 7. k k k 4··· 5 4 15 62k−3 ··· 2 1 0 17 k−1 k−2 0 ::: 0 4 k−2 k−3 5 2 2 ··· 1 0 ::: 0 2 2 ··· 2 1 0 2k−1 2k−2 ··· 2 1 (2) When n = 5, simple multiplications of A5 and Z5 show that 21 3 2 1 3 01 I2 : 0 0 −1 0 1 I : 0 0 Z−2A(2)= 6101 7 = and A(2) Z2= 6−1 0 1 7 = 2 , 5 5 4 5 d(2) 5 5 4 5 2 2101 A3 J3×2 −2−1 01 Zc3 J3×2 42101 −3−2−101 101 −1 0 1 because 210 = A(2) + J 2T = Ad(2) and −2−1 0 = −Z2 + J 2T = Zc2. 421 3 3 3 −3−2−1 3 3 3 2 (2) 2 hZ 0i (2) A 0 Now we write Z = n−1 and An = n−1 in n n; n − 1; ··· ; 2 1 2n−1; ··· 22; 2 1 block matrix forms, and assume an induction hypothesis that 2 ) 3 " # 0 I2 : 0 0 . Z2 = A(2) with J n−4 = 6 . 7. n A[(2) J n−4 n (n−2)×2 4 0 n−4 5 n−2 (n−2)×2 1 0 0 1 Then Theorem 2 says 21 321 3 21 3 2 1 0 1 2 1 3 2 1 61 0 1 7 622 2 1 7 44 3 2 1 542 1 0 1 5 = 423 22 2 1 5. .. n−3 n−4 .. n n − 1 . 2 1 2 2 . 2; 1; 0; 1 2n−1 2n−2 .. 2 1 h i But since the (n − 1) × (n + 1) block matrix [(2) n−3 equals An−1 J(n−1)×2 21 0 1 ::: 003 21 0 1 ::: 0 03 2 1 01 ::: 00 2 1 01::: 0 0 " # 6 ··· 007 6 ··· 0 07 A[(2) : J n−4 0 6 n−4 7 = 6 n−4 7 = n−2 (n−2)×2 , 2 ··· 2101 00 2 ··· 21010 0 n−2 42n−3 2n−4 ::: 210 105 42n−3 2n−4 ::: 2101 05 2 ··· 2: 1 0 1 2n−2 2n−3 ::: 21 01 2n−2 2n−3 ::: 210 1 the block matrix multiplications show 2 I : 0 : 0 03 " I : 0 0 # 2 2 hZ2 0i G H Z2 = n A[(2) : J n−4 0 = , n+1 A[(2) J n−3 n + 1; n; ··· ; 2 1 4 n−2 (n−2)×2 5 R S n−1 (n−1)×2 2n−2 ··· 2 : 1 0 1 " # I2 : 0 0 2 (2) where H = 0, S = 1 and G = Zn [(2) n−4 = An by hypothesis.