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 ∈ 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 i i i h1 i matrix form C(−1) = i2i i 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 1 −1 −1 1 {1, 2, ··· , n} and its inverse C(0) = −1 1 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−j i (−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 # 1 1 11 2 21 3 3 1 Note Z = 111 , Z = 321 and Z = 6 31 . For each 1111 4321 10631 k (k) (k) (k) (k) k ≥ 1, write Z = y0 | y1 | y2 | · · · where yj (j ≥ 0) are jth column vectors.
y1,1 y1,2 y1,3 ··· y y y ··· Lemma 2.1. Let Y = y(1)| y(2)| y(3)| · · · = 2,1 2,2 2,3 be a 0 0 0 y3,1 y3,2 y3,3 ··· ··· (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 1 1 (3) 3 (4) 4 satisfying yi,j = yi−1,j + yi,j−1 [3]. From y = 6 , y = 10 and 0 10 0 20 ··· ··· 1 5 (5) i(i+1) i+1 i(i+1)(i+2) i+2 y = 15, we notice yi,3 = = , yi,4 = = 0 35 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 1 1 2 1 3 1 P 2 = 4 4 1 , P 3 = 9 6 1 yield expansions of (2x + y)n and 8 12 6 1 27 27 9 1 ··· ··· (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). 1 1 2 1 3 1 Proof. Observe A(2) = 22 2 1 and A(3) = 32 3 1 . 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, 1 k 1 so A(k) = k2 k 1 = 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. 0 0) . 10 0 k . Let J = 10 = I . Then J = . 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 0010···0 0001···0 2T ···0 Proof. The transpose matrix Jk = ···1 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 [ˆai,j] = Ak + Jk and Zck = [ˆzi,j] = −Zk + Jk satisfya ˆi,j = ai,j, zˆi,j = −zi,j if i ≥ j, anda ˆi,j =z ˆi,j = 1 or 0 according to i = j + 1 or otherwise. Indeed, 1 0 1 1 0 0 1 0 2 1 0 1 0 0 0 1 2 (2) (2) T 2 1 2 2 1 0 1 Ad = A +J 2 = 22 2 1 + ... 0 = ··· . k k k ··· 1 2k−3 ··· 2 1 0 1 k−1 k−2 0 ... 0 k−2 k−3 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 1 1 01 I2 : 0 0 −1 0 1 I : 0 0 Z−2A(2)= 101 = and A(2) Z2= −1 0 1 = 2 , 5 5 d(2) 5 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 ) " # 0 I2 : 0 0 . Z2 = A(2) with J n−4 = . . n A[(2) J n−4 n (n−2)×2 0 n−4 n−2 (n−2)×2 1 0 0 1 Then Theorem 2 says 1 1 1 2 1 0 1 2 1 3 2 1 1 0 1 22 2 1 4 3 2 1 2 1 0 1 = 23 22 2 1 . .. 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 1 0 1 ... 00 1 0 1 ... 0 0 2 1 01 ... 00 2 1 01... 0 0 " # ··· 00 ··· 0 0 A[(2) : J n−4 0 n−4 = n−4 = n−2 (n−2)×2 , 2 ··· 2101 00 2 ··· 21010 0 n−2 2n−3 2n−4 ... 210 10 2n−3 2n−4 ... 2101 0 2 ··· 2: 1 0 1 2n−2 2n−3 ... 21 01 2n−2 2n−3 ... 210 1 the block matrix multiplications show I : 0 : 0 0 " 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 n−2 (n−2)×2 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. An−2 J(n−2)×2 And Zero commuting matrix with Pascal matrix 417
1 0 1 1 0 1 n−2 2 R = (n + 1, n, ··· , 3, 2)2 1 0 1 +(2 , ··· , 2 , 2, 1, 0) 2n−4 2n−5 ··· 1 0 1 0 2n−3 2n−4 ··· 2 1 0 1 n−3 n−4 n−5 = n + 1 + P (n − 1 − i)2i, n + P (n − 2 − i)2i, n − 1 + P (n − 3 − i)2i, i=0 i=0 i=0 ··· , 4 + P(2 − i)2i, 3, 2 +(2n−2, ··· , 2, 1, 0) i=0 n−2 n−3 n−4 = n + 1 + P (n − 1 − i)2i, n + P (n − 2 − i)2i, n − 1 + P (n − 3 − i)2i, i=0 i=0 i=0 1 ··· , 4 + P(2 − i)2i, 4, 2 . i=0 Thus the first few entries in R from right are 2, 22, 23, 24, ··· . So as an n−3 induction hypothesis, we assume n + P (n − 2 − i)2i = 2n−1. Then i=0 n−2 n−2 (n + 1) + P (n − 1 − i)2i = (n + 1) + (n − 1) + P (n − 1 − i)2i i=0 i=1 n−3 = 2n + 2 P (n − 2 − i)2i = 2n + 2(2n−1 − n) = i=0 2n. " # I2 : 0 0 n n−1 3 2 2 (2) So R = (2 , 2 , ··· , 2 , 2 , 2), thus Zn+1 [(2) n−3 = An+1. An−1 J(n−1)×2 And the second identity can be proved similarly.
We now go on Z3 and the arithmetic matrix A(3) of (3x + y)n with yx = 0.
(3) (3) 2T 3 Theorem 2.4. Let Ad = (3Ik − Jk)A + J and Zc = −(3Ik − k " k # k k" # I2 : 0 0 I2 : 0 0 3 2T (3) 3 3 (3) Jk)Zk +Jk . Then An = Zn [(3) n−4 and Zn = An [(3) n−4 An−2 J(n−2)×2 Zn−2 J(n−2)×2 for n ≥ 4.
(3) (3) Proof. When k = 5, Theorem 2 shows Z5A5 = A5 Z5 and 1 0 1 h i 3 0 1 −3 (3) I2 : 0 0 Z5 A5 = 3 0 1 = with X = 8 3 0 . 8 3 0 1 0 X J3×2 24 8 3 24 8 3 0 1 418 Eunmi Choi
3 0 0 0 0 1 3 0 0 2T Clearly X = 8 3 0 + 0 0 0 = Y + J3 with Y = 8 3 0 . But 24 8 3 0 0 0 24 8 3 3 (3) (3) −(3) since YA3 = A3 Y and A3 Y = −1 3 = 3I − J, we have −13 2T (3) 2T (3) 2T d(3) X = Y + J3 = A3 (3I − J) + J3 = (3I − J)A3 + J3 = A3 , I2 : 0 0 so Z−3A(3) = . For some n > 0, we assume 5 5 d(3) A3 J3×2 1 1 " # 3 1 0 1 I2 : 0 0 (3) 3 6 3 1 3 0 1 An = Zn [(3) n−4 = 10 6 3 1 8 3 0 1 . An−2 J (n−2)×2 . . n+1 n .. 3 n−4 3 n−5 .. 3 2 2 3 1 2 3 2 3 2 3 0 1 We note that d(3) (3) 2T An = (3I − J)An + Jn 1 0010···0 3 0 1 0 3 1 0001···0 8 3 0 1 2 ···0 ··· 0 0 = (3I −J) 3 3 1 + ···1 = . ··· ···0 233n−3 233n−4 ··· 3 0 1 0 3n−1 3n−2 31 ···0 233n−2 233n−3 ··· 23 3 0 1 h i so the block matrix [(3) n−3 equals An−1 J(n−1)×2 3 0 1 0 0 0 3 0 1 0 0 0 8 3 0 1 0 0 8 3 0 1 0 0 ··· = ··· 233n−4, 233n−5, ··· , 3, 0, 1, 0 1 0 233n−4, 233n−5, ··· , 3, 0, 1, 0 1 0 233n−3, 233n−4, ··· , 23, 3, 0, 1 0 1 233n−3, 233n−4, ··· , 23, 3, 0, 1 0 1 " # A[(3) : J n−4 0 = n−2 (n−2)×2 . 233n−3, ··· , 23, 3, 0 : 1, 0 1 Thus the block matrix multiplications yield " # I2 : 0 : 0 0 I2 : 0 0 3 3 Zn 0 [(3) n−4 Z (3) = n+2 n+1 A : J 0 n+1 A[ J n−3 , , ··· , 3 1 n−2 (n−2)×2 n−1 (n−1)×2 2 2 233n−3, ··· , 23, 3 : 1, 0 1 G H = , R S " # I2 : 0 0 3 (3) where G = Zn [(3) n−4 = An , H = 0n×1 and S = 1. And for An−2 J(n−2)×2 R, " # I2 : 0 0 n+2 n+1 3 n−3 3 R = ( 2 , 2 , ··· , 6, 3) [(3) n−4 + (2 3 , ··· , 2 , 3, 0, 1, 0) An−2 J(n−2)×2 Zero commuting matrix with Pascal matrix 419
1 0 1 3 0 1 = ( n+2, n+1, ··· , 6, 3) 8 3 0 1 + (233n−3, ··· , 3, 0, 1, 0) 2 2 ··· . 233n−4, 233n−5, .. 3, 0, 1 3 n−3 3 = (θ1, θ2, ··· , θn, θn+1) +(2 3 , ··· , 2 , 3, 0, 1, 0), where n+2 n 3 n−1 3 n−2 3 n−4 3 θ1 = 2 + 3 2 + 2 2 + 2 3 2 + ··· + 2 3 2 , n+1 n−1 3 n−2 3 n−3 3 n−5 3 θ2 = 2 + 3 2 + 2 2 + 2 3 2 + ··· + 2 3 2 , ··· , 6(5) 3 5(4) θn−2 = 2 + 6 · 3 + 3 · 2 , θn−1 = 2 + 3, θn = 6, θn+1 = 3. 2 3 3 So the last few entries in R are θn+1 = 3, θn + 3 = 3 , θn−1 + 2 = 3 3 4 n−1 3 n−4 and θn−2 + 2 · 3 = 3 . We now assume 3 = θ2 + 2 3 . Then n−1 n+1 n−1 3 n−2 3 n−3 3 n−5 3 3 = 2 + 3 2 + 2 2 + 2 3 2 + ··· + 2 3 2 + 233n−4. Hence 3 n−3 θ1 + 2 3 n+2 n 3 n−1 3 n−2 3 n−4 3 3 n−3 = 2 + 3 2 + 2 2 + 2 3 2 + ··· + 2 3 2 + 2 3 n+2 n 3 n−1 3 n−2 n−3 n−5 3 n−4 = 2 +3 2 +2 2 +3·2 2 +3 2 +···+3 2 +3 n+2 n 3 n−1 n−1 n+1 n−1 = 2 + 3 2 + 2 2 + 3 3 − 2 − 3 2 n n+2 n 3 n−1 n+1 2 n−1 n = 3 + 2 + 3 2 + 2 2 − 3 2 − 3 2 = 3 , by induction hypothesis. Thus R = (3n, 3n−1, ··· , 32, 3), so we have " # I2 : 0 0 G H (3) Z3 = = An 0 = A(3) . n+1 [(3) n−3 R S n n−1 n+1 An−1 J(n−1)×2 3 , 3 , ··· , 3 1 Similarly for the second identity, we have 1 0 1 " −3 0 1# −1 I : 0 0 A(3) Z3 = −3 0 1 = 2 with X = −8−3 0 . −8−3 01 X J −15−8−3 −15−8−301 ··· ··· " −3 0 0# "0 01# " 300# −8−3 0 0 00 2T 830 But since X = −15−8−3 + 0 00 = −Y + J with Y = 1583 , ··· ··· ··· " 3 # −3 −1 3 3 3 we have YZ = −13 = 3I − J. So Y = Z (3I − J) = (3I − J)Z ··· I2 : 0 0 and X = −(3I −J)Z3 +J 2T = Z3. Thus Z3 = A(3) . c n n [3 n−4 Zn−2 J(n−2)×2 420 Eunmi Choi
Connections of A(k) and Zk (k = 2, 3) are further explained by J. −1 −2 (2) 2 (2) (2) 2 2 2 Theorem 2.5. (1) Zn An = In +JnAn and An Zn = In −JnZn. −1 −3 (3) 2 (3) (3) 3 2 (2) Zn An = In + Jn(3In − Jn)An and An Zn = In − Jn(3In − 3 Jn)Zn. −1 −1 (2) −2 2 (3) −3 2 (3) An = Zn − Jn and An = Zn − Jn(3In − Jn).
d(2) (2) 2T Proof. With Ak = Ak + Jk , Theorem 3 shows 1 0 0 0 1 0 " # I2 : 0 0 0 Z−2A(2) = = 1 0 1 = I + 0 n n A[(2) J n−4 2 1 01 n (2) n−2 (n−2)×2 ··· 10 An−2 2n−3 2n−4 . .. 01 2 (2) = In + JnAn . 2 2 2T Similarly with Zck = −Zk + Jk , we also have 1 00 0 10 −1 I : 0 0 0 (2) 2 2 −1 01 A Z = n−4 = = I + 0 n n Z[2 J −2 −10 1 n 2 n−2 (n−2)×2 ... 10 −Zn−2 −en−2,1 −en−3,1 ... 01 2 2 = In − JnZn. d(3) (3) 2T On the other hand, with Ak = (3Ik − Jk)Ak + Jk , Theorem 4 shows 1 0 0 0 1 0 " # I2 : 0 0 3 0 1 0 Z−3A(3)= = 8 3 01 = I + 0 n n [(3) n−4 n (3) An−2 J(n−2)×2 24 8 301 00 (3I − J)A ··· 10 n−2 233n−4 232n−5 ... 01 2 (3) = In + Jn(3In − Jn)An . −1 I : 0 0 (3) 3 2 2 3 Moreover An Zn = [3 n−4 = In −Jn(3I −J)Zn. And also Zn−2 J(n−2)×2 −1 −1 −2 (2) 2 −3 (3) 2 Zn = An + Jn and Zn = An + Jn(3In − Jn) by (1) and (2). " 1 # "0 # " 1 # −2 2 −2 1 0 −2 1 (2)−1 In fact, Z − J = 1−2 1 − 100 = −2 1 = A . 1−21 100 −21
3. Zk as an arithmetic matrix
" 0 # −2 (2) 0 2 (2) Theorem 5 says Z An = In + = In + J An , n (2) n An−1 Zero commuting matrix with Pascal matrix 421
" 0 # −3 (3) 0 2 (3) Z An = In + = In + J (3In − Jn)An , n (3) n (3In−2 − Jn−2)An−2 −1 −1 (2) 2 2 2 (3) 3 2 3 An Zn = In − JnZn and An Zn = In − Jn(3In − Jn)Zn. In order to generalize the identities, first consider k = 4, 5, 6. Then we have " 0 # −k (k) 0 2 (k) Zn An = In + = In + JnS (1) S(k) 6 10 15 20 6 40 10 70 15 with S(4) = 81 206 , S(5) = 205 40 10 and S(6) = 435 70 15 . 32481206 10242054010 26044357015 ··· ··· ··· Also " # −1 0 (k) k 0 2 (k) An Zn = In − = In − JnT (2) T (k) 6 10 15 20 6 40 10 70 15 with T (4) = 45 206 , T (5) = 10540 10 and T (6) = 21070 15 . 84 45206 2241054010 5042107015 ··· ··· ··· −1 Now by means of the kth row (vk,0, ··· , vk,k) of P , we define a matrix k−2 (k) P k−j−2 X = vk,jJ . (3) j=0
(k) (k) (k) (k) (k) k Theorem 3.1. For 2 ≤ k ≤ 6, Xn An = Xn An and Xn Zn = −1 k (k) −k (k) 2 (k) (k) (k) k ZnXn . Moreover Zn An = In + JnXn An and An Zn = In − 2 (k) k JnXn Zn. Proof. Clearly X(2) = I and X(3) = 3I − J, so Theorem 5 implies 2 (2) (2) 2 (2) −2 (2) In + JnXn An = In + JnAn = Zn An 2 (3) (3) 2 (3) −3 (3) In + JnXn An = In + Jn(3In − Jn)An = Zn An −1 −1 2 (2) (2) (2) 2 2 (3) (3) (3) 3 In − JnXn An = An Zn and In + JnXn An = An Zn. −k (k) 2 (k) Now from Zn An = In + JnS in (1), we have 6 −1 −4 6 −1 A(4) S(4) = 1−4 6 = 6I − 4J + J 2 = S(4)A(4) 1−46 ··· 10 −10 10 (5)−1 (5) 5−10 10 2 3 (5) (5)−1 A S = −1 5−10 10 = 10I−10J+5J −J = S A −1 5−1010 ··· 422 Eunmi Choi
15 −1 −20 15 A(6) S(6) = 15−20 15 = 15I − 20J + 15J 2 − 6J 3 + J 4. −6 15−2015 ··· −1 Since each coefficients in expressions of A(k) S(k) correspond to el- −1 −1 ements in kth row of P −1, we have A(k) S(k) = S(k)A(k) = X(k), so (1) gives −k (k) 2 (k) 2 (k) (k) Zn An = In + JnS = In + JnXn An . −1 Similarly each T (k) in (2) satisfies Z−kT (k) = T (k)Z−k = A(k) S(k) = (k) −(k) k 2 (k) 2 (k) k X . So we have An Zn = In − JnT = In − JnXn Zn. k−2 P j (k) (k) Theorem 3.2. Let µ = k vk,j for k ≥ 2. Then X A is equal j=0 to vk,k−2 vk,k−3 + kvk,k−2,vk,k−2 ··· k−2 k−2 k−2 P j−1 P j−2 P j−3 k vk,j, k vk,j, k vk,j,··· j=1 j=2 j=3 k−2 k−2 P j−2 P j−2 µ k vk,j, k vk,j,··· j=1 j=2 k−2 kµ µ P kj−1v ,··· k,j j=1 k2µ kµ µ ··· ··· n−k+1 n−k k µ k µ ··· , µ, ··· , vk,k−3 + kvk,k−2,vk,k−2
Proof. We note that the first few X(k) in (3) are 0 0 "0 # "1 # " 6 # (4) 2 1 1 −4 6 X = v4,0J + v4,1J + v4,2I = 1 − 4 1 + 6 1 = 1−4 6 1 1 1 1−46 ...1 ...... ··· 10 −10 10 (5) 3 2 5−10 10 X = v5,0J + v5,1J + v5,2J + v5,3I = −1 5−10 10 and also, −1 5−10 10 −1 5−10 ··· vk,k−2 15 v v −20 15 k,k−3 ···k,k−2 15−20 15 vk,1 vk,2 ··· vk,k−2 X(6) = −6 15−20 15 . So X(k) = v v ··· 1 −6 15−20 15 k,0 vk,1 v v ···v 1 −6 15−20 k,0 vk,1 vk,2 ···vk,k−2 ··· k,0 k,1 ···k,k−3 1 3 1 (3) 2 for any k. Then since A = 3 3 1 and µ = v3,0 + 3v3,1, we have 33 32 31 ··· Zero commuting matrix with Pascal matrix 423
v3,1 v3,1 µ v3,1 v3,0 + 3v3,1 v3,1 (3) (3) 2 3µ µ v3,1 X An = 3v3,0 + 3 v3,1 v3,0 + 3v3,1 v3,1 = 2 . 2 3 µ 3µ µ v3,1 3 v3,0 + 3v3,1 ··· v3,1 ··· n−2 n−3 ··· 3 µ,3 µ,··· µ, v3,1 2 (4) P j Similarly A and µ = 4 v4,j imply j=0 v4,2 v4,1 + 4v4,2,v4,2 µ v4,1 + 4v4,2,v4,2 (4) (4) X A = 4µ µ v4,1 + 4v4,2,v4,2 . 42µ 4µ µ v + 4v ,v ··· 4,1 4,2 4,2 n−3 n−4 n−5 4 µ 4 µ 4 µ ··· µ v4,1 + 4v4,2,v4,2 Thus for any k > 1, it follows that (k) (k) k−2 k−3 (k) X A = (vk,0J + vk,1J + ··· + vk,k−3J + vk,k−2I)A 0) 0) . . . . h0 i (k) = vk,0 . + vk,1 . + ··· + vk,k−3 (k) + vk,k−2A . 0 k−2 0 k−3 A A(k) A(k) k−2 P j (k) (k) So with µ = k vk,j, X A holds the required form in Theorem j=0 7. We are ready to generalize Theorem 6 to Zk and A(k) for all k ≥ 2. −1 Theorem 3.3. Z−kA(k) = I + J 2X(k)A(k) and A(k) Zk = I − J 2X(k)Zk. (k) Proof. It is Theorem 6 if 2 ≤ k ≤ 6. For any k, consider An and (k) k−2 2 Xn = vk,0Jn + ··· + vk,k−4Jn + vk,k−3Jn + vk,k−2In. t Let n = 3. Since the nilpotent matrix Jn = 0 for t ≥ n, we have ek,k−2 0 0 (k) 2 X3 = vk,k−4Jn + vk,k−3Jn + vk,k−2I = −ek,k−3 ek,k−2 1 . ek,k−4 −ek,k−3 ek,k−2 "1 # "1 0 0# So with A(k) = k 1 and Zk = k 1 0 in Lemma 1, we 3 2 3 k k 1 ek+1,k−1 k 1 have 1 0 0 "1 # 2 (k) (k) k 2 (k) (k) I3 + J X A = 0 1 0 and Z (I3 + J X A ) = k 1 = 3 3 3 3 3 3 3 2 ek,k−2 0 1 k k 1 (k) A3 . Now for some n, as an induction hypothesis, we assume I : 0 0 (k) k 2 (k) (k) k 2 An = Zn(In + JnXn An ) = Zn (k) (k) 2 T n−4 . Xn−2An−2 + Jn−2 J(n−2)×2 By Lemma 1 we note that 424 Eunmi Choi
1 ek,k−1 1 k k Zn−1 0 Zn = ek+1,k−1 ek,k−1 1 = (4) ··· ek+(n−2),k−1, ··· , ek,k−1 1 ek+(n−2),k−1 ··· ek,k−1 1
Then
I : 0 0 k 2 (k) (k) k 2 Zn+1(In+1 + Jn+1Xn+1An+1) = Zn+1 (k) (k) 2 T n−3 Xn−1An−1 + Jn−1 J(n−1)×2
k " I2 : 0 0# Z 0 T = n X(k) A(k) + J 2 : J n−4 0 (5) Z0 1 n−2 n−2 n−2 (n−2)×2 X0 1 where bottom rows Z0 and X0 are from (4) and Theorem 7 that
0 Z = (ek+(n−1),k−1, ek+(n−2),k−1, ··· , ek+2,k−1, ek+1,k−1, ek,k−1),
k−2 k−2 0 n−k n−k−1 P j−1 P j−2 X = (k µ, k µ, ··· , kµ, µ, k vk,j, k vk,j, j=1 j=2
k−2 P j−(k−3) ··· , k vk,j, vk,k−2, 0). (6) j=k−3 Therefore the block matrix multiplications in (5) yield
G H Zk (I + J 2 X(k) A(k) ) = n+1 n+1 n+1 n+1 n+1 R S I : 0 0 k 2 (k) with H = 0, S = 1 and G = Zn (k) (k) 2 T n−4 = An by Xn−2An−2 + Jn−2 J(n−2)×2 induction hypothesis. Now we claim R = (kn, kn−1, ··· , k). If so, we get
G H (k) Zk (I +J 2 X(k) A(k) ) = = An 0 = A(k) , n+1 n+1 n+1 n+1 n+1 R S kn, kn−1, ··· , k 1 n+1 I : 0 0 0 2 0 as required. Now write R = Z (k) (k) 2 T n−4 + X by Xn−2An−2 + Jn−2 J(n−2)×2
0 R = (θ1, θ2, ··· , θn) + X = (χ1, χ2, ··· , χn), (7) with some θi, χi ∈ Z. Due to Theorem 7, the multiplication Zero commuting matrix with Pascal matrix 425
1 0 1 vk,k−2 0 1 vk,k−3 + kvk,k−2 vk,k−2 0 1 ··· k−2 k−2 P j−1 P j k vk,j k vk,j ··· j=1 j=2 (θ , ··· , θ ) = Z0 k−2 k−2 1 n P j P j µ k vk,j k vk,j ··· 0 1 j=1 j=2 k−2 P j kµ µ k vk,j ··· 0 1 j=1 k2µ kµ µ ··· 0 1 ··· n−k−1 n−k−2 k µ k µ ··· µ ··· vk,k−2 0 1 with Z0 in (6) satisfies
θ1 = ek+(n−1),k−1 +ek+(n−3),k−1vk,k−2 +ek+(n−4),k−1(vk,k−3 +kvk,k−2) k−2 k−2 P j−(k−4) P j−1 + ek+(n−5),k−1 k vk,j + ··· + en,k−1 k vk,j j=k−4 j=1 n−k−2 n−k−1 +en−1,k−1µ+en−2,k−1kµ+···+ek+1,k−1k µ+ek,k−1k µ, θ2 = ek+(n−2),k−1 +ek+(n−4),k−1vk,k−2 +ek+(n−5),k−1(vk,k−3 +kvk,k−2) k−2 k−2 P j−(k−4) P j−1 + ek+(n−6),k−1 k vk,j + ··· + en−1,k−1 k vk,j j=k−4 j=1 n−k−2 + en−2,k−1µ + en−3,k−1kµ + ··· + ek,k−1k µ, ··· , θn−2 = ek+2,k−1 + ek,k−1vk,k−2, θn−1 = ek+1,k−1, and θn = ek,k−1. 0 Then (χ1, ··· , χn) = R = (θ1, ··· , θn) + X in (6) and (7) satisfies χn = θn + 0 = ek,k−1 = k, k+1 k 2 χn−1 = θn−1 + vk,k−2 = ek+1,k−1 + ek,k−2 = k−1 + k−2 = k , 3 and χn−2 = (ek+2,k−1 + ek,k−1vk,k−2) + (vk,k−3 + kvk,k−2) = k . n−1 n Assume χ2 = k . Then the proof is complete if we show χ1 = k . n−k χ1 = θ1 + k µ = (ek+(n−1),k−1 + ek+(n−3),k−1vk,k−2 + ek+(n−4),k−1(vk,k−3 + kvk,k−2) 2 n−k−1 n−k + ··· + en−2,k−1kµ + en−3,k−1k µ + ··· + ek,k−1k µ) +k µ n−k−2 n−k−1 = A + k(en−2,k−1µ + en−3,k−1kµ + ··· + ek,k−1k µ + k µ), where A = ek+(n−1),k−1 + ek+(n−3),k−1vk,k−2 + ek+(n−4),k−1(vk,k−3 + k−2 P j−1 kvk,k−2) + ··· + en,k−1 k vk,j + en−1,k−1µ. j=1 On the other hand, the induction hypothesis shows n−1 n−k−1 k = χ2 = θ2 + k µ 426 Eunmi Choi
n−k−2 n−k−1 = B + (en−2,k−1µ + en−3,k−1kµ + ··· + ek,k−1k µ + k µ), where B = ek+(n−2),k−1 + ek+(n−4),k−1vk,k−2 + ek+(n−5),k−1(vk,k−3 + k−2 k−2 P j−2 P j−1 kvk,k−2) + ··· + en,k−1 k vk,j + en−1,k−1 k vk,j. j=2 j=1 Therefore n−k−2 n−k−1 χ1 = A + k(en−2,k−1µ + en−3,k−1kµ + ··· + ek,k−1k µ + k µ) = A + k(kn−1 − B) = kn + (A − kB). But we have A − kB = ek+(n−1),k−1 + ek+(n−3),k−1vk,k−2 + ek+(n−4),k−1(vk,k−3 + kvk,k−2) k−2 P j−1 + ··· + en,k−1 k vk,j + en−1,k−1µ j=1 −k ek+(n−2),k−1+ek+(n−4),k−1vk,k−2+ek+(n−5),k−1(vk,k−3+kvk,k−2) k−2 k−2 P j−2 P j−1 + ··· + en,k−1 k vk,j + en−1,k−1 k vk,j j=2 j=1 = ek+(n−1),k−1 − ek+(n−2),k−1k + ek+(n−3),k−1vk,k−2 + ek+(n−4),k−1((vk,k−3 + kvk,k−2) − kvk,k−2) + ··· k−2 k−2 k−2 P j−1 P j−1 P j + en,k−1( k vk,j − k vk,j) + en−1,k−1(µ − k vk,j) j=1 j=2 j=1 = ek+(n−1),k−1 − ek+(n−2),k−1k + ek+(n−3),k−1vk,k−2 + ek+(n−4),k−1vk,k−3 + ··· + en,k−1vk,1 + en−1,k−1vk,0 k P = vk,k−iek+(n−i−1),k−1. i=0 It means A − kB is a multiplication of kth row in P −1 and (k − 1)th n column in P , so A − kB = 0 and we have χ1 = k , as required.
−1 0 Therefore it follows that Z−k − A(k) = J 2X(k) = 0 . X(k)
References
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* Department of Mathematics Hannam University Daejon, Republic of Korea E-mail: [email protected]