# On the Determinants of the Square-Type Stirling Matrix and Bell Matrix

Authors: Eunmi Choi; Jiin Jo
Journal: International Journal of Mathematics and Mathematical Sciences
 (2021)

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## Abstract

We study determinants of the square-type Stirling matrixS∗ and the square-type Bell matrix B∗. For this purpose, we prove that S∗ and B∗ have LU factorizations S∗=LSUS and B∗=LBUB where the diagonal entries of US are kk−1, while those of UB are k! (k≥1).


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## Body

## 1. Introduction

The Stirling numberssm,n (m,n≥0) of second kind count the number of ways to partition an m element set into n subsets. The Stirling matrix S˜=sm,n satisfies a recurrence rule sm+1,n+1=sm,n+n+1sm,n+1 [1]. The sum ∑k=0msm,k of the mth row of S˜ is called the mth Bell number Bm, so Bm|m≥0=1,1,2,5,15,…. A triangular matrix B=bi,j having Bell numbers on both border and holding bi+1,j+1=bi,j+bi+1,ji,j≥1 is called the Bell matrix [2]. Since every entries in the first row and column of S˜ are zeros except s0,0=1, we denote by S the Stirling matrix deleted in the first row and column from S˜.(1)S˜=101011013101761⋯,S=si,j=111131176111525101⋯,B=bi,j=1122355710151520273752⋯,i,j≥1.Let(2)S∗=11111361017256511590350⋯,B∗=12515131037272711452087409⋯,be a square-type Stirling matrix and a square-type Bell matrix. In the work, we study the determinants of S∗ and B∗ by finding their LU factorizations. In fact, we prove that S∗ has an LU factorization S∗=LSUS, where LS=S and US has diagonal entries 1,2,32,43,… (Theorem 2), and B∗ has an LU factorization B∗=LBUB where LB=S˜P along with the Pascal matrix P and UB has diagonal entries 1,1,2!,3!,… (Theorem 12). This consideration is motivated by the square-type Pascal matrix(3)P∗=111123136⋯,where its LU factorization is P∗=PPT [3] so detP∗=1. Note that detBn∗ was discussed in [4] by means of Hankel transformation. Our feature in the work is to study detBn∗ by recurrence rules of Bell numbers over an LU factorization of Bn∗. For our notations, with a matrix M, Mk denotes the size k×k, and let riMk and cjMk be the ith row and jth column of Mk. Write x,…,x,a,b,… and a,b,…,x,…,x simply by x¯t;a,b,… and a,b,…;x¯t with the t copies x¯t of x. Therefore, x¯t;riMk means a row matrix having tx’s followed by a row matrix riMk, and similarly, x¯t;cjMk means a column matrix x⋮xcjMk. Let dia,b,… be a diagonal matrix having diagonal entries a,b,….


## 2. Square-Type Stirling Matrix

Fori,j≥1, let riS and cjS (resp., riS−1 and cjS−1) be the ith row and jth column of S (resp., S−1). Since S and S−1 are lower triangular matrices, riS and riS−1 can be considered as of size 1×i, while cjS and cjS−1 are of size ∞×1. But, if necessary, like the case of multiplication riScjS, we may regard riS filled with infinitely many zeros after the first i entries.Lemma 1.
LetS=si,j, S−1=si,j∨, and S∗=si,j∗ for i,j≥1. Let riS∗ and cjS∗ be the ith row and jth column of S∗.(1)
InS, si+1,j+1=si,j+j+1si,j+1 and cj+1S=0;cjS+j+10;cj+1S.(2)
InS−1, si+1,j+1∨=si,j∨−isi,j+1∨. Thus, si+1,1∨=−1ii! and ∑j=1isi,j∨=0 for i≥2.(3)
InS∗, si,j∗=si+j−1,j and si,j∗=si,j−1∗+jsi−1,j∗. And, cjS∗=cj−1S∗+j0;cjS∗=∑t=0k−1jt0¯t;cj−1S∗+jk0¯k;cjS∗=∑t=0∞jt0¯t;cj−1S∗.Proof.
The recurrence in (1) is well known [5]. The column cj+1S is(4)cj+1S=0,…,0,sj+1,j+1,sj+2,j+1,sj+3,j+1,…T=0¯j;sj,j,sj+1,j+j+1sj+1,j+1,sj+2,j+j+1sj+2,j+1,…T=0¯j;sj,j,sj+1,j,sj+2,j,…T+j+10¯j;0,sj+1,j+1,sj+2,j+1,…T=0;cjS+j+10;cj+1S.
The inverseS−1 is the signed Stirling matrix of first kind [6] satisfying the recurrence si+1,j+1∨=si,j∨−isi,j+1∨. From I=Sn−1Sn=riSn−1cjSn, we have 0=riS−1c1S=∑j=1isi,j∨ for i≥2, since c1S is composed of all 1s. Moreover, simple computation of S−1 shows si,1∨ (1≤i≤5) equals 1,−1,2,−3!,4!, respectively. Hence, if we assume si+1,1∨=−1ii! for some i, then si+2,1∨=si+1,0∨−i+1si+1,1∨=−i+1−1ii!=−1i+1i+1!.
Comparing(5)S∗=si,j∗=s1,1s2,2s3,3s2,1s3,2s4,3s3,1s4,2s5,3⋯with S=si,j, it is easy to see si,j∗=si+j−1,j and si,j∗=si,j−1∗+jsi−1,j∗. Now, for the jth column cjS∗, we have(6)cjS∗=cj−1S∗+j0;cjS∗=cj−1S∗+j0;cj−1S∗+j0;cjS∗=cj−1S∗+j0;cj−1S∗+j20¯2;cj−1S∗+j30¯3;cjS∗=⋯=∑t=0k−1jt0¯t;cj−1S∗+jk0¯k;cjS∗=⋯=∑t=0∞jt0¯t;cj−1S∗.Theorem 1.
riS−10¯j−t;cjS∗=0 for all 0≤t<j<i.Proof.
Note thatcjS∗=s1,j∗,s2,j∗,s3,j∗,…T=sj,j,sj+1,j,sj+2,j,…T and cjS=0¯j−1;cjS∗. Hence, we have riS−10¯j−1;cjS∗=riS−1cjS=0 from S−1S=I. When t=2, by Lemma 1 (3), we have(7)riS−10¯j−2;cjS∗=riS−10¯j−2;cj−1S∗+j0;cjS∗=riS−10¯j−2;cj−1S∗+jriS−10¯j−1;cjS∗=riS−1cj−1S+jriS−1cjS=0.Thus, by assuming riS−10¯j−t;cjS∗=0 for 1≤t≤j−2, we have riS−10¯;cjS∗=riS−10¯;cj−1S∗+j0;cjS∗=riS−10¯;cj−1S∗+jriS−10¯2;cjS∗=0.Theorem 2.
S∗ has an LU factorization SX, where X is an upper triangular matrix having diagonal entries ii−1 (i≥1). Therefore, detSk∗=∏i=1kii−1.Proof.
Letxi,j=X=S−1S∗. Then, Lemma 1 (2) shows xi,1=riS−1c1S∗=∑t=1isi,t∨=0 for all i>1, since all entries in c1S∗ are 1. And,(8)xi,2=riS−1c2S∗=riS−1c1S∗+20;c2S∗=xi,1+2riS−10;c2S∗=0,by Theorem 1. Thus, by assuming xi,i−2=0, we have(9)xi,i−1=riS−1ci−1S∗=riS−1ci−2S∗+i−10;ci−1S∗=xi,i−2+i−1riS−10;ci−2S∗=0,which shows X is an upper triangular matrix. Now, for xi,i, we have(10)xi,i=riS−1ciS∗=riS−1ci−1S∗+i0;ci−1S∗+⋯+ii−20¯i−2;ci−1S∗+ii−10¯i−1;ciS∗=riS−1ci−1S∗+iriS−10;ci−1S∗+⋯+ii−2riS−10¯i−2;ci−1S∗+ii−1riS−10¯i−1;ciS∗=ii−1riS−10¯i−1;1=ii−1.Indeed,(11)S∗=11111361017256511590350=11113117611111259323743.


## 3. Bell Matrix with the Pascal Matrix

LetriP and cjP be the ith row and jth column of the Pascal matrix P=pi,j (i,j≥1). Well-known recurrence rules of riP and cjP are(12)ri+1P=riP;0+0;riP,cj+1P=0;cjP+0;cj+1P.Theorem 3.
For any1≤j≤k, di0,1,…,k−10;cjPk−1=jcj+1Pk and rkPdi2k−1,…,2,1=rkPPk.Proof.
Due to the binomial identitypp−1q−1=qpq for p,q≥1, we have(13)jcj+1Pk=j0¯j,jj,j+1j,j+2j,…,k−1jT=0¯j,jj−1j−1,j+1jj−1,j+2j+1j−1,…,k−1k−2j−1T=di0,…,j−1;j,…,k−10,0¯j−1,j−1j−1,jj−1,j+1j−1,…,k−2j−1T=di0,1,…,k−10;cjPk−1.
Clearly,r3Pdi22,2,1=1,2,1P3. Assume rkPPk=rkPdi2k−1,…,2,1 for some k. Note that riP is the set of coefficients of x+1i−1 and riPPi equals riP2 which is the set of coefficients of 2x+1i−1 expanded in descending order. Thus, (12) with 2x+1k=2x2x+1k−1+2x+1k−1 implies(14)rk+1PPk+1=the set of coeff. of2x+1k=the set of coeff. of2x2x+1k−1+the set of coeff. of2x+1k−1=2rkPPk;0+0;rkPPk=2rkPdi2k−1,…,1;0+0;rkPdi2k−1,…,1=rkP;02k⋱21+0;rkP2k2k−1⋱1=rkP:0+0;rkPdi2k,…,1=rk+1Pdi2k,…,1.A matrixPkF=pi,jF is called a flipped matrix of Pk=pi,j if it is horizontally flipped sideways of Pk. Hence,(15)PkF=p1,k⋯p1,1p2,k⋯p2,1⋯pk,k⋯pk,1,pi,jF=pi,k−j+1,j≤k.Theorem 4.
LetriPkFPk be the ith row of PkFPk for 1≤i≤k. Then,(1)
r1PkFPk=rkPk and rkPkFPk=rkPkPk(2)
riPkFPk=ri−1PkFPk+ri−1Pk−1FPk−1;0=rkPk+∑t=1i−1rtPk−1FPk−1;0Proof.
Clearly,r1PkFPk=r1PkFPk=0¯k−1;1Pk=rkPk and rkPkFPk=rkPkFPk=rkPkPk. And, for 1≤i<k, we have(16)riPkFPk+riPk−1FPk−1;0=riPkF+riPk−1F;0Pk=ri+1PkFPk.Thus, it follows immediately that(17)riPkFPk=ri−1PkFPk+ri−1Pk−1FPk−1;0=ri−2PkFPk+ri−2Pk−1FPk−1;0+ri−1Pk−1FPk−1;0=ri−3PkFPk+ri−3Pk−1FPk−1;0+ri−2Pk−1FPk−1;0+ri−1Pk−1FPk−1;0=⋯=r1PkFPk+r1Pk−1FPk−1;0+r2Pk−1FPk−1;0+⋯+ri−1Pk−1FPk−1;0=rkPk+∑t=1i−1rtPk−1FPk−1;0.We now develop some interrelations of the Bell matrixB=bi,j, square-type Bell matrix B∗=bi,j∗, and Pascal matrix P.Theorem 5.
LetcjB∗ be the jth column of B∗ for i,j≤k. Then,(18)1rk+1Pbi,jbi+1,j⋯bi+k,j=bi+k,j+k,soPbi,jbi+1,jbi+2,j⋯=bi,jbi+1,j+1bi+2,j+2⋯.2PkcjB∗=bj+1,1bj+2,1⋯bj+k,1,soriPkcjB∗=bj+i,1.3Pk+1Fbm,1bm+1,1⋯bm+k,1=bm+k,1bm+k,2⋯bm+k,k+1,soriPk+1Fbm,1bm+1,1⋯bm+k,1=bm+k,i.Proof.
(19)bi+1,j+1=r2Pbi,jbi+1,j,bi+2,j+2=bi,j+2bi+1,j+bi+2,j=r3Pbi,jbi+1,jbi+2,j.So if we assume(20)rkPbi,jbi+1,j⋯bi+k−1,j=bi+k−1,j+k−1,for some k, then(21)rk+1Pbi,jbi+1,j⋯bi+k,j=rkPbi,j⋯bi+k−1,j+rkPbi+1,j⋯bi+k,j=bi+k−1,j+k−1+bi+k,j+k−1=bi+k,j+k,by recurrence (12). Comparing B∗=bi,j∗ with B=bi,j, we have(22)bi,j∗=bi+j−1,j,b1,j∗=bj+1,1∗,bi,j∗=bi,j−1∗+bi+1,j−1∗.Thus, with cjB∗=b1,j∗b2,j∗b3,j∗⋯=bj,jbj+1,jbj+2,j⋯, (1) implies(23)PkcjB∗=Pkbj,jbj+1,j⋯bj+k−1,j=bj,jbj+1,j+1⋯bj+k−1,j+k−1=bj+1,1bj+2,1⋯bj+k,1.
Moreover, (1) gives rise to (3) such that(24)Pk+1Fbm,1bm+1,1⋯bm+k,1=111⋯1k⋯k1bm,1bm+1,1⋯bm+k,1=bm+k,1,r2Pbm+k−1,1bm+k,1,r3Pbm+k−2,1bm+k−1,1bm+k,1,…,rk+1Pbm,1⋯bm+k,1T=bm+k,1,bm+k,2,bm+k,3,…,bm+k,k+1T.Theorem 6.
bi,j∗=bi,j−1∗+rj−1Pi−1FPi−1cjB∗,ifj>1,ri−1Pc1B∗,ifj=1.Proof.
Clearly,bi,1∗=bi,1=ri−1Pc1B∗ by Theorem 5 (2). And, observe b3,j∗ (1<j≤3) from B3∗=12513102727 such that b3,2∗=7=2+1,12,3T=b3,1∗+r1P2FP2b1,2∗,b2,2∗T and b3,3∗=27=7+2,15,10T=b3,2∗+r2P2FP2b1,3∗,b2,3∗T.
FromB4∗=15B3∗371145,20,87409, b4,j∗ (1<j≤4) satisfy b4,2∗=20=5+1,2,12,3,7T=b4,1∗+r1P3FP3b1,2∗,b2,2∗,b3,2∗T, b4,3∗=87=20+2,3,15,10,27T=b4,2∗+r2P3FP3b1,3∗,b2,3∗,b3,3∗T, and b4,4∗=409=87+4,4,115,37,114T=b4,3∗+r3P3FP3b1,4∗,b2,4∗,b3,4∗T.
Now, for somei, we assume bi,j∗=bi,j−1∗+rj−1Pi−1FPi−1cjB∗ for all 1<j<i. Then, bi+1,j∗=bi,j+1∗−bi,j∗ equals(25)bi+1,j∗=bi,j∗+rjPi−1FPi−1cj+1B∗−bi,j−1∗+rj−1Pi−1FPi−1cjB∗=bi,j∗−bi,j−1∗+rjPi−1FPi−1cj+1B∗−rj−1Pi−1FPi−1cjB∗.
But, since(26)rjPi−1FPi−1cj+1B∗=rjPi−1FPi−1cj+1B∗=rjPi−1Fbj+2,1bj+3,1⋯bj+i,1=bi+j,j,rj−1Pi−1FPi−1cjB∗=bi+j−1,j−1,by Theorem 5, we have(27)bi+1,j∗=bi,j∗−bi,j−1∗+bi+j,j−bi+j−1,j−1=bi+1,j−1∗+bi+j,j−1=bi+1,j−1∗+rj−1PiFbj,jbj+1,j+1bj+2,j+2⋯bj+i−1,j+i−1=bi+1,j−1∗+rj−1PiFPicjB∗=bi+1,j−1∗+rj−1PiFPicjB∗.


## 4. LU Factorization of the Square-Type Bell Matrix

We are ready to have an LU factorization ofBk∗=bi,j∗ with diagonal entries.Theorem 7.
Bk∗=LkUk (1≤k≤5) where the lower triangular matrix Lk=S˜kPk and the upper triangular matrix Uk has diagonal entries 1,1,2!,3!,4!.Proof.
LetUk=ui,j (1≤i,j≤k) be with(28)u1,j=b1,j∗,u2,j=b2,j∗−u1,j,for allj≥1.Then, B2∗=1213 yields U2=1201. And, B2∗U2−1=1011=S˜2P2 is a lower triangular matrix; denote it by L2.
FromB3∗=12513102727, clearly u1,3=b1,3∗ and u2,3=b2,3∗−u1,3=5 by (28). Let(29)u3,j=b3,j∗−r2Pb1,j∗,b2,j∗T−2u2,j,for allj≥1.Then, the identity bk+2,1∗=rk+1Pc1B∗ in Theorem 5 (2) implies b3,1∗=r2Pb1,1∗,b2,1∗T, so u3,1=0 because u2,1=0. Similarly, u3,2=b3,2∗−r2Pb1,2∗,b2,2∗T−2u2,2=2−2⋅1=0, u3,3=b3,3∗−r2Pb1,3∗,b2,3∗T−2u2,3=12−2⋅5=2, so(30)U3=U2u1,3u2,3u3,1,u3,2u3,3=125152,B3∗U3−1=111231=S˜3P3,is a lower triangular matrix and denote it by L3.
FromB4∗=15B3∗371145,20,87409, (28), and (29), we have u1,4=b1,4∗, u2,4=b2,4∗−u1,4=22, and u3,4=b3,4∗−r2Pb1,4∗,b2,4∗T−2u2,4=18. Now, let(31)u4,j=b4,j∗−r3PcjB∗−5,5u2,j,u3,jT,for allj≥1.
Sinceb4,1∗=r3Pc1B∗ and u2,1=u3,1=0, we have u4,1=0, u4,2=b4,2∗−r3Pc2B∗−5,5u2,2,u3,2T=5−5,51,0T=0, u4,3=b4,3∗−r3Pc3B∗−5,5u2,3,u3,3T=35−5,55,2T=0, and u4,4=b3,1∗−r3Pc4B∗−5,5u2,4,u3,4T=6. Therefore,(32)U4=15U322180¯36,B4∗U4−1=L305,10,61=11111311111211331=S˜4P4.Denote it by L4.
From(33)B5∗=52151B4∗523206615,67,322,16579089,(28), (29), and (31), give u2,5=b2,5∗−u1,5=99, u3,5=b3,5∗−r2Pc5B∗−2u2,5=122, and u4,5=b4,5∗−r3Pc5B∗−5,5u2,5,u3,5T=84. Now, let(34)W5,j=b5,j∗−r4PcjB∗−3b4,j∗−r3PcjB∗,and let u5,j=W5,j−7,6u3,j,u4,jT for j≥1.
Sinceu3,1=u4,1=0, b4,1∗=r3Pc1B∗, and b5,1∗=r4Pc1B∗, we have u5,1=0. Also, u3,2=u4,2=u4,3=0 and u3,3=2 imply u5,2=W5,2−7,6u3,2,u4,2T=15−3⋅5=0, u5,3=W5,3−7,6u3,3,u4,3T=0=W5,4−7,618,6T=u5,4, and u5,5=W5,5−7,6122,84T=4!. Hence,(35)U5=52U499122840¯44!,B5∗U5−1=L4015,37,31,101=S˜5P5.Denote it by L5.Noteuj,j=j−1! for 1≤j≤5. From (34), we let(36)Wt,j=bt,j∗−rt−1PcjB∗−3bt−1,j∗−rt−2PcjB∗,for allj≥1.Theorem 8.
Assume the matrixUk in Theorem 7 further satisfies u6,j=W6,j−Γ4u2,j,…,u5,jT and u7,j=W7,j−Γ5u2,j,…,u6,jT with row matrices Γ4=7,33,34,11 and Γ5=47,174,202,93,17. Then, for 1≤k≤7, Uk is an upper triangular matrix having uk,k=k−1! and Lk=Bk∗Uk−1 is a lower triangular matrix such that Lk=S˜kPk.Proof.
From(37)B6∗=203674B5∗2589111555292252,255,1335,7432,43833272947,(28), …, (34), show u1,6=b1,6∗, u2,6=471, u3,6=770, u4,6=810, and u5,6=480. Let(38)u6,j=W6,j−t1,t2,t3,t4u2,j,u3,j,u4,j,u5,jT,for someti∈ℤ.
Note from Theorem5 that u2,1=u3,1=u4,1=u5,1=0, b5,1∗=r4Pc1B∗, and b6,1∗=r5Pc1B∗. Thus, u6,1=0 by (38). And, we also observe u6,2=W6,2−t1,t2,t3,t4u2,2,…,u5,2T=52−3⋅15−t1,t2,t3,t41;0¯3T, so u6,2=0 if t1=7. Similarly, since W6,3=458−3⋅119, W6,4=3292−3⋅780, and W6,5=22686−3⋅4949 from (36), in order to be u6,3,u6,4, and u6,5 all zeros, the identities(39)u6,3=W6,3−t1,…,t4u2,3,…,u5,3T=101−7,t2,t3,t45,2;0¯2T,u6,4=W6,4−t1,…,t4u2,4,…,u5,4T=952−7,t2,t3,t422,18,6,0T,u6,5=W6,5−t1,…,t4u2,5,…,u5,5T=7839−7,t2,t3,t499,122,84,24T,yield t2=33, t3=34, and t4=11. Thus, with W6,6=156972−3⋅31775 in (36), we have u6,6=W6,6−7,33,34,11u2,6,…,u5,6T=5!.
Hence,(40)U6=203471U57708104800¯55!,L6=B6∗U6−1=L5052,151,160,75,151=S˜6P6.
Similarly, from(41)B7∗=8773263B6∗13744640773258691788850203,1080,6097,36401,229114,151590310515147,we get u1,7,…,u6,7=877,2386,4832,6840,6240,3240 by (28), …, (38). Now, let(42)u7,j=W7,j−t1,…,t5u2,j,…,u6,jT,for someti∈ℤ.
Sinceui,1=0 (1≤i≤6), b7,1∗=r6Pc1B∗, and b6,1∗=r5Pc1B∗, we also have u7,1=0. In order to get u7,j=02≤j≤6, the integers ti1≤i≤5 are determined as follows: Note that W7,2=203−3⋅52 from (36), so 0=u7,2=W7,2−t1,…,t51;0¯4T implies t1=47.
Analogously, withW7,3=1957−3⋅458, W7,4=15254−3⋅3292, W7,5=113139−3⋅22686, and W7,6=837333−3⋅156972 from (36), we also have(43)u7,3=W7,3−47,174,t3,t4,t55,2;0¯3T=0u7,4=W7,4−47,174,202,t4,t522,18,6;0¯2T=0u7,5=W7,5−47,174,202,93,t599,122,84,24,0T=0u7,6=W7,6−47,174,202,93,17471,770,810,480,120T=0.Thus, with t1,t2,t3,t4,t5=47,174,202,93,17 and W7,7=6301550−3⋅1110280 from (36), we have u7,7=W7,7−t1,t2,t3,t4,t5u2,7,u3,7,…,u6,7T=6!. Therefore,(44)U7=8772386U648326840624032400¯66!,L7=L60203,674,856,520,155,211=S˜7P7.Theorem 9.
The above upper triangular matrixUk=ui,jj≥i satisfies(45)u1,j=m1,1b1,j∗withm1,1=1,u2,j=m2,1,m2,2cjB2∗,withm2,t=−1,1,u3,j=m3,1,m3,2,m3,3cjB3∗withm3,t=1,−3,1,u4,j=m4,1,…,m4,4cjB4∗withm4,t=−1,8,−6,1,u5,j=m5,1,…,m5,5cjB5∗withm5,t=1,−24,29,−10,1,u6,j=m6,1,…,m6,6cjB6∗withm6,t=−1,89,−145,75,−15,1,u7,j=m7,1,…,m7,7cjB7∗withm7,t=1,−415,814,−545,160,−21,1.Proof.
u1,j=b1,j∗ and u2,j=b2,j∗−u1,j=−b1,j∗+b2,j∗=−1,1cjB2∗ by Theorems 7 and 8. Similarly, we have(46)u3,j=b3,j∗−r2Pb1,j∗,b2,j∗T−2u2,j=b3,j∗−1,1b1,j∗,b2,j∗T−2−1,1b1,j∗,b2,j∗T=1,−3,1b1,j∗,b2,j∗,b3,j∗T=1,−3,1cjB3∗u4,j=b4,j∗−r3Pb1,j∗,b2,j∗,b3,j∗T−5,5u2,j,u3,jT=−1,8,−6,1cjB4∗.
Moreover, withcjB∗=b1,j∗,b2,j∗,…T, we also have(47)u5,j=b5,j∗−r4PcjB4∗−3b4,j∗−r3PcjB3∗−7,6u3,j,u4,jT=1,−24,29,−10,1cjB5∗,and, similarly, the rest u6,j and u7,j follow immediately.With allmi,t’s in Theorem 9, write a lower triangular matrix(48)M7=mi,t=1−111−31−18−611−2429−101−189−14575−1511−415814−545160−211.Then,(49)U7=M7B7∗=12515522038771522994712386218122770483268481068402448062401203240720,gives an LU factorization B7∗=M7−1U7, where(50)L7=M7−1=111231510611537311015215116075151203674856520155211=S˜7P7.Observe that the matrix M7=mi,t is the exponential Riordan array (without signs) (refer to [7]) satisfying a recurrence rule(51)mi,t=mi−1,t+1,mi,t+1,mi+1,t+1i−1,i,1T.M is also known as the coefficients of the Charlier polynomial [8], and we may refer Table 3 in [9] for M−1=L.Theorem 10.
LetM=mi,t be a matrix satisfying recurrence (51) with mi,1=−1i−1 and M3=1−111−31. Then, the diagonal entries of the upper triangular matrix Uk=MkBk∗ are k−1! for all k>1.Proof.
LetU=ui,j. Then, U7=M7B7∗ has diagonal entries ui,i (1≤i≤7) as 1,1,2!,…,6! due to Theorem 9. We note the following identities.
SinceU=MB∗ is an upper triangular matrix, we have(52)riMcjB∗=ui,j=0,for all1≤j<i.
Recurrence (51) gives(53)mi,t=mi−1,t−1−i−2mi−2,t−i−1mi−1,t,over M. Hence, with (52), the i th row riM=mi,1,…,mi,i satisfies(54)riM=0,mi−1,1,…,mi−1,i−1−i−2mi−2,1,…,mi−2,i−2,0,0−i−1mi−1,1,…,mi−1,i−1,0=0;ri−1M−i−2ri−2M;0¯2−i−1ri−1M;0.
On the other hand, by (22), the i th column ciB∗ of B∗ satisfies(55)ciB∗=b1,i∗b2,i∗b3,i∗⋯=b1,i−1∗+b2,i−1∗b2,i−1∗+b3,i−1∗b3,i−1∗+b4,i−1∗⋯=ci−1B∗+b2,i−1∗b3,i−1∗b4,i−1∗⋯.Thus, (52) and (55) together show(56)riMb2,i−1∗b3,i−1∗⋯bi+1,i−1∗=riMciB∗−riMci−1B∗=ui,i−ui,i−1=ui,i,and similarly,(57)riMb2,i∗b3,i∗⋯bi,i∗=riMci+1B∗−ciB∗=ui,i+1−ui,i.
Moreover, by (28), (55), and (56), ui,i=riMciB∗ equals(58)ui,i=0;ri−1Mci−1B∗−i−2ri−2Mci−1B∗−i−1ri−1Mci−1B∗+riMb2,i−1∗b3,i−1∗⋯bi+1,i−1∗=ui−1,i−ui−1,i−1−i−2ui−2,i−1−i−1ui−1,i−1+ui,i,so we have ui−1,i−ui−1,i−1−i−2ui−2,i−1=i−1ui−1,i−1.
Hence, if we assumeuk,k=k−1! for all k≤i, then ui−1,i−ui−1,i−1−i−2ui−2,i−1=i−1!=ui,i, so the induction hypothesis with (58) yields(59)ui+1,i+1=ri+1Mci+1B∗=0;riMciB∗−i−1ri−1M;0¯2ciB∗−iriM;0ciB∗+ri+1Mb2,i∗b3,i∗b4,i∗⋯=0;riMciB∗−i−1ri−1MciB∗−iriMciB∗+ri+1Mb2,i∗b3,i∗b4,i∗⋯=i!−i−1i!+i!=i!.Theorem 11.
The lower triangular matrixLk=li,j=S˜kPk in Theorem 8 satisfies li+1,j+1=li,j,li,j+1,li,j+21,j+1,j+1T.Proof.
SinceLk=S˜kPk, for any 1≤i,j≤k, Theorem 3 shows(60)li+1,j+1=ri+1S˜cj+1P=0;riS˜+riS˜di0,1,…,i−10;cjP+0;cj+1P=0;riS˜0;cjP+0;riS˜0;cj+1P+riS˜di0,…,i−10;cjP+riS˜di0,…,i−10;cj+1P=li,j+li,j+1+riS˜di0,1,…0;cjP+riS˜di0,1,…0;cj+1P=li,j+li,j+1+jriS˜cj+1P+j+1riS˜cj+2P=li,j+li,j+1+jli,j+1+j+1li,j+2=li,j,li,j+1,li,j+21,j+1,j+1T.TheLk and Uk are obtained by the recurrence in Theorems 10 and 11. In fact, from M7 and L7, we get(61)L9=S˜7P7087732634802355614002872814140170072833724626119913290490361,M9=M70−12372−52434179−1575301−2811−1607238618−3486015659−3836518−361,so(62)M9B9∗=U7⋱07!,2217600,8!=U9,in which all diagonal entries are k−1!.Theorem 12.
detBk∗=∏i=1k−1i!=k−1!detBk−1∗.Clearly,detBk∗=detLkdetUk=detUk=∏i=1k−1i!=k−1!detBk−1∗.



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