More calculations on determinant evaluations

Electronic Journal of Linear Algebra Volume 16 Article 007 More calculations on determinant evaluations A. R. Moghaddamfar moghadam@kntu.ac.ir S. M. H. Pooya S. Navid Salehy S. Nima Salehy Follow this and additional works at: http://repository.uwyo.edu/ela Recommended Citation Moghaddamfar, A. R.; Pooya, S. M. H.; Salehy, S. Navid; and Salehy, S. Nima. (007), "More calculations on determinant evaluations", Electronic Journal of Linear Algebra, Volume 16. DOI: https://doi.org/10.13001/1081-3810.1179 This Article is brought to you for free and open access by Wyoming Scholars Repository. It has been accepted for inclusion in Electronic Journal of Linear Algebra by an authorized editor of Wyoming Scholars Repository. For more information, please contact scholcom@uwyo.edu.

MORE CALCULATIONS ON DETERMINANT EVALUATIONS A. R. MOGHADDAMFAR, S. M. H. POOYA, S. NAVID SALEHY, AND S. NIMA SALEHY Abstract. The purpose of this article is to prove several evaluations of determinants of matrices, the entries of which are given by the recurrence a i,j = a i 1,j 1 +a i 1,j, i, j, with various choices for the first row a 1,j and first column a i,1. Key words. Determinant, Matrix factorization, Recurrence relation. AMS subject classifications. 15A15, 11C0. 1. Introduction. Determinants have played a significant part in various areas in mathematics. For instance, they are quite useful in the analysis and solution of systems of linear equations. There are different perspectives on the study of determinants. One may notice several practical and effective instruments for calculating determinants in the nice survey articles [3]and [4]. Much attention has been paid to the symbolic evaluation of determinants of matrices, especially when their entries are given recursively (see Section 5.6 in [4]). Toward these matrices one usually introduces the first row and column as initial conditions for a recurrence relation used for constructing the other entries. Both this relation and the initial conditions play an important part in constructing the matrix, as well as in evaluating its determinant. Several relevant studies on evaluating determinants can be found in the literature; e.g., see [1,, 5, 7]. In this article we are interested in the sequences of determinants of matrices satisfying the recurrence relation a i,j = a i 1,j 1 + a i 1,j for i, j n, with various choices for the first row a 1,j and first column a i,1. To state our results we need to introduce some notation. The Fibonacci numbers F (n) satisfy { F (0) = 0, F (1) = 1 F (n +)=F (n +1)+F (n) (n 0). Throughout this article, we also use the following notation: Received by the editors 15 September 006. Accepted for publication 8 December 006. Handling Editor: Michael Neumann. Department of Mathematics, Faculty of Science, K. N. Toosi University of Technology, P.O. Box 16315-1594, Tehran, Iran. First author also affiliated with Institute for Studies in Theoretical Physics and Mathematics (IPM) (moghadam@kntu.ac.ir, moghadam@ipm.ir). This research was in part supported by a grant from IPM (No. 8500038). 19

0 A.R.Moghaddamfar,S.M.H.Pooya,S.NavidSalehy,andS.NimaSalehy ω k (n) =(ωi k) 1 i n =(1, 1,...,1, 0, 0,...,0, 1, 1,...,1), }{{}}{{}}{{} k times (n k) times k times ϖ k (n) =(ϖi k) 1 i n =(1, 1,...,1, 0, 0,...,0), }{{} (n k) times } {{ } k times χ k =(χ k i ) i 1 =(a,a,...,a, 0, 0, 0,...), }{{} k times ϑ k =(ϑ k i ) i 1 =(a, 0, 0,...,0,a,0, 0, 0,...). } {{ } k times Given a matrix A we denote by R i (A) andc j (A) therowi and the column j of A, respectively. We denote by e i,j the square matrix having 1 in the (i, j) position and 0 elsewhere. Now, it is easy to see that e i,j e k,l = δ jk e i,l. (1.1) In the case that the two matrices e i,j and A are to be multiplied together, we obtain R k (e i,j A) =δ ki R j (A) and C k (A e i,j )=δ kj C i (A). (1.) In fact, the matrix e i,j A is a matrix that all its rows except for the ith row are 0 and its ith row is the jth row of the matrix A. Similarly, the matrix A e i,j is a matrix that all its columns except for the jth column are 0 and its jth column is the ith column of the matrix A.. Main Results. Theorem.1. Let α =(α i ) i 1 be a given sequence and let A =(a i,j ) 1 i,j n be the doubly indexed sequence given by the recurrence a i,j = a i 1,j 1 + a i 1,j, i, j n, (.1) and the initial conditions a i,1 = α 1 +(i 1)d, a 1,j = α j, 1 i, j n. Then we have A = L B, wherel =(L i,j ) 1 i,j n is a lower triangular matrix given by the recurrence L i,j = L i 1,j 1 + L i 1,j, i, j n, (.) and the initial conditions L 1,1 =1, L 1,j =0, j n, andl i,1 =1, i n, and B =(B i,j ) 1 i,j n is a matrix given by the recurrence B i,j = B i 1,j 1, i, j n, (.3) and the initial conditions B 1,j = α j, 1 j n, B,1 = d and B i,1 =0, 3 i n. Note that B is a Hessenberg-Töplitz matrix. In particular, det(a) =det(b). Proof. For the proof of the claimed factorization we compute the (i, j) entry of L B, thatis (L B) i,j = n L i,k B k,j. k=1

More Calculations on Determinant Evaluations 1 In fact, so as to prove the theorem, we should establish R 1 (L B) =R 1 (A) =(α 1, α,..., α n ), and C 1 (L B) =C 1 (A) =(α 1, α 1 + d,..., α 1 +(n 1)d), (L B) i,j =(L B) i 1,j 1 +(L B) i 1,j, (.4) for i, j n. Let us do the required calculations. First, suppose that i =1. Then (L B) 1,j = n L 1,k B k,j = L 1,1 B 1,j = α j, k=1 and so R 1 (L B) =R 1 (A) =(α 1, α,..., α n ). Next, we suppose that j =1,andweobtain (L B) i,1 = n L i,k B k,1 = L i,1 B 1,1 + L i, B,1 = α 1 +(i 1)d, k=1 which implies that C 1 (L B) =C 1 (A) =(α 1,α 1 + d,..., α 1 +(n 1)d). Finally, we must establish Eq. (.4). At the moment, let us assume that i, j n. Inthiscasewehave (L B) i,j = P n k=1 L i,kb k,j = L i,1b 1,j + P n k= L i,kb k,j = L i,1b 1,j + P n k= (L i 1,k 1 + L i 1,k )B k,j (by (.)) = L i,1b 1,j + P n k= L i 1,k 1B k,j + P n k= L i 1,kB k,j = L i,1b 1,j + P n k= L i 1,k 1B k 1,j 1 + P n k=1 L i 1,kB k,j L i 1,1B 1,j (by (.3)) = (L i,1 L i 1,1)B 1,j + P n k=1 L i 1,kB k,j 1 + P n k=1 L i 1,kB k,j (it should be noticed that L i 1,n =0) = (L B) i 1,j 1 +(L B) i 1,j, which is Eq. (.4). Our proof is thus complete. An interesting corollary to Theorem.1 is the following. Corollary.. In Theorem.1, ifα = ω k (n), d =1and D(n) =det(a), then Furthermore, we have D(n) =( 1) k+1 D(n k 1), (n >3k). (.5)

A.R.Moghaddamfar,S.M.H.Pooya,S.NavidSalehy,andS.NimaSalehy (1) if k =,then D(n) = (resp. ) if n 6 0(resp. 3) 0 if n 6 1, 4 1(resp. 1) if n 6 (resp. 5) () if k 3 is even, then 1 if n k+ 0, 1, D(n) = 1 if n k +1,k+, 0 otherwise. Also, if k 3 is odd, then { D(n) = 1 if n k+1 0, 1 0 otherwise. Proof. Using Theorem.1, D(n) =det(b) whereb =(B i,j ) 1 i,j n is a matrix given by the recurrence B i,j = B i 1,j 1, i, j n, and the initial conditions B 1,j = ω k j,1 j n, B,1 =1andB i,1 =0,3 i n. To obtain the result we thus need to compute det(b). Put B = B k (n). We claim that B = U B L, where the matrices U =(U i,j ) 1 i,j n, B =( Bi,j ) 1 i,j n,andl =(L i,j ) 1 i,j n are defined as follows: 1 if i = j, U i,j = 1 if j = i + n k, 1 i k, 0 otherwise, 1 if 1 i k and n k + i 1 j n k + i 1, B i,j = L i,j = 0 if (i, j) =(n k, n k 1), B i,j otherwise, 1 if i = j or (i, j) =(n 1,n k 1), 1 if (i, j) =(n, n k 1), 0 otherwise. It is obvious that, the matrices U and L are the upper triangular matrix and lower triangular one, respectively, with 1 s on their diagonals. In addition, we can restate the matrices U and L as follows: U = I k e i,n k+i and L = I + e n 1,n k+1 e n,n k 1. i=1

More Calculations on Determinant Evaluations 3 Moreover, we can partition the matrix B in this way [ ] B1 B =, 0 B where B 1 = B k (n k 1), and 1 1 1... 1 0 1 1 1... 1 1 0 1 1... 1 1 B = 0 0 1... 1 1........... 0 0 0... 1 1 Since det(b )=( 1) k+1, it is obvious that the claimed factorization of B immediately implies the validity of Eq. (.5). For the proof of the claim, we observe that U B L = (I k i=1 e i,n k+i) B (I + e n 1,n k 1 e n,n k 1 ) = ( B k i=1 e i,n k+i B) (I + e n 1,n k 1 e n,n k 1 ) = (B e n 1,n k 1 + e k,n k 1 ) (I + e n 1,n k 1 e n,n k 1 ) = B + B e n 1,n k 1 B e n,n k 1 e n 1,n k 1 + e k,n k 1 (by (1.1)). Thus, it is enough to show that B e n 1,n k 1 B e n,n k 1 e n 1,n k 1 +e k,n k 1 = 0. To see this, from Eq. (1.), we obtain C l (B e n 1,n k 1 )=δ l,n k 1 C n 1 (B) and C l (B e n,n k 1 )=δ l,n k 1 C n (B), for l =1,,...,n. Hence, by the structure of B, it follows that B e n 1,n k 1 B e n,n k 1 = e n 1,n k 1 e k,n k 1. The rest of the proof is simple and left to the reader. Second Proof of Corollary.. Here, we compute det(b) directly. We will assume that k>1, since the case k =1iseasy. PutR i =R i (B) andc j =C j (B). We first subtract C 1 from C through C k and from C n k+ through C n. Then we subtract R from R 1. So this leaves us with a matrix B 1 = 0 0... 0 1... 1 0... 0 1 0... 0 where the boxed entries 1 and 1 are in positions (1,k+1)and(1,n k +1) respectively, and the part is unchanged from B. The goal now is to push the,

4 A.R.Moghaddamfar,S.M.H.Pooya,S.NavidSalehy,andS.NimaSalehy entries 1 and 1 all the way to position (1,n), with no other changes in the matrix. Let us first deal with 1. Unlessk =, we can get rid of it by replacing R 1 R 1 + R n k+1 R n k+. If k =,wemaker 1 R 1 R n and we then have 1at(1,n). Next we treat 1. Wecanmoveitk + 1 positions to the right by doing R 1 R 1 + R k+ R k+3. We can repeatedly move it to the right k + 1 places by adding and subtracting the appropriate consecutive rows. This is done until we are at (1,r)forr>n k. There are three cases: (i) If n 0(modk + 1), we bring the 1 to (1,n). (ii) If n 1(modk+1), then 1 is brought to (1,n 1) and then R 1 R 1 +R n brings it to (1,n) by changing sign. (iii) Otherwise we bring 1 to (1,r)forn k<r<n 1andthenwemake it disappear by R 1 R 1 + R r+1 R r+. At this point we have the matrix B = 0... 0 c 1 0... 0, where c = a n + b n for a n = { 1 k = 0 k> and b n = 1 n 0 (mod k +1) 1 n 1 (mod k +1) 0 otherwise. Expanding det(b ) by the first row we see that det(b) =det(b )=( 1) n 1 c,and the corollary follows from here immediately. Corollary.3. Suppose that in Theorem.1, D(n) =det(a). Then (i) If α = χ k,then D(n) =a k ( d) i 1 D(n i), n k +1. i=1 In particular, in case that k =and a =1, we have D(n) = 1 [( 1+η ) n+1 ( 1 η ) n+1 ]. η where η = 1 4d. (ii) If α = ϖ k (n) and d =1, then we have ( 1) n if n 0(modn k +1), det(a) = ( 1) n+1 if n 1(modn k +1), 0 otherwise. (iii) If α = ϑ k,then a n if 1 n k +1, D(n) = a n +( 1) n 1 ad n 1 if n = k +, ad(n 1) + a( d) k+1 D(n k ) if n k +3.

More Calculations on Determinant Evaluations 5 Proof. First of all, by Theorem.1, D(n) =det(b) whereb =(B i,j ) 1 i,j n is a matrix given by the recurrence B i,j = B i 1,j 1, i, j n, and the initial conditions B 1,j = α j,1 j n, B,1 = d and B i,1 =0,3 i n. (i). Suppose α = χ k. If we expand the determinant det(b) alongthefirstrow, then the (1,j) cofactor is of the form det(di j 1 B n j ), where B n j equals the principal submatrix of B taking the first n j rows and columns. Now, we easily get the desired recursion, that is D(n) =a k ( d) i 1 D(n i), n k +1. i=1 Now, assume k =anda = 1. In this case, we have D(1) = 1, D() = 1 d, and D(n) =D(n 1) dd(n ), n 3. To solve this recurrence relation, we see that its characteristic equation as follows x x + d =0, and its characteristic roots are x 1 = 1 η and x = 1+η, where η = 1 4d. Therefore, the general solution is given by D(n) =λ 1 ( 1 η ) n ( 1+η ) n, + λ (.6) where λ 1 and λ are constants. The initial conditions D(1) = 1 and D() = 1 d imply that λ 1 = 1 η ( 1 η )andλ = 1 η ( 1+η ), and so D(n) = 1 η [( 1+η ) n+1 ( 1 η ) n+1 ]. (ii). Assume α = ϖ k (n) andd = 1. We put n k = p. Similar arguments as in part (i) withk replaced by p, show that and D(1) = 1, D() = D(3) = = D(p) =0, D(n) = p ( 1) i+1 D(n i), n p +1. i=1

6 A.R.Moghaddamfar,S.M.H.Pooya,S.NavidSalehy,andS.NimaSalehy Hence, we observe that D(m + p +1) = p i=1 ( 1)i+1 D(m + p +1 i) = D(m + p)+ p i= ( 1)i+1 D(m + p +1 i) = p i=1 ( 1)i+1 D(m + p i)+ p i= ( 1)i+1 D(m + p +1 i) = ( 1) p+1 D(m), (m 1). (.7) Now, using the division algorithm we find integers r and s such that n = s(p +1)+r, and 0 r<p+1. First, assume that r = 0. In this case, from Eq. (.7) we obtain D(n) =D(s(p +1))=( 1) (s 1)(p+1) D(p +1)=( 1) (s 1)(p+1) ( 1) (p+1) =( 1) n. Next assume that 1 r<p+ 1. In this case, again by Eq. (.7) it follows that { 0 if 1 <r<p+1 D(n) =D(s(p +1)+r) =( 1) s(p+1) D(r) = ( 1) n 1 if r =1. This completes the proof of part (ii). (iii) The proof is similar to the previous parts. Theorem.4. Let α =(α i ) i 1 be a given sequence and let β =(β i ) i 1 be a sequence satisfying β 1 = α 1, β = α and the linear recursion β i = β i + β i 1, i 3. Let (a i,j ) i,j 1 be the doubly indexed sequence given by the recurrence det (a i,j) = 1 i,j n a i,j = a i 1,j 1 + a i 1,j, i,j, (.8) and the initial conditions a i,1 = β i and a 1,j = α j,wherei, j 1. Then α 1 if n =1, α 1 α + α 1α if n =, (α 1 α + α 1α )(α 1 + α α 3 ) n if n 3. Proof. Forn 3 the result is straightforward. Hence, we assume that n 4. Let A denote the matrix (a i,j ) 1 i,j n. We claim that A = L B, where L =(L i,j ) 1 i,j n is a lower triangular matrix by the recurrence L i,j = L i 1,j 1 + L i 1,j, i,j 3, (.9)

More Calculations on Determinant Evaluations 7 and the initial conditions L i,1 = F (i), L 1,i =0,i, and L i, = F (i 1), i, and where B =(B i,j ) 1 i,j n with α j if i =1,j 1, α α 1 if i =,j =1, B i,j = B,j 1 + B,j B 1,j if i =3,j, (.10) B i 1,j 1 if i 1, 3, j, 0 if i 3,j =1. For instance, when n =4thematrixB is hence given by α 1 α α 3 α 4 B = α α 1 α 1 α α 3 0 0 α 1 + α α 3 α + α 3 α 4 0 0 0 α 1 + α α 3 Notice that, by the structure of matrices L and B, for every i, it follows that L i,3 L i 1,3 = L i 1,, L i, = L i 1,1, B,i = B 1,i 1. (.11) By expanding, in terms of the first column, we easily see that α 1 if n =1, det(b) = α 1 α + α 1α if n =, (α 1 α + α 1α )(α 1 + α α 3 ) n if n 3, and so the claimed factorization of A immediately implies the validity of the theorem. As before, the proof of the claim requires again some calculations. In fact, it suffices to show that R 1 (L B) =R 1 (A), C 1 (L B) =C 1 (A) and. (L B) i,j =(L B) i 1,j 1 +(L B) i 1,j, (.1) for i, j n. First, suppose that i =1. Then n (L B) 1,j = L 1,k B k,j = L 1,1 B 1,j = α j, k=1 and so R 1 (L B) =R 1 (A) =(α 1, α,..., α n ). Next, we assume that i andj =1. Inthiscasewehave (L B) i,1 = n L i,k B k,1 = L i,1 B 1,1 + L i, B,1 k=1 = F (i)α 1 + F (i 1)(α α 1 )=F (i )α 1 + F (i 1)α, and we get C 1 (L B) =C 1 (A) =(α 1, α,..., F(n )α 1 + F (n 1)α ). Finally, we must establish Eq. (.1). Therefore, we assume that i, j n. In this case we have

8 A.R.Moghaddamfar,S.M.H.Pooya,S.NavidSalehy,andS.NimaSalehy (L B) i,j = P n k=1 L i,kb k,j = P 3 k=1 L i,kb k,j + P n k=4 L i,kb k,j P P 3 = P k=1 L n i,kb k,j + P k=4 (L i 1,k 1 + L i 1,k P )B k,j (by (.9)) 3 = k=1 L n i,kb k,j + k=4 L n i 1,k 1B k,j + k=4 L i 1,kB k,j = P 3 k=1 L i,kb k,j + P n k=4 L i 1,k 1B k 1,j 1 + P n k=4 L i 1,kB k,j (by (.10)) = P 3 k=1 L i,kb k,j + P n k=3 L i 1,kB k,j 1 + P n k=4 L i 1,kB k,j (it should be noticed that L i 1,n P =0) 3 = (L B) i 1,j 1 +(L B) i 1,j + k=1 L i,kb k,j P k=1 L i 1,kB k,j 1 P 3 k=1 L i 1,kB k,j, = (L B) i 1,j 1 +(L B) i 1,j +(L i,1 L i 1,1)B 1,j +(L i 1, +L i 1,3)B 3,j L i 1,B,j 1 L i 1,B,j L i 1,3B 3,j, (by (.11)) = (L B) i 1,j 1 +(L B) i 1,j + L i 1,(B 1,j + B 3,j B,j 1 B,j), (because L i,1 L i 1,1 = L i,1 = L i 1,) = (L B) i 1,j 1 +(L B) i 1,j (by (.10)), which is Eq. (.1), and the proof is completed. As an immediate consequence of Theorem.4, we have the following corollary. Corollary.5. In Theorem.4, ifα = ω k (n) with k and β =(β i ) 1 i n = (1, 1,,...,F(n)), then det (a i,j) = 1 i,j n 1 if k =, n =4, n if k =, n 5, 1 if k 3, n k. Theorem.6. Let (a i,j ) 1 i,j n be the doubly indexed sequence given by the recurrence a i,j = a i 1,j 1 + a i 1,j, i,j, (.13) and the initial conditions a i,1 = F (i) and a 1,j = ωj 1,where1 i, j n. Then { if n even, det (a i,j) = 1 i,j n 0 if n odd; where n 3. Proof. Let us denote the matrix (a i,j ) 1 i,j n by A. We claim that A = L B,

More Calculations on Determinant Evaluations 9 where L =(L i,j ) 1 i,j n is a matrix by the recurrence L i,j = L i 1,j 1 + L i 1,j, i,j 3, (.14) and the initial conditions L i,1 = F (i), L 1,i =0,andL i, = F (i 1), i, and where B =(B i,j ) 1 i,j n with ωj 1 if i =1,j 1, B B i,j =,j 1 + B,j B 1,j if i =3,j, (.15) B i 1,j 1 if i 1, 3, j, 0 if i,j =1. The proof of the claim is quite similar to the proof of Theorem.4 and we leave it to the reader. Notice that, the matrix L is a lower triangular one with 1 s on the diagonal, and hence det(a) =det(b). Through a short computation one gets det(b) =1+( 1) n, and so the theorem follows now immediately. Acknowledgement We would like to express our sincere gratitude to the referee who read the article thoroughly and made many outstanding comments and suggestions. Moreover, the second proof of Corollary. is due to the referee. REFERENCES [1] R. Bacher. Determinants of matrices related to the Pascal triangle. J. Theorie Nombres Bordeaux, 14:19-41, 00. [] C. Krattenthaler. Evaluations of some determinants of matrices related to the Pascal triangle. Séminaire Lotharingien Combin., Article b47g, 19 pp, 00. [3] C. Krattenthaler. Advanced determinant calculus, Séminaire Lotharingien Combin., Article b4q, 67 pp, 1999. [4] C. Krattenthaler. Advanced determinant calculus: a complement. Linear Algebra Appl., 411:68-166, 005. [5] A. R. Moghaddamfar, S. Navid Salehy, and S. Nima Salehy. Evaluating some determinants of matrices with recursive entries, submitted. [6] E. Neuwirth. Treeway Galton arrays and generalized Pascal-like determinants. Adv. Appl. Math., to appear. [7] H. Zakrajšek and M. Petkovšek. Pascal-like determinants are recursive. Adv. Appl. Math., 33:431-450, 004.