Spec. Matrices 2019; 7:20–26
Research Article Open Access
S. Takahira*, T. Sogabe, and T.S. Usuda Bidiagonalization of (k, k + 1)-tridiagonal matrices https://doi.org/10.1515/spma-2019-0002 Received April 2, 2018; accepted September 28, 2018
Abstract: In this paper, we present the bidiagonalization of n-by-n (k, k+1)-tridiagonal matrices when n ≤ 2k. Moreover, we show that the determinant of an n-by-n (k, k+1)-tridiagonal matrix is the product of the diagonal elements and the eigenvalues of the matrix are the diagonal elements. This paper is related to the fast block diagonalization algorithm using the permutation matrix from [T. Sogabe and M. El-Mikkawy, Appl. Math. Comput., 218, (2011), 2740-2743] and [A. Ohashi, T. Sogabe, and T. S. Usuda, Int. J. Pure and App. Math., 106, (2016), 513-523].
Keywords: (k, k + 1)-tridiagonal matrix, bidiagonalization, eigenvalues
MSC: Primary 65F15; Secondary 65F30
1 Introduction
(k,k+1) We consider an n-by-n (k, k + 1)-tridiagonal matrix Tn de ned by
1 2 ··· k + 1 ······ n 1 d1 a1 . 2 d .. 2 . .. .. . . . T(k,k+1) . n := . dn k an k . (1) − − .. k + 2 b1 . . . .. .. . . . n bn−k−1 dn
(k,k+1) The numbers of elements di , ai and bi are n, n − k and n − k − 1, respectively. The matrix Tn can also be described as follows:
n n−k n−k−1 (k,k+1) X T X T X T Tn = di ei ei + ai ei ek+i + bi ei+k+1ei , (2) i=1 i=1 i=1
T where ei denotes the i-th canonical n dimensional basis vector and ei denotes the transpose of ei.
*Corresponding Author: S. Takahira: Graduate School of Information Science & Technology, Aichi Prefectural University, Japan, E-mail: [email protected] T. Sogabe: Graduate School of Engineering, Nagoya University, Japan, E-mail: [email protected] T.S. Usuda: Graduate School of Information Science & Technology, Aichi Prefectural University, Japan, E-mail: [email protected]
Open Access. © 2019 S. Takahira et al., published by De Gruyter. his work is licensed under the Creative Commons Attribution alone 4.0 License. Bidiagonalization of (k, k + 1)-tridiagonal matrices Ë 21
(k, k+1)-tridiagonal matrices arise in applications such as the discrete hungry Lotka-Volterra system and (k,k+1) linear systems (see, e.g., [3, 4]). These matrices are related to k-Hessenberg matrices Hn(k) [2], in that Tn with k = 1 corresponds to Hn(k) with k = 2, and can be regarded as a variant of k-tridiagonal matrices (see, 0 e.g., [1, 7]). Moreover, a (k, k + 1)-tridiagonal matrix is a special case of a (k, k )-pentadiagonal matrix (see, 0 0 0 e.g., [6]). In fact, if k = k + 1, and the k-th superdiagonal and k -th subdiagonal are all zero, then a (k, k )- 0 pentadiagonal matrix is a (k, k + 1)-tridiagonal matrix. For k-tridiagonal matrices and (k, k )-pentadiagonal matrices, the fast block diagonalization algorithm using a permutation matrix is discussed in [6, 7]. However, the (k, k + 1)-tridiagonal matrices are irreducible in the sense that the block size of the matrix produced by the block diagonalization algorithm in [6] is one. In this paper, we consider another type of diagonalization, that is, bidiagonalization using a permuta- (k,k+1) 0 tion matrix. The bidiagonalization of Tn is the process of nding a lower bidiagonal matrix T and a permutation matrix P such that 0 T (k,k+1) T = P Tn P, (3) where P is the n-by-n permutation matrix de ned as n X T P := eσ(i)ei , (4) i=1 and σ is a permutation in the symmetric group of order n. Moreover, we give an explicit representation of 0 the bidiagonal matrix T and the permutation matrix P. We consider only bidiagonalizations of the form T (k,k+1) T (k,k+1) P Tn P, and in particular do not consider bidiagonalizations of the form P Tn Q for dierent per- mutation matrices P and Q. One reason is that, as we will see in Corollary 4, the eigenvalues of an n-by-n (k,k+1) (k, k + 1)-tridiagonal matrix Tn are the diagonal elements d1, d2, ... , dn. (k,k+1) We now give a necessary condition for bidiagonalization of n-by-n (k, k + 1)-tridiagonal matrices Tn (k,k+1) when the elements di , ai , bi are all nonzero. Then, the number of nonzero elements of Tn is 3n − 2k − 1. Thus, we have the necessary condition given in the following remark.
(k,k+1) Remark 1. Suppose that the elements di , ai , bi of an n-by-n (k, k + 1)-tridiagonal matrices Tn are all (k,k+1) nonzero. Then n ≤ 2k is a necessary condition for bidiagonalization of Tn using a permutation matrix.
(k,k+1) Proof. The number of elements of di , ai , bi of Tn is 3n − 2k − 1. The position of nonzero elements of 0 the bidiagonal matrix T must be on the diagonal or on the subdiagonal. The number of nonzero elements (k,k+1) 0 of Tn is equal to the number of nonzero elements of T because we consider the bidiagonalization by a permutation matrix. Thus, we have 3n − 2k − 1 ≤ n + n − 1, that is n ≤ 2k.
Therefore, we consider only the case n ≤ 2k.
2 Main Result
(k,k+1) We present the bidiagonalization of the n-by-n (k, k+1)-tridiagonal matrix Tn when n ≤ 2k and an explicit representation of the permutation matrix as a theorem. In addition, we give the determinant and eigenvalues of the n-by-n (k, k + 1)-tridiagonal matrix when n ≤ 2k as corollaries of the theorem.
(k,k+1) Theorem 2. Let Tn be the n-by-n (k, k + 1)-tridiagonal matrix in Eq.(1) with n ≤ 2k. Then there is a bidi- 0 (k,k+1) agonalization T of Tn of the form n−k k 0 T (k,k+1) X T T X T T = P Tn P = (di e2i e2i + dk+i e2i−1e2i−1) + di en−k+i en−k+i i=1 i=n−k+1 n−k n−k−1 X T X T + ai e2i e2i−1 + bi e2i+1e2i , (5) i=1 i=1 22 Ë S. Takahira, T. Sogabe, and T.S. Usuda where P is the permutation matrix
n−k k X T T X T P = (ek+j e2j−1 + ej e2j) + ej en−k+j . (6) j=1 j=n−k+1
(k,k+1) Proof. For an n-by-n (k, k + 1)-tridiagonal matrix Tn , we de ne
n n−k n−k−1 X T X T X T Dn := di ei ei , An := ai ei ek+i , Bn := bi ei+k+1ei , (7) i=1 i=1 i=1
T T T T (k,k+1) and compute each of P Dn P, P An P, P Bn P to obtain P Tn P. T The vector P ei can be written n−k n−k k T X T X T X T P ei = e2j−1ek+j + e2j ej + en−k+j ej ei , j=1 j=1 j=n−k+1 n k n k k X− X− X = e2j−1δk+j,i + e2j δj,i + en−k+j δj,i , j=1 j=1 j=n−k+1 e i 1 ≤ i ≤ n − k 2 = en−k+i n − k + 1 ≤ i ≤ k , (8) e2(i−k)−1 k + 1 ≤ i ≤ n where δi,j is the Kronecker delta, that is, δi,i = 1 and δi,j = 0 for i ≠ j. Note that n − k ≤ k < k + j for j > 0 T because n ≤ 2k. By transposing Eq.(8), we also obtain ei P. T We compute P Dn P as follows:
n−k k n ! T T X T X T X T P Dn P = P di ei ei + di ei ei + di ei ei P, i=1 i=n−k+1 i=k+1 n−k k n X T T X T T X T T = di P ei ei P + di P ei ei P + di P ei ei P, i=1 i=n−k+1 i=k+1 n−k k n X T X T X T = di e2i e2i + di en−k+i en−k+i + di e2(i−k)−1e2(i−k)−1, i=1 i=n−k+1 i=k+1 n−k k X T T X T = di e2i e2i + di+k e2i−1e2i−1 + di en−k+i en−k+i . (9) i=1 i=n−k+1 T Next, we compute P An P as follows:
n−k n−k T X T T X T P An P = ai P ei ek+i P = ai e2i e2i−1. (10) i=1 i=1 T Finally, we compute P Bn P as follows:
n−k−1 n−k−1 T X T T X T P Bn P = bi P ei+k+1ei P = bi e2i+1e2i . (11) i=1 i=1 Therefore, we have a bidiagonalization of the form (5). This completes the proof of Theorem 2.
Note that the case n = 2k is essential for bidiagonalization using a permutation matrix. Because, as we will see in the appendix, the permutation matrix when n = 2k is uniquely derived as
n−k X T T P = (ek+j e2j−1 + ej e2j). (12) j=1 Bidiagonalization of (k, k + 1)-tridiagonal matrices Ë 23
In addition, the permutation matrix when n < 2k can be obtained from the permutation matrix when n = 2k. Moreover, bidiagonalization is not uniquely determined when n < 2k. For the case n < 2k, the matrix (5) is the block diagonal matrix that has one bidiagonal block and some 1-by-1 blocks (for example, see Example 0 6). Therefore, choosing an appropriate permutation matrix P to exchange blocks, we can obtain another 00 0 T (k,k+1) 0 bidiagonalization of the form T = (PP ) Tn (PP ). However, this is not essential for bidiagonalization 0 because such a permutation matrix P only exchanges blocks. From Theorem 2, we have the following corollary giving the determinant and eigenvalues of an n-by-n (k,k+1) (k, k + 1)-tridiagonal matrix Tn when n ≤ 2k.
(k,k+1) Corollary 3. The determinant of the n-by-n (k, k + 1)-tridiagonal matrix Tn in Eq.(1) with n ≤ 2k is the (k,k+1) product of the elements on the main diagonal, that is, det(Tn ) = d1d2 ··· dn.
(k,k+1) Corollary 4. The eigenvalues λ of the n-by-n (k, k + 1)-tridiagonal matrix Tn in Eq.(1) with n ≤ 2k are the elements on the main diagonal, that is, λ = d1, d2, ... , dn.
(k,k+1) Corollary 4 relates to the result of Theorem 1 in [5]. If diagonal elements d1, d2, ... , dn of Tn are all zero (k,k+1) and ai , bi are all positive real numbers, then the (k, k+1)-tridiagonal matrix Tn coincides with the double band matrix A with b = k + 1 in [5]. When n ≤ 2k, the eigenvalues of the matrix are d1 = d2 = ··· = dn = 0 from Corollary 4, and moreover the matrix has the zero eigenvalue of multiplicity n from Theorem 1 in [5].
3 Examples
In this section, we present two examples based on the result in the previous section.
Example 5 n k . Let T(5,6) be the following -tridiagonal matrix: ( = 10, = 5) 10 (5, 6) d1 0 0 0 0 a1 0 0 0 0 0 d 0 0 0 0 a 0 0 0 2 2 d a 0 0 3 0 0 0 0 3 0 0 0 0 0 d4 0 0 0 0 a4 0 0 0 0 0 d 0 0 0 0 a T(5,6) = 5 5 . (13) 10 d 0 0 0 0 0 6 0 0 0 0 b1 0 0 0 0 0 d7 0 0 0 0 b2 0 0 0 0 0 d8 0 0 0 0 b3 0 0 0 0 0 d9 0 0 0 0 b4 0 0 0 0 0 d10 Applying Theorem 2, we have the permutation matrix
P = [e6, e1, e7, e2, e8, e3, e9, e4, e10, e5], (14) and the bidiagonal matrix T0 is d6 0 0 0 0 0 0 0 0 0 a d 0 0 0 0 0 0 0 0 1 1 b d 0 1 7 0 0 0 0 0 0 0 0 0 a2 d2 0 0 0 0 0 0 0 0 0 b d 0 0 0 0 0 PTT(5,6)P = 2 8 . (15) 10 a d 0 0 0 0 3 3 0 0 0 0 0 0 0 0 0 b3 d9 0 0 0 0 0 0 0 0 0 a4 d4 0 0 0 0 0 0 0 0 0 b4 d10 0 0 0 0 0 0 0 0 0 a5 d5 24 Ë S. Takahira, T. Sogabe, and T.S. Usuda
This example corresponds to the case n = 2k. The total number of values ai and bi is 9 = n − 1. We can see that some ai or bi appears in each position of the subdiagonal.
Example 6 n k . Let T(6,7) be the following -tridiagonal matrix: ( = 10, = 6) 10 (6, 7) d1 0 0 0 0 0 a1 0 0 0 0 d 0 0 0 0 0 a 0 0 2 2 d a 0 0 3 0 0 0 0 0 3 0 0 0 0 d4 0 0 0 0 0 a4 0 0 0 0 d 0 0 0 0 0 T(6,7) = 5 . (16) 10 d 0 0 0 0 0 6 0 0 0 0 0 0 0 0 0 0 d7 0 0 0 b1 0 0 0 0 0 0 d8 0 0 0 b2 0 0 0 0 0 0 d9 0 0 0 b3 0 0 0 0 0 0 d10 Applying Theorem 2, we have the permutation matrix
P = [e7, e1, e8, e2, e9, e3, e10, e4, e5, e6], (17) and the bidiagonal matrix T0 is d7 0 0 0 0 0 0 0 0 0 a d 0 0 0 0 0 0 0 0 1 1 b d 0 1 8 0 0 0 0 0 0 0 0 0 a2 d2 0 0 0 0 0 0 0 0 0 b d 0 0 0 0 0 PTT(6,7)P = 2 9 . (18) 10 a d 0 0 0 0 3 3 0 0 0 0 0 0 0 0 0 b3 d10 0 0 0 0 0 0 0 0 0 a4 d4 0 0 0 0 0 0 0 0 0 0 d5 0 0 0 0 0 0 0 0 0 0 d6
This example corresponds to the case n < 2k. The total number of values ai and bi is 7 < (n − 1). Thus, there are two zero elements in the subdiagonal. In particular, both the (9, 8) element and the (10, 9) element are zero.
Acknowledgement: We thank Peter Humphries, PhD, from Edanz Group (www.edanzediting.com/ac) for editing a draft of this manuscript. We also thank the editor and the anonymous referees for the comments for the original manuscript of this paper. This work has been supported in part by JSPS KAKENHI Grant Numbers JP16H04367, JP18H05392.
Appendix A : Derivation of the bidiagonalization
(k,k+1) In this appendix, we derive the bidiagonalization of an n-by-n (k, k + 1)-tridiagonal matrix Tn when n = 2k. Although the general case is n ≤ 2k from Remark 1, we consider n = 2k because the permutation matrix when n < 2k can be described using the permutation matrix when n = 2k. Moreover, the case n = 2k is essential for the bidiagonalization as described in Section 2. 0 We rst show a relationship between the elements of the matrices X and X = PTXP for an n-by-n matrix X. We then derive the bidiagonalization of an n-by-n (k, k + 1)-tridiagonal matrix and an explicit representation of the permutation matrix P when n = 2k. Bidiagonalization of (k, k + 1)-tridiagonal matrices Ë 25
A.1 Preliminaries
0 We present a relationship between the elements of X and X in the following lemma. Throughout the paper, −1 Xi,j denotes the (i, j) element of a matrix X and σ denotes the inverse of the permutation σ.
0 Lemma 7. Let X be an n-by-n matrix and P be an n-by-n permutation matrix. The (i, j) element of X = PTXP 0 −1 −1 is the (σ(i), σ(j)) element of X, that is, Xi,j = Xσ(i),σ(j). Alternatively, the (i, j) element of X is the (σ (i), σ (j)) element of X0, that is, X X0 . i,j = σ−1(i),σ−1(j)
Proof. We prove this lemma by direct calculation: n ! n ! n n 0 T X T X T X X T T X = P XP = ei eσ(i) X eσ(j)ej = eσ(i)Xeσ(j)ei ej , i=1 j=1 i=1 j=1 n n X X T = Xσ(i),σ(j)ei ej . (19) i=1 j=1
A.2 The nonzero structure of the bidiagonal matrix
0 T (k,k+1) We describe the nonzero structure of T = P Tn P by considering submatrices. Using Lemma 7, we can determine the submatrices for which the lower left element is ai or bi as follows.
(k,k+1) Lemma 8. Let Tn be the n-by-n (k, k + 1)-tridiagonal matrix in Eq.(1) and let P be the n-by-n permutation (k,k+1) 0 T (k,k+1) matrix in Eq.(4). If n = 2k and P bidiagonalizes Tn , then the matrix T = P Tn P has the following 2-by-2 submatrices whose lower left element is ai: 0 ! ! T −1 −1 0 dk i 0 σ (i)−1,σ (i)−1 = + , i = 1, 2, ... , n − k. (20) T0 T0 a d σ−1(i),σ−1(i)−1 σ−1(i),σ−1(i) i i
−1 (k,k+1) 0 Proof. We write τ and T for σ and Tn , respectively. For the main diagonal of the matrix T , we focus on T0 T0 the 2-by-2 submatrix whose lower right element is τ(1),τ(1). Because is a bidiagonal matrix, a submatrix 0 of T is a lower bidiagonal matrix. From Lemma 2, the form of the submatrix is 0 ! ! T 0 T 0 τ(1)−1,τ(1)−1 = σ(τ(1)−1),σ(τ(1)−1) , T0 T0 T T τ(1),τ(1)−1 τ(1),τ(1) σ(τ(1)),σ(τ(1)−1) σ(τ(1)),σ(τ(1)) ! T 0 = σ(τ(1)−1),σ(τ(1)−1) . (21) T1,σ(τ(1)−1) T1,1
The nonzero elements in the rst row of T are d1 and a1. From T1,1 = d1, it follows that T1,σ(τ(1)−1) = a1 = 0 T1,k+1 because the subdiagonals of T are all nonzero when n = 2k. Thus, we have σ(τ(1) − 1) = k + 1. Therefore, we can rewrite equation (21) as 0 ! ! T 0 dk 0 τ(1)−1,τ(1)−1 = +1 . (22) T0 T0 a d τ(1),τ(1)−1 τ(1),τ(1) 1 1 The cases i = 2, 3, ... , n − k follow similarly, which proves the lemma.
(k,k+1) Lemma 9. Let Tn be the n-by-n (k, k + 1)-tridiagonal matrix in Eq.(1) and let P be the n-by-n permutation (k,k+1) 0 T (k,k+1) matrix as in Eq.(4). If n = 2k and P bidiagonalizes Tn , then the matrix T = P Tn P has the following 2-by-2 submatrices whose lower left element is bi: 0 ! ! T −1 −1 0 di 0 σ (i),σ (i) = , i = 1, 2, ... , n − k − 1. (23) T0 T0 b d σ−1(i)+1,σ−1(i) σ−1(i)+1,σ−1(i)+1 i k+i+1 26 Ë S. Takahira, T. Sogabe, and T.S. Usuda
Proof. The proof is similar to that of Lemma 8. 0 We now show the nonzero structure of the bidiagonal matrix T in Lemma 8 and Lemma 9.
(k,k+1) Proposition 10. Let Tn be the n-by-n (k, k + 1)-tridiagonal matrix in Eq.(1) and let P be the permutation (k,k+1) 0 T (k,k+1) matrix in Eq.(4). If n = 2k and P bidiagonalizes Tn , then the matrix T = P Tn P is a bidiagonal matrix of the form
n−k n−k n−k−1 0 T (k,k+1) X T T X T X T T = P Tn P = (di+k e2i−1e2i−1 + di e2i e2i) + ai e2i e2i−1 + bi e2i+1e2i . (24) i=1 i=1 i=1
0 Proof. From Eqs.(20) and (23), we have the following 3-by-3 bidiagonal submatrices of T : T0 τ(i)−1,τ(i)−1 0 0 di+k 0 0 T0 T0 a d i ... n k τ(i)−1,τ(i) τ(i),τ(i) 0 = i i 0 , = 1, 2, , − − 1. (25) T0 T0 b d 0 τ(i),τ(i)+1 τ(i)+1,τ(i)+1 0 i i+1+k (k,k+1) The elements of Tn are not divided into two parts by the transformation. For example, d1 is the only 0 element of T . Therefore, we have τ(i) + 1 = τ(i + 1) − 1 for i = 1, 2, ... , n − k − 1. Considering the submatrix 0 (20) for i = n − k, the sequence on the main diagonal of T is d1+k , d1, d2+k , d2, ... , dn−k , dn. Moreover, the 0 sequence on the subdiagonal of T is a1, b1, a2, b2, ... , an−k−1, bn−k−1, an−k.
A.3 Derivation of the permutation matrix
We now derive an explicit representation of the permutation matrix P when n = 2k. Proposition 10 suggests P Pn e eT σ a permutation matrix of the form = i=1 σ(i) i , where is the permutation. From the proof of the propo- 0 sition, the sequence on the main diagonal of T when n = 2k is dk+1, d1, dk+2, d2, ... , dn , dn−k. Moreover, 0 (k,k+1) from Lemma 7, the (i, i) element of T is the (σ(i), σ(i)) element of Tn . Thus, we have k + 1 = σ(1), 1 = σ(2), k + 2 = σ(3), 2 = σ(4), ... , n = σ(n − 1), n − k = σ(n). Namely, when n = 2k, we have the permutation matrix n−k X T T P = (ek+j e2j−1 + ej e2j). (26) j=1 We summarize this in the following proposition.
(k,k+1) Proposition 11. The permutation matrix P that bidiagonalizes the n-by-n (k, k + 1)-tridiagonal matrix Tn when n = 2k is the matrix in Eq.(26).
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