Abstract
Abstract In this paper,we present the bidiagonalization of n-by-n (k, k+1)-tridiagonal matriceswhen 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].
Highlights
We consider an n-by-n (k, k +)-tridiagonal matrix T (k,k+ n ) de ned by··· k+ ··· ··· n d a d Tn(k,k+ ) := ... dn−k an−k (1) k+ b n bn−k− dnThe numbers of elements di, ai and bi are n, n − k and n − k −, respectively
We present the bidiagonalization of the n-by-n (k, k+ )-tridiagonal matrix Tn(k,k+ ) when n ≤ k and an explicit representation of the permutation matrix as a theorem
We present a relationship between the elements of X and X in the following lemma
Summary
The position of nonzero elements of the bidiagonal matrix T must be on the diagonal or on the subdiagonal. ) is equal to the number of nonzero elements of T because we consider the bidiagonalization by a permutation matrix. We present the bidiagonalization of the n-by-n (k, k+ )-tridiagonal matrix Tn(k,k+ ) when n ≤ k and an explicit representation of the permutation matrix as a theorem. Choosing an appropriate permutation matrix P to exchange blocks, we can obtain another bidiagonalization of the form T = (PP )TTn(k,k+ )(PP ). Let T( , ) be the following ( , )-tridiagonal matrix: This example corresponds to the case n < k.
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