再谈初等变换法在矩阵计算中的应用(英翻中)

谁有 这篇文章的中文版的？急需翻译中文
Usually, for an n×n matrix over a number field F , eigenvalues are obtained by solving its eigenequation.
That is to say, we need to solve an equation of order n in one variable, whose processing is very trouble.
Similarly, if we diagonalize a matrix A , we need to judge whether A has n linear independent eigenvectors;
if similar transformation matrix is required, we have to find basic solutions of a system of n linear equations in
n variables.
The QR decomposition of a matrix is one of the useful methods in matrix computation. But its computing
process is very complex, which makes us trouble. In this paper, we give an elementary transformation method for
finding a QR decomposition of any full-column-rank matrix.
2 Basic theory
Theorem 2[1] If A ∈ Rm×nis of full column rank，then AT Ais symmetric positive definite. In addition,
AT Ahas the unique triangular decomposition

AT A= LDLT,（1）
where L is a low triangular matrix of all
diagonal elements 1, D is a diagonal matrix with
positive diagonal elements，and A Tis the transpose
of A .
Theorem 3 If A ∈ Rm×nis of full column
rank，then A has QR decomposition
A = QR（2）
where Q = A(L ?1)T D?1/2has orthogonal
normal columns, and R = D1/2LTis nonsingular
upper triangular.
Proof By (1), we have
n
( D ? 1 /2L?1AT )(D?1/2L?1AT)T=I, which
implies that ( D ? 1 /2L?1AT)Thas orthogonal
normal columns. Let Q = A( L?1)T D?1/2and
R = D1/2LT, then the proof is complete.
From Theorems 2 and 3, calculating QR
decomposition (2) of a full-column-rank matrix
A ∈ Rm×nis transformed into two steps: first
calculate triangular decomposition (1) of AT A, and
then calculate Q and R according to
Q = A(L ?1)T D?1/2and R = D1/2LT,
respectively. The first step can be completed by
using approach provided in textbooks of numerical
algebra (see, e.g., [4]), while the second step involves
only the inverses and multiplications of matrices.
However, the following algorithm provides an
elementary transformation approach, which is simpler
than those in the textbooks of numerical algebra and
simultaneously avoid to calculating inverse of
Matrices.
Algorithm
Input: A full-column-rank matrix A ∈ Rm×n.
Output: Two matrices Q ∈ Rm×nand
R ∈ Rn×nsuch that A = QR, where Q
has orthogonal normal columns and R is upper
triangular.

通常，为在数字域F的一个n×n矩阵，本征值通过解决它的eigenequation获得。
也就是说，我们需要解决秩序n的等式在一可变物的，处理是非常麻烦。
Similarly，如果我们斜向移动矩阵A，我们需要判断A是否有n线性独立特征向量; 相似的变革矩阵需要的if，我们必须发现n线性方程系统的基本解法在的n可变物。 矩阵的The QR分解是其中一个在矩阵计算的有用的方法。 但是它计算
process是非常复杂的，造成我们困难。 在本文，我们给的一个基本的变革方法finding QR分解任何充分专栏排列矩阵。
2基本的理论
Theorem 2 [1]，如果∈充分的专栏rank，then Rm×nis在Ais相称正面确定的。 另外，
AT Ahas独特的三角分解

AT A= LDLT， （1）
where L是所有的一个低三角形矩阵
diagonal元素1， D是一个对角矩阵与
positive对角elements，and Tis移置
of A。
Theorem 3，如果∈充分的专栏Rm×nis
rank，then A有QR分解
A = QR（2） WHERe Q = A (L ？1) T D ？正交的1/2has
normal专栏和非奇R =的D1/2LTis 三角的upper。 由(1)的Proof，我们有
n
(D ？ 1 /2L ？1AT) (D ？1/2L ？1AT) T=I，
implies (D ？ 1 /2L ？1AT)正交的Thas
normal专栏。 让Q = A (L ？1) T D ？1/2and
R = D1/2LT，然后证明是完全的。
From定理2和3，计算QR
decomposition (2)充分专栏排列矩阵
A ∈ Rm×nis被变换成二步： 首先 在A的calculate三角分解(1)，和
then计算Q和R根据
Q = A (L ？1) T D ？1/2and R = D1/2LT，
respectively. 第一步可以完成
using方法在课本提供了数字
algebra (参见，即， [4])，而第二步介入
only矩阵的反面和增殖。
However，以下算法提供
elementary变革方法，是更加简单的
than那些在数字代数课本和
simultaneously避免对计算反面
Matrices.
Algorithm
Input ： 充分专栏排列矩阵A ∈ Rm×n。
Output ： 二个矩阵Q ∈ Rm×nand
R ∈ Rn×nsuch A = QR， Q
has正交正常专栏和R上部
triangular.

通常，在数域F中的n*n矩阵，通过解其特征方程得到特征值。 也就是说，我们要解一元n次方程，这是很复杂的。 如果把矩阵A对角化，需要保证A有n个线性无关的特征值 如果需要相似变换，首先要计算出n个n元线性方程组的解。 矩阵的QR分解是矩阵运算的重要方法之一