## 澳洲代写｜MTH3320｜Computational linear algebra计算线性代数 蒙纳士大学

statistics-labTM为您提供蒙纳士大学（Monash University）Computational linear algebra计算线性代数澳洲代写代考辅导服务！

The overall aim of this unit is to study the numerical methods for matrix computations that lie at the core of a wide variety of large-scale computations and innovations in the sciences, engineering, technology and data science. You will receive an introduction to the mathematical theory of numerical methods for linear algebra (with derivations of the methods and some proofs). This will broadly include methods for solving linear systems of equations, least-squares problems, eigenvalue problems, and other matrix decompositions. Special attention will be paid to conditioning and stability, dense versus sparse problems, and direct versus iterative solution techniques. You will learn to implement the computational methods efficiently, and will learn how to thoroughly test their implementations for accuracy and performance. You will work on realistic matrix models for applications in a variety of fields. Applications may include, for example: computation of electrostatic potentials and heat conduction problems; eigenvalue problems for electronic structure calculation; ranking algorithms for webpages; algorithms for movie recommendation, classification of handwritten digits, and document clustering; and principal component analysis in data science.

## Computational linear algebra计算线性代数问题集

Let $\mathbf{A}=\left(\begin{array}{ll}1 & 2 \ 3 & 4\end{array}\right)$ and $\mathbf{B}=\left(\begin{array}{ll}5 & 6 \ 7 & 8\end{array}\right)$. Determine (a) $(\mathbf{A}-\mathbf{B})(\mathbf{A}+\mathbf{B})(\mathrm{b}) \mathbf{A}^2-\mathbf{B}^2$ Explain why $\mathbf{A}^2-\mathbf{B}^2 \neq(\mathbf{A}-\mathbf{B})(\mathbf{A}+\mathbf{B})$.
University of Hertfordshire, UK
Let $\mathbf{A}$ and $\mathbf{B}$ be invertible (non-singular) $n$ by $n$ matrices. Find the errors, if any, in the following derivation:
\begin{aligned} \mathbf{A B}(\mathbf{A B})^{-1} & =\mathbf{A B A}^{-1} \mathbf{B}^{-1} \ & =\mathbf{A A}^{-1} \mathbf{B B}^{-1} \ & =\mathbf{I} \times \mathbf{I}=\mathbf{I} \end{aligned}
You need to explain why you think there is an error.
University of Hertfordshire, UK
Given the matrix
$$\mathbf{A}=\frac{1}{7}\left(\begin{array}{rrr} 3 & -2 & -6 \ -2 & 6 & -3 \ -6 & -3 & -2 \end{array}\right)$$
(a) Compute $\mathbf{A}^2$ and $\mathbf{A}^3$.
(b) Based on these results, determine the matrices $\mathbf{A}^{-1}$ and $\mathbf{A}^{2004}$.

Give an example of the following, or state that no such example exists: $2 \times 2$ matrix $\mathbf{A}$ and $2 \times 1$ non-zero vectors $\mathbf{u}$ and $\mathbf{v}$ such that $\mathbf{A u}=\mathbf{A v}$ yet $\mathbf{u} \neq \mathbf{v}$.

Illinois State University, USA (part question)
(a) If $\mathbf{A}=\left(\begin{array}{ll}1 & 2 \ 3 & 4\end{array}\right)$ and $\mathbf{B}=\left(\begin{array}{rr}0 & 1 \ -1 & 0\end{array}\right)$, compute $\mathbf{A}^2, \mathbf{B}^2, \mathbf{A B}$ and $\mathbf{B A}$.
(b) If $\mathbf{A}=\left(\begin{array}{ll}a & b \ c & d\end{array}\right)$ and $\mathbf{B}=\left(\begin{array}{ll}e & f \ g & h\end{array}\right)$, compute $\mathbf{A B}-\mathbf{B A}$.
Queen Mary, University of London, UK
Let $\mathbf{M}=\left(\begin{array}{ll}1 & 1 \ 1 & 1\end{array}\right)$. Compute $\mathbf{M}^n$ for $n=2,3,4$. Find a function $c(n)$ such that $\mathbf{M}^n=c(n) \mathbf{M}$ for all $n \in \mathbb{Z}, n \geq 1$. (You are not required to prove any of your results.) Queen Mary, University of London, UK (part question)
Let $\mathbf{A}=\left(\begin{array}{cc}\frac{1}{3} & \frac{1}{3} \ \frac{1}{3} & \frac{1}{3}\end{array}\right)$. Determine (i) $\mathbf{A}^2$ (ii) $\mathbf{A}^3$ Prove that $\mathbf{A}^n=\frac{1}{2}\left(\frac{2}{3}\right)^n \mathbf{A}$.
University of Hertfordshire, UK How many rows does $\mathbf{B}$ have if $\mathbf{B C}$ is a $4 \times 6$ matrix? Explain.

Prove (give a clear reason): If $\mathbf{A}$ is a symmetric invertible matrix then $\mathbf{A}^{-1}$ is also symmetric.

Massachusetts Institute of Technology USA
If $\mathbf{A}$ is a matrix such that $\mathbf{A}^2-\mathbf{A}+\mathbf{I}=\mathbf{O}$ show that $\mathbf{A}$ is invertible with inverse I – A.

(part question)
(a) Define what is meant by a square matrix $\mathbf{A}$ being invertible. Show that the inverse of $\mathbf{A}$, if it exists, is unique.
(b) Show that the product of any finite number of invertible matrices is invertible.
(c) Find the inverse of the matrix
$$\mathbf{A}=\left[\begin{array}{rrr} 1 & 0 & 1 \ -1 & 1 & 1 \ 0 & 1 & 0 \end{array}\right]$$
University of Sussex, UK
Let $\mathbf{A}$ and $\mathbf{B}$ be $n \times n$ invertible matrices, with $\mathbf{A X A}^{-1}=\mathbf{B}$. Explain why $\mathbf{X}$ is invertible and calculate $\mathbf{X}^{-1}$ in terms of $\mathbf{A}$ and $\mathbf{B}$.

\begin{prob}

Show that, for any non-zero vector $\mathbf{u}$ in $\mathbb{R}^n$, we have $\left|\frac{1}{|\mathbf{u}|} \mathbf{u}\right|=1$.
Let $\mathbf{u}$ and $\mathbf{v}$ be vectors in $\mathbb{R}^n$. Disprove the following propositions:
(a) If $\mathbf{u} \cdot \mathbf{v}=0$ then $\mathbf{u}=\mathbf{O}$ or $\mathbf{v}=\mathbf{O}$.
(b) $|\mathbf{u}+\mathbf{v}|=|\mathbf{u}|+|\mathbf{v}|$
Let $\mathbf{u}_1, \mathbf{u}_2, \mathbf{u}_3, \ldots, \mathbf{u}_n$ be orthogonal vectors in $\mathbb{R}^n$. Prove
(i) $\left|\mathbf{u}_1+\mathbf{u}_2\right|^2=\left|\mathbf{u}_1\right|^2+\left|\mathbf{u}_2\right|^2$
(ii) $\left|\mathbf{u}_1+\mathbf{u}_2+\cdots+\mathbf{u}_n\right|^2=\left|\mathbf{u}_1\right|^2+\left|\mathbf{u}_2\right|^2+\cdots+\left|\mathbf{u}_n\right|^2$
For part (ii) use mathematical induction.

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

## 数学代写|计算线性代数代写Computational Linear Algebra代考|MATH4076

statistics-lab™ 为您的留学生涯保驾护航 在代写计算线性代数Computational Linear Algebra方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写计算线性代数Computational Linear Algebra代写方面经验极为丰富，各种代写计算线性代数Computational Linear Algebra相关的作业也就用不着说。

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 数学代写|计算线性代数代写Computational Linear Algebra代考|Determinant of a Matrix

Definition 2.20. Let us consider $n$ objects. We will call permutation every grouping of these objects. For example, if we consider three objects $a, b$, and $c$, we could group them as $a-b-c$ or $a-c-b$, or $c-b-a$ or $b-a-c$ or $c-a-b$ or $b-c-a$ . In this case, there are totally six possible permutations. More generally, it can be checked that for $n$ objects there are $n$ ! ( $n$ factorial) permutations where $n !=$ $(n)(n-1)(n-2) \ldots(2)(1)$ with $n \in \mathbb{N}$ and $(0) !=1$.

We could fix a reference sequence (e.g. $a-b-c$ ) and name it fundamental permutation. Every time two objects in a permutation follow each other in a reverse order with respect to the fundamental we will call it inversion. Let us define even class permutation a permutation undergone to an even number of inversions and odd class permutation a permutation undergone to an odd number of inversions, see also [1].

In other words, a sequence is an even class permutation if an even number of swaps is necessary to obtain the fundamental permutation. Analogously, a sequence is an odd class permutation if an odd number of swaps is necessary to obtain the fundamental permutation.

Example 2.19. Let us consider the fundamental permutation $a-b-c-d$ associated with the objects $a, b, c, d$. The permutation $d-a-c-b$ is of even class since two swaps are required to reconstruct the fundamental permutation. At first we swap $a$ and $d$ to obtain $a-d-c-b$ and then we swap $d$ and $b$ to obtain the fundamental permutation $a-b-c-d$.

On the contrary, the permutation $d-c-a-b$ is of odd class since three swaps are necessary to reconstruct the fundamental permutation. Let us reconstruct the fundamental permutation step-by-step. At first we swap $d$ and $b$ and obtain $b-c-$ $a-d$. Then, let us swap $b$ and $a$ to obtain $a-c-b-d$. Eventually, we swap $c$ and $b$ to obtain the fundamental permutation $a-b-c-d$.

## 数学代写|计算线性代数代写Computational Linear Algebra代考|Linear Dependence of Row and Column Vectors of a Matrix

Definition 2.24. Let A be a matrix. The $i^{t h}$ row is said linear combination of the other rows if each of its element $a_{i, j}$ can be expressed as weighted sum of the other elements of the $j^{t h}$ column by means of the same scalars $\lambda_{1}, \lambda_{2}, \ldots, \lambda_{i-1}, \lambda_{i+1}, \ldots \lambda_{n}$ :
$$\mathbf{a}{\mathbf{i}}=\lambda{1} \mathbf{a}{1}+\lambda{2} \mathbf{a}{2}+\cdots+\lambda{i-1} \mathbf{a}{\mathbf{i}-\mathbf{1}}+\lambda{i+1} \mathbf{a}{\mathbf{i}+\mathbf{1}}+\ldots+\lambda{n} \mathbf{a}{\mathbf{n}}$$ Equivalently, we may express the same concept by considering each row element: \begin{aligned} &\forall j: \exists \lambda{1}, \lambda_{2}, \ldots, \lambda_{i-1}, \lambda_{i+1}, \ldots \lambda_{n} \mid \ &a_{i, j}=\lambda_{1} a_{1, j}+\lambda_{2} a_{2, j}+\ldots \lambda_{i-1} a_{i-1, j}+\lambda_{i+1} a_{i+1, j}+\ldots \lambda_{n} a_{n, j} . \end{aligned}

Example 2.27. Let us consider the following matrix:
$$\mathbf{A}=\left(\begin{array}{lll} 0 & 1 & 1 \ 3 & 2 & 1 \ 6 & 5 & 3 \end{array}\right)$$
The third row is a linear combination of the first two by means of scalars $\lambda_{1}, \lambda_{2}=$ 1,2 , the third row is equal to the weighted sum obtained by multiplying the first row by 1 and summing to it the second row multiplied by 2 :
$$(6,5,3)=(0,1,1)+2(3,2,1)$$
that is
$$\mathbf{a}{3}=\mathbf{a}{1}+2 \mathbf{a}{2} .$$ Definition 2.25. Let A be a matrix. The $j^{t h}$ column is said linear combination of the other column if each of its element $a{i, j}$ can be expressed as weighted sum of the other elements of the $i^{t h}$ row by means of the same scalars $\lambda_{1}, \lambda_{2}, \ldots, \lambda_{j-1}$, $\lambda_{j+1}, \ldots \lambda_{n}$ :
$$\mathbf{a}^{\mathbf{j}}=\lambda_{1} \mathbf{a}^{\mathbf{1}}+\lambda_{2} \mathbf{a}^{2}+\cdots+\lambda_{j-1} \mathbf{a}^{\mathbf{j}-\mathbf{1}}+\lambda_{j+1} \mathbf{a}^{\mathbf{j}+\mathbf{1}}+\ldots+\lambda_{n} \mathbf{a}^{\mathbf{n}}$$
Equivalently, we may express the same concept by considering each row element:
\begin{aligned} &\forall i: \exists \lambda_{1}, \lambda_{2}, \ldots, \lambda_{j-1}, \lambda_{j+1}, \ldots \lambda_{n} \mid \ &a_{i, j}=\lambda_{1} a_{i, 1}+\lambda_{2} a_{i, 2}+\ldots \lambda_{i-1} a_{i, j-1}+\lambda_{i+1} a_{i, j+1}+\ldots \lambda_{n} a_{i, n} \end{aligned}

## 数学代写|计算线性代数代写Computational Linear Algebra代考|Laplace Theorems on Determinants

Theorem 2.2. I Laplace Theorem Let $\mathbf{A} \in \mathbb{R}{n, n}$. The determinant of $\mathbf{A}$ can be computed as the sum of each row (element) multiplied by the corresponding cofactor: $\operatorname{det} \mathbf{A}=\sum{j=1}^{n} a_{i, j} A_{i, j}$ for any arbitrary $i$ and
$\operatorname{det} \mathbf{A}=\sum_{i=1}^{n} a_{i, j} A_{i, j}$ for any arbitrary $j$.
The I Laplace Theorem can be expressed in the equivalent form: the determinant of a matrix is equal to scalar product of a row (column) vector by the corresponding vector of cofactors.
Example 2.46. Let us consider the following $\mathbf{A} \in \mathbb{R}{3,3}$ : $$\mathbf{A}=\left(\begin{array}{ccc} 2 & -1 & 3 \ 1 & 2 & -1 \ -1 & -2 & 1 \end{array}\right)$$ The determinant of this matrix is $\operatorname{det} \mathbf{A}=4-1-6+6+1-4=0$. Hence, the matrix is singular. Let us now calculate the determinant by applying the I Laplace Theorem. If we consider the first row, it follows that det $\mathbf{A}=a{1,1} A_{1,1}+a_{1,2}(-1) A_{1,2}+$ $a_{1,3} A_{1,3}$, $\operatorname{det} \mathbf{A}=2(0)+1(0)+3(0)=0$. We arrive to the same conclusion.
Example 2.47. Let us consider the following $\mathbf{A} \in \mathbb{R}{3,3}$ : $$\mathbf{A}=\left(\begin{array}{lll} 1 & 2 & 1 \ 0 & 1 & 1 \ 4 & 2 & 0 \end{array}\right)$$ The determinant of this matrix is $\operatorname{det} \mathbf{A}=8-4-2=2$. Hence, the matrix is nonsingular. Let us now calculate the determinant by applying the I Laplace Theorem. If we consider the second row, it follows that $\operatorname{det} \mathbf{A}=a{2,1}(-1) A_{2,1}+a_{2,2} A_{2,2}+$ $a_{2,3}(-1) A_{2,3}$, $\operatorname{det} \mathbf{A}=0(-1)(-2)+1(-4)+1(-1)(-6)=2$. The result is the same.

## 数学代写|计算线性代数代写Computational Linear Algebra代考|Linear Dependence of Row and Column Vectors of a Matrix

$$\mathbf{a i}=\lambda 1 \mathbf{a} 1+\lambda 2 \mathbf{a} 2+\cdots+\lambda i-1 \mathbf{a} \mathbf{i}-\mathbf{1}+\lambda i+1 \mathbf{a} \mathbf{i}+\mathbf{1}+\ldots+\lambda n \mathbf{a n}$$

$$\forall j: \exists \lambda 1, \lambda_{2}, \ldots, \lambda_{i-1}, \lambda_{i+1}, \ldots \lambda_{n} \mid \quad a_{i, j}=\lambda_{1} a_{1, j}+\lambda_{2} a_{2, j}+\ldots \lambda_{i-1} a_{i-1, j}+\lambda_{i+1} a_{i+1, j}+\ldots \lambda_{n} a_{n, j}$$

$$\mathbf{A}=\left(\begin{array}{lllllll} 0 & 1 & 13 & 2 & 16 & 5 & 3 \end{array}\right)$$

$$(6,5,3)=(0,1,1)+2(3,2,1)$$

$$\mathbf{a} 3=\mathbf{a} 1+2 \mathbf{a} 2 .$$

$$\mathbf{a}^{\mathbf{j}}=\lambda_{1} \mathbf{a}^{\mathbf{1}}+\lambda_{2} \mathbf{a}^{2}+\cdots+\lambda_{j-1} \mathbf{a}^{\mathbf{j}-\mathbf{1}}+\lambda_{j+1} \mathbf{a}^{\mathbf{j}+\mathbf{1}}+\ldots+\lambda_{n} \mathbf{a}^{\mathbf{n}}$$

$$\forall i: \exists \lambda_{1}, \lambda_{2}, \ldots, \lambda_{j-1}, \lambda_{j+1}, \ldots \lambda_{n} \mid \quad a_{i, j}=\lambda_{1} a_{i, 1}+\lambda_{2} a_{i, 2}+\ldots \lambda_{i-1} a_{i, j-1}+\lambda_{i+1} a_{i, j+1}+\ldots \lambda_{n} a_{i, n}$$

## 数学代写|计算线性代数代写Computational Linear Algebra代考|Laplace Theorems on Determinants

$\operatorname{det} \mathbf{A}=\sum j=1^{n} a_{i, j} A_{i, j}$ 对于任何任意 $i$ 和
$\operatorname{det} \mathbf{A}=\sum_{i=1}^{n} a_{i, j} A_{i, j}$ 对于任何任意 $j$.
। 拉普拉斯定理可以表示为等价形式: 矩阵的行列式等于行 (列) 向量与相应的辅因子向量的标量积。 例 2.46。让我们考虑以下内容 $\mathbf{A} \in \mathbb{R} 3,3$ :

$$\mathbf{A}=\left(\begin{array}{llllllll} 1 & 2 & 10 & 1 & 14 & 2 & 0 \end{array}\right)$$

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

## 数学代写|计算线性代数代写Computational Linear Algebra代考|MAST10007

statistics-lab™ 为您的留学生涯保驾护航 在代写计算线性代数Computational Linear Algebra方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写计算线性代数Computational Linear Algebra代写方面经验极为丰富，各种代写计算线性代数Computational Linear Algebra相关的作业也就用不着说。

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 数学代写|计算线性代数代写Computational Linear Algebra代考|A Preliminary Introduction to Algebraic Structures

If a set is a primitive concept, on the basis of a set, algebraic structures are sets that allow some operations on their elements and satisfy some properties. Although an in depth analysis of algebraic structures is out of the scopes of this chapter, this section gives basic definitions and concepts. More advanced concepts related to algebraic structures will be given in Chap. $7 .$

Definition 1.32. An operation is a function $f: A \rightarrow B$ where $A \subset X_{1} \times X_{2} \times \ldots \times X_{k}$, $k \in \mathbb{N}$. The $k$ value is said arity of the operation.

Definition 1.33. Let us consider a set $A$ and an operation $f: A \rightarrow B$. If $A$ is $X \times X \times$ $\ldots \times X$ and $B$ is $X$, i.e. the result of the operation is still a member of the set, the set is said to be closed with respect to the operation $f$.

Definition 1.34. Ring. A ring $R$ is a set equipped with two operations called sum and product. The sum is indicated with $\mathrm{a}+$ sign while the product operator is simply omitted (the product of $x_{1}$ by $x_{2}$ is indicated as $x_{1} x_{2}$ ). Both these operations process two elements of $R$ and return an element of $R(R$ is closed with respect to these two operations). In addition, the following properties must be valid.

• commutativity (sum): $x_{1}+x_{2}=x_{2}+x_{1}$
• associativity (sum): $\left(x_{1}+x_{2}\right)+x_{3}=x_{1}+\left(x_{2}+x_{3}\right)$
• neutral element (sum): $\exists$ an element $0 \in R$ such that $\forall x \in R: x+0=x$
• inverse element (sum): $\forall x \in R: \exists(-x) \mid x+(-x)=0$
• associativity (product): $\left(x_{1} x_{2}\right) x_{3}=x_{1}\left(x_{2} x_{3}\right)$
• distributivity $1: x_{1}\left(x_{2}+x_{3}\right)=x_{1} x_{2}+x_{1} x_{3}$
• distributivity $2:\left(x_{2}+x_{3}\right) x_{1}=x_{2} x_{1}+x_{3} x_{1}$
• neutral element (product): $\exists$ an element $1 \in R$ such that $\forall x \in R x 1=1 x=x$
The inverse element with respect to the sum is also named opposite element.

## 数学代写|计算线性代数代写Computational Linear Algebra代考|Numeric Vectors

Although this chapter intentionally refers to the set of real numbers $\mathbb{R}$ and its sum and multiplication operations, all the concepts contained in this chapter can be easily extended to the set of complex numbers $\mathbb{C}$ and the complex field. This fact is further remarked in Chap. 5 after complex numbers and their operations are introduced.
Definition 2.1. Numeric Vector. Let $n \in \mathbb{N}$ and $n>0$. The set generated by the Cartesian product of $\mathbb{R}$ by itself $n$ times $(\mathbb{R} \times \mathbb{R} \times \mathbb{R} \times \mathbb{R} \ldots)$ is indicated with $\mathbb{R}^{n}$ and is a set of ordered $n$-tuples of real numbers. The generic element $\mathbf{a}=\left(a_{1}, a_{2}, \ldots, a_{n}\right)$ of this set is named numeric vector or simply vector of order $n$ on the real field and the generic $a_{i} \forall i$ from 1 to $n$ is said the $i^{t h}$ component of the vector a.
Example 2.1. The $n$-tuple
$$\mathbf{a}=(1,0,56.3, \sqrt{2})$$
is a vector of $\mathbb{R}^{4}$.
Definition 2.2. Scalar. A numeric vector $\lambda \in \mathbb{R}^{1}$ is said scalar.
Definition 2.3. Let $\mathbf{a}=\left(a_{1}, a_{2}, \ldots, a_{n}\right)$ and $\mathbf{b}=\left(b_{1}, b_{2}, \ldots, b_{n}\right)$ be two numeric vectors $\in \mathbb{R}^{n}$. The sum of these two vectors is the vector $\mathbf{c}=\left(a_{1}+b_{1}, a_{2}+b_{2}, \ldots, a_{n}\right.$ $\left.+b_{n}\right)$ generated by the sum of the corresponding components.
Example 2.2. Let us consider the following vectors of $\mathbb{R}^{3}$
\begin{aligned} &\mathbf{a}=(1,0,3) \ &\mathbf{b}=(2,1,-2) \end{aligned}

## 数学代写|计算线性代数代写Computational Linear Algebra代考|Basic Definitions About Matrices

Definition 2.6. Matrix. Let $m, n \in \mathbb{N}$ and both $m, n>0$. A matrix $(m \times n) \mathbf{A}$ is a generic table of the kind:
$$\boldsymbol{\Lambda}=\left(\begin{array}{cccc} a_{1,1} & a_{1,2} & \ldots & a_{1, n} \ a_{2,1} & a_{2,2} & \ldots & a_{2, n} \ \ldots & \ldots & \ldots & \ldots \ a_{m, 1} & a_{m, 2} & \ldots & a_{m, n} \end{array}\right)$$
where each matrix element $a_{i, j} \in \mathbb{R}$. If $m=n$ the matrix is said square while it is said rectangular otherwise.

The numeric vector $\mathbf{a}{\mathbf{i}}=\left(a{i, 1}, a_{i, 2}, \ldots, a_{i, n}\right)$ is said generic $i^{t h}$ row vector while $\mathbf{a}^{\mathbf{j}}=\left(a_{1, j}, a_{2, j}, \ldots, a_{m, j}\right)$ is said generic $j^{\text {th }}$ column vector.

The set containing all the matrices of real numbers having $m$ rows and $n$ columns is indicated with $\mathbb{R}{m, n}$. Definition 2.7. A matrix is said null $\mathbf{O}$ if all its elements are zeros. Example 2.5. The null matrix of $\mathbb{R}{2,3}$ is
$$\mathbf{O}=\left(\begin{array}{lll} 0 & 0 & 0 \ 0 & 0 & 0 \end{array}\right)$$
Definition 2.8. Let $\mathbf{A} \in \mathbb{R}{m, n}$. The transpose matrix of $\mathbf{A}$ is a matrix $\mathbf{A}^{\mathbf{T}}$ whose elements are the same of $\mathbf{A}$ but $\forall i, j: a{j, i}=a_{i, j}^{T}$.
Example 2.6.
$\mathbf{A}=\left(\begin{array}{cccc}2 & 7 & 3.4 & \sqrt{2} \ 5 & 0 & 4 & 1\end{array}\right)$
$\mathbf{A}^{\mathbf{T}}=\left(\begin{array}{cc}2 & 5 \ 7 & 0 \ 3.4 & 4 \ \sqrt{2} & 1\end{array}\right)$
It can be easily proved that the transpose of the transpose of a matrix is the matrix itself: $\left(\mathbf{A}^{\mathbf{T}}\right)^{\mathbf{T}}$.
Definition 2.9. A matrix $\mathbf{A} \in \mathbb{R}_{n, n}$ is said n order square matrix.

## 数学代写|计算线性代数代写Computational Linear Algebra代考|A Preliminary Introduction to Algebraic Structures

• 交换性 (总和) : $x_{1}+x_{2}=x_{2}+x_{1}$
• 关联性 (总和) : $\left(x_{1}+x_{2}\right)+x_{3}=x_{1}+\left(x_{2}+x_{3}\right)$
• 中性元素 (总和) : $\exists$ 个个元素 $0 \in R$ 这样 $\forall x \in R: x+0=x$
• 逆元素 (总和) : $\forall x \in R: \exists(-x) \mid x+(-x)=0$
• 关联性 (产品) : $\left(x_{1} x_{2}\right) x_{3}=x_{1}\left(x_{2} x_{3}\right)$
• 分配性 $1: x_{1}\left(x_{2}+x_{3}\right)=x_{1} x_{2}+x_{1} x_{3}$
• 分配性 $2:\left(x_{2}+x_{3}\right) x_{1}=x_{2} x_{1}+x_{3} x_{1}$
• 中性元素 (产品) : $\exists 一$ 个元素 $1 \in R$ 这样 $\forall x \in R x 1=1 x=x$ 关于和的逆元也称为逆元。

## 数学代写|计算线性代数代写Computational Linear Algebra代考|Numeric Vectors

$$\mathbf{a}=(1,0,56.3, \sqrt{2})$$

$$\mathbf{a}=(1,0,3) \quad \mathbf{b}=(2,1,-2)$$

## 数学代写|计算线性代数代写Computational Linear Algebra代考|Basic Definitions About Matrices

$$\mathbf{O}=\left(\begin{array}{llllll} 0 & 0 & 0 & 0 & 0 \end{array}\right)$$

$\mathbf{A}=\left(\begin{array}{lllllll}2 & 7 & 3.4 & \sqrt{2} 5 & 0 & 4 & 1\end{array}\right)$
$\mathbf{A}^{\mathbf{T}}=\left(\begin{array}{lllll}2 & 5 & 7 & 0 & 3.4\end{array}\right.$

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

## 数学代写|计算线性代数代写Computational Linear Algebra代考|МАTH 1014

statistics-lab™ 为您的留学生涯保驾护航 在代写计算线性代数Computational Linear Algebra方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写计算线性代数Computational Linear Algebra代写方面经验极为丰富，各种代写计算线性代数Computational Linear Algebra相关的作业也就用不着说。

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 数学代写|计算线性代数代写Computational Linear Algebra代考|Axiomatic System

A concept is said to be primitive when it cannot be rigorously defined since its meaning is intrinsically clear. An axiom or postulate is a premise or a starting point for reasoning. Thus, an axiom is a statement which appears unequivocally true and that does not require any proof to be verified but cannot be, in any way, falsified.
Primitive concepts and axioms compose the axiomatic system. The axiomatic system is the ground onto the entire mathematics is built. On the basis of this ground, a definition is a statement that introduces a new concept/object by using previously known concepts (and thus primitive concepts are necessary for defining new ones). When the knowledge can be extended on the basis of previously established statements, this knowledge extension is named theorem. The previously known statements are the hypotheses while the extension is the thesis. A theorem can be expressed in the form: “if the hypotheses are verified then the thesis occurs”. In some cases, the theorem is symmetric, i.e. besides being true that “if the hypotheses are verified then the thesis occurs” it is also true that “if the thesis is verified then the hypotheses occur”. More exactly, if $A$ and $B$ are two statements, a theorem of this kind can be expressed as “if $A$ is verified than $B$ occurs and if $B$ is verified then A occurs”. In other words the two statements are equivalent since the truth of one of them automatically causes the truth of the other. In this book, theorems of this kind will be expressed in the form ” $A$ is verified if and only if $B$ is verified”.

The set of logical steps to deduce the thesis on the basis of the hypotheses is here referred as mathematical proof or simply proof. A large number of proof strategies exist. In this book, we will use only the direct proof, i.e. from the hypotheses we will logically arrive to the thesis or by contradiction (or reductio ad absurdum), i.e. the negated thesis will be new hypothesis that will lead to a paradox. A successful proof is indicated with the symbol $\quad \square$. It must be remarked that a theorem that states the equivalence of two facts requires two proofs. More specifically, a theorem of the kind ‘ $A$ is verified if and only if $B$ is verified” is essentially two theorems in one. Hence, the statements “if $A$ is verified than $B$ occurs” and “if $B$ is verified then $A$ occurs” require two separate proofs.

A theorem that enhances the knowledge by achieving a minor result that is then usable to prove a major result is called lemma while a minor result that uses a major theorem to be proved is called corollary. A proved result that is not as important as a theorem is called proposition.

## 数学代写|计算线性代数代写Computational Linear Algebra代考|Order and Equivalence

Definition 1.15. Order Relation. Let us consider a set $A$ and a relation $\mathscr{R}$ on $A$. This relation is said order relation and is indicated with $\preceq$ if the following properties are verified.

• reflexivity: $\forall x \in A: x \preceq x$
• transitivity: $\forall x, y, z \in A:$ if $x \preceq y$ and $y \preceq z$ then $x \preceq z$
• antisymmetry: $\forall x, y \in A:$ if $x \preceq y$ then $y \not x$
The set $A$, on which the order relation $\preceq$ is valid, is said totally ordered set.
Example 1.4. If we consider a group of people we can always sort them according theirs age. Hence the relation “to not be older than” (i.e. to be younger or to have the same age) with a set of people is a totally ordered set since every group of people can be fully sorted on the basis of their age.

From the definition above, the order relation can be interpreted as a predicate to be defined over the elements of a set. Although this is not wrong, we must recall that, rigorously, a relation is a set and an order relation is a set with some properties. In order to emphasise this fact, let us give again the definition of order relation by using a different notation.

Definition 1.16. Order Relation (Set Notation). Let us consider a set $A$ and the Cartesian product $A \times A=A^{2}$. Let $\mathscr{R}$ be a relation on $A$, that is $\mathscr{R} \subseteq A^{2}$. This relation is said order relation if the following properties are verified for the set $\mathscr{R}$.

## 数学代写|计算线性代数代写Computational Linear Algebra代考|Functions

Definition 1.23. Function. A relation is said to be a mapping or function when it relates to any element of a set a unique element of another. Let $A$ and $B$ be two sets, a mapping $f: A \rightarrow B$ is a relation $\mathscr{R} \subseteq A \times B$ such that $\forall x \in A, \forall y_{1}$ and $y_{2} \in B$ it follows that

• $\left(x, y_{1}\right) \in f$ and $\left(x, y_{2}\right) \in f \Rightarrow y_{1}=y_{2}$
• $\forall x \in A: \exists y \in B \mid(x, y) \in f$
where the symbol : $A \rightarrow B$ indicates that the mapping puts into relationship the set $A$ and the set $B$ and should be read “from $A$ to $B$ “, while $\Rightarrow$ indicates the material implications and should be read “it follows that”. In addition, the concept $(x, y) \in f$ can be also expressed as $y=f(x)$.
An alternative definition of function is the following.
Definition 1.24. Let $A$ and $B$ be two sets, a mapping $f: A \rightarrow B$ is a relation $\mathscr{R} \subseteq$ $A \times B \mid$ that satisfies the following property: $\forall x \in A$ it follows that $\exists ! y \in B$ such that $(x, y) \in \mathscr{R}$ (or, equivalently $y=f(x)$ ).

Example 1.12. The latter two definitions tell us that for example $(2,3)$ and $(2,6)$ cannot be both element of a function. We can express the same concept by stating that if $f(2)=3$ then it cannot happen that $f(2)=6$. In other words, if we fix $x=2$ then we can have only one $y$ value such that $y=f(x)$.

Thus, although functions are often interpreted as “laws” that connect two sets, mathematically, a function is any set (subset of a Cartesian product) for which the property in Definition $1.24$ is valid.

## 数学代写|计算线性代数代写Computational Linear Algebra代考|Order and Equivalence

-反身性: $\forall x \in A: x \preceq x$

• 传递性: $\forall x, y, z \in A$ :如果 $x \preceq y$ 和 $y \preceq z$ 然后 $x \preceq z$
• 反对称: $\forall x, y \in A$ :如果 $x \preceq y$ 然后 $y \not x$
套装 $A$ ，其上的顺序关系了是有效的，就是说全序集。
例 1.4。如果我们考虑一组人，我们总是可以根据他们的年龄对他们进行排序。因此，与一组人的“不比”（即 年轻或具有相同年龄) 的关系是一个完全有序的集合，因为每组人都可以根据他们的年龄进行完全排序。
根据上面的定义，顺序关系可以解释为要在集合的元素上定义的谓词。虽然这没有错，但我们必须记住，严格地 说，关系是一个集合，而顺序关系是一个具有某些属性的集合。为了强调这一事实，让我们使用不同的符号再次给 出顺序关系的定义。
定义 1.16。顺序关系 (集合符号)。让我们考虑一个集合 $A$ 和笛卡尔积 $A \times A=A^{2}$. 让 $\mathscr{R}$ 成为关系 $A$ ，那是 $\mathscr{R} \subseteq A^{2}$. 如果为集合验证了以下属性，则该关系称为顺序关系 $\mathscr{R}$.

## 数学代写|计算线性代数代写Computational Linear Algebra代考|Functions

• $\left(x, y_{1}\right) \in f$ 和 $\left(x, y_{2}\right) \in f \Rightarrow y_{1}=y_{2}$
• $\forall x \in A: \exists y \in B \mid(x, y) \in f$ 读“它遵循”。此外，概念 $(x, y) \in f$ 也可以表示为 $y=f(x)$.
函数的另一种定义如下。
定义 1.24。让 $A$ 和 $B$ 是两个集合，一个映射 $f: A \rightarrow B$ 是关系 $\mathscr{R} \subseteq A \times B \mid$ 满足以下性质: $\forall x \in A$ 它遵 循 $\exists ! y \in B$ 这样 $(x, y) \in \mathscr{R}$ (或者，等效地 $y=f(x)$ ).
示例 1.12。后两个定义告诉我们，例如 $(2,3)$ 和 $(2,6)$ 不能同时是函数的元素。我们可以表达同样的概念，如果 $f(2)=3$ 那么它不可能发生 $f(2)=6$. 换句话说，如果我们修复 $x=2$ 那么我们只能有一个 $y$ 值使得 $y=f(x)$.
因此，尽管函数通常被解释为连接两个集合的“定律”，但在数学上，函数是定义中的属性的任何集合（笛卡尔积的 子集） $1.24$ 已验证。

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

## 数学代写|计算线性代数代写Computational Linear Algebra代考|MATH4076

statistics-lab™ 为您的留学生涯保驾护航 在代写计算线性代数Computational Linear Algebra方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写计算线性代数Computational Linear Algebra代写方面经验极为丰富，各种代写计算线性代数Computational Linear Algebra相关的作业也就用不着说。

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 数学代写|计算线性代数代写Computational Linear Algebra代考|Real and Complex Inner Products

Definition 5.1 (Inner Product) An inner product in a complex vector space $\mathcal{V}$ is a function $\mathcal{V} \times \mathcal{V} \rightarrow \mathrm{C}$ satisfying for all $\boldsymbol{x}, \boldsymbol{y}, z \in \mathcal{V}$ and all $a, b \in \mathrm{C}$ the following conditions:

1. $\langle\boldsymbol{x}, \boldsymbol{x}\rangle \geq 0$ with equality if and only if $\boldsymbol{x}=\mathbf{0}$.
(positivity)
2. $\langle\boldsymbol{x}, \boldsymbol{y}\rangle=\overline{\langle\boldsymbol{y}, \boldsymbol{x}\rangle}$
(skew symmetry)
3. $(a \boldsymbol{x}+b \boldsymbol{y}, z\rangle=a\langle\boldsymbol{x}, z\rangle+b\langle\boldsymbol{y}, z\rangle$.
(linearity)
The pair $(\mathcal{V},\langle\cdot, \cdot))$ is called an inner product space.
Note the complex conjugate in 2 . Since
$$\langle\boldsymbol{x}, a \boldsymbol{y}+b z\rangle=\overline{\langle a \boldsymbol{y}+b z, \boldsymbol{x}\rangle}=\overline{a\langle\boldsymbol{y}, \boldsymbol{x}\rangle+b\langle z, \boldsymbol{x}\rangle}=\bar{a} \overline{\langle\boldsymbol{y}, \boldsymbol{x}\rangle}+\overline{b\langle z, \boldsymbol{x}\rangle}$$
we find
$$\langle\boldsymbol{x}, a \boldsymbol{y}+b z\rangle=\bar{a}\langle\boldsymbol{x}, \boldsymbol{y}\rangle+\bar{b}\langle\boldsymbol{x}, z\rangle, \quad\langle a \boldsymbol{x}, a \boldsymbol{y}\rangle=|a|^{2}\langle\boldsymbol{x}, \boldsymbol{y}\rangle$$
An inner product in a real vector space $\mathcal{V}$ is real valued function satisfying Properties $1,2,3$ in Definition $5.1$, where we can replace skew symmetry by symmetry
$$\langle\boldsymbol{x}, \boldsymbol{y}\rangle=\langle\boldsymbol{y}, \boldsymbol{x}\rangle \quad(\text { symmetry }) .$$
In the real case we have linearity in both variables since we can remove the complex conjugates in (5.1).
Recall that (cf. (1.10)) the standard inner product in $\mathbb{C}^{n}$ is given by
$$\langle\boldsymbol{x}, \boldsymbol{y}\rangle:=\boldsymbol{y}^{} x=\boldsymbol{x}^{T} \bar{y}=\sum_{j=1}^{n} x_{j} \overline{y_{j}}$$ Note the complex conjugate on $\boldsymbol{y}$. It is clearly an inner product in $\mathbb{C}^{n}$. The function $$|\cdot|: \mathcal{V} \rightarrow \mathbb{R}, \quad x \longmapsto|x|:=\sqrt{\langle x, x\rangle}$$ is called the inner product norm. The inner product norm for the standard inner product is the Euclidian norm $$|x|=|x|_{2}=\sqrt{x^{} x}$$

## 数学代写|计算线性代数代写Computational Linear Algebra代考|Orthogonality

Definition $5.2$ (Orthogonality) Two vectors $\boldsymbol{x}, \boldsymbol{y}$ in a real or complex inner product space are orthogonal or perpendicular, denoted as $\boldsymbol{x} \perp \boldsymbol{y}$, if $\langle\boldsymbol{x}, \boldsymbol{y}\rangle=0$. The vectors are orthonormal if in addition $|x|=|y|=1$.

From the definitions (5.6), (5.20) of angle $\theta$ between two nonzero vectors in $\mathbb{R}^{n}$ or $\mathbb{C}^{n}$ it follows that $\boldsymbol{x} \perp \boldsymbol{y}$ if and only if $\theta=\pi / 2$.
Theorem $5.3$ (Pythagoras) For a real or complex inner product space
$$|x+y|^{2}=|x|^{2}+|y|^{2}, \quad \text { if } \quad x \perp y .$$
Proof We set $a=1$ in (5.5) and use the orthogonality.
Definition $5.3$ (Orthogonal- and Orthonormal Bases) A set of nonzero vectors $\left{v_{1}, \ldots, v_{k}\right}$ in a subspace $\mathcal{S}$ of a real or complex inner product space is an orthogonal basis for $\mathcal{S}$ if it is a basis for $\mathcal{S}$ and $\left\langle v_{i}, v_{j}\right\rangle=0$ for $i \neq j$. It is an orthonormal basis for $\mathcal{S}$ if it is a basis for $\mathcal{S}$ and $\left\langle\boldsymbol{v}{i}, \boldsymbol{v}{j}\right\rangle=\delta_{i j}$ for all $i, j$.

A basis for a subspace of an inner product space can be turned into an orthogonalor orthonormal basis for the subspace by the following construction (Fig. 5.1).

## 数学代写|计算线性代数代写Computational Linear Algebra代考|Sum of Subspaces and Orthogonal Projections

Suppose $\mathcal{S}$ and $\mathcal{T}$ are subspaces of a real or complex vector space $\mathcal{V}$ endowed with an inner product $\langle\boldsymbol{x}, \boldsymbol{y}\rangle$. We define

• Sum: $\mathcal{S}+\mathcal{T}:={s+t: s \in \mathcal{S}$ and $t \in \mathcal{T}}$,
• direct sum $\mathcal{S} \oplus \mathcal{T}:$ a sum where $\mathcal{S} \cap \mathcal{T}={0}$,
• orthogonal sum $\mathcal{S} \oplus \mathcal{T}$ : a sum where $\langle s, t\rangle=0$ for all $s \in \mathcal{S}$ and $t \in \mathcal{T}$.
We note that
• $\mathcal{S}+\mathcal{T}$ is a vector space, a subspace of $\mathcal{V}$ which in this book will be $\mathbb{R}^{n}$ or $\mathcal{C}^{n}$ (cf. Example 1.2).
• Every $v \in \mathcal{S} \oplus \mathcal{T}$ can be decomposed uniquely in the form $v=s+t$, where $s \in \mathcal{S}$ and $t \in \mathcal{T}$. For if $v=s_{1}+t_{1}=s_{2}+t_{2}$ for $s_{1}, s_{2} \in \mathcal{S}$ and $t_{1}, t_{2} \in \mathcal{T}$, then $0=s_{1}-s_{2}+t_{1}-t_{2}$ or $s_{1}-s_{2}=t_{2}-t_{1}$. It follows that $s_{1}-s_{2}$ and $t_{2}-t_{1}$ belong to both $\mathcal{S}$ and $\mathcal{T}$ and hence to $\mathcal{S} \cap \mathcal{T}$. But then $s_{1}-s_{2}=t_{2}-t_{1}=0$ so $s_{1}=s_{2}$ and $t_{2}=t_{1}$.
By $(1.8)$ in the introduction chapter we have
$$\operatorname{dim}(\mathcal{S} \oplus \mathcal{T})=\operatorname{dim}(\mathcal{S})+\operatorname{dim}(\mathcal{T})$$
The subspaces $\mathcal{S}$ and $\mathcal{T}$ in a direct sum are called complementary subspaces.
• An orthogonal sum is a direct sum. For if $v \in \mathcal{S} \cap \mathcal{T}$ then $v$ is orthogonal to itself, $\langle v, v\rangle=0$, which implies that $v=0$. We often write $\mathcal{T}:=\mathcal{S}^{\perp}$.
• Suppose $v=s_{0}+t_{0} \in \mathcal{S} \oplus \mathcal{T}$, where $s_{0} \in \mathcal{S}$ and $t_{0} \in \mathcal{T}$. The vector $s_{0}$ is called the oblique projection of $v$ into $\mathcal{S}$ along $\mathcal{T}$. Similarly, The vector $t_{0}$ is called the oblique projection of $v$ into $\mathcal{T}$ along $\mathcal{S}$. If $\mathcal{S} \oplus \mathcal{T}$ is an orthogonal sum then $s_{0}$ is called the orthogonal projection of $v$ into $\mathcal{S}$. Similarly, $t_{0}$ is called the orthogonal projection of $v$ in $\mathcal{T}=\mathcal{S}^{\perp}$. The orthogonal projections are illustrated in Fig. 5.2.

## 数学代写|计算线性代数代写Computational Linear Algebra代考|Real and Complex Inner Products

1. ⟨X,X⟩≥0当且仅当X=0.
（积极性）
2. ⟨X,是⟩=⟨是,X⟩¯
（斜对称）
3. (一个X+b是,和⟩=一个⟨X,和⟩+b⟨是,和⟩.
（线性）
对(在,⟨⋅,⋅))称为内积空间。
注意 2 中的复共轭。自从
⟨X,一个是+b和⟩=⟨一个是+b和,X⟩¯=一个⟨是,X⟩+b⟨和,X⟩¯=一个¯⟨是,X⟩¯+b⟨和,X⟩¯
我们发现
⟨X,一个是+b和⟩=一个¯⟨X,是⟩+b¯⟨X,和⟩,⟨一个X,一个是⟩=|一个|2⟨X,是⟩
实向量空间中的内积在是满足属性的实值函数1,2,3在定义5.1, 我们可以用对称性代替斜对称
⟨X,是⟩=⟨是,X⟩( 对称 ).
在实际情况下，我们在两个变量中都有线性，因为我们可以去除（5.1）中的复共轭。
回想一下（参见（1.10））中的标准内积Cn是（谁）给的
⟨X,是⟩:=是X=X吨是¯=∑j=1nXj是j¯注意复共轭是. 它显然是一个内积Cn. 功能|⋅|:在→R,X⟼|X|:=⟨X,X⟩称为内积范数。标准内积的内积范数是欧几里得范数|X|=|X|2=XX

## 数学代写|计算线性代数代写Computational Linear Algebra代考|Orthogonality

|X+是|2=|X|2+|是|2, 如果 X⊥是.

## 数学代写|计算线性代数代写Computational Linear Algebra代考|Sum of Subspaces and Orthogonal Projections

• 和：小号+吨:=s+吨:s∈小号$一个nd$吨∈吨,
• 直接和小号⊕吨:总和小号∩吨=0,
• 正交和小号⊕吨: 总和⟨s,吨⟩=0对所有人s∈小号和吨∈吨.
我们注意到
• 小号+吨是一个向量空间，一个子空间在在本书中将是Rn或者Cn（参见示例 1.2）。
• 每一个在∈小号⊕吨可以唯一地分解为形式在=s+吨， 在哪里s∈小号和吨∈吨. 如果在=s1+吨1=s2+吨2为了s1,s2∈小号和吨1,吨2∈吨， 然后0=s1−s2+吨1−吨2或者s1−s2=吨2−吨1. 它遵循s1−s2和吨2−吨1属于两者小号和吨并因此小号∩吨. 但是之后s1−s2=吨2−吨1=0所以s1=s2和吨2=吨1.
经过(1.8)在介绍章节中，我们有
暗淡⁡(小号⊕吨)=暗淡⁡(小号)+暗淡⁡(吨)
子空间小号和吨在直接和中称为互补子空间。
• 正交和是直接和。如果在∈小号∩吨然后在与自身正交，⟨在,在⟩=0，这意味着在=0. 我们经常写吨:=小号⊥.
• 认为在=s0+吨0∈小号⊕吨， 在哪里s0∈小号和吨0∈吨. 向量s0称为斜投影在进入小号沿着吨. 同样，向量吨0称为斜投影在进入吨沿着小号. 如果小号⊕吨那么是正交和s0称为正交投影在进入小号. 相似地，吨0称为正交投影在在吨=小号⊥. 正交投影如图 5.2 所示。

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

## 数学代写|计算线性代数代写Computational Linear Algebra代考|LDL* Factorization and Positive Definite

statistics-lab™ 为您的留学生涯保驾护航 在代写计算线性代数Computational Linear Algebra方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写计算线性代数Computational Linear Algebra代写方面经验极为丰富，各种代写计算线性代数Computational Linear Algebra相关的作业也就用不着说。

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 数学代写|计算线性代数代写Computational Linear Algebra代考|The LDL* Factorization

There are special versions of the LU factorization for Hermitian and positive definite matrices which takes advantage of the special properties of such matrices. The most important ones are

1. the LDL* factorization which is an LDU factorization with $\boldsymbol{U}=\boldsymbol{L}^{*}$ and $\boldsymbol{D}$ a diagonal matrix with real diagonal elements
2. the $\mathrm{LL}^{}$ factorization which is an LU factorization with $\boldsymbol{U}=\boldsymbol{L}^{}$ and $l_{i i}>0$ all $i$.

A matrix $A$ having an LDL factorization must be Hermitian since $D$ is real so that $A^{}=\left(\boldsymbol{L} \boldsymbol{D} L^{}\right)^{}=\boldsymbol{L} D^{} L^{}=A$. The LL factorization is called a Cholesky factorization .

Example $4.1$ (LDL* of $2 \times 2$ Hermitian Matrix) Let $a, d \in \mathbb{R}$ and $b \in \mathrm{C}$. An LDL factorization of a $2 \times 2$ Hermitian matrix must satisfy the equations
$$\left[\begin{array}{ll} a & \bar{b} \ b & d \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \ l_{1} & 1 \end{array}\right]\left[\begin{array}{cc} d_{1} & 0 \ 0 & d_{2} \end{array}\right]\left[\begin{array}{ll} 1 & \overline{l_{1}} \ 0 & 1 \end{array}\right]=\left[\begin{array}{cc} d_{1} & d_{1} \overline{l_{1}} \ d_{1} l_{1} & d_{1}\left|l_{1}\right|^{2}+d_{2} \end{array}\right]$$ for the unknowns $l_{1}$ in $\boldsymbol{L}$ and $d_{1}, d_{2}$ in $\boldsymbol{D}$. They are determined from
$$d_{1}=a . \quad a l_{1}=b, \quad d_{2}=d-a\left|l_{1}\right|^{2}$$
There are essentially three cases

1. $a \neq 0$ : The matrix has a unique LDL* factorization. Note that $d_{1}$ and $d_{2}$ are real.
2. $a=b=0$ : The LDL* factorization exists, but it is not unique. Any value for $l_{1}$ can be used.
3. $a=0, b \neq 0$ : No LDL* factorization exists.
Lemma 3. I carries over to the Hermitian case.

## 数学代写|计算线性代数代写Computational Linear Algebra代考|Positive Definite and Semidefinite Matrices

Given $A \in \mathbb{C}^{n \times n}$. The function $f: \mathbb{C}^{n} \rightarrow \mathbb{R}$ given by
$$f(x)=x^{*} A x=\sum_{i=1}^{n} \sum_{j=1}^{n} a_{i j} \bar{x}{i} x{j}$$

is called a quadratic form. Note that $f$ is real valued if $A$ is Hermitian. Indeed, $\overline{f(x)}=\overline{x^{} A x}=\left(x^{} A x\right)^{}=x^{} A^{} x=f(x)$ Definition 4.1 (Positive Definite Matrix) We say that a matrix $A \in \mathbb{C}^{n \times n}$ is (i) positive definite if $A^{}=A$ and $x^{} A x>0$ for all nonzero $x \in \mathbb{C}^{n}$; (ii) positive semidefinite if $A^{}=A$ and $x^{*} A x \geq 0$ for all $x \in \mathbb{C}^{n}$;
(iii) negative (semi)definite if $-A$ is positive (semi)definite.
We observe that

1. The zero-matrix is positive semidefinite, while the unit matrix is positive definite.
2. The matrix $A$ is positive definite if and only if it is positive semidefinite and $x^{*} A x=0 \Longrightarrow x=0$.
3. A positive definite matrix $A$ is nonsingular. For if $A x=0$ then $x^{*} A x=0$ and this implies that $\boldsymbol{x}=\mathbf{0}$.
4. It follows from Lemma $4.6$ that a nonsingular positive semidefinite matrix is positive definite.
5. If $A$ is real then it is enough to show definiteness for real vectors only. Indeed, if $\boldsymbol{A} \in \mathbb{R}^{n \times n}, \boldsymbol{A}^{T}=\boldsymbol{A}$ and $\boldsymbol{x}^{T} \boldsymbol{A} \boldsymbol{x}>0$ for all nonzero $\boldsymbol{x} \in \mathbb{R}^{n}$ then $z^{} \boldsymbol{A} z>0$ for all nonzero $z \in \mathbb{C}^{n}$. For if $z=x+i y \neq 0$ with $x, y \in \mathbb{R}^{n}$ then \begin{aligned} z^{} \boldsymbol{A} z &=(\boldsymbol{x}-i \boldsymbol{y})^{T} \boldsymbol{A}(\boldsymbol{x}+i \boldsymbol{y})=\boldsymbol{x}^{T} \boldsymbol{A} \boldsymbol{x}-i \boldsymbol{y}^{T} \boldsymbol{A} \boldsymbol{x}+i \boldsymbol{x}^{T} \boldsymbol{A} \boldsymbol{y}-i^{2} \boldsymbol{y}^{T} \boldsymbol{A} \boldsymbol{y} \ &=\boldsymbol{x}^{T} \boldsymbol{A} \boldsymbol{x}+\boldsymbol{y}^{T} \boldsymbol{A} \boldsymbol{y} \end{aligned}
and this is positive since at least one of the real vectors $\boldsymbol{x}, \boldsymbol{y}$ is nonzero.

## 数学代写|计算线性代数代写Computational Linear Algebra代考|The Cholesky Factorization

Recall that a principal submatrix $\boldsymbol{B}=\boldsymbol{A}(\boldsymbol{r}, \boldsymbol{r}) \in \mathbb{C}^{k \times k}$ of a matrix $A \in \mathbb{C}^{n \times n}$ has elements $b_{i, j}=a_{r i, r j}$ for $i, j=1, \ldots, k$, where $1 \leq r_{1}<\cdots<r_{k} \leq n$. It is a leading principal submatrix, denoted $A_{[k]}$ if $\boldsymbol{r}=[1,2, \ldots, k]^{T}$. We have
$$\boldsymbol{A}(\boldsymbol{r}, \boldsymbol{r})=\boldsymbol{X}^{*} \boldsymbol{A} \boldsymbol{X}, \quad \boldsymbol{X}:=\left[\boldsymbol{e}{r{1}}, \ldots, \boldsymbol{e}{r{k}}\right] \in \mathbb{C}^{n \times k}$$
Lemma 4.4 (Submatrices) Any principal submatrix of a positive (semi)definite matrix is positive (semi)definite.

Proof Let $\boldsymbol{X}$ and $\boldsymbol{B}:=\boldsymbol{A}(\boldsymbol{r}, \boldsymbol{r})$ be given by (4.5). If $\boldsymbol{A}$ is positive semidefinite then $B$ is positive semidefinite since
$$\boldsymbol{y}^{} \boldsymbol{B} \boldsymbol{y}=\boldsymbol{y}^{} \boldsymbol{X}^{} \boldsymbol{A} \boldsymbol{X} \boldsymbol{y}=\boldsymbol{x}^{} \boldsymbol{A} \boldsymbol{x} \geq 0, \quad \boldsymbol{y} \in \mathbb{C}^{k}, \quad \boldsymbol{x}:=\boldsymbol{X} \boldsymbol{y}$$
Suppose $\boldsymbol{A}$ is positive definite and $\boldsymbol{y}^{*} \boldsymbol{B} \boldsymbol{y}=0$. By (4.6) we have $\boldsymbol{x}=\mathbf{0}$ and since $\boldsymbol{X}$ has linearly independent columns it follows that $\boldsymbol{y}=\mathbf{0}$. We conclude that $\boldsymbol{B}$ is positive definite.

## 数学代写|计算线性代数代写Computational Linear Algebra代考|The LDL* Factorization

Hermitian 和正定矩阵有特殊版本的 LU 分解，它利用了这些矩阵的特殊性质。最重要的是

1. LDL* 分解，它是一个 LDU 分解在=大号∗和D具有实对角元素的对角矩阵
2. 这大号大号因式分解，这是一个 LU 因式分解在=大号和l一世一世>0全部一世.

[一个b¯ bd]=[10 l11][d10 0d2][1l1¯ 01]=[d1d1l1¯ d1l1d1|l1|2+d2]对于未知数l1在大号和d1,d2在D. 它们由

d1=一个.一个l1=b,d2=d−一个|l1|2

1. 一个≠0：矩阵具有唯一的 LDL* 分解。注意d1和d2是真实的。
2. 一个=b=0：存在 LDL* 分解，但它不是唯一的。任何价值l1可以使用。
3. 一个=0,b≠0：不存在 LDL* 分解。
引理 3. 我继续讨论 Hermitian 案例。

## 数学代写|计算线性代数代写Computational Linear Algebra代考|Positive Definite and Semidefinite Matrices

F(X)=X∗一个X=∑一世=1n∑j=1n一个一世jX¯一世Xj

(iii) 否定（半）定如果−一个是正（半）定的。

1. 零矩阵是半正定的，而单位矩阵是正定的。
2. 矩阵一个是正定的当且仅当它是半正定的并且X∗一个X=0⟹X=0.
3. 一个正定矩阵一个是非奇异的。如果一个X=0然后X∗一个X=0这意味着X=0.
4. 它遵循引理4.6非奇异半正定矩阵是正定矩阵。
5. 如果一个是真实的，那么仅显示实向量的确定性就足够了。确实，如果一个∈Rn×n,一个吨=一个和X吨一个X>0对于所有非零X∈Rn然后和一个和>0对于所有非零和∈Cn. 如果和=X+一世是≠0和X,是∈Rn然后和一个和=(X−一世是)吨一个(X+一世是)=X吨一个X−一世是吨一个X+一世X吨一个是−一世2是吨一个是 =X吨一个X+是吨一个是
这是肯定的，因为至少有一个实向量X,是是非零的。

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

## 数学代写|计算线性代数代写Computational Linear Algebra代考|MATHS 7104

statistics-lab™ 为您的留学生涯保驾护航 在代写计算线性代数Computational Linear Algebra方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写计算线性代数Computational Linear Algebra代写方面经验极为丰富，各种代写计算线性代数Computational Linear Algebra相关的作业也就用不着说。

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 数学代写|计算线性代数代写Computational Linear Algebra代考|Algorithms for Triangular Systems

A nonsingular triangular linear system $\boldsymbol{A x}=\boldsymbol{b}$ is easy to solve. By Lemma $2.5 \boldsymbol{A}$ has nonzero diagonal elements. Consider first the lower triangular case. For $n=3$ the system is
$$\left[\begin{array}{ccc} a_{11} & 0 & 0 \ a_{21} & a_{22} & 0 \ a_{31} & a_{32} & a_{33} \end{array}\right]\left[\begin{array}{l} x_{1} \ x_{2} \ x_{3} \end{array}\right]=\left[\begin{array}{l} b_{1} \ b_{2} \ b_{3} \end{array}\right]$$
From the first equation we find $x_{1}=b_{1} / a_{11}$. Solving the second equation for $x_{2}$ we obtain $x_{2}=\left(b_{2}-a_{21} x_{1}\right) / a_{22}$. Finally the third equation gives $x_{3}=\left(b_{3}-a_{31} x_{1}-\right.$ $\left.a_{32} x_{2}\right) / a_{33}$. This process is known as forward substitution. In general
$$x_{k}=\left(b_{k}-\sum_{j=1}^{k-1} a_{k, j} x_{j}\right) / a_{k k}, \quad k=1,2, \ldots, n .$$
When $A$ is a lower triangular band matrix the number of arithmetic operations necessary to find $\boldsymbol{x}$ can be reduced. Suppose $\boldsymbol{A}$ is a lower triangular $d$-banded, so that $a_{k, j}=0$ for $j \notin\left{l_{k}, l_{k}+1, \ldots, k\right.$ for $k=1,2, \ldots, n$, and where $l_{k}:=\max (1, k-d)$, see Fig. 3.2. For a lower triangular $d$-band matrix the calculation in (3.7) can be simplified as follows
$$x_{k}=\left(b_{k}-\sum_{j=l_{k}}^{k-1} a_{k, j} x_{j}\right) / a_{k k}, \quad k=1,2, \ldots, n .$$
Note that (3.8) reduces to $(3.7)$ if $d=n$. Letting $A\left(k, l_{k}:(k-1)\right) * x\left(l_{k}:(k-1)\right)$ denote the sum $\sum_{j=l_{k}}^{k-1} a_{k j} x_{j}$ we arrive at the following algorithm, where the initial ” $\mathrm{r} “$ in the name signals that this algorithm is row oriented. The algorithm takes a nonsingular lower triangular $d$-banded matrix $A \in \mathbb{C}^{n \times n}$, and $\boldsymbol{b} \in \mathbb{C}^{n}$, as input, and returns an $\boldsymbol{x} \in \mathbb{C}^{n}$ so that $\boldsymbol{A} \boldsymbol{x}=\boldsymbol{b}$. For each $k$ we take the inner product of a part of a row with the already computed unknowns.

## 数学代写|计算线性代数代写Computational Linear Algebra代考|Counting Operations

It is useful to have a number which indicates the amount of work an algorithm requires. In this book we measure this by estimating the total number of (complex) arithmetic operations. We count both additions, subtractions, multiplications and divisions, but not work on indices. As an example we show that the LU factorization of a full matrix of order $n$ using Gaussian elimination requires exactly
$$N_{L U}:=\frac{2}{3} n^{3}-\frac{1}{2} n^{2}-\frac{1}{6} n$$
operations. Let $M, D, A, S$ be the number of (complex) multiplications, divisions, additions, and subtractions. In (3.2) the multiplications and subtractions occur in the calculation of $a_{i j}^{k+1}=a_{i j}^{(k)}-l_{i k}^{(k)} a_{k j}^{(k)}$ which is carried out $(n-k)^{2}$ times. Moreover,

each calculation involves one subtraction and one multiplication. Thus we find $M+$ $S=2 \sum_{k=1}^{n-1}(n-k)^{2}=2 \sum_{m=1}^{n-1} m^{2}=\frac{2}{3} n(n-1)\left(n-\frac{1}{2}\right)$. For each $k$ there are $n-k$ divisions giving a sum of $\sum_{k=1}^{n-1}(n-k)=\frac{1}{2} n(n-1)$. Since there are no additions we obtain the total
$$M+D+A+S=\frac{2}{3} n(n-1)\left(n-\frac{1}{2}\right)+\frac{1}{2} n(n-1)=N_{L U}$$
given by $(3.9)$
We are only interested in $N_{L U}$ when $n$ is large and for such $n$ the term $\frac{2}{3} n^{3}$ dominates. We therefore regularly ignore lower order terms and use number of operations both for the exact count and for the highest order term. We also say more loosely that the number of operations is $O\left(n^{3}\right)$. We will use the number of operations counted in one of these ways as a measure of the complexity of an algorithm and say that the complexity of LU factorization of a full matrix is $O\left(n^{3}\right)$ or more precisely $\frac{2}{3} n^{3}$.

We will compare the number of arithmetic operations of many algorithms with the number of arithmetic operations of Gaussian elimination and define for $n \in \mathbb{N}$ the number $G_{n}$ as follows:

## 数学代写|计算线性代数代写Computational Linear Algebra代考|Pivoting

Interchanging two rows (and/or two columns) during Gaussian elimination is known as pivoting. The element which is moved to the diagonal position $(k, k)$ is called the pivot element or pivot for short, and the row containing the pivot is called the pivot row. Gaussian elimination with row pivoting can be described as follows.

1. Choose $r_{k} \geq k$ so that $a_{r_{k}, k}^{(k)} \neq 0$.
2. Interchange rows $r_{k}$ and $k$ of $A^{(k)}$.
3. Eliminate by computing $l_{i k}^{(k)}$ and $a_{i j}^{(k+1)}$ using (3.2).
To show that Gaussian elimination can always be carried to completion by using suitable row interchanges suppose by induction on $k$ that $A^{(k)}$ is nonsingular. Since $A^{(1)}=A$ this holds for $k=1$. By Lemma $2.4$ the lower right diagonal block in $A^{(k)}$ is nonsingular. But then at least one element in the first column of that block must be nonzero and it follows that $r_{k}$ exists so that $a_{r_{k}, k}^{(k)} \neq 0$. But then $A^{(k+1)}$ is nonsingular since it is computed from $A^{(k)}$ using row operations preserving the nonsingularity. We conclude that $A^{(k)}$ is nonsingular for $k=1, \ldots, n$.

## 数学代写|计算线性代数代写Computational Linear Algebra代考|Algorithms for Triangular Systems

[一个1100 一个21一个220 一个31一个32一个33][X1 X2 X3]=[b1 b2 b3]

Xķ=(bķ−∑j=1ķ−1一个ķ,jXj)/一个ķķ,ķ=1,2,…,n.

## 数学代写|计算线性代数代写Computational Linear Algebra代考|Counting Operations

ñ大号在:=23n3−12n2−16n

## 数学代写|计算线性代数代写Computational Linear Algebra代考|Pivoting

1. 选择rķ≥ķ以便一个rķ,ķ(ķ)≠0.
2. 交换行rķ和ķ的一个(ķ).
3. 通过计算消除l一世ķ(ķ)和一个一世j(ķ+1)使用（3.2）。
为了证明高斯消元总是可以通过使用适当的行交换来完成，假设通过归纳ķ那一个(ķ)是非奇异的。自从一个(1)=一个这适用于ķ=1. 引理2.4右下角块一个(ķ)是非奇异的。但是，该块的第一列中至少有一个元素必须是非零的，并且它遵循rķ存在使得一个rķ,ķ(ķ)≠0. 但是之后一个(ķ+1)是非奇异的，因为它是从一个(ķ)使用保留非奇异性的行操作。我们得出结论一个(ķ)是非奇异的ķ=1,…,n.

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

## 数学代写|计算线性代数代写Computational Linear Algebra代考|Block Multiplication and Triangular Matrices

statistics-lab™ 为您的留学生涯保驾护航 在代写计算线性代数Computational Linear Algebra方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写计算线性代数Computational Linear Algebra代写方面经验极为丰富，各种代写计算线性代数Computational Linear Algebra相关的作业也就用不着说。

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 数学代写|计算线性代数代写Computational Linear Algebra代考|Block Multiplication

A rectangular matrix $A$ can be partitioned into submatrices by drawing horizontal lines between selected rows and vertical lines between selected columns. For example, the matrix
$$A=\left[\begin{array}{lll} 1 & 2 & 3 \ 4 & 5 & 6 \ 7 & 8 & 9 \end{array}\right]$$ can be partitioned as
(i) $\left[\begin{array}{ll}A_{11} & A_{12} \ A_{21} & A_{22}\end{array}\right]=\left[\begin{array}{l|ll}1 & 2 & 3 \ \hline 4 & 5 & 6 \ 7 & 8 & 9\end{array}\right]$,
(ii) $\left[\boldsymbol{a}: 1, \boldsymbol{a}{: 2}, \boldsymbol{a}: 3\right]=\left[\begin{array}{c|c|c}1 & 2 & 3 \ 4 & 5 & 6 \ 7 & 8 & 9\end{array}\right]$, (iii) $\left[\begin{array}{c}a{1}^{T} \ a_{2:}^{T} \ a_{3:}^{T}\end{array}\right]=\left[\begin{array}{lll}\frac{1}{2} & 3 \ \frac{4}{}\end{array}\right.$
(iv) $\left[A_{11}, A_{12}\right]=\left[\begin{array}{c|ll}1 & 2 & 3 \ 4 & 5 & 6 \ 7 & 8 & 9\end{array}\right]$.
In ( $i$ ) the matrix $A$ is divided into four submatrices
$$A_{11}=[1], \quad A_{12}=[2,3], A_{21}=\left[\begin{array}{l} 4 \ 7 \end{array}\right] \text {, and } A_{22}=\left[\begin{array}{ll} 5 & 6 \ 8 & 9 \end{array}\right] \text {, }$$
while in (ii) and (iii) A has been partitioned into columns and rows, respectively. The submatrices in a partition are often referred to as blocks and a partitioned matrix is sometimes called a block matrix.

In the following we assume that $\boldsymbol{A} \in \mathbb{C}^{m \times p}$ and $\boldsymbol{B} \in \mathbb{C}^{p \times n}$. Here are some rules and observations for block multiplication.

1. If $\boldsymbol{B}=\left[\boldsymbol{b}{: 1}, \ldots, \boldsymbol{b}{: n}\right]$ is partitioned into columns then the partition of the product $\boldsymbol{A B}$ into columns is
$$A B=\left[A b_{: 1}, A b_{: 2}, \ldots, A b_{: n}\right]$$
In particular, if $\boldsymbol{I}$ is the identity matrix of order $p$ then
$$A=A I=A\left[e_{1}, e_{2}, \ldots, e_{p}\right]=\left[A e_{1}, A e_{2}, \ldots, A e_{p}\right]$$
and we see that column $j$ of $A$ can be written $A e_{j}$ for $j=1, \ldots, p$.
2. Similarly, if $\boldsymbol{A}$ is partitioned into rows then
$$\boldsymbol{A} \boldsymbol{B}=\left[\begin{array}{c} a_{1:}^{T} \ a_{2:}^{T} \ \vdots \ a_{m:}^{T} \end{array}\right] \boldsymbol{B}=\left[\begin{array}{c} a_{1:}^{T} \boldsymbol{B} \ a_{2:}^{T} \boldsymbol{B} \ \vdots \ a_{m:}^{T} \boldsymbol{B} \end{array}\right]$$
and taking $\boldsymbol{A}=\boldsymbol{I}$ it follows that row $i$ of $\boldsymbol{B}$ can be written $\boldsymbol{e}_{i}^{T} \boldsymbol{B}$ for $i=1, \ldots, m$.

## 数学代写|计算线性代数代写Computational Linear Algebra代考|Triangular Matrices

Lemma $2.4$ (Inverse of a Block Triangular Matrix) Suppose
$$A=\left[\begin{array}{cc} \boldsymbol{A}{11} & \boldsymbol{A}{12} \ \mathbf{0} & \boldsymbol{A}{22} \end{array}\right]$$ where $\boldsymbol{A}, \boldsymbol{A}{11}$ and $\boldsymbol{A}{22}$ are square matrices. Then $\boldsymbol{A}$ is nonsingular if and only if both $A{11}$ and $A_{22}$ are nonsingular. In that case
$$\boldsymbol{A}^{-1}=\left[\begin{array}{cc} \boldsymbol{A}{11}^{-1} & \boldsymbol{C} \ \mathbf{0} & \boldsymbol{A}{22}^{-1} \end{array}\right]$$
for some matrix $\boldsymbol{C}$.
Proof Suppose $A$ is nonsingular. We partition $B:=A^{-1}$ conformally with $A$ and have
$$B A=\left[\begin{array}{ll} B_{11} & B_{12} \ B_{21} & B_{22} \end{array}\right]\left[\begin{array}{cc} A_{11} & A_{12} \ \mathbf{0} & A_{22} \end{array}\right]=\left[\begin{array}{ll} I & 0 \ \mathbf{0} & I \end{array}\right]=I$$
Using block-multiplication we find
$$B_{11} A_{11}=I, B_{21} A_{11}=\mathbf{0}, B_{21} A_{12}+B_{22} A_{22}=I, \quad B_{11} A_{12}+B_{12} A_{22}=\mathbf{0}$$

The first equation implies that $A_{11}$ is nonsingular, this in turn implies that $\boldsymbol{B}{21}=$ $\mathbf{0} \boldsymbol{A}{11}^{-1}=\mathbf{0}$ in the second equation, and then the third equation simplifies to $\boldsymbol{B}{22} \boldsymbol{A}{22}=\boldsymbol{I}$. We conclude that also $\boldsymbol{A}{22}$ is nonsingular. From the fourth equation we find $$B{12}=C=-A_{11}^{-1} A_{12} A_{22}^{-1}$$
Conversely, if $\boldsymbol{A}{11}$ and $\boldsymbol{A}{22}$ are nonsingular then
$$\left[\begin{array}{cc} A_{11}^{-1} & -A_{11}^{-1} A_{12} A_{22}^{-1} \ \mathbf{0} & A_{22}^{-1} \end{array}\right]\left[\begin{array}{cc} A_{11} & A_{12} \ \mathbf{0} & A_{22} \end{array}\right]=\left[\begin{array}{ll} I & 0 \ 0 & I \end{array}\right]=I$$
and $A$ is nonsingular with the indicated inverse.
Consider now a triangular matrix.

## 数学代写|计算线性代数代写Computational Linear Algebra代考|3 by 3 Example

Gaussian elimination with row interchanges is the classical method for solving $n$ linear equations in $n$ unknowns. ${ }^{1}$ We first recall how it works on a $3 \times 3$ system.
Example $3.1$ (Gaussian Elimination on a $3 \times 3$ System) Consider a nonsingular system of three equations in three unknowns:
$a_{11}^{(1)} x_{1}+a_{12}^{(1)} x_{2}+a_{13}^{(1)} x_{3}=b_{1}^{(1)}, \quad \mathbf{I}$
$a_{21}^{(1)} x_{1}+a_{22}^{(1)} x_{2}+a_{23}^{(1)} x_{3}=b_{2}^{(1)}, \quad$ II
$a_{31}^{(1)} x_{1}+a_{32}^{(1)} x_{2}+a_{33}^{(1)} x_{3}=b_{3}^{(1)}$. III.

To solve this system by Gaussian elimination suppose $a_{11}^{(1)} \neq 0$. We subtract $l_{21}^{(1)}:=$ $a_{21}^{(1)} / a_{11}^{(1)}$ times equation I from equation II and $l_{31}^{(1)}:=a_{31}^{(1)} / a_{11}^{(1)}$ times equation I from equation III. The result is
$a_{11}^{(1)} x_{1}+a_{12}^{(1)} x_{2}+a_{13}^{(1)} x_{3}=b_{1}^{(1)}, \quad \mathrm{I}$
$a_{22}^{(2)} x_{2}+a_{23}^{(2)} x_{3}=b_{2}^{(2)}, \quad \mathbf{I I}^{\prime}$
$a_{32}^{(2)} x_{2}+a_{33}^{(2)} x_{3}=b_{3}^{(2)}, \quad \mathrm{III}^{\prime}$,
where $b_{i}^{(2)}=b_{i}^{(1)}-l_{i 1}^{(1)} b_{i}^{(1)}$ for $i=2,3$ and $a_{i j}^{(2)}=a_{i j}^{(1)}-l_{i, 1}^{(1)} a_{1 j}^{(1)}$ for $i, j=2,3$. If $a_{11}^{(1)}=0$ and $a_{21}^{(1)} \neq 0$ we first interchange equation I and equation II. If $a_{11}^{(1)}=$ $a_{21}^{(1)}=0$ we interchange equation I and III. Since the system is nonsingular the first column cannot be zero and an interchange is always possible.

If $a_{22}^{(2)} \neq 0$ we subtract $l_{32}^{(2)}:=a_{32}^{(2)} / a_{22}^{(2)}$ times equation $\mathrm{II}^{\prime}$ from equation $\mathrm{III}^{\prime}$ to obtain
$a_{11}^{(1)} x_{1}+a_{12}^{(1)} x_{2}+a_{13}^{(1)} x_{3}=b_{1}^{(1)}, \quad \mathbf{I}$
$a_{22}^{(2)} x_{2}+a_{23}^{(2)} x_{3}=b_{2}^{(2)}, \quad$ II $^{\prime}$
$a_{33}^{(3)} x_{3}=b_{3}^{(3)}, \quad \mathrm{III}^{\prime \prime}$,
where $a_{33}^{(3)}=a_{33}^{(2)}-l_{32}^{(2)} a_{23}^{(2)}$ and $b_{3}^{(3)}=b_{3}^{(2)}-l_{32}^{(2)} b_{2}^{(2)}$. If $a_{22}^{(2)}=0$ then $a_{32}^{(2)} \neq 0$ (cf. Sect. 3.4) and we first interchange equation $\mathrm{II}^{\prime}$ and equation $\mathrm{III}^{\prime}$. The reduced system is easy to solve since it is upper triangular. Starting from the bottom and moving upwards we find
\begin{aligned} &x_{3}=b_{3}^{(3)} / a_{33}^{(3)} \ &x_{2}=\left(b_{2}^{(2)}-a_{23}^{(2)} x_{3}\right) / a_{22}^{(2)} \ &x_{1}=\left(b_{1}^{(1)}-a_{12}^{(1)} x_{2}-a_{13}^{(1)} x_{3}\right) / a_{11}^{(1)} \end{aligned}

## 数学代写|计算线性代数代写Computational Linear Algebra代考|Block Multiplication

(i)\left[\begin{array}{ll}A_{11} & A_{12} \A_{21} & A_{22}\end{array}\right]=\left[\begin{array}{l| ll}1 & 2 & 3 \ \hline 4 & 5 & 6 \ 7 & 8 & 9\end{array}\right]\left[\begin{array}{ll}A_{11} & A_{12} \A_{21} & A_{22}\end{array}\right]=\left[\begin{array}{l| ll}1 & 2 & 3 \ \hline 4 & 5 & 6 \ 7 & 8 & 9\end{array}\right],
(ii)[一个:1,一个:2,一个:3]=[123 456 789], (iii)[一个1吨 一个2:吨 一个3:吨]=[123 4
(四)[一个11,一个12]=[123 456 789].

1. 如果乙=[b:1,…,b:n]被划分为列，然后是产品的划分一个乙成列是
一个乙=[一个b:1,一个b:2,…,一个b:n]
特别是，如果我是阶单位矩阵p然后
一个=一个我=一个[和1,和2,…,和p]=[一个和1,一个和2,…,一个和p]
我们看到那一栏j的一个可以写一个和j为了j=1,…,p.
2. 同样，如果一个然后被划分为行
一个乙=[一个1:吨 一个2:吨 ⋮ 一个米:吨]乙=[一个1:吨乙 一个2:吨乙 ⋮ 一个米:吨乙]
并采取一个=我它跟随那一行一世的乙可以写和一世吨乙为了一世=1,…,米.

## 数学代写|计算线性代数代写Computational Linear Algebra代考|Triangular Matrices

[一个11−1−一个11−1一个12一个22−1 0一个22−1][一个11一个12 0一个22]=[我0 0我]=我

## 数学代写|计算线性代数代写Computational Linear Algebra代考|3 by 3 Example

X3=b3(3)/一个33(3) X2=(b2(2)−一个23(2)X3)/一个22(2) X1=(b1(1)−一个12(1)X2−一个13(1)X3)/一个11(1)

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

## 数学代写|计算线性代数代写Computational Linear Algebra代考|MATHS 2104

statistics-lab™ 为您的留学生涯保驾护航 在代写计算线性代数Computational Linear Algebra方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写计算线性代数Computational Linear Algebra代写方面经验极为丰富，各种代写计算线性代数Computational Linear Algebra相关的作业也就用不着说。

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 数学代写|计算线性代数代写Computational Linear Algebra代考|A Two Point Boundary Value Problem

Consider the simple two point boundary value problem
$$-u^{\prime \prime}(x)=f(x), \quad x \in[0,1], \quad u(0)=0, u(1)=0$$
where $f$ is a given continuous function on $[0,1]$ and $u$ is an unknown function. This problem is also known as the one-dimensional (1D) Poisson problem. In principle it is easy to solve $(2.20)$ exactly. We just integrate $f$ twice and determine the two integration constants so that the homogeneous boundary conditions $u(0)=u(1)=$ 0 are satisfied. For example, if $f(x)=1$ then $u(x)=x(x-1) / 2$ is the solution.
Suppose $f$ cannot be integrated exactly. Problem $(2.20)$ can then be solved approximately using the finite difference method. We need a difference approximation to the second derivative. If $g$ is a function differentiable at $x$ then
$$g^{\prime}(x)=\lim _{h \rightarrow 0} \frac{g\left(x+\frac{h}{2}\right)-g\left(x-\frac{h}{2}\right)}{h}$$ and applying this to a function $u$ that is twice differentiable at $x$
\begin{aligned} u^{\prime \prime}(x) &=\lim {h \rightarrow 0} \frac{u^{\prime}\left(x+\frac{h}{2}\right)-u^{\prime}\left(x-\frac{h}{2}\right)}{h}=\lim {h \rightarrow 0} \frac{\frac{u(x+h)-u(x)}{h}-\frac{u(x)-u(x-h)}{h}}{h} \ &=\lim {h \rightarrow 0} \frac{u(x+h)-2 u(x)+u(x-h)}{h^{2}} \end{aligned} To define the points where this difference approximation is used we choose a positive integer $m$, let $h:=1 /(m+1)$ be the discretization parameter, and replace the interval $[0,1]$ by grid points $x{j}:=j h$ for $j=0,1, \ldots, m+1$. We then obtain approximations $v_{j}$ to the exact solution $u\left(x_{j}\right)$ for $j=1, \ldots, m$ by replacing the differential equation by the difference equation
$$\frac{-v_{j-1}+2 v_{j}-v_{j+1}}{h^{2}}=f(j h), \quad j=1, \ldots, m, \quad v_{0}=v_{m+1}=0$$
Moving the $h^{2}$ factor to the right hand side this can be written as an $m \times m$ linear system

The matrix $T$ is called the second derivative matrix and will occur frequently in this book. It is our second example of a tridiagonal matrix, $T=\operatorname{tridiag}\left(a_{i}, d_{i}, c_{i}\right) \in$ $\mathbb{R}^{m \times m}$, where in this case $a_{i}=c_{i}=-1$ and $d_{i}=2$ for all $i$.

## 数学代写|计算线性代数代写Computational Linear Algebra代考|Diagonal Dominance

We want to show that $(2.21)$ has a unique solution. Note that $T$ is not strictly diagonally dominant. However, $T$ is weakly diagonally dominant in accordance with the following definition.

Definition $2.3$ (Diagonal Dominance) The matrix $A=\left[a_{i j}\right] \in \mathbb{C}^{n \times n}$ is weakly diagonally dominant if
$$\left|a_{i i}\right| \geq \sum_{j \neq i}\left|a_{i j}\right|, i=1, \ldots, n$$

We showed in Theorem $2.2$ that a strictly diagonally dominant matrix is nonsingular. This is in general not true in the weakly diagonally dominant case. Consider the 3 matrices
$$\boldsymbol{A}{1}=\left[\begin{array}{lll} 1 & 1 & 0 \ 1 & 2 & 1 \ 0 & 1 & 1 \end{array}\right], \quad \boldsymbol{A}{2}=\left[\begin{array}{lll} 1 & 0 & 0 \ 0 & 0 & 0 \ 0 & 0 & 1 \end{array}\right], \quad \boldsymbol{A}{3}=\left[\begin{array}{rrr} 2 & -1 & 0 \ -1 & 2 & -1 \ 0 & -1 & 2 \end{array}\right]$$ They are all weakly diagonally dominant, but $A{1}$ and $A_{2}$ are singular, while $A_{3}$ is nonsingular. Indeed, for $A_{1}$ column two is the sum of columns one and three, $A_{2}$ has a zero row, and $\operatorname{det}\left(\boldsymbol{A}_{3}\right)=4 \neq 0$. It follows that for the nonsingularity and existence of an LU factorization of a weakly diagonally dominant matrix we need some additional conditions. Here are some sufficient conditions.

Theorem 2.4 (Weak Diagonal Dominance) Suppose $\boldsymbol{A}=\operatorname{tridiag}\left(a_{i}, d_{i}, c_{i}\right) \in$ $\mathbb{C}^{n \times n}$ is tridiagonal and weakly diagonally dominant. If in addition $\left|d_{1}\right|>\left|c_{1}\right|$ and $a_{i} \neq 0$ for $i=1, \ldots, n-2$, then $\boldsymbol{A}$ has a unique $L U$ factorization (2.15). If in addition $d_{n} \neq 0$, then $\boldsymbol{A}$ is nonsingular.

Proof The proof is similar to the proof of Theorem 2.2. The matrix $\boldsymbol{A}$ has an LU factorization if the $u_{k}$ ‘s in (2.16) are nonzero for $k=1, \ldots, n-1$. For this it is sufficient to show by induction that $\left|u_{k}\right|>\left|c_{k}\right|$ for $k=1, \ldots, n-1$. By assumption $\left|u_{1}\right|=\left|d_{1}\right|>\left|c_{1}\right|$. Suppose $\left|u_{k}\right|>\left|c_{k}\right|$ for some $1 \leq k \leq n-2$. Then $\left|c_{k}\right| /\left|u_{k}\right|<1$ and by (2.16) and since $a_{k} \neq 0$ $$\left|u_{k+1}\right|=\left|d_{k+1}-l_{k} c_{k}\right|=\left|d_{k+1}-\frac{a_{k} c_{k}}{u_{k}}\right| \geq\left|d_{k+1}\right|-\frac{\left|a_{k}\right|\left|c_{k}\right|}{\left|u_{k}\right|}>\left|d_{k+1}\right|-\left|a_{k}\right| .$$
This also holds for $k=n-1$ if $a_{n-1} \neq 0$. By (2.23) and weak diagonal dominance $\left|u_{k+1}\right|>\left|d_{k+1}\right|-\left|a_{k}\right| \geq\left|c_{k+1}\right|$ and it follows by induction that an LU factorization exists. It is unique since any LU factorization must satisfy (2.16). For the nonsingularity we need to show that $u_{n} \neq 0$. For then by Lemma $2.5$, both $\boldsymbol{L}$ and $\boldsymbol{U}$ are nonsingular, and this is equivalent to $\boldsymbol{A}=\boldsymbol{L} \boldsymbol{U}$ being nonsingular. If $a_{n-1} \neq 0$ then by (2.16) $\left|u_{n}\right|>\left|d_{n}\right|-\left|a_{n-1}\right| \geq 0$ by weak diagonal dominance, while if $a_{n-1}=0$ then again by (2.23) $\left|u_{n}\right| \geq\left|d_{n}\right|>0$.

## 数学代写|计算线性代数代写Computational Linear Algebra代考|The Buckling of a Beam

Consider a horizontal beam of length $L$ located between 0 and $L$ on the $x$-axis of the plane. We assume that the beam is fixed at $x=0$ and $x=L$ and that a force $F$ is applied at $(L, 0)$ in the direction towards the origin. This situation can be modeled by the boundary value problem
$$R y^{\prime \prime}(x)=-F y(x), \quad y(0)=y(L)=0,$$
where $y(x)$ is the vertical displacement of the beam at $x$, and $R$ is a constant defined by the rigidity of the beam. We can transform the problem to the unit interval $[0,1]$ by considering the function $u:[0,1] \rightarrow \mathbb{R}$ given by $u(t):=y(t L)$. Since $u^{\prime \prime}(t)=$ $L^{2} y^{\prime \prime}(t L)$, the problem $(2.24)$ then becomes
$$u^{\prime \prime}(t)=-K u(t), \quad u(0)=u(1)=0, \quad K:=\frac{F L^{2}}{R} .$$
Clearly $u=0$ is a solution, but we can have nonzero solutions corresponding to certain values of the $\mathrm{K}$ known as eigenvalues. The corresponding function $u$ is called an eigenfunction. If $F=0$ then $K=0$ and $u=0$ is the only solution, but if the force is increased it will reach a critical value where the beam will buckle and maybe break. This critical value corresponds to the smallest eigenvalue of (2.25). With $u(t)=\sin (\pi t)$ we find $u^{\prime \prime}(t)=-\pi^{2} u(t)$ and this $u$ is a solution if $K=\pi^{2}$. It can be shown that this is the smallest eigenvalue of (2.25) and solving for $F$ we find $F=\frac{\pi^{2} R}{L^{2}}$.

We can approximate this eigenvalue numerically. Choosing $m \in \mathbb{N}, h:=1 /(m+$
1) and using for the second derivative the approximation
$$u^{\prime \prime}(j h) \approx \frac{u((j+1) h)-2 u(j h)+u((j-1) h)}{h^{2}}, \quad j=1, \ldots, m,$$
(this is the same finite difference approximation as in Sect. 2.2) we obtain
$$\frac{-v_{j-1}+2 v_{j}-v_{j+1}}{h^{2}}=K v_{j}, \quad j=1, \ldots, m, h=\frac{1}{m+1}, \quad v_{0}=v_{m+1}=0$$

where $v_{j} \approx u(j h)$ for $j=0, \ldots, m+1$. If we define $\lambda:=h^{2} K$ then we obtain the equation
$$T v=\lambda v, \text { with } v=\left[v_{1}, \ldots, v_{m}\right]^{T}$$
and
The problem now is to determine the eigenvalues of $T$. Normally we would need a numerical method to determine the eigenvalues of a matrix, but for this simple problem the eigenvalues can be determined exactly. We show in the next subsection that the smallest eigenvalue of $(2.26)$ is given by $\lambda=4 \sin ^{2}(\pi h / 2)$. Since $\lambda=$ $h^{2} K=\frac{h^{2} F L^{2}}{R}$ we can solve for $F$ to obtain
$$F=\frac{4 \sin ^{2}(\pi h / 2) R}{h^{2} L^{2}}$$
For small $h$ this is a good approximation to the value $\frac{\pi^{2} R}{L^{2}}$ we computed above.

## 数学代写|计算线性代数代写Computational Linear Algebra代考|A Two Point Boundary Value Problem

−在′′(X)=F(X),X∈[0,1],在(0)=0,在(1)=0

G′(X)=林H→0G(X+H2)−G(X−H2)H并将其应用于函数在在X

−在j−1+2在j−在j+1H2=F(jH),j=1,…,米,在0=在米+1=0

## 数学代写|计算线性代数代写Computational Linear Algebra代考|Diagonal Dominance

|一个一世一世|≥∑j≠一世|一个一世j|,一世=1,…,n

|在ķ+1|=|dķ+1−lķCķ|=|dķ+1−一个ķCķ在ķ|≥|dķ+1|−|一个ķ||Cķ||在ķ|>|dķ+1|−|一个ķ|.

## 数学代写|计算线性代数代写Computational Linear Algebra代考|The Buckling of a Beam

R是′′(X)=−F是(X),是(0)=是(大号)=0,

1) 并使用二阶导数的近似值

（这与第 2.2 节中的有限差分近似相同）我们得到

−在j−1+2在j−在j+1H2=ķ在j,j=1,…,米,H=1米+1,在0=在米+1=0

F=4罪2⁡(圆周率H/2)RH2大号2

## 有限元方法代写

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## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

## 数学代写|计算线性代数代写Computational Linear Algebra代考|Diagonally Dominant Tridiagonal Matrices; Three Examples

statistics-lab™ 为您的留学生涯保驾护航 在代写计算线性代数Computational Linear Algebra方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写计算线性代数Computational Linear Algebra代写方面经验极为丰富，各种代写计算线性代数Computational Linear Algebra相关的作业也就用不着说。

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 数学代写|计算线性代数代写Computational Linear Algebra代考|Piecewise Linear and Cubic Spline Interpolation

To avoid oscillations like the one in Fig. $2.1$ piecewise linear interpolation can be used. An example is shown in Fig. 2.2. The interpolant $g$ approximates the original function quite well, and for some applications, like plotting, the linear interpolant using many points is what is used. Note that $g$ is a piecewise polynomial of the form
$$g(x):= \begin{cases}p_{1}(x), & \text { if } x_{1} \leq x<x_{2} \ p_{2}(x), & \text { if } x_{2} \leq x<x_{3} \ \vdots & \ p_{n-1}(x), & \text { if } x_{n-1} \leq x<x_{n} \ p_{n}(x), & \text { if } x_{n} \leq x \leq x_{n+1}\end{cases}$$

where each $p_{i}$ is a polynomial of degree $\leq 1$. In particular, $p_{1}$ is given in (2.3) and the other polynomials $p_{i}$ are given by similar expressions.

The piecewise linear interpolant is continuous, but the first derivative will usually have jumps at the interior sites. We can obtain a smoother approximation by letting $g$ be a piecewise polynomial of higher degree. With degree 3 (cubic) we obtain continuous derivatives of order $\leq 2\left(C^{2}\right)$. We consider here the following functions giving examples of $C^{2}$ cubic spline interpolants.

Definition 2.1 (The $D_{2}-$ Spline Problem) Given $n \in \mathbb{N}$, an interval $[a, b], y \in$ $\mathbb{R}^{n+1}$, knots (sites) $x_{1}, \ldots, x_{n+1}$ given by $(2.1)$ and numbers $\mu_{1}, \mu_{n+1}$. The problem is to find a function $g:[a, b] \rightarrow \mathbb{R}$ such that

• piecewise cubic polynomial: $g$ is of the form (2.4) with each $p_{i}$ a cubic polynomial,
• smoothness: $g \in C^{2}[a, b]$, i.e., derivatives of order $\leq 2$ are continuous on $\mathbb{R}$,
• interpolation: $g\left(x_{i}\right)=y_{i}, \quad i=1,2, \ldots, n+1$,
• $D_{2}$ boundary conditions: $g^{\prime \prime}(a)=\mu_{1}, \quad g^{\prime \prime}(b)=\mu_{n+1}$.
We call $g$ a $D_{2}$-spline. It is called an $N$-spline or natural spline if $\mu_{1}=\mu_{n+1}=0$.

## 数学代写|计算线性代数代写Computational Linear Algebra代考|Give Me a Moment

Existence and uniqueness of a solution of the $D_{2}$-spline problem hinges on the nonsingularity of a linear system of equations that we now derive. The unknowns are derivatives at the knots. Here we use second derivatives which are sometimes called moments. We start with the following lemma.

Lemma $2.1$ (Representing Each $p_{i}$ Using $(0,2)$ Interpolation) Given $a<b$, $h=(b-a) / n$ with $n \geq 2, x_{i}=a+(i-1) h$, and numbers $y_{i}, \mu_{i}$ for $i=1, \ldots, n+1$. For $i=1, \ldots, n$ there are unique cubic polynomials $p_{i}$ such that
$$p_{i}\left(x_{i}\right)=y_{i}, p_{i}\left(x_{i+1}\right)=y_{i+1}, \quad p_{i}^{\prime \prime}\left(x_{i}\right)=\mu_{i}, p_{i}^{\prime \prime}\left(x_{i+1}\right)=\mu_{i+1}$$
Moreover,
$$p_{i}(x)=c_{i, 1}+c_{i, 2}\left(x-x_{i}\right)+c_{i, 3}\left(x-x_{i}\right)^{2}+c_{i, 4}\left(x-x_{i}\right)^{3} \quad i=1, \ldots, n$$
where
$$c_{i 1}=y_{i}, c_{i 2}=\frac{y_{i+1}-y_{i}}{h}-\frac{h}{3} \mu_{i}-\frac{h}{6} \mu_{i+1}, c_{i, 3}=\frac{\mu_{i}}{2}, c_{i, 4}=\frac{\mu_{i+1}-\mu_{i}}{6 h} .$$ Proof Consider $p_{i}$ in the form (2.7) for some $1 \leq i \leq n$. Evoking (2.6) we find $p_{i}\left(x_{i}\right)=c_{i, 1}=y_{i}$. Since $p_{i}^{\prime \prime}(x)=2 c_{i, 3}+6 c_{i, 4}\left(x-x_{i}\right)$ we obtain $c_{i, 3}$ from $p_{i}^{\prime \prime}\left(x_{i}\right)=$ $2 c_{i, 3}=\mu_{i}$ (a moment), and then $c_{i, 4}$ from $p_{i}^{\prime \prime}\left(x_{i+1}\right)=\mu_{i}+6 h c_{i, 4}=\mu_{i+1}$. Finally we find $c_{i, 2}$ by solving $p_{i}\left(x_{i+1}\right)=y_{i}+c_{i, 2} h+\frac{\mu_{i}}{2} h^{2}+\frac{\mu_{i+1}-\mu_{i}}{6 h} h^{3}=y_{i+1}$. For $j=0,1,2,3$ the shifted powers $\left(x-x_{i}\right)^{j}$ constitute a basis for cubic polynomials and the formulas (2.8) are unique by construction. It follows that $p_{i}$ is unique.

## 数学代写|计算线性代数代写Computational Linear Algebra代考|LU Factorization of a Tridiagonal System

To find the $D^{2}$-spline $g$ we have to solve the triangular system (2.11). Consider solving a general tridiagonal linear system $A x=b$ where $A=\operatorname{tridiag}\left(a_{i}, d_{i}, c_{i}\right) \in$ $\mathbb{C}^{n \times n}$. Instead of using Gaussian elimination directly, we can construct two matrices $\boldsymbol{L}$ and $\boldsymbol{U}$ such that $\boldsymbol{A}=\boldsymbol{L} \boldsymbol{U}$. Since $\boldsymbol{A} \boldsymbol{x}=\boldsymbol{L} \boldsymbol{U} \boldsymbol{x}=\boldsymbol{b}$ we can find $\boldsymbol{x}$ by solving two systems $\boldsymbol{L z}=\boldsymbol{b}$ and $\boldsymbol{U} \boldsymbol{x}=z$. Moreover $\boldsymbol{L}$ and $\boldsymbol{U}$ are both triangular and bidiagonal, and if in addition they are nonsingular the two systems can be solved easily without using elimination.
In our case we write the product $\boldsymbol{A}=\boldsymbol{L U}$ in the form
$$\left[\begin{array}{lllll} d_{1} & c_{1} & & & \ a_{1} & d_{2} & c_{2} & & \ & \ddots & \ddots & \ddots & \ & & a_{n-2} & d_{n-1} & c_{n-1} \ & & & a_{n-1} & d_{n} \end{array}\right]=\left[\begin{array}{cccc} 1 & & \ l_{1} & 1 & \ & \ddots & \ddots & \ & & l_{n-1} & 1 \end{array}\right]\left[\begin{array}{cccc} u_{1} & c_{1} & & \ & \ddots & \ddots & \ & & u_{n-1} & c_{n-1} \ & & & u_{n} \end{array}\right]$$
To find $\boldsymbol{L}$ and $\boldsymbol{U}$ we first consider the case $n=3$. Equation (2.15) takes the form
$$\left[\begin{array}{lll} d_{1} & c_{1} & 0 \ a_{1} & d_{2} & c_{2} \ 0 & a_{2} & d_{3} \end{array}\right]=\left[\begin{array}{ccc} 1 & 0 & 0 \ l_{1} & 1 & 0 \ 0 & l_{2} & 1 \end{array}\right]\left[\begin{array}{ccc} u_{1} & c_{1} & 0 \ 0 & u_{2} & c_{2} \ 0 & 0 & u_{3} \end{array}\right]=\left[\begin{array}{ccc} u_{1} & c_{1} & 0 \ l_{1} u_{1} l_{1} c_{1}+u_{2} & c_{2} \ 0 & l_{2} u_{2} & l_{2} c_{2}+u_{3} \end{array}\right],$$
and the systems $\boldsymbol{L z}=\boldsymbol{b}$ and $\boldsymbol{U} \boldsymbol{x}=z$ can be written
$$\left[\begin{array}{lll} 1 & 0 & 0 \ l_{1} & 1 & 0 \ 0 & l_{2} & 1 \end{array}\right]\left[\begin{array}{l} z_{1} \ z_{2} \ z_{3} \end{array}\right]=\left[\begin{array}{l} b_{1} \ b_{2} \ b_{3} \end{array}\right],\left[\begin{array}{ccc} u_{1} & c_{1} & 0 \ 0 & u_{2} & c_{2} \ 0 & 0 & u_{3} \end{array}\right]\left[\begin{array}{l} x_{1} \ x_{2} \ x_{3} \end{array}\right]=\left[\begin{array}{l} z_{1} \ z_{2} \ z_{3} \end{array}\right]$$
Comparing elements we find
\begin{aligned} &u_{1}=d_{1}, \quad l_{1}=a_{1} / u_{1}, \quad u_{2}=d_{2}-l_{1} c_{1}, \quad l_{2}=a_{2} / u_{2}, \quad u_{3}=d_{3}-l_{2} c_{2}, \ &z_{1}=b_{1}, \quad z_{2}=b_{2}-l_{1} z_{1}, \quad z_{3}=b_{3}-l_{2} z_{2} \ &x_{3}=z_{3} / u_{3}, \quad x_{2}=\left(z_{2}-c_{2} x_{3}\right) / u_{2}, \quad x_{1}=\left(z_{1}-c_{1} x_{2}\right) / u_{1} \end{aligned}
In general, if
$$u_{1}=d_{1}, \quad l_{k}=a_{k} / u_{k}, \quad u_{k+1}=d_{k+1}-l_{k} c_{k}, \quad k=1,2, \ldots, n-1,$$
then $\boldsymbol{A}=\boldsymbol{L} \boldsymbol{U}$. If $u_{1}, u_{2}, \ldots, u_{n-1}$ are nonzero then (2.16) is well defined. If in addition $u_{n} \neq 0$ then we can solve $\boldsymbol{L} z=\boldsymbol{b}$ and $\boldsymbol{U} \boldsymbol{x}=z$ for $z$ and $\boldsymbol{x}$. We formulate this as two algorithms. In trifactor, vectors $l \in \mathbb{C}^{n-1}, \boldsymbol{u} \in \mathbb{C}^{n}$ are computed from $a, c \in \mathbb{C}^{n-1}, \boldsymbol{d} \in \mathbb{C}^{n}$. This implements the LU factorization of a tridiagonal matrix:

## 数学代写|计算线性代数代写Computational Linear Algebra代考|Piecewise Linear and Cubic Spline Interpolation

G(X):={p1(X), 如果 X1≤X<X2 p2(X), 如果 X2≤X<X3 ⋮ pn−1(X), 如果 Xn−1≤X<Xn pn(X), 如果 Xn≤X≤Xn+1

• 分段三次多项式：G是 (2.4) 的形式，每个p一世三次多项式，
• 光滑度：G∈C2[一个,b]，即阶导数≤2是连续的R,
• 插值：G(X一世)=是一世,一世=1,2,…,n+1,
• D2边界条件：G′′(一个)=μ1,G′′(b)=μn+1.
我们称之为G一个D2-样条。它被称为ñ-spline 或自然样条 ifμ1=μn+1=0.

## 数学代写|计算线性代数代写Computational Linear Algebra代考|Give Me a Moment

p一世(X一世)=是一世,p一世(X一世+1)=是一世+1,p一世′′(X一世)=μ一世,p一世′′(X一世+1)=μ一世+1

p一世(X)=C一世,1+C一世,2(X−X一世)+C一世,3(X−X一世)2+C一世,4(X−X一世)3一世=1,…,n

C一世1=是一世,C一世2=是一世+1−是一世H−H3μ一世−H6μ一世+1,C一世,3=μ一世2,C一世,4=μ一世+1−μ一世6H.证明考虑p一世以 (2.7) 的形式对某些1≤一世≤n. 唤起 (2.6) 我们发现p一世(X一世)=C一世,1=是一世. 自从p一世′′(X)=2C一世,3+6C一世,4(X−X一世)我们获得C一世,3从p一世′′(X一世)= 2C一世,3=μ一世（片刻），然后C一世,4从p一世′′(X一世+1)=μ一世+6HC一世,4=μ一世+1. 最后我们发现C一世,2通过解决p一世(X一世+1)=是一世+C一世,2H+μ一世2H2+μ一世+1−μ一世6HH3=是一世+1. 为了j=0,1,2,3转移的权力(X−X一世)j构成三次多项式的基础，并且公式（2.8）在构造上是唯一的。它遵循p一世是独特的。

## 数学代写|计算线性代数代写Computational Linear Algebra代考|LU Factorization of a Tridiagonal System

[d1C1 一个1d2C2 ⋱⋱⋱ 一个n−2dn−1Cn−1 一个n−1dn]=[1 l11 ⋱⋱ ln−11][在1C1 ⋱⋱ 在n−1Cn−1 在n]

[d1C10 一个1d2C2 0一个2d3]=[100 l110 0l21][在1C10 0在2C2 00在3]=[在1C10 l1在1l1C1+在2C2 0l2在2l2C2+在3],

[100 l110 0l21][和1 和2 和3]=[b1 b2 b3],[在1C10 0在2C2 00在3][X1 X2 X3]=[和1 和2 和3]

## 有限元方法代写

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## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。