### 统计代写|数值分析和优化代写numerical analysis and optimazation代考|Conjugate Gradients

statistics-lab™ 为您的留学生涯保驾护航 在代写数值分析和优化numerical analysis and optimazation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。我们的专家在代写数值分析和优化numerical analysis and optimazation方面经验极为丰富，各种代写数值分析和优化numerical analysis and optimazation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

## 统计代写|数值分析和优化代写numerical analysis and optimazation代考|Conjugate Gradients

We have already seen how the positive definiteness of $A$ can be used to define a norm. This can be extended to have a concept similar to orthogonality.
Definition 2.8. The vectors $\mathbf{u}$ and $\mathbf{v}$ are conjugate with respect to the positive definite matrix $A$, if they are nonzero and satisfy $\mathbf{u}^{T} A \mathbf{v}=0$.

Conjugacy plays such an important role, because through it search directions can be constructed such that the new gradient $\mathbf{g}^{(k+1)}$ is not just orthogonal to the current search direction $\mathbf{d}^{(k)}$ but to all previous search directions. This avoids revisiting search directions as in Figure 2.1.

Specifically, the first search direction is chosen as $\mathbf{d}^{(0)}=-\mathbf{g}^{(0)}$. The following search directions are then constructed by
$$\mathbf{d}^{(k)}=-\mathbf{g}^{(k)}+\beta^{(k)} \mathbf{d}^{(k-1)}, \quad k=1,2, \ldots,$$
where $\beta^{(k)}$ is chosen such that the conjugacy condition $\mathbf{d}^{(k)^{T}} A \mathbf{d}^{(k-1)}=0$ is satisfied. This yields
$$\beta^{(k)}=\frac{\mathbf{g}^{(k)^{T}} A \mathbf{d}^{(k-1)}}{\mathbf{d}^{(k-1)^{T}} A \mathbf{d}^{(k-1)}}, \quad k=1,2, \ldots$$
This gives the conjugate gradient method.
The values of $\mathbf{x}^{(k+1)}$ and $\omega^{(k)}$ are calculated as before in (2.7) and $(2.9)$.
These search directions satisfy the descent condition. From Equation (2.8) we have seen that the descent condition is equivalent to $\mathbf{d}^{(k)^{T}} \mathbf{g}^{(k)}<0$. Using the formula for $\mathbf{d}^{(k)}$ above and the fact that the new gradient is orthogonal to the previous search direction, i.e., $\mathbf{d}^{(k-1)^{T}} \mathbf{g}^{(k)}=0$, we see that
$$\mathbf{d}^{(k)^{T}} \mathbf{g}^{(k)}=-\left|\mathbf{g}^{(k)}\right|^{2}<0$$

## 统计代写|数值分析和优化代写numerical analysis and optimazation代考|Krylov Subspaces and Pre-Conditioning

Definition 2.9. Let $A$ be an $n \times n$ matrix, $\mathbf{b} \in \mathbb{R}^{n}$ a non-zero vector, then for a number $m$ the space spanned by $A^{j} \mathbf{b}, j=0, \ldots, m-1$ is the $m^{\text {th }}$ Krylov subspace of $\mathbb{R}^{n}$ and is denoted by $K_{m}(A, \mathbf{b})$.

In our analysis of the conjugate gradient method we saw that in the $k^{\text {th }}$ iteration the space spanned by the search directions $\mathbf{d}^{(j)}$ and the space spanned by the gradients $\mathbf{g}^{(j)}, j=0, \ldots, k$, are the same.

Lemma 2.3. The space spanned by $\mathbf{g}^{(j)}\left(\right.$ or $\left.\mathbf{d}^{(j)}\right), j=0, \ldots, k$, is the same as the $k+1^{\text {th }}$ Krylov subspace.
Proof. For $k=0$ we have $\mathbf{d}^{(0)}=-\mathbf{g}^{(0)}=\mathbf{b} \in K_{1}(A, \mathbf{b})$.
We assume that the space spanned by $\mathbf{g}^{(j)}, j=0, \ldots, k$, is the same as the space spanned by $\mathbf{b}, A \mathbf{b}, \ldots, A^{k} \mathbf{b}$ and increase $k$ by one.

In the formula $\mathbf{g}^{(k+1)}=\mathbf{g}^{(k)}+\omega^{(k)} A \mathbf{d}^{(k)}$ both $\mathbf{g}^{(k)}$ and $\mathbf{d}^{(k)}$ can be expressed as linear combinations of $\mathbf{b}, A \mathbf{b}, \ldots, A^{k} \mathbf{b}$ by the inductive hypothesis. Thus $\mathbf{g}^{(k+1)}$ can be expressed as a linear combination of $\mathbf{b}, A \mathbf{b}, \ldots, A^{k+1} \mathbf{b}$. Equally, using $\mathbf{d}^{(k+1)}=-\mathbf{g}^{(k+1)}+\beta^{(k)} \mathbf{d}^{(k)}$ we show that $\mathbf{d}^{(k+1)}$ lies in the space spanned by b, $A \mathbf{b}, \ldots, A^{k+1} \mathbf{b}$, which is $K_{k+2}(A, \mathbf{b})$. This completes the proof.

In the following lemma we show some properties of the Krylov subspaces.
Lemma 2.4. Given $A$ and b. Then $K_{m}(A, \mathbf{b})$ is a subspace of $K_{m+1}(A, \mathbf{b})$ and there exists a positive integer $s \leq n$ such that for every $m \geq s K_{m}(A, \mathbf{b})=$ $K_{s}(A, \mathbf{b})$. Furthermore, if we express $\mathbf{b}$ as $\mathbf{b}=\sum_{i=1}^{t} c_{i} \mathbf{v}{i}$, where $\mathbf{v}{1}, \ldots, \mathbf{v}{t}$ are eigenvectors of $A$ corresponding to distinct eigenvalues and all coefficients $c{i}, i=1, \ldots, t$, are non-zero, then $s=t$.

Proof. Clearly, $K_{m}(A, \mathbf{b}) \subseteq K_{m+1}(A, \mathbf{b})$. The dimension of $K_{m}(A, \mathbf{b})$ is less than or equal to $m$, since it is spanned by $m$ vectors. It is also at most $n$ since

$K_{m}(A, \mathbf{b})$ is a subspace of $\mathbb{R}^{n}$. The first Krylov subspace has dimension 1 . We let $s$ be the greatest integer such that the dimension of $K_{s}(A, \mathbf{b})$ is $s$. Then the dimension of $K_{s+1}(A, \mathbf{b})$ cannot be $s+1$ by the choice of $s$. It has to be $s$, since $K_{s}(A, \mathbf{b}) \subseteq K_{s+1}(A, \mathbf{b})$. Hence the two spaces are the same. This means that $A^{s} \mathbf{b} \in K_{s}(A, \mathbf{b})$, i.e., $A^{s} \mathbf{b}=\sum_{j=0}^{s-1} a_{j} A^{j} \mathbf{b}$. But then
$$A^{s+r} \mathbf{b}=\sum_{j=0}^{s-1} a_{j} A^{j+r} \mathbf{b}$$
for any positive $r$. This means that also the spaces $K_{s+r+1}(A$, b $)$ and $K_{s+r}(A, \mathbf{b})$ are the same. Therefore, for every $m \geq s K_{m}(A, \mathbf{b})=K_{s}(A, \mathbf{b})$.
Suppose now that $\mathbf{b}=\sum_{i=1}^{t} c_{i} \mathbf{v}{i}$, where $\mathbf{v}{1}, \ldots, \mathbf{v}{t}$ are eigenvectors of $A$ corresponding to distinct eigenvalues $\lambda{i}$. Then for every $j=1, \ldots, s$
$$A^{j} \mathbf{b}=\sum_{i=1}^{t} c_{i} \lambda_{i}^{j} \mathbf{v}_{i}$$

## 统计代写|数值分析和优化代写numerical analysis and optimazation代考|Eigenvalues and Eigenvectors

So far we only considered eigenvalues when analyzing the properties of numerical methods. In the following sections we look at how to determine eigenvalues and eigenvectors. Let $A$ be a real $n \times n$ matrix. The eigenvalue equation is given by
$$A \mathbf{v}=\lambda \mathbf{v} \text {, }$$
where $\lambda$ is scalar. It may be complex if $A$ is not symmetric. There exists $\mathbf{v} \in \mathbb{R}^{n}$ satisfying the eigenvalue equation if and only if the determinant $\operatorname{det}(A-\lambda I)$ is zero. The function $p(\lambda)=\operatorname{det}(A-\lambda I), \lambda \in \mathbb{C}$, is a polynomial of degree $n$. However, calculating the eigenvalues by finding the roots of $p$ is generally unsuitable because of loss of accuracy due to rounding errors. In Chapter 1 , Fundamentals, we have seen how even finding the roots of a quadratic can be difficult due to loss of significance.

If the polynomial has some multiple roots and if $A$ is not symmetric, then $A$ might have less than $n$ linearly independent eigenvalues. However, there are always $n$ mutually orthogonal real eigenvectors when $A$ is symmetric. In the following we assume that $A$ has $n$ linearly independent eigenvectors $\mathbf{v}{j}$ for each eigenvalue $\lambda{j}, j=1, \ldots, n$. This can be achieved by perturbing $A$ slightly if necessary. The task is now to find $\mathbf{v}{j}$ and $\lambda{j}, j=1, \ldots, n$.

## 统计代写|数值分析和优化代写numerical analysis and optimazation代考|Conjugate Gradients

We have already seen how the positive definiteness of A can be used to define a norm. This can be extended to have a concept similar to orthogonality.
Definition 2.8. The vectors u and v are conjugate with respect to the positive definite matrix A, if they are nonzero and satisfy uTAv=0.

Conjugacy plays such an important role, because through it search directions can be constructed such that the new gradient g(k+1) is not just orthogonal to the current search direction d(k) but to all previous search directions. This avoids revisiting search directions as in Figure 2.1.

Specifically, the first search direction is chosen as d(0)=−g(0). The following search directions are then constructed by
d(k)=−g(k)+β(k)d(k−1),k=1,2,…,
where β(k) is chosen such that the conjugacy condition d(k)TAd(k−1)=0 is satisfied. This yields
This gives the conjugate gradient method.
The values of x(k+1) and ω(k) are calculated as before in (2.7) and (2.9).
These search directions satisfy the descent condition. From Equation (2.8) we have seen that the descent condition is equivalent to d(k)Tg(k)<0. Using the formula for d(k) above and the fact that the new gradient is orthogonal to the previous search direction, i.e., d(k−1)Tg(k)=0, we see that
d(k)Tg(k)=−|g(k)|2<0

## 统计代写|数值分析和优化代写numerical analysis and optimazation代考|Krylov Subspaces and Pre-Conditioning

Definition 2.9. Let A be an n×n matrix, b∈Rn a non-zero vector, then for a number m the space spanned by Ajb,j=0,…,m−1 is the mth  Krylov subspace of Rn and is denoted by Km(A,b).

In our analysis of the conjugate gradient method we saw that in the kth  iteration the space spanned by the search directions d(j) and the space spanned by the gradients g(j),j=0,…,k, are the same.

Lemma 2.3. The space spanned by g(j)( or d(j)),j=0,…,k, is the same as the k+1th  Krylov subspace.
Proof. For k=0 we have d(0)=−g(0)=b∈K1(A,b).
We assume that the space spanned by g(j),j=0,…,k, is the same as the space spanned by b,Ab,…,Akb and increase k by one.

In the formula g(k+1)=g(k)+ω(k)Ad(k) both g(k) and d(k) can be expressed as linear combinations of b,Ab,…,Akb by the inductive hypothesis. Thus g(k+1) can be expressed as a linear combination of b,Ab,…,Ak+1b. Equally, using d(k+1)=−g(k+1)+β(k)d(k) we show that d(k+1) lies in the space spanned by b, Ab,…,Ak+1b, which is Kk+2(A,b). This completes the proof.

In the following lemma we show some properties of the Krylov subspaces.
Lemma 2.4. Given A and b. Then Km(A,b) is a subspace of Km+1(A,b) and there exists a positive integer s≤n such that for every m≥sKm(A,b)= Ks(A,b). Furthermore, if we express b as b=∑i=1tcivi, where v1,…,vt are eigenvectors of A corresponding to distinct eigenvalues and all coefficients ci,i=1,…,t, are non-zero, then s=t.

Proof. Clearly, Km(A,b)⊆Km+1(A,b). The dimension of Km(A,b) is less than or equal to m, since it is spanned by m vectors. It is also at most n since

Km(A,b) is a subspace of Rn. The first Krylov subspace has dimension 1 . We let s be the greatest integer such that the dimension of Ks(A,b) is s. Then the dimension of Ks+1(A,b) cannot be s+1 by the choice of s. It has to be s, since Ks(A,b)⊆Ks+1(A,b). Hence the two spaces are the same. This means that Asb∈Ks(A,b), i.e., Asb=∑j=0s−1ajAjb. But then
As+rb=∑j=0s−1ajAj+rb
for any positive r. This means that also the spaces Ks+r+1(A, b ) and Ks+r(A,b) are the same. Therefore, for every m≥sKm(A,b)=Ks(A,b).
Suppose now that b=∑i=1tcivi, where v1,…,vt are eigenvectors of A corresponding to distinct eigenvalues λi. Then for every j=1,…,s
Ajb=∑i=1tciλijvi

## 统计代写|数值分析和优化代写numerical analysis and optimazation代考|Eigenvalues and Eigenvectors

So far we only considered eigenvalues when analyzing the properties of numerical methods. In the following sections we look at how to determine eigenvalues and eigenvectors. Let A be a real n×n matrix. The eigenvalue equation is given by
Av=λv,
where λ is scalar. It may be complex if A is not symmetric. There exists v∈Rn satisfying the eigenvalue equation if and only if the determinant det⁡(A−λI) is zero. The function p(λ)=det⁡(A−λI),λ∈C, is a polynomial of degree n. However, calculating the eigenvalues by finding the roots of p is generally unsuitable because of loss of accuracy due to rounding errors. In Chapter 1 , Fundamentals, we have seen how even finding the roots of a quadratic can be difficult due to loss of significance.

If the polynomial has some multiple roots and if A is not symmetric, then A might have less than n linearly independent eigenvalues. However, there are always n mutually orthogonal real eigenvectors when A is symmetric. In the following we assume that A has n linearly independent eigenvectors vj for each eigenvalue λj,j=1,…,n. This can be achieved by perturbing A slightly if necessary. The task is now to find vj and λj,j=1,…,n.

## 广义线性模型代考

statistics-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。