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• 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

## 计算机代写|机器学习代写machine learning代考|Systems of Linear Equations and Determinants

We now consider some applications of determinants to linear independence and to solving systems of linear equations. Although these results hold for matrices over certain rings, their proofs require more sophisticated methods. ‘Therefore, we assume again that $K$ is a field (usually, $K=\mathbb{R}$ or $K=\mathbb{C}$ ).

Let $A$ be an $n \times n$-matrix, $x$ a column vectors of variables, and $b$ another column vector, and let $A^1, \ldots, A^n$ denote the columns of $A$. Observe that the system of equations $A x=b$,
$$\left(\begin{array}{cccc} a_{11} & a_{12} & \ldots & a_{1 n} \ a_{21} & a_{22} & \ldots & a_{2 n} \ \vdots & \vdots & \ddots & \vdots \ a_{n 1} & a_{n 2} & \ldots & a_{n n} \end{array}\right)\left(\begin{array}{c} x_1 \ x_2 \ \vdots \ x_n \end{array}\right)=\left(\begin{array}{c} b_1 \ b_2 \ \vdots \ b_n \end{array}\right)$$
is equivalent to
$$x_1 A^1+\cdots+x_j A^j+\cdots+x_n A^n=b,$$
since the equation corresponding to the $i$-th row is in both cases
$$a_{i 1} x_1+\cdots+a_{i j} x_j+\cdots+a_{i n} x_n=b_i .$$
First we characterize linear independence of the column vectors of a matrix $A$ in terms of its determinant.

Proposition 7.11. Given an $n \times n$-matrix $A$ over a field $K$, the columns $A^1, \ldots, A^n$ of A are linearly dependent iff $\operatorname{det}(A)=\operatorname{det}\left(A^1, \ldots, A^n\right)=0$. Equivalently, A has rank $n$ iff $\operatorname{det}(A) \neq 0$

Proof. First assume that the columns $A^1, \ldots, A^n$ of $A$ are linearly dependent. Then there are $x_1, \ldots, x_n \in K$, such that
$$x_1 A^1+\cdots+x_j A^j+\cdots+x_n A^n=0,$$

where $x_j \neq 0$ for some $j$. If we compute
$$\operatorname{det}\left(A^1, \ldots, x_1 A^1+\cdots+x_j A^j+\cdots+x_n A^n, \ldots, A^n\right)=\operatorname{det}\left(A^1, \ldots, 0, \ldots, A^n\right)=0,$$
where 0 occurs in the $j$-th position. By multilinearity, all terms containing two identical columns $A^k$ for $k \neq j$ vanish, and we get
$$\operatorname{det}\left(A^1, \ldots, x_1 A^1+\cdots+x_j A^j+\cdots+x_n A^n, \ldots, A^n\right)=x_j \operatorname{det}\left(A^1, \ldots, A^n\right)=0 .$$
Since $x_j \neq 0$ and $K$ is a field, we must have $\operatorname{det}\left(A^1, \ldots, A^n\right)=0$.

## 计算机代写|机器学习代写machine learning代考|Determinant of a Linear Map

Given a vector space $E$ of finite dimension $n$, given a basis $\left(u_1, \ldots, u_n\right)$ of $E$, for every linear map $f: E \rightarrow E$, if $M(f)$ is the matrix of $f$ w.r.t. the basis $\left(u_1, \ldots, u_n\right)$, we can define $\operatorname{det}(f)=\operatorname{det}(M(f))$. If $\left(v_1, \ldots, v_n\right)$ is any other basis of $E$, and if $P$ is the change of basis matrix, by Corollary $4.5$, the matrix of $f$ with respect to the basis $\left(v_1, \ldots, v_n\right)$ is $P^{-1} M(f) P$. By Proposition $7.9$, we have
$\operatorname{det}\left(P^{-1} M(f) P\right)=\operatorname{det}\left(P^{-1}\right) \operatorname{det}(M(f)) \operatorname{det}(P)=\operatorname{det}\left(P^{-1}\right) \operatorname{det}(P) \operatorname{det}(M(f))=\operatorname{det}(M(f))$.
Thus, $\operatorname{det}(f)$ is indeed independent of the basis of $E$.
Definition 7.8. Given a vector space $E$ of finite dimension, for any linear map $f: E \rightarrow E$, we define the determinant $\operatorname{det}(f)$ of $f$ as the determinant $\operatorname{det}(M(f))$ of the matrix of $f$ in any basis (since, from the discussion just before this definition, this determinant does not depend on the basis).
”hen we have the following proposition.
Proposition 7.13. Given any vector space $E$ of finite dimension $n$, a linear map $f: E \rightarrow E$ is invertible iff $\operatorname{det}(f) \neq 0$.

Proof. The linear map $f: E \rightarrow E$ is invertible iff its matrix $M(f)$ in any basis is invertible (by Proposition $4.2$ ), iff $\operatorname{det}(M(f)) \neq 0$, by Proposition $7.10$.

Given a vector space of finite dimension $n$, it is easily seen that the set of bijective linear maps $f: E \rightarrow E$ such that $\operatorname{det}(f)=1$ is a group under composition. This group is a subgroup of the general linear group $\mathrm{GL}(E)$. It is called the special linear group (of $E$ ), and it is denoted by $\mathbf{S L}(E)$, or when $E=K^n$, by $\mathbf{S L}(n, K)$, or even by $\mathbf{S L}(n)$.

# 机器学习代考

## 计算机代写|机器学习代写machine learning代考|Systems of Linear Equations and Determinants

$$x_1 A^1+\cdots+x_j A^j+\cdots+x_n A^n=b,$$

$$a_{i 1} x_1+\cdots+a_{i j} x_j+\cdots+a_{i n} x_n=b_i .$$

$$x_1 A^1+\cdots+x_j A^j+\cdots+x_n A^n=0,$$

$$\operatorname{det}\left(A^1, \ldots, x_1 A^1+\cdots+x_j A^j+\cdots+x_n A^n, \ldots, A^n\right)=\operatorname{det}\left(A^1, \ldots, 0, \ldots, A^n\right)=0,$$

$$\operatorname{det}\left(A^1, \ldots, x_1 A^1+\cdots+x_j A^j+\cdots+x_n A^n, \ldots, A^n\right)=x_j \operatorname{det}\left(A^1, \ldots, A^n\right)=0 .$$

## 计算机代写|机器学习代写machine learning代考|Determinant of a Linear Map

$\operatorname{det}\left(P^{-1} M(f) P\right)=\operatorname{det}\left(P^{-1}\right) \operatorname{det}(M(f)) \operatorname{det}(P)=\operatorname{det}\left(P^{-1}\right) \operatorname{det}(P) \operatorname{det}(M(f))=\operatorname{det}(M(f))$.

“那么我们有以下命题。

## 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。

assignmentutor™您的专属作业导师
assignmentutor™您的专属作业导师