### 分类： 高等线性代数代写

• 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|高等线性代数代写Advanced Linear Algebra代考|Subspaces and more examples of vector spaces

Given a vector space $V$ over a field $\mathbb{F}$, and $W \subseteq V$. When $\mathbf{w}, \mathbf{y} \in W$, then as $W$ is a subset of $V$, we also have $\mathbf{w}, \mathbf{y} \in V$, and thus $\mathbf{w}+\mathbf{y}$ is well-defined. In addition, when $c \in \mathbb{F}$ and $\mathbf{w} \in W \subseteq V$, then $c \mathbf{w}$ is well-defined. Thus we can consider the question whether $W$ with the operations as defined on $V$, is itself a vector space. If so, we call $W$ a subspace of $V$.

Proposition 2.3.1 Given a vector space $V$ over a field $\mathbb{F}$, and $W \subseteq V$, then $W$ is a subspace of $V$ if and only if
(i) $\mathbf{0} \in W$.
(ii) $W$ is closed under addition: for all $\mathbf{w}, \mathbf{y} \in W$, we have $\mathbf{w}+\mathbf{y} \in W$.
(iii) $W$ is closed under scalar multiplication: for all $c \in \mathbb{F}$ and $\mathbf{w} \in W$, we have that $c \mathbf{w} \in W$.
Proof. If $W$ is a vector space, then (i), (ii) and (iii) are clearly satisfied.
For the converse, we need to check that when $W$ satisfies (i), (ii) and (iii), it satisfies all ten axioms in the definition of a vector space. Clearly properties (i), (ii) and (iii) above take care of axioms 1,4 and 6 in the definition of a vector space. Axiom 5 follows from (iii) in combination with Lemma 2.1.1(ii). The other properties (associativity, commutativity, distributivity, unit multiplication) are satisfied as they hold for all elements of $V$, and thus also for elements of $W$.
In Proposition 2.3.1 one may replace (i) by

(i) $’ W \neq \emptyset$.
Clearly, if (i) holds then (i)’ holds.
For the other direction, note that if $\mathbf{w} \in W$ (existence of such $\mathbf{w}$ is guaranteed by (i)’) then by (iii) and Lemma 2.1.1(i), we get that $\mathbf{0}=0 \mathbf{w} \in W$. Thus (i)’ and (iii) together imply (i).

Given two subspaces $U$ and $W$ of a vector space $V$, we introduce $U+W:={\mathbf{v} \in V:$ there exist $\mathbf{u} \in U$ and $\mathbf{w} \in W$ so that $\mathbf{v}=\mathbf{u}+\mathbf{w}}$ $U \cap W:={\mathbf{v} \in V: \mathbf{v} \in U$ and $\mathbf{v} \in W}$

## 数学代写|高等线性代数代写Advanced Linear Algebra代考|Vector spaces of polynomials

We let $\mathbb{F}[X]$ be the set of all polynomials in $X$ with coefficients in $\mathbb{F}$. Thus a typical element of $\mathbb{F}[X]$ has the form
$$p(X)=\sum_{j=0}^n p_j X^j=p_0 X^0+p_1 X+p_2 X^2+\cdots+p_n X^n,$$
where $n \in \mathbb{N}$ and $p_0, \ldots, p_n \in \mathbb{F}$. Here $X$ is merely a symbol and so are its powers $X^i$, with the understanding that $X^i X^j=X^{i+j}$. Often $X^0$ is omitted, as when we specify $X$ we will have that $X^0$ is a multiplicative neutral element (as for instance the equality $X^0 X^i=X^i$ suggests).
When we have two polynomials $p(X)=\sum_{j=0}^n p_j X^j$ and $q(X)=\sum_{j=0}^m q_j X^j$, it is often convenient to have $m=n$. We do this by introducing additional terms with a zero coefficient. For instance, if we want to view
$$p(X)=1+X \text { and } q(X)=1+2 X^2-X^5$$
as having the same number of terms we may view them as
$$p(X)=1+X+0 X^2+0 X^3+0 X^4+0 X^5, q(X)=1+0 X+2 X^2+0 X^3+0 X^4-X^5 .$$
Notice that the term $X$ is really $1 X$, and $-X^5$ is $(-1) X^5$.
Two polynomials $p(X)=\sum_{j=0}^n p_j X^j$ and $q(X)=\sum_{j=0}^n q_j X^j$ are equal exactly when all their coefficients are equal: $p_j=q_j, j=0, \ldots, n$.

The sum of two polynomials $p(X)=\sum_{j=0}^n p_j X^j$ and $q(X)=\sum_{j=0}^n q_j X^j$ is given by
$$(p+q)(X)=\sum_{j=0}^n\left(p_j+q_j\right) X^j$$
When $c \in \mathbb{F}$ and $p(X)=\sum_{j=0}^n p_j X^j$ are given, we define the polynomial $(c p)(X)$ via
$$(c p)(X)=\sum_{j=0}^n\left(c p_j\right) X^j$$

## 数学代写|高等线性代数代写高级线性代数代考|子空间和更多向量空间的例子

(i) $\mathbf{0} \in W$.
(ii) $W$ 在加法下是封闭的:for all $\mathbf{w}, \mathbf{y} \in W$，我们有 $\mathbf{w}+\mathbf{y} \in W$.
(iii) $W$ 在标量乘法下是封闭的:for all $c \in \mathbb{F}$ 和 $\mathbf{w} \in W$我们有这个 $c \mathbf{w} \in W$.

(i) $’ W \neq \emptyset$ .

## 数学代写|高等线性代数代写高级线性代数代考|多项式的向量空间

$$p(X)=\sum_{j=0}^n p_j X^j=p_0 X^0+p_1 X+p_2 X^2+\cdots+p_n X^n,$$
where $n \in \mathbb{N}$ 和 $p_0, \ldots, p_n \in \mathbb{F}$。这里 $X$ 只是一个象征，它的力量也是吗 $X^i$，但我们理解 $X^i X^j=X^{i+j}$。经常 $X^0$ 省略，如指定 $X$ 我们会得到的 $X^0$ 乘法的中性元素(例如等式? $X^0 X^i=X^i$ )。

$$p(X)=1+X \text { and } q(X)=1+2 X^2-X^5$$

$$p(X)=1+X+0 X^2+0 X^3+0 X^4+0 X^5, q(X)=1+0 X+2 X^2+0 X^3+0 X^4-X^5 .$$
$X$ 真的是 $1 X$，以及 $-X^5$ 是 $(-1) X^5$

$$(p+q)(X)=\sum_{j=0}^n\left(p_j+q_j\right) X^j$$

$$(c p)(X)=\sum_{j=0}^n\left(c p_j\right) X^j$$ 定义多项式$(c p)(X)$

## MATLAB代写

• 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|高等线性代数代写Advanced Linear Algebra代考|Matrix algebra over different fields

All the matrix algebra techniques that you learned in the first Linear Algebra course carry over to any field. Indeed, these algebra techniques were based on elementary algebraic operations, which work exactly the same in another field. In this section we illustrate these techniques by going through several examples with different fields. You will be reminded of matrix multiplication, row reduction, pivots, solving systems of linear equations, checking whether a vector is a linear combination of other vectors, finding a basis of a nullspace, column space, row space, eigenspace, computing determinants, finding inverses, Cramer’s rule, etc., but now we do these techniques in other fields.
One notable exception where $\mathbb{R}$ differs from the other fields we are considering, is that $\mathbb{R}$ is an ordered field (that is, $\geq$ defines an order relation on pairs of real numbers, that satisfies $x \geq y \Rightarrow x+z \geq z+y$ and $x, y \geq 0 \Rightarrow x y \geq 0)$. So anytime we want to use $\leq,<, \geq$ or $>$, we will have to make sure we are dealing with real numbers. We will do this when we talk about inner products and related concepts in Chapter 5.

Let $\mathbb{F}$ be a field, and the $n \times n$ matrix $A \in \mathbb{F}^{n \times n}$ and vector $\mathbf{b} \in \mathbb{F}^n$ be given. Let $\mathbf{a}_i$ denote the $i$ th column of $A$. Now we define

Thus $A_i(\mathbf{b})$ is the matrix obtained from $A$ by replacing its $i$ th column by $\mathbf{b}$.
We now have the following result.
Theorem 1.4.9 (Cramer’s rule) Let $A \in \mathbb{F}^{n \times n}$ be invertible. For any $\mathbf{b} \in \mathbb{F}^n$, the unique solution $\mathbf{x}=\left(x_i\right)_{i=1}^n$ to the equation $A \mathbf{x}=\mathbf{b}$ has entries given by
$$x_i=\operatorname{det} A_i(\mathbf{b})(\operatorname{det} A)^{-1}, i=1, \ldots, n .$$
Proof. We denote the columns of the $n \times n$ identity matrix $I$ by $\mathbf{e}_1, \ldots, \mathbf{e}_n$.
Let us compute
But then, using the multiplicativity of the determinant, we get $\operatorname{det} A \operatorname{det} I_i(\mathbf{x})=\operatorname{det} A_i(\mathbf{b})$. It is easy to see that $\operatorname{det} I_i(\mathbf{x})=x_i$, and (1.8) follows.

## 数学代写|高等线性代数代写Advanced Linear Algebra代考|Vector spaces of functions

The set of all functions from a set $X$ to a field $\mathbb{F}$ is denoted by $\mathbb{F}^X$. Thus $\mathbb{F}^X:={f: X \rightarrow \mathbb{F}: f$ is a function $}$.
When $f, g: X \rightarrow \mathbb{F}$ we can define the sum of $f$ and $g$ as the function
$$f+g: X \rightarrow \mathbb{F}, \quad(f+g)(x)=f(x)+g(x) .$$
Thus, by virtue that $\mathbb{F}$ has a well-defined addition, the set $\mathbb{F}^X$ now also has a well-defined addition. It is a fine point, but it is important to recognize that in the equation
$$(f+g)(x)=f(x)+g(x)$$
the first $+$ sign represents addition between functions, while the second $+$ sign represents addition in $\mathbb{F}$, so really the two $+\mathrm{s}$ are different. We still choose to use the same $+$ sign for both, although technically we could have made them different $\left(+{\mathbb{F}^x}\right.$ and $+{\mathbb{F}}$, say) and written
$$\left(f+{\mathbb{F}^X} g\right)(x)=f(x)+{\mathbb{F}} g(x) .$$
Next, it is also easy to define the scalar multiplication on $\mathbb{F}^X$ as follows. Given $c \in \mathbb{F}$ and $f: X \rightarrow \mathbb{F}$, we define the function $c f$ via
$$c f: X \rightarrow \mathbb{F}, \quad(c f)(x)=c(f(x)) .$$
Again, let us make the fine point that there are two different multiplications here, namely the multiplication of a scalar (i.e., an element of $\mathbb{F}$ ) with a function and the multiplication of two scalars. Again, if we want to highlight this difference, one would write this for instance as
$$\left(c \cdot F^X f\right)(x)=c \cdot{ }_F f(x) .$$
We now have the following claim.

## 高等线性代数代考

. . . . .

$$x_i=\operatorname{det} A_i(\mathbf{b})(\operatorname{det} A)^{-1}, i=1, \ldots, n .$$

$$f+g: X \rightarrow \mathbb{F}, \quad(f+g)(x)=f(x)+g(x) .$$因此，凭借 $\mathbb{F}$ 有一个定义良好的附加，集合 $\mathbb{F}^X$ 现在也有一个定义良好的附加项。这是一个很好的观点，但重要的是要认识到，在
$$(f+g)(x)=f(x)+g(x)$$

$$\left(f+{\mathbb{F}^X} g\right)(x)=f(x)+{\mathbb{F}} g(x) .$$接下来，定义上的标量乘法也很容易 $\mathbb{F}^X$ 如下所示。给定 $c \in \mathbb{F}$ 和 $f: X \rightarrow \mathbb{F}$，我们定义函数 $c f$ via
$$c f: X \rightarrow \mathbb{F}, \quad(c f)(x)=c(f(x)) .$$再次强调，这里有两种不同的乘法，即一个标量的乘法(即，的元素) $\mathbb{F}$ )，用一个函数和两个标量的乘法。同样，如果我们想要突出显示这个差异，可以将其写成
$$\left(c \cdot F^X f\right)(x)=c \cdot{ }_F f(x) .$$

## MATLAB代写

• 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|高等线性代数代写Advanced Linear Algebra代考|The field Z3

Let us consider the set $\mathbb{Z}_3={0,1,2}$, and use the following tables to define addition and multiplication:

So, in other words, $1+1=2,2+1=0,2 \cdot 2=1,0 \cdot 1=0$, etc. In fact, to take the sum of two elements we take the usual sum, and then take the remainder after division by 3 . For example, to compute $2+2$ we take the remainder of 4 after division by 3 , which is 1 . Similarly for multiplication.
What you notice in the table is that when you add 0 to any number, it does not change that number (namely, $0+0=0,0+1=1,1+0=1,0+2=2$, $2+0=2$ ). We say that 0 is the neutral element for addition. Analogously, 1 is the neutral element for multiplication, which means that when we multiply a number in this field by 1 , it does not change that number $(0 \cdot 1=0$, $1 \cdot 2=2$, etc.). Every field has these neutral elements, and they are typically denoted by 0 and 1 , although there is no rule that you have to denote them this way.

Another important observation is that in the core part of the addition table
$$\begin{array}{c|lll} • & & & \ \hline & 0 & 1 & 2 \ 1 & 2 & 0 \ & 2 & 0 & 1 \end{array}$$
the 0 appears exactly once in every row and column. What this means is that whatever $x$ we choose in $\mathbb{Z}_3={0,1,2}$, we can always find exactly one $y \in \mathbb{Z}_3$ so that
$$x+y=0 .$$
We are going to call $y$ the additive inverse of $x$, and we are going to write $y=-x$.

The complex numbers are defined as
$$\mathbb{C}={a+b i ; a, b \in \mathbb{R}},$$
with addition and multiplication defined by
$$\begin{gathered} (a+b i)+(c+d i):=(a+c)+(b+d) i \ (a+b i)(c+d i):=(a c-b d)+(a d+b c) i . \end{gathered}$$
Notice that with these rules, we have that $(0+1 i)(0+1 i)=-1+0 i$, or in shorthand $i^2=-1$. Indeed, this is how to remember the multiplication rule:
$$(a+b i)(c+d i)=a c+b d i^2+(a d+b c) i=a c-b d+(a d+b c) i,$$
where in the last step we used that $i^2=-1$. It may be obvious, but we should state it clearly anyway: two complex numbers $a+b i$ and $c+d i$, with $a, b, c, d \in \mathbb{R}$ are equal if and only if $a=c$ and $b=d$. A typical complex number may be denoted by $z$ or $w$. When
$$z=a+b i \text { with } a, b \in \mathbb{R},$$
we say that the real part of $z$ equals $a$ and the imaginary part of $z$ equals $b$. The notation for this is,
$$\operatorname{Re} z=a, \quad \operatorname{Im} z=b \text {. }$$
It is quite laborious, but in principle elementary, to prove that $\mathbb{C}$ satisfies all the field axioms. In fact, in doing so one needs to use that $\mathbb{R}$ satisfies the field axioms, as addition and multiplication in $\mathbb{C}$ are defined via addition and multiplication in $\mathbb{R}$. As always, it is important to realize what the neutral elements are:
$$0=0+0 i, \quad 1=1+0 i .$$
Another tricky part of this is the multiplicative inverse, for instance,
$$(1+i)^{-1},(2-3 i)^{-1}$$

## 高等线性代数代考

$$\begin{array}{c|lll} • & & & \ \hline & 0 & 1 & 2 \ 1 & 2 & 0 \ & 2 & 0 & 1 \end{array}$$
的核心部分，0在每一行和每一列中恰好出现一次。这意味着，无论我们在$\mathbb{Z}_3={0,1,2}$中选择什么$x$，我们总是能找到一个准确的$y \in \mathbb{Z}_3$，因此
$$x+y=0 .$$
我们将$y$称为$x$的加性逆，我们将写入$y=-x$。

$$\mathbb{C}={a+b i ; a, b \in \mathbb{R}},$$
，加法和乘法定义为
$$\begin{gathered} (a+b i)+(c+d i):=(a+c)+(b+d) i \ (a+b i)(c+d i):=(a c-b d)+(a d+b c) i . \end{gathered}$$

$$(a+b i)(c+d i)=a c+b d i^2+(a d+b c) i=a c-b d+(a d+b c) i,$$
，而在上一步中我们使用了$i^2=-1$。这可能是显而易见的，但我们无论如何都应该说清楚:两个复数$a+b i$和$c+d i$，其中$a, b, c, d \in \mathbb{R}$当且仅当$a=c$和$b=d$相等。一个典型的复数可以用$z$或$w$表示。当
$$z=a+b i \text { with } a, b \in \mathbb{R},$$

$$\operatorname{Re} z=a, \quad \operatorname{Im} z=b \text {. }$$

$$0=0+0 i, \quad 1=1+0 i .$$

$$(1+i)^{-1},(2-3 i)^{-1}$$

## 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。