数学代写|高等线性代数代写Advanced Linear Algebra代考|Math4571
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- Statistical Inference 统计推断
- Statistical Computing 统计计算
- Advanced Probability Theory 高等概率论
- Advanced Mathematical Statistics 高等数理统计学
- (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
$$
(c p)(X)=\sum_{j=0}^n\left(c p_j\right) X^j
$$

高等线性代数代考
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
给定一个向量空间$V$在一个字段$\mathbb{F}$和$W \subseteq V$上。当$\mathbf{w}, \mathbf{y} \in W$时,由于$W$是$V$的一个子集,我们也有$\mathbf{w}, \mathbf{y} \in V$,因此$\mathbf{w}+\mathbf{y}$是定义良好的。此外,当$c \in \mathbb{F}$和$\mathbf{w} \in W \subseteq V$时,那么$c \mathbf{w}$是定义良好的。因此,我们可以考虑这样一个问题:带有$V$上定义的操作的$W$本身是否是一个向量空间。如果是,我们称$W$为$V$的子空间 命题2.3.1给定向量空间 $V$ 越过田野 $\mathbb{F}$,以及 $W \subseteq V$,那么 $W$ 是的子空间 $V$ 当且仅当
(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 $c \mathbf{w} \in W$.
证明。如果 $W$ 是一个向量空间,则(i), (ii)和(iii)明显满足。
对于相反的,我们需要检查当 $W$ 满足(i) (ii)和(iii),它满足向量空间定义中的所有十个公理。显然,上述性质(i)、(ii)和(iii)涉及向量空间定义中的公理1、4和6。公理5由(iii)与引理2.1.1(ii)结合而来。其他的性质(结合性,交换性,分配性,单位乘法)满足,因为它们适用于的所有元素 $V$,因此也为的元素 $W$在命题2.3.1中,可以用
代替(i)
(i) $’ W \neq \emptyset$ .
显然,如果(i)成立,那么(i)’成立。对于另一个方向,注意如果$\mathbf{w} \in W$(这样的$\mathbf{w}$的存在由(i)’保证)那么由(iii)和引理2.1.1(i),我们得到$\mathbf{0}=0 \mathbf{w} \in W$。因此(i)’和(iii)一起隐含(i).
给定两个子空间 $U$ 和 $W$ 一个向量空间的 $V$,我们介绍 $U+W:={\mathbf{v} \in V:$ 存在 $\mathbf{u} \in U$ 和 $\mathbf{w} \in W$ 所以 $\mathbf{v}=\mathbf{u}+\mathbf{w}}$ $U \cap W:={\mathbf{v} \in V: \mathbf{v} \in U$ 和 $\mathbf{v} \in W}$
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
我们让 $\mathbb{F}[X]$ 是所有多项式的集合 $X$ 加上系数 $\mathbb{F}$。因此一个典型的元素 $\mathbb{F}[X]$ 是否有
$$
p(X)=\sum_{j=0}^n p_j X^j=p_0 X^0+p_1 X+p_2 X^2+\cdots+p_n X^n,
$$
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? $X^0 X^i=X^i$ )。
当我们有两个多项式 $p(X)=\sum_{j=0}^n p_j X^j$ 和 $q(X)=\sum_{j=0}^m q_j X^j$,它往往是方便的 $m=n$。我们通过引入系数为零的附加项来做到这一点。例如,如果我们想查看
$$
p(X)=1+X \text { and } q(X)=1+2 X^2-X^5
$$
因为有相同数量的项,我们可以把它们看作
$$
p(X)=1+X+0 X^2+0 X^3+0 X^4+0 X^5, q(X)=1+0 X+2 X^2+0 X^3+0 X^4-X^5 .
$$
$X$ 真的是 $1 X$,以及 $-X^5$ 是 $(-1) X^5$
两个多项式 $p(X)=\sum_{j=0}^n p_j X^j$ 和 $q(X)=\sum_{j=0}^n q_j X^j$ 当它们的系数都相等时正好相等: $p_j=q_j, j=0, \ldots, n$.
两个多项式$p(X)=\sum_{j=0}^n p_j X^j$和$q(X)=\sum_{j=0}^n q_j X^j$的和由
$$
(p+q)(X)=\sum_{j=0}^n\left(p_j+q_j\right) X^j
$$
给出,当给出$c \in \mathbb{F}$和$p(X)=\sum_{j=0}^n p_j X^j$时,我们通过
$$
(c p)(X)=\sum_{j=0}^n\left(c p_j\right) X^j
$$ 定义多项式$(c p)(X)$