# 线性代数网课代修|同调代数代写homological algebra代考|MAST90068

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• 数值分析
• 高等线性代数
• 矩阵论
• 优化理论
• 线性规划
• 逼近论

## 线性代数作业代写linear algebra代考|Modules

Many properties of vector spaces and of abelian groups are also enjoyed by modules. We assume that much of this section is familiar to most readers, and so our account is written to refresh one’s memory. All rings $R$ in this book are assumed to have an identity element 1 (or unit) (where $r 1=r=1 r$ for all $r \in R$ ). We do not insist that $1 \neq 0$; however, should $1=0$, then $R$ is the zero ring having only one element. If $f: R \rightarrow S$ is a ring homomorphism, we assume that $f(1)=1$; that is, $f(1)$ is the identity element of $S$.

We can view modules as a tool for studying rings. If $M$ is an abelian group, then we saw, in Exercise $1.13$ on page 35, that
$$\operatorname{End}{\mathbb{Z}}(M)={\text { homomorphisms } f: M \rightarrow M}$$ is a ring under pointwise addition $[f+g: m \mapsto f(m)+g(m)]$ and composition as multiplication. A representation of a ring $R$ is a ring homomorphism $\varphi: R \rightarrow \operatorname{End}{\mathbb{Z}}(M)$ for some abelian group $M$.

## 线性代数作业代写linear algebra代考|Tensor Products

One of the most compelling reasons to introduce tensor products comes from Algebraic Topology. The homology groups of a space are interesting (for example, computing the homology groups of spheres enables us to prove the Jordan Curve Theorem), and the homology groups of the cartesian product $X \times Y$ of two topological spaces are computed (by the Kunneth formula) in terms of the tensor product of the homology groups of the factors $X$ and $Y$.
Here is a second important use of tensor products. We saw, in Example 2.2, that if $k$ is a field, then every $k$-representation $\varphi: H \rightarrow \operatorname{Mat}_{n}(k)$ of a group $H$ to $n \times n$ matrices makes the vector space $k^{n}$ into a left $k H$-module;

conversely, every such module gives a representation of $H$. If $H$ is a subgroup of a group $G$, can we obtain a $k$-representation of $G$ from a $k$-representation of $H$; that is, can we construct a $k G$-module from a $k H$-module? Now $k H$ is a subring of $k G$; can we “adjoin more scalars” to form a $k G$-module from the $k H$-module? Tensor products will give a very simple construction, induced modules, which does exactly this.

More generally, if $S$ is a subring of a ring $R$ and $M$ is a left $S$-module, can we adjoin more scalars to form a left $R$-module $M^{\prime}$ that contains $M$ ? If a left $S$-module $M$ is generated by a set $X$ (so that each $m \in M$ has an expression of the form $m=\sum_{i} s_{i} x_{i}$ for $s_{i} \in S$ and $x_{i} \in X$ ), can we define a left $R$-module $M^{\prime}$ containing $M$ as the set of all expressions of the form $\sum_{i} r_{i} x_{i}$ for $r_{i} \in R$ ? Recall that if $V$ is a vector space over a field $k$ and $q v=0$ in $V$, where $q \in k$ and $v \in V$, then either $q=0$ or $v=0$. Now suppose that $M=\langle a\rangle$ is a cyclic $\mathbb{Z}$-module (abelian group) of order 2 ; if $M$ could be imbedded in a $\mathbb{Q}$-module (i.e., a vector space $V$ over $\mathbb{Q}$ ), then $2 a=0$ in $V$ and yet neither factor is 0 . Thus, our goal of extending scalars has merit, but we cannot be so cavalier about its solution. We must consider two problems: given a left $S$-module $M$, can we extend scalars to obtain a left $R$-module $M^{\prime}$ (always); if we can extend scalars, does $M$ imbed in $M^{\prime}$ (sometimes).

## 线性代数作业代写linear algebra代考|Modules

End $\mathbb{Z}(M)=$ homomorphisms $f: M \rightarrow M$

# 计量经济学代写

## 在这种情况下，如何学好线性代数？如何保证线性代数能获得高分呢？

1.1 mark on book

【重点的误解】划重点不是书上粗体，更不是每个定义，线代概念这么多，很多朋友强迫症似的把每个定义整整齐齐用荧光笔标出来，然后整本书都是重点，那期末怎么复习呀。我认为需要标出的重点为

A. 不懂，或是生涩，或是不熟悉的部分。这点很重要，有的定义浅显，但证明方法很奇怪。我会将晦涩的定义，证明方法标出。在看书时，所有例题将答案遮住，自己做，卡住了就说明不熟悉这个例题的方法，也标出。

B. 老师课上总结或强调的部分。这个没啥好讲的，跟着老师走就对了

C. 你自己做题过程中，发现模糊的知识点

1.2 take note

1.3 understand the relation between definitions