# 线性代数网课代修|同调代数代写homological algebra代考|MATH6350

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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|Semisimple Rings

If $k$ is a field, then $k$-modules are vector spaces. It follows that all $k$-modules are projective (even free, for every vector space has a basis). Indeed, every $k$-module is injective and flat as well. We now describe all rings for which this is true.

Definition. Let $R$ be a ring. A left $R$-module $M$ is simple (or irreducible) if $M \neq{0}$ and if $M$ has no proper nonzero submodules; we say that $M$ is semisimple (or completely reducible) if it is a direct sum of (possibly infinitely many) simple modules.

The zero module is not simple, but it is semisimple, for ${0}=\bigoplus_{i \in \varnothing} S_{i}$.

Corollary 4.2. Every submodule and every quotient module of a semisimple module $M$ is semisimple.

Proof. Let $N$ be a submodule of $M$. Every submodule of $N$ is a direct summand of $M$, by Proposition 4.1, so that Corollary $2.24$ shows that every submodule of $N$ is a direct summand of $N$; therefore, $N$ is semisimple. A quotient $M / N$ is semisimple, for $M=N \oplus Q$ for some submodule $Q$ of $M$. But $M / N \cong Q$, and $Q$ is semisimple, as we have just seen.

Lemma 4.3. If a ring $R$ is a direct sum of left ideals, say, $R=\bigoplus_{i \in I} L_{i}$, then only finitely many $L_{i}$ are nonzero.

Proof. Each element in a direct sum has finite support; in particular, the unit element can be written as $1=e_{1}+\cdots+e_{n}$, where $e_{i} \in L_{i}$. If $a \in L_{j}$ for some $j \neq 1, \ldots, n$, then
$$a=a 1=a e_{1}+\cdots+a e_{n} \in L_{j} \cap\left(L_{1} \oplus \cdots \oplus L_{n}\right)={0} .$$
Therefore, $L_{j}={0}$, and $R=L_{1} \oplus \cdots \oplus L_{n}$.

## 线性代数作业代写linear algebra代考|von Neumann Regular Rings

We have just seen that every $R$-module is projective (or injective) if and only if $R$ is semisimple. What if every $R$-module is flat?

Definition. A ring $R$ is von Neumann regular if, for each $r \in R$, there is $r^{\prime} \in R$ with $r r^{\prime} r=r$.
Informally, one may think of $r^{\prime}$ as a generalized inverse of $r$.
Example 4.7.
(i) A ring $R$ is a Boolean ring if every element $r \in R$ is idempotent; that is, $r^{2}=r$. Boolean rings are von Neumann regular: if $r \in R$, define $r^{\prime}=r$. Boolean rings are commutative.
(ii) Here is a proof that if $V$ is a (possibly infinite-dimensional) vector space over a field $k$, then $R=\operatorname{End}_{k}(V)$ is von Neumann regular. Given a linear transformation $\varphi: V \rightarrow V$, we have $V=\operatorname{ker} \varphi \oplus W$, for every subspace of a vector space is a direct summand. Let $X$ be a basis of $\operatorname{ker} \varphi$ and let $Y$ be a basis of $W$, so that $X \cup Y$ is a basis of $V$. Now $\varphi(Y)$ is a linearly independent subset (because $W \cap \operatorname{ker} \varphi={0}$ ), and so it can be extended to a basis $\varphi(Y) \cup Z$ of $V$. If we detine $\varphi^{\prime}: V \rightarrow V$ by $\varphi^{\prime}(\varphi(y))=y$ for all $y \in Y$ and $\varphi^{\prime}(z)=0$ for all $z \in Z$, then $\varphi \varphi^{\prime} \varphi=\varphi$. (Example $2.36$ shows that von Neumann regular rings may not have IBN; on the other hand, the uniqueness part of the Wedderburn-Arlin Theorem shows that semisimple rings do have IBN.)

## 线性代数作业代写linear algebra代考|Semisimple Rings

$$a=a 1=a e_{1}+\cdots+a e_{n} \in L_{j} \cap\left(L_{1} \oplus \cdots \oplus L_{n}\right)=0 .$$

## 线性代数作业代写linear algebra代考|von Neumann Regular Rings

$(一)$ 戒指 $R$ 如果每个元素都是布尔环 $r \in R$ 是幂等的；那是， $r^{2}=r$. 布尔环是冯诺依曼正 则: 如果 $r \in R$ ，定义 $r^{\prime}=r$. 布尔环是可交换的。
(ii) 这是一个证明，如果 $V$ 是场上的 (可能是无限维的) 向量空间 $k$ ，然后 $R=\operatorname{End}_{k}(V)$ 是 冯诺依曼正则。给定一个线性变换 $\varphi: V \rightarrow V$ ，我们有 $V=\operatorname{ker} \varphi \oplus W$ ，因为向量空间的 每个子空间都是直接和。让 $X$ 成为基础 $\operatorname{ker} \varphi$ 然后让 $Y$ 成为基础 $W$ ，以便 $X \cup Y$ 是一个基础 $V$. 现在 $\varphi(Y)$ 是一个线性独立的子集（因为 $W \cap \operatorname{ker} \varphi=0$ ），因此它可以扩展到一个基 $\varphi(Y) \cup Z$ 的 $V$. 如果我们确定 $\varphi^{\prime}: V \rightarrow V$ 经过 $\varphi^{\prime}(\varphi(y))=y$ 对所有人 $y \in Y$ 和 $\varphi^{\prime}(z)=0$ 对 所有人 $z \in Z$ ，然后 $\varphi \varphi^{\prime} \varphi=\varphi$. (例子 $2.36$ 表明冯诺依曼正则环可能没有 IBN；另一方 面，Wedderburn-Arlin 定理的唯一性部分表明半单环确实有 IBN。)

# 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