# 线性代数网课代修|张量代数代写Tensor algebra辅导|MATH489

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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|Inner products of exterior algebras

The space $T_{p}(V)=T^{p}\left(V^{}\right)$ of covariant tensors of degree $p$ can be regarded naturally as the dual space of $T^{p}(V)$ (see Example 2.3, §2.1). In this case, the inner product (a nondegenerate bilinear form by which these two spaces can be regarded as dual space of each other) is given by, for $T_{p}(V) \ni \varphi_{1} \otimes \cdots \otimes \varphi_{p}$ and $T^{p}(V) \ni v_{1} \otimes \cdots \otimes v_{p}$ $$\left\langle v_{1} \otimes \cdots \otimes v_{p}, \varphi_{1} \otimes \cdots \otimes \varphi_{p}\right\rangle=\varphi_{1}\left(v_{1}\right) \cdots \varphi_{p}\left(v_{p}\right) \quad\left(v_{i} \in V, \varphi_{j} \in V^{}\right) .$$
Using this, we show that the exterior algebras $A(V)$ and $A\left(V^{}\right)$ can also be regarded as dual spaces of each other. First we define a bilinear form $\langle\mid\rangle_{p}$ on $A^{p}(V) \times A^{p}\left(V^{}\right)$ by
$$\langle z \mid \xi\rangle_{p}=p !\langle z, \xi\rangle \quad\left(z \in A^{p}(V), \xi \in A^{p}\left(V^{}\right)\right) .$$ PROPOSITION 3.5. (1) For $z=v_{1} \wedge \cdots \wedge v_{p}\left(v_{i} \in V\right), \xi=\varphi_{1} \wedge \cdots \wedge \varphi_{p}$ $\left(\varphi_{i} \in V^{}\right)$, we have
$$\langle z \mid \xi\rangle_{p}=\operatorname{det}\left(\varphi_{i}\left(v_{j}\right)\right) .$$
(2) $A^{p}(V)$ and $A^{p}\left(V^{}\right)$ are dual spaces of each other with respect to the inner product $\langle\mid\rangle_{p} .\left{e_{H} \mid H \subseteq I, #(H)=p\right}$ and $\left{f^{K} \mid K \subseteq I, #(K)=\right.$ $p}$ are dual bases of each other for $A^{p}(V)$ and $A^{p}\left(V^{}\right)$ respectively, where $e_{H}$ and $f^{K}$ are standard basis elements of $A(V)$ and $A\left(V^{}\right)$ corresponding to a basis $\mathscr{E}=\left(e_{1}, \ldots, e_{n}\right)$ of $V$ and its dual basis $\mathscr{F}=\left(f^{1}, \ldots, f^{n}\right)$ respectively. (3) By using the sum of $\langle\mid\rangle_{p}$ as the inner product, $A(V)$ and $A\left(V^{}\right)$ are dual spaces of each other. That is, we define a bilinear form $\langle\mid\rangle$ on $A(V) \times A\left(V^{}\right)$ as follows, for $z=\sum_{p=0}^{n} z_{p}\left(z_{p} \in A^{p}(V)\right)$ and $\xi=\sum_{p=0}^{n} \xi_{p}$ $\left(\xi_{p} \in A\left(V^{}\right)\right)$,
$$\langle z \mid \xi\rangle=\sum_{p=0}^{n}\left\langle z_{p} \mid \xi_{p}\right\rangle_{p} .$$

## 线性代数作业代写linear algebra代考|Applications to geometry

Let $V$ be an $n$-dimensional $k$-vector space and $A^{p}(V)$ the space of alternating tensors of degree $p(1 \leq p \leq n)$. In this section, we shall examine the correspondence between elements of $A^{p}(V)$ and subspaces of $V$.

For $v_{1}, \ldots, v_{p} \in V$, the exterior product $v_{1} \wedge \cdots \wedge v_{p}$ is an element of $A^{p}(V)$, and every element of $A^{p}(V)$ can be expressed as a linear combination of several elements of this type. For example, taking the standard basis of $A(V)$ corresponding to a basis $\mathscr{E}$ of $V$, we have such an expression.
First of all, we have the following easy lemma.
LEMMA 3.4. For $v_{1}, \ldots, v_{p} \in V, v_{1} \wedge \cdots \wedge v_{p} \neq 0$ if and only if $v_{1}, \ldots, v_{p}$ are linearly independent.

Proof. If $v_{1}, \ldots, v_{p}$ are linearly independent, then there exists a basis for $V$ containing all $v_{i}(1 \leq i \leq p)$. Then $v_{1} \wedge \cdots \wedge v_{p}$ is a standard basis element of $A^{p}(V)$. In particular, $v_{1} \wedge \cdots \wedge v_{p} \neq 0$. Conversely, if $v_{1}, \ldots, v_{p}$ are linearly dependent, e.g., if $v_{1}=\sum_{i=2}^{p} \alpha_{i} v_{i}$, then we have
\begin{aligned} v_{1} \wedge \cdots \wedge v_{p} &=\left(\sum_{i=2}^{p} \alpha_{i} v_{i}\right) \wedge v_{2} \wedge \cdots \wedge v_{p} \ &=\sum_{i=2}^{p} \alpha_{i}\left(v_{i} \wedge v_{2} \wedge \cdots \wedge v_{p}\right)=0 . \end{aligned}
DEFINITION 3.4. An element $z \neq 0$ of $A^{p}(V)$ which can be expressed in the form $z=v_{1} \wedge \cdots \wedge v_{p}$ with $v_{1}, \ldots, v_{p} \in V$, is called decomposable or pure.

## 线性代数作业代写linear algebra代考|Inner products of exterior algebras

$2.1$ 节) 。在这种情况下，内积（一种非退化双线性形式，这两个空间可以被视为彼此的对

$$\left\langle v_{1} \otimes \cdots \otimes v_{p}, \varphi_{1} \otimes \cdots \otimes \varphi_{p}\right\rangle=\varphi_{1}\left(v_{1}\right) \cdots \varphi_{p}\left(v_{p}\right) \quad\left(v_{i} \in V, \varphi_{j} \in V\right) .$$

$$\langle z \mid \xi\rangle_{p}=p !\langle z, \xi\rangle \quad\left(z \in A^{p}(V), \xi \in A^{p}(V)\right) .$$

(2) $A^{p}(V)$ 和 $A^{p}(V)$ 是彼此关于内积的对偶空间 和 $f^{K}$ 是标准的基本元素 $A(V)$ 和 $A(V)$ 对应一个基 $\mathscr{E}=\left(e_{1}, \ldots, e_{n}\right)$ 的 $V$ 及其双重基础 $\mathscr{F}=\left(f^{1}, \ldots, f^{n}\right)$ 分别。(3) 通过使用总和 $\langle\mid\rangle_{p}$ 作为内积， $A(V)$ 和 $A(V)$ 是彼此的对偶空 间。也就是说，我们定义了一个双线性形式 $\langle\mid\rangle$ 上 $A(V) \times A(V)$ 如下，对于 $z=\sum_{p=0}^{n} z_{p}\left(z_{p} \in A^{p}(V)\right)$ 和 $\xi=\sum_{p=0}^{n} \xi_{p}\left(\xi_{p} \in A(V)\right)$,
$$\langle z \mid \xi\rangle=\sum_{p=0}^{n}\left\langle z_{p} \mid \xi_{p}\right\rangle_{p}$$

## 线性代数作业代写linear algebra代考|Applications to geometry

$$v_{1} \wedge \cdots \wedge v_{p}=\left(\sum_{i=2}^{p} \alpha_{i} v_{i}\right) \wedge v_{2} \wedge \cdots \wedge v_{p} \quad=\sum_{i=2}^{p} \alpha_{i}\left(v_{i} \wedge v_{2} \wedge \cdots \wedge v_{p}\right)=0 .$$

# 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