# 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|Tensor algebras and their properties

(a) Definition of tensor algebra. Collecting all contravariant tensors of different degree, we define the (infinite) direct sum of $T^{p}(V), p=0,1, \ldots$,
$$T(V)=\bigoplus_{p=0}^{\infty} T^{p}(V) .$$
In general, suppose that, to each element $i$ of $I={0,1, \ldots,$,$} there$ corresponds a $k$-vector space $V_{i}$. Then we define the infinite direct sum $\tilde{V}=\bigoplus_{i=0}^{\infty} V_{i}$ as follows. As a set, $\tilde{V}$ consists of all infinite sequences
$$v=\left(v_{0}, v_{1}, \ldots\right) \quad\left(v_{i} \in V_{i}\right)$$
such that the number of indices $i$ with $v_{i} \neq 0$ is finite. If we define, for $v=\left(v_{0}, v_{1}, \ldots\right) \in \widetilde{V}$ and $v^{\prime}=\left(v_{0}^{\prime}, v_{1}^{\prime}, \ldots\right) \in \widetilde{V}$, the sum $v+v^{\prime}$ and the scalar multiple $\alpha v(\alpha \in k)$ by the following formulas, then $\tilde{V}$ has the structure of a $k$-vector space (not necessarily finite dimensional).
$$v+v^{\prime}=\left(v_{0}+v_{0}^{\prime}, v_{1}+v_{1}^{\prime}, \ldots\right), \quad \alpha v=\left(\alpha v_{0}, \alpha v_{1}, \ldots\right) .$$
If we identify $v_{i} \in V_{i}$ with the element
$$\left(0, \ldots, 0, v_{i}, 0, \ldots\right)$$
of $\tilde{V}$, then $V_{i}$ can be considered as a subset of $\tilde{V}$. Let $v=\left(v_{0}, v_{1}, \ldots\right) \in$ $\tilde{V}$. Then there exists an $i_{0} \in I$ such that $v_{j}=0$ for all $j>i_{0}$. Thus we can write
$$v=\sum_{j=0}^{i_{0}} v_{j} \quad\left(v_{j} \in V_{j}\right) .$$
We write this expression formally as $v=\sum_{j=0}^{\infty} v_{j}$.

## 线性代数作业代写linear algebra代考|Symmetric algebras and their properties

We shall define a multiplication on the (infinite) direct sum of the spaces $S^{p}(V)$ of symmetric tensors of degree $p \quad(p=0,1,2, \ldots)$
$$S(V)=\bigoplus_{p=0}^{\infty} S^{p}(V) \quad\left(\text { where } S^{0}(V)=T^{0}(V)=k\right) .$$
Then we show that, with respect to this product, $S(V)$ becomes an associative algebra and we discuss its properties.

To define a multiplication, let $t \in S^{p}(V)$ and $t^{\prime} \in S^{q}(V) . t \otimes t^{\prime} \in T^{p+q}(V)$ is not necessarily symmetric. So we apply the symmetrizer $\mathscr{S}{p+q}$ to $t \otimes t^{\prime}$. The image is called the product of $t$ and $t^{\prime}$ and is denoted by $t \cdot t^{\prime}$ : $$t \cdot t^{\prime}=\mathscr{S}{p+q}\left(t \otimes t^{\prime}\right) .$$
EXAMPLE 2.19. (1) Let $t=v_{1}$ and $t^{\prime}=v_{2}\left(v_{1}, v_{2} \in V\right)$.
$$t \cdot t^{\prime}=\mathscr{S}{2}\left(v{1} \otimes v_{2}\right)=\frac{1}{2}\left(v_{1} \otimes v_{2}+v_{2} \otimes v_{1}\right) \text {. }$$
(2) Let $t=\left(v_{1} \otimes v_{2}+v_{2} \otimes v_{1}\right) / 2 \in S^{2}(V)$ and $t^{\prime}=v_{3} \in S^{1}(V)=V$, then
\begin{aligned} t \cdot t^{\prime}=& \mathscr{S}{3}\left(t \otimes t^{\prime}\right)=\mathscr{S}{3}\left(\frac{1}{2}\left(v_{1} \otimes v_{2}+v_{2} \otimes v_{1}\right) \otimes v_{3}\right) \ =& \frac{1}{6}\left(v_{1} \otimes v_{2} \otimes v_{3}+v_{2} \otimes v_{1} \otimes v_{3}+v_{1} \otimes v_{3} \otimes v_{2}\right.\ &\left.\quad+v_{3} \otimes v_{1} \otimes v_{2}+v_{2} \otimes v_{3} \otimes v_{1}+v_{3} \otimes v_{2} \otimes v_{1}\right) \ =& \mathscr{S}{3}\left(v{1} \otimes v_{2} \otimes v_{3}\right) . \end{aligned}

## 线性代数作业代写linear algebra代考|Tensor algebras and their properties

(a) 张量代数的定义。收集所有不同程度的逆变张量，我们定义 (无限) 直接和 $T^{p}(V), p=0,1, \ldots$
$$T(V)=\bigoplus_{p=0}^{\infty} T^{p}(V)$$

v=\left(v_{0}, v_{1}, \ldots\right) \quad\left(v_{i} \in V_{i}\right)
$$这样索引的数量 i 和 v_{i} \neq 0 是有限的。如果我们定义，对于 v=\left(v_{0}, v_{1}, \ldots\right) \in \widetilde{V} 和 v^{\prime}=\left(v_{0}^{\prime}, v_{1}^{\prime}, \ldots\right) \in \widetilde{V} ，总和 v+v^{\prime} 和标量倍数 \alpha v(\alpha \in k) 由以下公式，则 \tilde{V} 有一个结构 k 向量空间 (不一定是有限维)。$$
v+v^{\prime}=\left(v_{0}+v_{0}^{\prime}, v_{1}+v_{1}^{\prime}, \ldots\right), \quad \alpha v=\left(\alpha v_{0}, \alpha v_{1}, \ldots\right) .
$$如果我们确定 v_{i} \in V_{i} 与元素$$
\left(0, \ldots, 0, v_{i}, 0, \ldots\right)
$$的 \tilde{V} ，然后 V_{i} 可以看作是一个子集 \tilde{V}. 让 v=\left(v_{0}, v_{1}, \ldots\right) \in \tilde{V}. 那么存在一个 i_{0} \in I 这样 v_{j}=0 对所有人 j>i_{0}. 因此我们可以写$$
v=\sum_{j=0}^{i_{0}} v_{j} \quad\left(v_{j} \in V_{j}\right) .
$$我们将这个表达式正式写为 v=\sum_{j=0}^{\infty} v_{j}. ## 线性代数作业代写linear algebra代考|Symmetric algebras and their properties 我们将在空间的 (无限) 直接和上定义一个乘法 S^{p}(V) 度的对称张量 p \quad(p=0,1,2, \ldots)$$
S(V)=\bigoplus_{p=0}^{\infty} S^{p}(V) \quad\left(\text { where } S^{0}(V)=T^{0}(V)=k\right) .
$$然后我们证明，关于这个产品， S(V) 变成结合代数，我们讨论它的性质。 要定义一个乘法，让 t \in S^{p}(V) 和 t^{\prime} \in S^{q}(V) \cdot t \otimes t^{\prime} \in T^{p+q}(V) 不一定是对称的。所以我 们应用对称器 \mathscr{S} p+q 至 t \otimes t^{\prime}. 图像被称为产品 t 和 t^{\prime} 并表示为 t \cdot t^{\prime} :$$
t \cdot t^{\prime}=\mathscr{S} p+q\left(t \otimes t^{\prime}\right) .
$$例 2.19。(1) 让 t=v_{1} 和 t^{\prime}=v_{2}\left(v_{1}, v_{2} \in V\right).$$
t \cdot t^{\prime}=\mathscr{S} 2\left(v 1 \otimes v_{2}\right)=\frac{1}{2}\left(v_{1} \otimes v_{2}+v_{2} \otimes v_{1}\right) .
$$(2) 让 t=\left(v_{1} \otimes v_{2}+v_{2} \otimes v_{1}\right) / 2 \in S^{2}(V) 和 t^{\prime}=v_{3} \in S^{1}(V)=V ，然后$$
t \cdot t^{\prime}=\mathscr{S} 3\left(t \otimes t^{\prime}\right)=\mathscr{S} 3\left(\frac{1}{2}\left(v_{1} \otimes v_{2}+v_{2} \otimes v_{1}\right) \otimes v_{3}\right)=\frac{1}{6}\left(v_{1} \otimes v_{2} \otimes v_{3}+v_{2} \otimes\right.


# 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