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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

Let $U, V$ be finite-dimensional vector spaces over a field $\mathbb{F}$ and $U^{\prime}, V^{\prime}$ their dual spaces. For $T \in L(U, V)$ and $v^{\prime} \in V^{\prime}$, we see that
$$\left\langle T(u), v^{\prime}\right\rangle, \quad \forall u \in U,$$
defines a linear functional over $U$. Hence there is a unique vector $u^{\prime} \in U^{\prime}$ such that
$$\left\langle u, u^{\prime}\right\rangle=\left\langle T(u), v^{\prime}\right\rangle, \quad \forall u \in U .$$
Of course, $u^{\prime}$ depends on $T$ and $v^{\prime}$. So we may write this relation as
$$u^{\prime}=T^{\prime}\left(v^{\prime}\right) .$$
Under such a notion, we can rewrite (2.3.2) as
$$\left\langle u, T^{\prime}\left(v^{\prime}\right)\right\rangle=\left\langle T(u), v^{\prime}\right\rangle, \quad \forall u \in U, \quad \forall v^{\prime} \in V^{\prime} .$$
Thus, in this way we have constructed a mapping $T^{\prime}: V^{\prime} \rightarrow U^{\prime}$. We now show that $T^{\prime}$ is linear.
In fact, let $v_{1}^{\prime}, v_{2}^{\prime} \in V^{\prime}$. Then (2.3.4) gives us
$$\left\langle u, T^{\prime}\left(v_{i}^{\prime}\right)\right\rangle=\left\langle T(u), v_{i}^{\prime}\right\rangle, \quad \forall u \in U, \quad i=1,2 .$$
Thus
$$\left\langle u, T^{\prime}\left(v_{1}^{\prime}\right)+T^{\prime}\left(v_{2}^{\prime}\right)\right\rangle=\left\langle T(u), v_{1}^{\prime}+v_{2}^{\prime}\right\rangle, \quad \forall u \in U .$$
In view of (2.3.4) and (2.3.6), we arrive at $T^{\prime}\left(v_{1}^{\prime}+v_{2}^{\prime}\right)=T\left(v_{1}^{\prime}\right)+T^{\prime}\left(v_{2}^{\prime}\right)$. Besides, for any $a \in \mathbb{F}$, we have from (2.3.4) that
\begin{aligned} \left\langle u, T^{\prime}\left(a v^{\prime}\right)\right\rangle &=\left\langle T(u), a v^{\prime}\right\rangle=a\left\langle T(u), v^{\prime}\right\rangle=a\left\langle u, T^{\prime}\left(v^{\prime}\right)\right\rangle \ &=\left\langle u, a T^{\prime}\left(v^{\prime}\right)\right\rangle, \quad \forall u \in U, \quad \forall v^{\prime} \in V^{\prime}, \end{aligned}
which yields $T^{\prime}\left(a v^{\prime}\right)=a T^{\prime}\left(v^{\prime}\right)$. Thus the linearity of $T^{\prime}$ is established.

## 线性代数作业代写linear algebra代考|Quotient mappings

Let $U, V$ be two vector spaces over a field $\mathbb{F}$ and $T \in L(U, V)$. Suppose that $X, Y$ are subspaces of $U, V$, respectively, which satisfy the property $T(X) \subset Y$ or $T \in L(X, Y)$. We now show that such a property allows us to generate a linear mapping
from $T$ naturally.
As before, we use [.] to denote a coset in $U / X$ or $V / Y$. Define $\tilde{T}: U /$ $X \rightarrow V / Y$ by setting
$$\tilde{T}([u])=[T(u)], \quad \forall[u] \in U / X .$$
We begin by showing that this definition does not suffer any ambiguity by verifying
$$\tilde{T}\left(\left[u_{1}\right]\right)=\tilde{T}\left(\left[u_{2}\right]\right) \quad \text { whenever }\left[u_{1}\right]=\left[u_{2}\right] .$$
In fact, if $\left[u_{1}\right]=\left[u_{2}\right]$, then $u_{1}-u_{2} \in X$. Thus $T\left(u_{1}\right)-T\left(u_{2}\right)=T\left(u_{1}-u_{2}\right) \in Y$, which implies $\left[T\left(u_{1}\right)\right]=\left[T\left(u_{2}\right)\right]$. So $(2.4 .3)$ follows.
The linearity of $\tilde{T}$ can now be checked directly.
First, let $u_{1}, u_{2} \in U$. Then, by (2.4.2), we have
\begin{aligned} \tilde{T}\left(\left[u_{1}\right]+\left[u_{2}\right]\right) &=\tilde{T}\left(\left[u_{1}+u_{2}\right]\right)=\left[T\left(u_{1}+u_{2}\right)\right]=\left[T\left(u_{1}\right)+T\left(u_{2}\right)\right] \ &=\left[T\left(u_{1}\right)\right]+\left[T\left(u_{2}\right)\right]=\tilde{T}\left(\left[u_{1}\right]\right)+\tilde{T}\left(\left[u_{2}\right]\right) . \end{aligned}
Next, let $a \in \mathbb{F}$ and $u \in U$. Then, again by (2.4.2), we have
$$\tilde{T}(a[u])=\tilde{T}([a u])=[T(a u)]=a[T(u)]=a \tilde{T}([u]) .$$

$$\left\langle T(u), v^{\prime}\right\rangle, \quad \forall u \in U,$$

$$\left\langle u, u^{\prime}\right\rangle=\left\langle T(u), v^{\prime}\right\rangle, \quad \forall u \in U .$$

$$u^{\prime}=T^{\prime}\left(v^{\prime}\right)$$

$$\left\langle u, T^{\prime}\left(v^{\prime}\right)\right\rangle=\left\langle T(u), v^{\prime}\right\rangle, \quad \forall u \in U, \quad \forall v^{\prime} \in V^{\prime} .$$

$$\left\langle u, T^{\prime}\left(v_{i}^{\prime}\right)\right\rangle=\left\langle T(u), v_{i}^{\prime}\right\rangle, \quad \forall u \in U, \quad i=1,2 .$$

$$\left\langle u, T^{\prime}\left(v_{1}^{\prime}\right)+T^{\prime}\left(v_{2}^{\prime}\right)\right\rangle=\left\langle T(u), v_{1}^{\prime}+v_{2}^{\prime}\right\rangle, \quad \forall u \in U .$$

$$\left\langle u, T^{\prime}\left(a v^{\prime}\right)\right\rangle=\left\langle T(u), a v^{\prime}\right\rangle=a\left\langle T(u), v^{\prime}\right\rangle=a\left\langle u, T^{\prime}\left(v^{\prime}\right)\right\rangle \quad=\left\langle u, a T^{\prime}\left(v^{\prime}\right)\right\rangle, \quad \forall u \in U, \quad \forall v^{\prime} \in V^{\prime},$$

## 线性代数作业代写linear algebra代考|Quotient mappings

$$\tilde{T}([u])=[T(u)], \quad \forall[u] \in U / X .$$

$$\tilde{T}\left(\left[u_{1}\right]\right)=\tilde{T}\left(\left[u_{2}\right]\right) \quad \text { whenever }\left[u_{1}\right]=\left[u_{2}\right] .$$

$$\tilde{T}\left(\left[u_{1}\right]+\left[u_{2}\right]\right)=\tilde{T}\left(\left[u_{1}+u_{2}\right]\right)=\left[T\left(u_{1}+u_{2}\right)\right]=\left[T\left(u_{1}\right)+T\left(u_{2}\right)\right] \quad=\left[T\left(u_{1}\right)\right]+\left[T\left(u_{2}\right)\right]=\tilde{T}\left(\left[u_{1}\right]\right)+\tilde{T}$$

$$\tilde{T}(a[u])=\tilde{T}([a u])=[T(a u)]=a[T(u)]=a \tilde{T}([u]) .$$

# 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