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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|Null-space, range, nullity, and rank

For $T \in L(U, V)$, the subset
$$N(T)={x \in U \mid T(x)=0}$$
in $U$ is a subspace of $U$ called the null-space of $T$, which is sometimes called the kernel of $T$, and denoted as $\operatorname{Kernel}(T)$ or $T^{-1}(0)$. Here and in the sequel, note that, in general, $T^{-1}(v)$ denotes the set of preimages of $v \in V$ under $T$. That is,
$$T^{-1}(v)={x \in U \mid T(x)=v} .$$
Besides, the subset
$$R(T)={v \in V \mid v=T(x) \text { for some } x \in U}$$
is a subspace of $V$ called the range of $T$, which is sometimes called the image of $U$ under $T$, and denoted as Image $(T)$ or $T(U)$.

For $T \in L(U, V)$, we say that $T$ is one-to-one, or 1-1, or injective, if $T(x) \neq$ $T(y)$ whenever $x, y \in U$ with $x \neq y$. It is not hard to show that $T$ is 1-1 if and only if $N(T)={0}$. We say that $T$ is onto or surjective if $R(T)=V$.
A basic problem in linear algebra is to investigate whether the equation
$$T(x)=v$$
has a solution $x$ in $U$ for a given vector $v \in V$. From the meaning of $R(T)$, we see that (2.1.17) has a solution if and only if $v \in R(T)$, and the solution is unique if and only if $N(T)={0}$. More generally, if $v \in R(T)$ and $u \in U$ is any particular solution of (2.1.17), then the set of all solutions of (2.1.17) is simply the coset
$$[u]={u}+N(T)=u+N(T) .$$
Thus, in terms of $N(T)$ and $R(T)$, we may understand the structure of the set of solutions of (2.1.17) completely.

## 线性代数作业代写linear algebra代考|Change of basis

It will be important to investigate how the associated matrix changes with respect to a change of basis for a linear mapping.

Let $\left{u_{1}, \ldots, u_{n}\right}$ and $\left{\tilde{u}{1}, \ldots, \tilde{u}{n}\right}$ be two bases of the $n$-dimensional vector space $U$ over a field $\mathbb{F}$. A change of bases from $\left{u_{1}, \ldots, u_{n}\right}$ to $\left{\tilde{u}{1}, \ldots, \tilde{u}{n}\right}$ is simply a linear mapping $R \in L(U, U)$ such that
$$R\left(u_{j}\right)=\tilde{u}{j}, \quad j=1, \ldots, n .$$ Following the study of the previous section, we know that $R$ is invertible. Moreover, if we rewrite $(2.2 .1)$ in a matrix form, we have $$\tilde{u}{k}=\sum_{j=1}^{n} b_{j k} u_{j}, \quad k=1, \ldots, n,$$
where the matrix $B=\left(b_{j k}\right) \in \mathbb{F}(n, n)$ is called the basis transition matrix from the basis $\left{u_{1}, \ldots, u_{n}\right}$ to the basis $\left{\tilde{u}{1}, \ldots, \tilde{u}{n}\right}$, which is necessarily invertible.

Let $V$ be an $m$-dimensional vector space over $\mathbb{F}$ and take $T \in L(U, V)$. Let $A=\left(a_{i j}\right)$ and $\tilde{A}=\left(\tilde{a}{i j}\right)$ be the $m \times n$ matrices of the linear mapping $T$ associated with the pairs of the bases $$\left{u{1}, \ldots, u_{n}\right} \quad \text { and } \quad\left{v_{1}, \ldots, v_{m}\right}, \quad\left{\tilde{u}{1}, \ldots, \tilde{u}{n}\right} \quad \text { and } \quad\left{v_{1}, \ldots, v_{m}\right},$$

## 线性代数作业代写linear algebra代考|Null-space, range, nullity, and rank

$$N(T)=x \in U \mid T(x)=0$$

$$T^{-1}(v)=x \in U \mid T(x)=v .$$

$$R(T)=v \in V \mid v=T(x) \text { for some } x \in U$$

$$T(x)=v$$

$$[u]=u+N(T)=u+N(T) .$$

## 线性代数作业代写linear algebra代考|Change of basis

lleftntilde{u}{1}, Ildots, Itilde{u}{n}}right} 只是一个线性映射 $R \in L(U, U)$ 这样
$$R\left(u_{j}\right)=\tilde{u} j, \quad j=1, \ldots, n .$$

$$\tilde{u} k=\sum_{j=1}^{n} b_{j k} u_{j}, \quad k=1, \ldots, n,$$ Mleft{tilde{u}{1}, Iddots, Itilde{u}{n}|right}，这必然是可逆的。

# 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