# 线性代数网课代修|高阶线性代数代写Advanced Linear Algebra代考|MATH270

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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|Subspaces, span, and linear dependence

Let $U$ be a vector space over a field $\mathbb{F}$ and $V \subset U$ a non-empty subset of $U$. We say that $V$ is a subspace of $U$ if $V$ is a vector space over $\mathbb{F}$ with the inherited addition and scalar multiplication from $U$. It is worth noting that, when checking whether a subset $V$ of a vector space $U$ becomes a subspace, one only needs to verify the closure axiom (1) in the definition of a vector space since the rest of the axioms follow automatically as a consequence of (1).

The two trivial subspaces of $U$ are those consisting only of zero vector, ${0}$, and $U$ itself. A nontrivial subspace is also called a proper subspace. Consider the subset $\mathcal{P}{n}(n \in \mathbb{N})$ of $\mathcal{P}$ defined by $$\mathcal{P}{n}=\left{a_{0}+a_{1} t+\cdots+a_{n} t^{n} \mid a_{0}, a_{1}, \ldots, a_{n} \in \mathbb{F}\right} .$$ It is clear that $\mathcal{P}{n}$ is a subspace of $\mathcal{P}$ and $\mathcal{P}{m}$ is subspace of $\mathcal{P}{n}$ when $m \leq n$. Consider the set $S{a}$ of all vectors $\left(x_{1}, \ldots, x_{n}\right)$ in $\mathbb{F}^{n}$ satisfying the equation $$x_{1}+\cdots+x_{n}=a,$$ where $a \in \mathbb{F}$. Then $S_{a}$ is a subspace of $\mathbb{F}^{n}$ if and only if $a=0$. Let $u_{1}, \ldots, u_{k}$ be vectors in $U$. The linear span of $\left{u_{1}, \ldots, u_{k}\right}$, denoted by $\operatorname{Span}\left{u_{1}, \ldots, u_{k}\right}$, is the subspace of $U$ defined by $$\operatorname{Span}\left{u_{1}, \ldots, u_{k}\right}=\left{u \in U \mid u=a_{1} u_{1}+\cdots+a_{k} u_{k}, a_{1}, \ldots, a_{k} \in \mathbb{F}\right} .$$ Thus, if $u \in \operatorname{Span}\left{u_{1}, \ldots, u_{k}\right}$, then there are $a_{1}, \ldots, a_{k} \in \mathbb{F}$ such that $$u=a_{1} u_{1}+\cdots+a_{k} u_{k} .$$

## 线性代数作业代写linear algebra代考|Bases, dimensionality, and coordinates

Let $U$ be a vector space over a field $\mathbb{F}$, take $u_{1}, \ldots, u_{n} \in U$, and set $V=\operatorname{Span}\left{u_{1}, \ldots, u_{n}\right}$. Eliminating linearly dependent vectors from the set $\left{u_{1}, \ldots, u_{n}\right}$ if necessary, we can certainly assume that the vectors $u_{1}, \ldots, u_{n}$ are already made linearly independent. Thus, any vector $u \in V$ may take the form $$u=a_{1} u_{1}+\cdots+a_{n} u_{n}, \quad a_{1}, \ldots, a_{n} \in \mathbb{F} .$$ It is not hard to see that the coefficients $a_{1}, \ldots, a_{n}$ in the above representation must be unique. In fact, if we also have $$u=b_{1} u_{1}+\cdots+b_{n} u_{n}, \quad b_{1}, \ldots, b_{n} \in \mathbb{F},$$ then, combining the above two relations, we have $\left(a_{1}-b_{1}\right) u_{1}+\cdots+$ $\left(a_{n}-b_{n}\right) u_{n}=0$. Since $u_{1}, \ldots, u_{n}$ are linearly independent, we obtain $a_{1}=b_{1}, \ldots, a_{n}=b_{n}$ and the uniqueness follows. Furthermore, if there is another set of vectors $v_{1}, \ldots, v_{m}$ in $U$ such that $$\operatorname{Span}\left{v_{1}, \ldots, v_{m}\right}=\operatorname{Span}\left{u_{1}, \ldots, u_{n}\right},$$ then $m \geq n$ in view of Theorem 1.5. As a consequence, if $v_{1}, \ldots, v_{m}$ are also linearly independent, then $m=n$. This observation leads to the following.

## 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