# 线性代数网课代修|计算机图形学代写Computer Graphics代考|MS-C1342

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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|The Cross Product Reexamined

In Section $1.5$ we observed that $\mathbf{R}^{3}$ has not only a dot product but also a cross product. Note that the cross product produces another vector, whereas the dot product was a real number. Various identities involving the dot and cross product are known. The cross product is a “product” that behaves very much like the product in the case of real numbers except that it is not commutative. The two operations of vector addition and the cross product make $\mathbf{R}^{3}$ into a (noncommutative) ring. Is there a similar product in other dimensions? Unfortunately not, but the cross product does arise from a general construction that applies to all dimensions and that is worth looking at because it will give us additional insight into the cross product. 1.10.1. Theorem. Let $\mathbf{v}{1}, \mathbf{v}{2}, \ldots, \mathbf{v}{n-1} \in \mathbf{R}^{n}$. Define a map $T: \mathbf{R}^{n} \rightarrow \mathbf{R}$ by $$\mathrm{T}(\mathbf{w})=\operatorname{det}\left(\begin{array}{c} \mathbf{v}{1} \ \vdots \ \mathbf{v}_{\mathbf{n}-1} \ \mathbf{w} \end{array}\right)$$ Then there is a unique $\mathbf{u} \in \mathbf{R}^{\mathrm{n}}$ such that $\mathrm{T}(\mathbf{w})=\mathbf{u} \bullet \mathbf{w}$ for all $\mathbf{w}$. Proof. This theorem is an immediate corollary to Theorem 1.8.2 because properties of the determinant function show that $\mathrm{T}$ is a linear functional.

Definition. Using the notation of Theorem 1.10.1, the vector $\mathbf{u}$ is called the (generalized) cross product of the vectors $\mathbf{v}{1}, \mathbf{v}{2}, \ldots \mathbf{v}{\mathrm{n}-1}$ and is denoted by $\mathbf{v}{1} \times \mathbf{v}{2} \times \cdots \times$ $\mathbf{v}{\mathrm{n}-1}$.

## 线性代数作业代写linear algebra代考|The Generalized Inverse Matrix

Let $$\mathrm{T}: \mathbf{R}^{\mathrm{m}} \rightarrow \mathbf{R}^{\mathrm{n}}$$ be a linear transformation. Now normally one would not expect this arbitrary map T to have an inverse, especially if $\mathrm{m}>\mathrm{n}$, but it turns out that it is possible to define something close to that that is useful. Define a map $$\mathrm{T}^{+}: \mathbf{R}^{\mathrm{II}} \rightarrow \mathbf{R}^{\mathrm{II}}$$ as follows: See Figure 1.19. Let $\mathbf{b} \in \mathbf{R}^{\mathrm{n}}$. The point $\mathbf{b}$ may not be in the image of $T$, $\operatorname{im}(\mathrm{T})$, since we are not assuming that $\mathrm{T}$ is onto, but $\operatorname{im}(\mathrm{T})$ is a plane in $\mathbf{R}^{\mathrm{n}}$. Therefore, there is a unique point $\mathbf{c} \in \operatorname{im}(\mathrm{T})$ that is closest to $\mathbf{b}$ (Theorem 4.5.12). If the transformation $\mathrm{T}$ is onto, then obviously $\mathbf{c}=\mathbf{b}$. It is easy to show that $\mathrm{T}^{-1}(\mathbf{c})$ is a plane in $\mathbf{R}^{\mathrm{m}}$ that is parallel to the kernel of $\mathrm{T}$, $\operatorname{ker}(\mathrm{T})$. This plane will meet the orthogonal complement of the kernel of $T, \operatorname{ker}(T)^{\perp}$, in a unique point a. For an alternative definition of the point a write $\mathbf{R}^{\mathrm{m}}$ in the form $$\mathbf{R}^{\mathrm{m}}=\operatorname{ker}(\mathrm{T}) \oplus \operatorname{ker}(\mathrm{T})^{\perp}$$ and let $$\varphi=\mathrm{T} \mid \operatorname{ker}(\mathrm{T})^{\perp}: \operatorname{ker}(\mathrm{T})^{\perp} \rightarrow \operatorname{im}(\mathrm{T}) .$$ It is easy to show that $\varphi$ is an isomorphism and $\mathbf{a}=\varphi^{-1}(\mathbf{c})$. In either case, we define $\mathrm{T}^{+}(\mathbf{b})=\mathbf{a}$.

## 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