# 线性代数网课代修|计算机图形学代写Computer Graphics代考|MTH309

linearalgebra.me 为您的留学生涯保驾护航 在线性代数linear algebra作业代写方面已经树立了自己的口碑, 保证靠谱, 高质且原创的线性代数linear algebra代写服务。我们的专家在线性代数linear algebra代写方面经验极为丰富，各种线性代数linear algebra相关的作业也就用不着 说。

• 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|Reflections in the Plane

Another important type of motion is a reflection. Such a motion can be defined in several ways. After giving our definition we shall discuss some of these other characterizations.

Definition. Let $\mathbf{L}$ be a line in the plane. Define a map $\mathbf{S}: \mathbf{R}^{2} \rightarrow \mathbf{R}^{2}$, called the reflection about the line $\boldsymbol{L}$, as follows: Choose a point $\mathbf{A}$ on $\mathbf{L}$ and a unit normal vector $\mathbf{N}$ for $\mathbf{L}$. If $\mathbf{P}$ is any point in $\mathbf{R}^{2}$, then $$\mathrm{S}(\mathbf{P})=\mathbf{P}^{\prime}=\mathbf{P}+2(\mathbf{P A} \cdot \mathbf{N}) \mathbf{N} .$$ The line $\mathbf{L}$ is called the axis for the reflection $\mathrm{S}$. The reader will find Figure $2.4$ helpful as we discuss the geometry behind reflections. First, note that $\mathbf{W}=(\mathbf{P A} \cdot \mathbf{N}) \mathbf{N}$ is just the orthogonal projection of the vector $\mathbf{P A}$ onto $\mathbf{N}$. Define a point $\mathbf{Q}$ by the equation $$\mathbf{P Q}=\mathbf{W}=(\mathbf{P A} \cdot \mathbf{N}) \mathbf{N} .$$ Intuitively, it should be clear that $\mathbf{Q}$ is the point on $\mathbf{L}$ as shown in Figure 2.4. This does not follow from the definition however and must be proved. The following string of equalities: $$\mathbf{A Q} \cdot \mathbf{N}=(\mathbf{P Q}+\mathbf{A P}) \cdot \mathbf{N}=[(\mathbf{P A} \cdot \mathbf{N}) \mathbf{N}+\mathbf{A P}] \cdot \mathbf{N}=\mathbf{P A} \cdot \mathbf{N}+\mathbf{A P} \cdot \mathbf{N}=0$$ shows that $\mathbf{Q}$ satisfies the point-normal form of the equation $\mathbf{A X} \cdot \mathbf{N}=0$ for the points $\mathbf{X}$ on the line (or hyperplane) $\mathbf{L}$, so that $\mathbf{Q}$ does indeed lie on $\mathbf{L}$. Furthermore, it is easy to check that $\mathbf{A Q}$ is the orthogonal projection of $\mathbf{A P}$ on $\mathbf{L}$. This means that, if $\mathbf{V}$ is a unit direction vector for $\mathbf{L}$, then $\mathbf{A Q}=(\mathbf{A P} \bullet \mathbf{V}) \mathbf{V}$ and we could have defined the reflection $\mathrm{S}$ by $$\mathrm{S}(\mathbf{P})=\mathbf{P}+2(\mathbf{P A}+\mathbf{A Q})$$

## 线性代数作业代写linear algebra代考|Motions Preserve the Dot Product

Proof. We shall show that $\mathrm{M}$ is a linear transformation in two steps. Claim 1. $\mathrm{M}(\mathbf{u}+\mathbf{v})=\mathrm{M}(\mathbf{u})+\mathrm{M}(\mathbf{v})$. Define a vector $\mathbf{w}$ by the equation $$\mathbf{u}+\mathbf{v}=2 \mathbf{w}$$ This equation can be rewritten as $$\mathbf{w}=\mathbf{u}+\frac{1}{2}(\mathbf{v}-\mathbf{u})$$ See Figure 2.6. Since $M(\mathbf{0})=\mathbf{0}$ (which implies that $|\mathrm{M}(\mathbf{p})|=|\mathbf{p}|$ for any vector $\mathbf{p}$ ), we can use equation (2.15) and Lemma 2.2.2 to conclude that $$M(\mathbf{u}+\mathbf{v})=2 \mathrm{M}(\mathbf{w}) .$$ Similarly, equation (2.16) and Lemma 2.2.2 implies that $$\mathrm{M}(\mathbf{w})=\mathrm{M}(\mathbf{u})+\frac{1}{2}(\mathbf{M}(\mathbf{v})-\mathrm{M}(\mathbf{u}))$$ Substituting the expression for $\mathbf{M}(\mathbf{w})$ in equation (2.18) into equation (2.17) and simplifying the result proves Claim $1 .$ Claim 2. $\mathrm{M}(\mathrm{cv})=\mathrm{cM}(\mathbf{v})$, for any real number $\mathrm{c}$. This follows from Lemma 2.2.2 (let $\mathbf{A}=\mathbf{0}, \mathbf{B}=\mathbf{v}$, and $\mathrm{t}=\mathrm{c}$ in that Lemma). Theorem $2.2 .4 .1$ is proved.

## 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