### 数学代写|线性代数代写linear algebra代考|MATH1051

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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 数学代写|线性代数代写linear algebra代考|Perspective Projections

A three-dimensional object is represented on the two-dimensional computer screen by projecting the object onto a viewing plane. (We ignore other important steps, such as selecting the portion of the viewing plane to display on the screen.) For simplicity, let the $x y$-plane represent the computer screen, and imagine that the eye of a viewer is along the positive $z$-axis, at a point $(0,0, d)$. A perspective projection maps each point $(x, y, z)$ onto an image point $\left(x^, y^, 0\right)$ so that the two points and the eye position, called the center of projection, are on a line. See Figure 6(a).

The triangle in the $x z$-plane in Figure 6(a) is redrawn in part (b) showing the lengths of line segments. Similar triangles show that
$$\frac{x^}{d}=\frac{x}{d-z} \quad \text { and } \quad x^=\frac{d x}{d-z}=\frac{x}{1-z / d}$$
Similarly,
$$y^*=\frac{y}{1-z / d}$$
Using homogeneous coordinates, we can represent the perspective projection by a matrix, say, $P$. We want $(x, y, z, 1)$ to map into $\left(\frac{x}{1-z / d}, \frac{y}{1-z / d}, 0,1\right)$. Scaling these coordinates by $1-z / d$, we can also use $(x, y, 0,1-z / d)$ as homogeneous coordinates for the image. Now it is easy to display $P$. In fact,
$$P\left[\begin{array}{l} x \ y \ z \ 1 \end{array}\right]=\left[\begin{array}{cccc} 1 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 \ 0 & 0 & 0 & 0 \ 0 & 0 & -1 / d & 1 \end{array}\right]\left[\begin{array}{c} x \ y \ z \ 1 \end{array}\right]=\left[\begin{array}{c} x \ y \ 0 \ 1-z / d \end{array}\right]$$
EXAMPLE 8 Let $S$ be the box with vertices $(3,1,5),(5,1,5),(5,0,5),(3,0,5)$, $(3,1,4),(5,1,4),(5,0,4)$, and $(3,0,4)$. Find the image of $S$ under the perspective projection with center of projection at $(0,0,10)$.

## 数学代写|线性代数代写linear algebra代考|Column Space and Null Space of a Matrix

Subspaces of $\mathbb{R}^n$ usually occur in applications and theory in one of two ways. In both cases, the subspace can be related to a matrix.
The column space of a matrix $A$ is the set $\operatorname{Col} A$ of all linear combinations of the columns of $A$.
If $A=\left[\begin{array}{lll}\mathbf{a}_1 & \cdots & \mathbf{a}_n\end{array}\right]$, with the columns in $\mathbb{R}^m$, then $\operatorname{Col} A$ is the same as Span $\left{\mathbf{a}_1, \ldots, \mathbf{a}_n\right}$. Example 4 shows that the column space of an $\boldsymbol{m} \times \boldsymbol{n}$ matrix is a subspace of $\mathbb{R}^m$. Note that $\operatorname{Col} A$ equals $\mathbb{R}^m$ only when the columns of $A$ span $\mathbb{R}^m$. Otherwise, $\operatorname{Col} A$ is only part of $\mathbb{R}^m$.

EXAMPLE 4 Let $A=\left[\begin{array}{rrr}1 & -3 & -4 \ -4 & 6 & -2 \ -3 & 7 & 6\end{array}\right]$ and $\mathbf{b}=\left[\begin{array}{r}3 \ 3 \ -4\end{array}\right]$. Determine whether $\mathbf{b}$ is in the column space of $A$.

SOLUTION The vector $\mathbf{b}$ is a linear combination of the columns of $A$ if and only if $\mathbf{b}$ can be written as $A \mathbf{x}$ for some $\mathbf{x}$, that is, if and only if the equation $A \mathbf{x}=\mathbf{b}$ has a solution. Row reducing the augmented matrix $\left[A \begin{array}{ll}A & \mathbf{b}\end{array}\right]$,
$$\left[\begin{array}{rrrr} 1 & -3 & -4 & 3 \ -4 & 6 & -2 & 3 \ -3 & 7 & 6 & -4 \end{array}\right] \sim\left[\begin{array}{rrrr} 1 & -3 & -4 & 3 \ 0 & -6 & -18 & 15 \ 0 & -2 & -6 & 5 \end{array}\right] \sim\left[\begin{array}{rrrr} 1 & -3 & -4 & 3 \ 0 & -6 & -18 & 15 \ 0 & 0 & 0 & 0 \end{array}\right]$$
we conclude that $A \mathbf{x}=\mathbf{b}$ is consistent and $\mathbf{b}$ is in $\operatorname{Col} A$.

The solution of Example 4 shows that when a system of linear equations is written in the form $A \mathbf{x}=\mathbf{b}$, the column space of $A$ is the set of all $\mathbf{b}$ for which the system has a solution.
The null space of a matrix $A$ is the set $\mathrm{Nul} A$ of all solutions of the homogeneous equation $A \mathbf{x}=\mathbf{0}$
When $A$ has $n$ columns, the solutions of $A \mathbf{x}=\mathbf{0}$ belong to $\mathbb{R}^n$, and the null space of $A$ is a subset of $\mathbb{R}^n$. In fact, $\mathrm{Nul} A$ has the properties of a subspace of $\mathbb{R}^n$.
The null space of an $m \times n$ matrix $A$ is a subspace of $\mathbb{R}^n$. Equivalently, the set of all solutions of a system $A \mathbf{x}=\mathbf{0}$ of $m$ homogeneous linear equations in $n$ unknowns is a subspace of $\mathbb{R}^n$.
PROOF The zero vector is in $\operatorname{Nul} A$ (because $A 0=0$ ). To show that $\mathrm{Nul} A$ satisfies the other two properties required for a subspace, take any $\mathbf{u}$ and $\mathbf{v}$ in $\mathrm{Nul} A$. That is, suppose $A \mathbf{u}=\mathbf{0}$ and $A \mathbf{v}=\mathbf{0}$. Then, by a property of matrix multiplication,
$$A(\mathbf{u}+\mathbf{v})=A \mathbf{u}+A \mathbf{v}=\mathbf{0}+\mathbf{0}=\mathbf{0}$$
Thus $\mathbf{u}+\mathbf{v}$ satisfies $A \mathbf{x}=\mathbf{0}$, and so $\mathbf{u}+\mathbf{v}$ is in $\operatorname{Nul} A$. Also, for any scalar $c, A(c \mathbf{u})=$ $c(A \mathbf{u})=c(0)=\mathbf{0}$, which shows that $c \mathbf{u}$ is in $\mathrm{Nul} A$.

To test whether a given vector $\mathbf{v}$ is in $\operatorname{Nul} A$, just compute $A \mathbf{v}$ to see whether $A \mathbf{v}$ is the zero vector. Because $\mathrm{Nul} A$ is described by a condition that must be checked for each vector, we say that the null space is defined implicitly. In contrast, the column space is defined explicitly, because vectors in Col A can be constructed (by linear combinations) from the columns of $A$. To create an explicit description of $\mathrm{Nul} A$, solve the equation $A \mathbf{x}=\mathbf{0}$ and write the solution in parametric vector form. (See Example 6 , below.) ${ }^2$

# 线性代数代考

## 数学代写|线性代数代写linear algebra代考|Perspective Projections

y^*=\frac{y}{1-z / d}
$$使用齐次坐标，我们可以用矩阵表示透视投影，比如说， P. 我们想要 (x, y, z, 1) 映射到 \left(\frac{x}{1-z / d}, \frac{y}{1-z / d}, 0,1\right). 缩放这些坐标 1-z / d ，我们也可以使用 (x, y, 0,1-z / d) 作为图像的齐次坐 标。现在很容易显示 P. 实际上，$$
P[x y z 1]=\left[\begin{array}{llllllllllllllll}
1 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 / d & 1
\end{array}\right]\left[\begin{array}{ll}
x y z & -1
\end{array}\right]=\left[\begin{array}{lll}
x & 0 & 1
\end{array}\right.
$$例 8 让 S 是有顶点的盒子 (3,1,5),(5,1,5),(5,0,5),(3,0,5) ，(3,1,4),(5,1,4),(5,0,4) ， 和 (3,0,4). 找到图像 S 在投影中心位于的透视投影下 (0,0,10). ## 数学代写|线性代数代写linear algebra代考|Column Space and Null Space of a Matrix 通过将物体投影到观察平面上，三维物体在二维计算机屏幕上呈现。（我们忽略其他重要步㡜，例如选择 要在屏幕上显示的视图平面部分。) 为简单起见，让 x y-plane 代表计算机屏幕，并想象观众的眼睛沿着 正面 z-轴，在一点 (0,0, d). 透视投影映射每个点 (x, y, z) 到图像点 \ V \mathrm{eft}\left(\mathrm{X}^{\wedge}, y^{\wedge} ， 0 \backslash r i g h t\right) \$$ 上，这样两 个点和眼睛位置 (称为投影中心) 在一条线上。见图 6(a)。

$\left\langle f r a c\left{x^{\wedge}\right}{d}=|f r a c{x}{d z} \backslash q u a d| t e x t{\right.$ 和 $\left.} \backslash q u a d x^{\wedge}=\right| f r a c{d x}{d z}=\backslash f r a c{x}{1-z / d}$

$$y^*=\frac{y}{1-z / d}$$

$$P[x y z 1]=\left[\begin{array}{llllllllllllllll} 1 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 / d & 1 \end{array}\right]\left[\begin{array}{ll} x y z & -1 \end{array}\right]=\left[\begin{array}{lll} x & 0 & 1 \end{array}\right.$$

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## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。