Posted on Categories:Linear algebra, 数学代写, 线性代数

数学代写|线性代数代写Linear algebra代考|KMA254 Eigenvectors and Difference Equations

avatest™

avatest™帮您通过考试

avatest™的各个学科专家已帮了学生顺利通过达上千场考试。我们保证您快速准时完成各时长和类型的考试，包括in class、take home、online、proctor。写手整理各样的资源来或按照您学校的资料教您，创造模拟试题，提供所有的问题例子，以保证您在真实考试中取得的通过率是85%以上。如果您有即将到来的每周、季考、期中或期末考试，我们都能帮助您！

•最快12小时交付

•200+ 英语母语导师

•70分以下全额退款

数学代写|线性代数代写Linear algebra代考|Eigenvectors and Difference Equations

This section concludes by showing how to construct solutions of the first-order difference equation discussed in the chapter introductory example:
$$\mathbf{x}_{k+1}=A \mathbf{x}_k \quad(k=0,1,2, \ldots)$$
If $A$ is an $n \times n$ matrix, then (8) is a recursive description of a sequence $\left{\mathbf{x}_k\right}$ in $\mathbb{R}^n$. A solution of (8) is an explicit description of $\left{\mathbf{x}_k\right}$ whose formula for each $\mathbf{x}_k$ does not depend directly on $A$ or on the preceding terms in the sequence other than the initial term $\mathbf{x}_0$.

The simplest way to build a solution of (8) is to take an eigenvector $\mathbf{x}0$ and its corresponding eigenvalue $\lambda$ and let $$\mathbf{x}_k=\lambda^k \mathbf{x}_0 \quad(k=1,2, \ldots)$$ This sequence is a solution because $$A \mathbf{x}_k=A\left(\lambda^k \mathbf{x}_0\right)=\lambda^k\left(A \mathbf{x}_0\right)=\lambda^k\left(\lambda \mathbf{x}_0\right)=\lambda^{k+1} \mathbf{x}_0=\mathbf{x}{k+1}$$
Linear combinations of solutions in the form of equation (9) are solutions, too! See Exercise $33 .$

数学代写|线性代数代写Linear algebra代考|THE CHARACTERISTIC EQUATION

Useful information about the eigenvalues of a square matrix $A$ is encoded in a special scalar equation called the characteristic equation of $A$. A simple example will lead to the general case.
EXAMPLE 1 Find the eigenvalues of $A=\left[\begin{array}{rr}2 & 3 \ 3 & -6\end{array}\right]$.
SOLUTION We must find all scalars $\lambda$ such that the matrix equation
$$(A-\lambda I) \mathbf{x}=\mathbf{0}$$
has a nontrivial solution. By the Invertible Matrix Theorem in Section 2.3, this problem is equivalent to finding all $\lambda$ such that the matrix $A-\lambda I$ is not invertible, where
$$A-\lambda I=\left[\begin{array}{rr} 2 & 3 \ 3 & -6 \end{array}\right]-\left[\begin{array}{ll} \lambda & 0 \ 0 & \lambda \end{array}\right]=\left[\begin{array}{cc} 2-\lambda & 3 \ 3 & -6-\lambda \end{array}\right]$$
By Theorem 4 in Section 2.2, this matrix fails to be invertible precisely when its determinant is zero. So the eigenvalues of $A$ are the solutions of the equation
$$\operatorname{det}(A-\lambda I)=\operatorname{det}\left[\begin{array}{cc} 2-\lambda & 3 \ 3 & -6-\lambda \end{array}\right]=0$$
Recall that
$$\operatorname{det}\left[\begin{array}{ll} a & b \ c & d \end{array}\right]=a d-b c$$
So
\begin{aligned} \operatorname{det}(A-\lambda I) &=(2-\lambda)(-6-\lambda)-(3)(3) \ &=-12+6 \lambda-2 \lambda+\lambda^2-9 \ &=\lambda^2+4 \lambda-21 \ &=(\lambda-3)(\lambda+7) \end{aligned}
If $\operatorname{det}(A-\lambda I)=0$, then $\lambda=3$ or $\lambda=-7$. So the eigenvalues of $A$ are 3 and $-7$.
The determinant in Example 1 transformed the matrix equation $(A-\lambda I) \mathbf{x}=\mathbf{0}$, which involves two unknowns $(\lambda$ and $\mathbf{x})$, into the scalar equation $\lambda^2+4 \lambda-21=0$, which involves only one unknown. The same idea works for $n \times n$ matrices. However, before turning to larger matrices, we summarize the properties of determinants needed to study eigenvalues.

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。