# 线性代数作业代写linear algebra代考|Definitions and examples

my-assignmentexpert™ 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|Eigenvalue, eigenvector

Let $A$ be a complex square matrix. Then if $\lambda$ is a complex number and $X$ a non-zero complex column vector satisfying $A X=\lambda X$, we call $X$ an eigenvector of $A$, while $\lambda$ is called an eigenvalue of $A$. We also say that $X$ is an eigenvector corresponding to the eigenvalue $\lambda$.

So in the above example $P_{1}$ and $P_{2}$ are eigenvectors corresponding to $\lambda_{1}$ and $\lambda_{2}$, respectively. We shall give an algorithm which starts from the eigenvalues of $A=\left[\begin{array}{ll}a & h \ h & b\end{array}\right]$ and constructs a rotation matrix $P$ such that $P^{t} A P$ is diagonal.

As noted above, if $\lambda$ is an eigenvalue of an $n \times n$ matrix $A$, with corresponding eigenvector $X$, then $\left(A-\lambda I_{n}\right) X=0$, with $X \neq 0$, so $\operatorname{det}\left(A-\lambda I_{n}\right)=0$ and there are at most $n$ distinct eigenvalues of $A$.

Conversely if $\operatorname{det}\left(A-\lambda I_{n}\right)=0$, then $\left(A-\lambda I_{n}\right) X=0$ has a non-trivial solution $X$ and so $\lambda$ is an eigenvalue of $A$ with $X$ a corresponding eigenvector.

## 线性代数作业代写linear algebra代考|Characteristic equation, polynomial

The equation $\operatorname{det}\left(A-\lambda I_{n}\right)=0$ is called the characteristic equation of $A$, while the polynomial $\operatorname{det}\left(A-\lambda I_{n}\right)$ is called the characteristic polynomial of $A$. The characteristic polynomial of $A$ is often denoted by $\operatorname{ch}_{A}(\lambda)$.

Hence the eigenvalues of $A$ are the roots of the characteristic polynomial of $A$.

For a $2 \times 2$ matrix $A=\left[\begin{array}{ll}a & b \ c & d\end{array}\right]$, it is easily verified that the characteristic polynomial is $\lambda^{2}-(\operatorname{trace} A) \lambda+\operatorname{det} A$, where trace $A=a+d$ is the sum of the diagonal elements of $A$.

## 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