# 线性代数作业代写linear algebra代考|EIGENVALUES AND EIGENVECTORS

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|Motivation

equal to 1 is a rotation matrix.
We can also solve for the new coordinates in terms of the old ones:
$$\left[\begin{array}{l} x_{1} \ y_{1} \end{array}\right]=Y=P^{t} X=\left[\begin{array}{rr} \cos \theta & \sin \theta \ -\sin \theta & \cos \theta \end{array}\right]\left[\begin{array}{l} x \ y \end{array}\right] \text {, }$$
so $x_{1}=x \cos \theta+y \sin \theta$ and $y_{1}=-x \sin \theta+y \cos \theta$. Then
$$X^{t} A X=(P Y)^{t} A(P Y)=Y^{t}\left(P^{t} A P\right) Y \text {. }$$
Now suppose, as we later show, that it is possible to choose an angle $\theta$ so that $P^{t} A P$ is a diagonal matrix, say $\operatorname{diag}\left(\lambda_{1}, \lambda_{2}\right)$. Then
$$X^{t} A X=\left[\begin{array}{ll} x_{1} & y_{1} \end{array}\right]\left[\begin{array}{cc} \lambda_{1} & 0 \ 0 & \lambda_{2} \end{array}\right]\left[\begin{array}{l} x_{1} \ y_{1} \end{array}\right]=\lambda_{1} x_{1}^{2}+\lambda_{2} y_{1}^{2}$$
and relative to the new axes, the equation $a x^{2}+2 h x y+b y^{2}=c$ becomes $\lambda_{1} x_{1}^{2}+\lambda_{2} y_{1}^{2}=c$, which is quite easy to sketch. This curve is symmetrical about the $x_{1}$ and $y_{1}$ axes, with $P_{1}$ and $P_{2}$, the respective columns of $P$, giving the directions of the axes of symmetry.
Also it can be verified that $P_{1}$ and $P_{2}$ satisfy the equations
$$A P_{1}=\lambda_{1} P_{1} \text { and } A P_{2}=\lambda_{2} P_{2} .$$
These equations force a restriction on $\lambda_{1}$ and $\lambda_{2}$. For if $P_{1}=\left[\begin{array}{l}u_{1} \ v_{1}\end{array}\right]$, the first equation becomes
$$\left[\begin{array}{ll} a & h \ h & b \end{array}\right]\left[\begin{array}{l} u_{1} \ v_{1} \end{array}\right]=\lambda_{1}\left[\begin{array}{l} u_{1} \ v_{1} \end{array}\right] \text { or }\left[\begin{array}{cc} a-\lambda_{1} & h \ h & b-\lambda_{1} \end{array}\right]\left[\begin{array}{l} u_{1} \ v_{1} \end{array}\right]=\left[\begin{array}{l} 0 \ 0 \end{array}\right]$$
Hence we are dealing with a homogeneous system of two linear equations in two unknowns, having a non-trivial solution $\left(u_{1}, v_{1}\right)$. Hence
$$\left|\begin{array}{cc} a-\lambda_{1} & h \ h & b-\lambda_{1} \end{array}\right|=0$$
Similarly, $\lambda_{2}$ satisfies the same equation. In expanded form, $\lambda_{1}$ and $\lambda_{2}$ satisfy
$$\lambda^{2}-(a+b) \lambda+a b-h^{2}=0$$
This equation has real roots
$$\lambda=\frac{a+b \pm \sqrt{(a+b)^{2}-4\left(a b-h^{2}\right)}}{2}=\frac{a+b \pm \sqrt{(a-b)^{2}+4 h^{2}}}{2}$$
(The roots are distinct if $a \neq b$ or $h \neq 0$. The case $a=b$ and $h=0$ needs no investigation, as it gives an equation of a circle.)

The equation $\lambda^{2}-(a+b) \lambda+a b-h^{2}=0$ is called the eigenvalue equation of the matrix $A$.

## 线性代数作业代写linear algebra代考|Motivation

[X1 和1]=和=磷吨X=[某物⁡θ没有⁡θ −没有⁡θ某物⁡θ][X 和],

X吨一种X=(磷和)吨一种(磷和)=和吨(磷吨一种磷)和.

X吨一种X=[X1和1][λ10 0λ2][X1 和1]=λ1X12+λ2和12

[一种H Hb][你1 v1]=λ1[你1 v1] 要么 [一种−λ1H Hb−λ1][你1 v1]=[0 0]

|一种−λ1H Hb−λ1|=0

λ2−(一种+b)λ+一种b−H2=0

λ=一种+b±(一种+b)2−4(一种b−H2)2=一种+b±(一种−b)2+4H22
（根是不同的，如果一种≠b要么H≠0. 案子一种=b和H=0不需要研究，因为它给出了一个圆的方程。）

# 计量经济学代写

## 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