## 线性代数作业代写linear algebra代考|A classification algorithm

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|Rotate the

Rotate the $\left(x_{1}, y_{1}\right)$ axes with the new positive $x_{2}$-axis in the direction of
$$\begin{gathered} {[(b-a} \ +4 h^{2} \end{gathered}$$
where $R=\sqrt{(a-b)^{2}+4 h^{2}}$. Then equation $7.12$ becomes
$$\lambda_{1} x_{2}^{2}+\lambda_{2} y_{2}^{2}=-\frac{\Delta}{C}$$
where
$$\lambda_{1}=(a+b-R) / 2, \lambda_{2}=(a+b+R) / 2$$
Here $\lambda_{1} \lambda_{2}=C$.
(a) $C<0$ : Hyperbola. Here $\lambda_{2}>0>\lambda_{1}$ and equation $7.13$ becomes
$$\frac{x_{2}^{2}}{u^{2}}-\frac{y_{2}^{2}}{v^{2}}=\frac{-\Delta}{|\Delta|},$$
where
$$u=\sqrt{\frac{|\Delta|}{C \lambda_{1}}}, v=\sqrt{\frac{|\Delta|}{-C \lambda_{2}}}$$
(b) $C>0$ and $a \Delta>0$ : Empty set.
(c) $C>0$ and $a \Delta<0$ : Ellipse.
Here $\lambda_{1}, \lambda_{2}, a, b$ have the same sign and $\lambda_{1} \neq \lambda_{2}$ and equation $7.13$ becomes
$$\frac{x_{2}^{2}}{u^{2}}+\frac{y_{2}^{2}}{v^{2}}=1$$
where
$$u=\sqrt{\frac{\Delta}{-C \lambda_{1}}}, v=\sqrt{\frac{\Delta}{-C \lambda_{2}}}$$

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

Identify the curve
$$x^{2}+2 x y+y^{2}++2 x+2 y+1=0 .$$
Solution. Here
$$\Delta=\left|\begin{array}{lll} 1 & 1 & 1 \ 1 & 1 & 1 \ 1 & 1 & 1 \end{array}\right|=0 .$$
Let $x=x_{1}+\alpha, y=y_{1}+\beta$ and substitute in equation $7.16$ to get $\left(x_{1}+\alpha\right)^{2}+2\left(x_{1}+\alpha\right)\left(y_{1}+\beta\right)+\left(y_{1}+\beta\right)^{2}+2\left(x_{1}+\alpha\right)+2\left(y_{1}+\beta\right)+1=0 .$
Then equating the coefficients of $x_{1}$ and $y_{1}$ to 0 gives the same equation
$$2 \alpha+2 \beta+2=0 .$$
Take $\alpha=0, \beta=-1$. Then equation $7.17$ simplifies to
$$x_{1}^{2}+2 x_{1} y_{1}+y_{1}^{2}=0=\left(x_{1}+y_{1}\right)^{2},$$
and in terms of $x, y$ coordinates, equation $7.16$ becomes
$$(x+y+1)^{2}=0, \text { or } x+y+1=0 .$$

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

Rotate the (x1,y1) axes with the new positive x2-axis in the direction of
[(b−a +4h2
where R=(a−b)2+4h2. Then equation 7.12 becomes
λ1×22+λ2y22=−ΔC
where
λ1=(a+b−R)/2,λ2=(a+b+R)/2
Here λ1λ2=C.
(a) C<0 : Hyperbola. Here λ2>0>λ1 and equation 7.13 becomes
x22u2−y22v2=−Δ|Δ|,
where
u=|Δ|Cλ1,v=|Δ|−Cλ2
(b) C>0 and aΔ>0 : Empty set.
(c) C>0 and aΔ<0 : Ellipse.
Here λ1,λ2,a,b have the same sign and λ1≠λ2 and equation 7.13 becomes
x22u2+y22v2=1
where
u=Δ−Cλ1,v=Δ−Cλ2

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

Identify the curve
x2+2xy+y2++2x+2y+1=0.
Solution. Here
Δ=|111 111 111|=0.
Let x=x1+α,y=y1+β and substitute in equation 7.16 to get (x1+α)2+2(x1+α)(y1+β)+(y1+β)2+2(x1+α)+2(y1+β)+1=0.
Then equating the coefficients of x1 and y1 to 0 gives the same equation
2α+2β+2=0.
Take α=0,β=−1. Then equation 7.17 simplifies to
x12+2x1y1+y12=0=(x1+y1)2,
and in terms of x,y coordinates, equation 7.16 becomes
(x+y+1)2=0, or x+y+1=0.

# 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

## 线性代数作业代写linear algebra代考|Identifying second degree equations

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|The eigenvalue method

In this section we apply eigenvalue methods to determine the geometrical nature of the second degree equation
$$a x^{2}+2 h x y+b y^{2}+2 g x+2 f y+c=0,$$
where not all of $a, h, b$ are zero.
Let $A=\left[\begin{array}{ll}a & h \ h & b\end{array}\right]$ be the matrix of the quadratic form $a x^{2}+2 h x y+b y^{2}$. We saw in section 6.1, equation $6.2$ that $A$ has real eigenvalues $\lambda_{1}$ and $\lambda_{2}$, given by
$$\lambda_{1}=\frac{a+b-\sqrt{(a-b)^{2}+4 h^{2}}}{2}, \lambda_{2}=\frac{a+b+\sqrt{(a-b)^{2}+4 h^{2}}}{2} .$$
We show that it is always possible to rotate the $x, y$ axes to $x_{1}, x_{2}$ axes whose positive directions are determined by eigenvectors $X_{1}$ and $X_{2}$ corresponding to $\lambda_{1}$ and $\lambda_{2}$ in such a way that relative to the $x_{1}, y_{1}$ axes, equation $7.1$ takes the form
$$a^{\prime} x^{2}+b^{\prime} y^{2}+2 g^{\prime} x+2 f^{\prime} y+c=0 .$$
Then by completing the square and suitably translating the $x_{1}, y_{1}$ axes, to new $x_{2}, y_{2}$ axes, equation $7.2$ can be reduced to one of several standard forms, each of which is easy to sketch. We need some preliminary definitions.

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

An $n \times n$ real matrix $P$ is called orthogonal if
$$P^{t} P=I_{n} .$$
It follows that if $P$ is orthogonal, then $\operatorname{det} P=\pm 1$. For
$$\operatorname{det}\left(P^{t} P\right)=\operatorname{det} P^{t} \operatorname{det} P=(\operatorname{det} P)^{2},$$
so $(\operatorname{det} P)^{2}=\operatorname{det} I_{n}=1$. Hence $\operatorname{det} P=\pm 1$.
If $P$ is an orthogonal matrix with $\operatorname{det} P=1$, then $P$ is called a proper orthogonal matrix.

THEOREM 7.1.1 If $P$ is a $2 \times 2$ orthogonal matrix with $\operatorname{det} P=1$, then
$$P=\left[\begin{array}{rr} \cos \theta & -\sin \theta \ \sin \theta & \cos \theta \end{array}\right]$$
for some $\theta$.
REMARK 7.1.1 Hence, by the discusssion at the beginning of Chapter 6 , if $P$ is a proper orthogonal matrix, the coordinate transformation
$$\left[\begin{array}{l} x \ y \end{array}\right]=P\left[\begin{array}{l} x_{1} \ y_{1} \end{array}\right]$$
represents a rotation of the axes, with new $x_{1}$ and $y_{1}$ axes given by the repective columns of $P$.

## 线性代数作业代写linear algebra代考|The eigenvalue method

λ1=一种+b−(一种−b)2+4H22,λ2=一种+b+(一种−b)2+4H22.

[X 和]=磷[X1 和1]

# 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

## 线性代数作业代写linear algebra代考|PROBLEMS

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|singular matrix

1. Let $A=\left[\begin{array}{rr}4 & -3 \ 1 & 0\end{array}\right]$. Find a non-singular matrix $P$ such that $P^{-1} A P=$ $\operatorname{diag}(1,3)$ and hence prove that
$$A^{n}=\frac{3^{n}-1}{2} A+\frac{3-3^{n}}{2} I_{2} .$$
2. If $A=\left[\begin{array}{ll}0.6 & 0.8 \ 0.4 & 0.2\end{array}\right]$, prove that $A^{n}$ tends to a limiting matrix
$$\left[\begin{array}{ll} 2 / 3 & 2 / 3 \ 1 / 3 & 1 / 3 \end{array}\right]$$
as $n \rightarrow \infty$.

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

1. Solve the system of differential equations
\begin{aligned} &\frac{d x}{d t}=3 x-2 y \ &\frac{d y}{d t}=5 x-4 y, \end{aligned}
given $x=13$ and $y=22$ when $t=0$.
[Answer: $x=7 e^{t}+6 e^{-2 t}, y=7 e^{t}+15 e^{-2 t}$.]
2. Solve the system of recurrence relations
\begin{aligned} x_{n+1} &=3 x_{n}-y_{n} \ y_{n+1} &=-x_{n}+3 y_{n}, \end{aligned}
given that $x_{0}=1$ and $y_{0}=2$.
[Answer: $x_{n}=2^{n-1}\left(3-2^{n}\right), y_{n}=2^{n-1}\left(3+2^{n}\right)$.]
3. Let $A=\left[\begin{array}{ll}a & b \ c & d\end{array}\right]$ be a real or complex matrix with distinct eigenvalues $\lambda_{1}, \lambda_{2}$ and corresponding eigenvectors $X_{1}, X_{2}$. Also let $P=\left[X_{1} \mid X_{2}\right]$.
(a) Prove that the system of recurrence relations
\begin{aligned} &x_{n+1}=a x_{n}+b y_{n} \ &y_{n+1}=c x_{n}+d y_{n} \end{aligned}
has the solution
$$\left[\begin{array}{l} x_{n} \ y_{n} \end{array}\right]=\alpha \lambda_{1}^{n} X_{1}+\beta \lambda_{2}^{n} X_{2},$$
where $\alpha$ and $\beta$ are determined by the equation
$$\left[\begin{array}{l} \alpha \ \beta \end{array}\right]=P^{-1}\left[\begin{array}{l} x_{0} \ y_{0} \end{array}\right] .$$
(b) Prove that the system of differential equations
\begin{aligned} &\frac{d x}{d t}=a x+b y \ &\frac{d y}{d t}=c x+d y \end{aligned}
has the solution
$$\left[\begin{array}{l} x \ y \end{array}\right]=\alpha e^{\lambda_{1} t} X_{1}+\beta e^{\lambda_{2} t} X_{2},$$

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

1. 让一种=[4−3 10]. 找到一个非奇异矩阵磷这样磷−1一种磷= 诊断⁡(1,3)并因此证明
一种n=3n−12一种+3−3n2一世2.
2. 如果一种=[0.60.8 0.40.2]， 证明一种n趋于极限矩阵
[2/32/3 1/31/3]
作为n→∞.

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

1. 求解微分方程组
dXd吨=3X−2和 d和d吨=5X−4和,
给定X=13和和=22什么时候吨=0.
[回答：X=7和吨+6和−2吨,和=7和吨+15和−2吨.]
2. 解决递归关系系统
Xn+1=3Xn−和n 和n+1=−Xn+3和n,
鉴于X0=1和和0=2.
[回答：Xn=2n−1(3−2n),和n=2n−1(3+2n).]
3. 让一种=[一种b Cd]是具有不同特征值的实数或复数矩阵λ1,λ2和相应的特征向量X1,X2. 也让磷=[X1∣X2].
(a) 证明递推关系系统
Xn+1=一种Xn+b和n 和n+1=CXn+d和n
有解决办法
[Xn 和n]=一种λ1nX1+bλ2nX2,
在哪里一种和b由等式确定
[一种 b]=磷−1[X0 和0].
(b) 证明微分方程组
dXd吨=一种X+b和 d和d吨=CX+d和
有解决办法
[X 和]=一种和λ1吨X1+b和λ2吨X2,

# 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

## 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 $A$.

# 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

## 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

## 线性代数作业代写linear algebra代考|PROBLEMS

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|subspaces

1. Which of the following subsets of $\mathbb{R}^{2}$ are subspaces?
(a) $[x, y]$ satisfying $x=2 y$;
(b) $[x, y]$ satisfying $x=2 y$ and $2 x=y$;
(c) $[x, y]$ satisfying $x=2 y+1$;
(d) $[x, y]$ satisfying $x y=0$;
CHAPTER 3. SUBSPACES
(e) $[x, y]$ satisfying $x \geq 0$ and $y \geq 0$.
2. If $X, Y, Z$ are vectors in $\mathbb{R}^{n}$, prove that
$$\langle X, Y, Z\rangle=\langle X+Y, X+Z, Y+Z\rangle$$
3. Determine if $X_{1}=\left[\begin{array}{l}1 \ 0 \ 1 \ 2\end{array}\right], X_{2}=\left[\begin{array}{l}0 \ 1 \ 1 \ 2\end{array}\right]$ and $X_{3}=\left[\begin{array}{l}1 \ 1 \ 1 \ 3\end{array}\right]$ are linearly independent in $\mathbb{R}^{4}$.
4. For which real numbers $\lambda$ are the following vectors linearly independent in $\mathbb{R}^{3}$ ?
$$X_{1}=\left[\begin{array}{r} \lambda \ -1 \ -1 \end{array}\right], \quad X_{2}=\left[\begin{array}{r} -1 \ \lambda \ -1 \end{array}\right], \quad X_{3}=\left[\begin{array}{r} -1 \ -1 \ \lambda \end{array}\right]$$
5. Find bases for the row, column and null spaces of the following matrix over $\mathbb{Q}$ :
$$A=\left[\begin{array}{ccccc} 1 & 1 & 2 & 0 & 1 \ 2 & 2 & 5 & 0 & 3 \ 0 & 0 & 0 & 1 & 3 \ 8 & 11 & 19 & 0 & 11 \end{array}\right]$$
6. Find bases for the row, column and null spaces of the following matrix over $\mathbb{Z}_{2}$ :
$$A=\left[\begin{array}{lllll} 1 & 0 & 1 & 0 & 1 \ 0 & 1 & 0 & 1 & 1 \ 1 & 1 & 1 & 1 & 0 \ 0 & 0 & 1 & 1 & 0 \end{array}\right]$$
7. Find bases for the row, column and null spaces of the following matrix over $\mathbb{Z}_{5}$ :
$$A=\left[\begin{array}{llllll} 1 & 1 & 2 & 0 & 1 & 3 \ 2 & 1 & 4 & 0 & 3 & 2 \ 0 & 0 & 0 & 1 & 3 & 0 \ 3 & 0 & 2 & 4 & 3 & 2 \end{array}\right]$$

## 线性代数作业代写linear algebra代考|Find bases for the row

1. Find bases for the row, column and null spaces of the matrix $A$ defined in section 1.6, Problem 17. (Note: In this question, $F$ is a field of four elements.)
2. If $X_{1}, \ldots, X_{m}$ form a basis for a subspace $S$, prove that
$$X_{1}, X_{1}+X_{2}, \ldots, X_{1}+\cdots+X_{m}$$
also form a basis for $S$.
3. Let $A=\left[\begin{array}{ccc}a & b & c \ 1 & 1 & 1\end{array}\right]$. Find conditions on $a, b, c$ such that (a) rank $A=$ 1; (b) $\operatorname{rank} A=2$.
[Answer: (a) $a=b=c ;$ (b) at least two of $a, b, c$ are distinct.]
4. Let $S$ be a subspace of $F^{n}$ with $\operatorname{dim} S=m$. If $X_{1}, \ldots, X_{m}$ are vectors in $S$ with the property that $S=\left\langle X_{1}, \ldots, X_{m}\right\rangle$, prove that $X_{1} \ldots, X_{m}$ form a basis for $S$.
5. Find a basis for the subspace $S$ of $\mathbb{R}^{3}$ defined by the equation
$$x+2 y+3 z=0$$
Verify that $Y_{1}=[-1,-1,1]^{t} \in S$ and find a basis for $S$ which includes $Y_{1}$.
6. Let $X_{1}, \ldots, X_{m}$ be vectors in $F^{n}$. If $X_{i}=X_{j}$, where $i<j$, prove that $X_{1}, \ldots X_{m}$ are linearly dependent.
7. Let $X_{1}, \ldots, X_{m+1}$ be vectors in $F^{n}$. Prove that
$$\operatorname{dim}\left\langle X_{1}, \ldots, X_{m+1}\right\rangle=\operatorname{dim}\left\langle X_{1}, \ldots, X_{m}\right\rangle$$
if $X_{m+1}$ is a linear combination of $X_{1}, \ldots, X_{m}$, but
$$\operatorname{dim}\left\langle X_{1}, \ldots, X_{m+1}\right\rangle=\operatorname{dim}\left\langle X_{1}, \ldots, X_{m}\right\rangle+1$$
if $X_{m+1}$ is not a linear combination of $X_{1}, \ldots, X_{m}$.
Deduce that the system of linear equations $A X=B$ is consistent, if and only if
$$\operatorname{rank}[A \mid B]=\operatorname{rank} A$$

## 线性代数作业代写linear algebra代考|subspaces

1. 以下哪个子集R2是子空间？
（一种）[X,和]令人满意的X=2和;
(二)[X,和]令人满意的X=2和和2X=和;
（C）[X,和]令人满意的X=2和+1;
(d)[X,和]令人满意的X和=0;
第 3 章子空间
(e)[X,和]令人满意的X≥0和和≥0.
[答案：(a) 和 (b)。]
2. 如果X,和,和是向量Rn， 证明
⟨X,和,和⟩=⟨X+和,X+和,和+和⟩
3. 确定是否X1=[1 0 1 2],X2=[0 1 1 2]和X3=[1 1 1 3]是线性独立的R4.
4. 对于哪些实数λ以下向量是线性独立的R3?
X1=[λ −1 −1],X2=[−1 λ −1],X3=[−1 −1 λ]
5. 查找以下矩阵的行、列和零空间的基数问:
一种=[11201 22503 00013 81119011]
6. 查找以下矩阵的行、列和零空间的基数和2:
一种=[10101 01011 11110 00110]
7. 查找以下矩阵的行、列和零空间的基数和5:
一种=[112013 214032 000130 302432]

## 线性代数作业代写linear algebra代考|Find bases for the row

1. 查找矩阵的行、列和零空间的基一种在第 1.6 节，问题 17 中定义。（注意：在这个问题中，F是四个元素的字段。）
2. 如果X1,…,X米形成子空间的基础小号， 证明
X1,X1+X2,…,X1+⋯+X米
也形成了基础小号.
3. 让一种=[一种bC 111]. 查找条件一种,b,C使得 (a) 排名一种=1个；(二)秩⁡一种=2.
[答案：（一）一种=b=C;(b) 至少两个一种,b,C是不同的。]
4. 让小号是一个子空间Fn和不⁡小号=米. 如果X1,…,X米是向量小号与财产小号=⟨X1,…,X米⟩， 证明X1…,X米形成一个基础小号.
5. 找到子空间的基础小号的R3由等式定义
X+2和+3和=0
验证和1=[−1,−1,1]吨∈小号并找到依据小号包括和1.
6. 让X1,…,X米成为向量Fn. 如果X一世=Xj， 在哪里一世<j， 证明X1,…X米是线性相关的。
7. 让X1,…,X米+1成为向量Fn. 证明
不⁡⟨X1,…,X米+1⟩=不⁡⟨X1,…,X米⟩
如果X米+1是一个线性组合X1,…,X米， 但
不⁡⟨X1,…,X米+1⟩=不⁡⟨X1,…,X米⟩+1
如果X米+1不是的线性组合X1,…,X米.
推导出线性方程组一种X=乙是一致的，当且仅当
秩⁡[一种∣乙]=秩⁡一种

# 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

## 线性代数作业代写linear algebra代考|Rank and nullity of a matrix

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|Every linearly independent

Every linearly independent family of vectors in a subspace $S$ can be extended to a basis of $S$.

Proof. Suppose $S$ has basis $X_{1}, \ldots, X_{m}$ and that $Y_{1}, \ldots, Y_{r}$ is a linearly independent family of vectors in $S$. Then
$$S=\left\langle X_{1}, \ldots, X_{m}\right\rangle=\left\langle Y_{1}, \ldots, Y_{r}, X_{1}, \ldots, X_{m}\right\rangle$$
as each of $Y_{1}, \ldots, Y_{r}$ is a linear combination of $X_{1}, \ldots, X_{m}$.
Then applying the left-to-right algorithm to the second spanning family for $S$ will yield a basis for $S$ which includes $Y_{1}, \ldots, Y_{r}$.

## 线性代数作业代写linear algebra代考|Given that the reduced row–echelon form of

Given that the reduced row-echelon form of
$$A=\left[\begin{array}{rrrrr} 1 & 1 & 5 & 1 & 4 \ 2 & -1 & 1 & 2 & 2 \ 3 & 0 & 6 & 0 & -3 \end{array}\right]$$
equal to
$$B=\left[\begin{array}{rrrrr} 1 & 0 & 2 & 0 & -1 \ 0 & 1 & 3 & 0 & 2 \ 0 & 0 & 0 & 1 & 3 \end{array}\right]$$
find bases for $R(A), C(A)$ and $N(A)$.
Solution. $[1,0,2,0,-1],[0,1,3,0,2]$ and $[0,0,0,1,3]$ form a basis for $R(A)$. Also
$$A_{* 1}=\left[\begin{array}{l} 1 \ 2 \ 3 \end{array}\right], A_{* 2}=\left[\begin{array}{r} 1 \ -1 \ 0 \end{array}\right], A_{* 4}=\left[\begin{array}{l} 1 \ 2 \ 0 \end{array}\right]$$
form a basis for $C(A)$.
Finally $N(A)$ is given by
$$\left[\begin{array}{l} x_{1} \ x_{2} \ x_{3} \ x_{4} \ x_{5} \end{array}\right]=\left[\begin{array}{c} -2 x_{3}+x_{5} \ -3 x_{3}-2 x_{5} \ x_{3} \ -3 x_{5} \ x_{5} \end{array}\right]=x_{3}\left[\begin{array}{r} -2 \ -3 \ 1 \ 0 \ 0 \end{array}\right]+x_{5}\left[\begin{array}{r} 1 \ -2 \ 0 \ -3 \ 1 \end{array}\right]=x_{3} X_{1}+x_{5}$$
where $x_{3}$ and $x_{5}$ are arbitrary. Hence $X_{1}$ and $X_{2}$ form a basis for $N(A)$. Here $\operatorname{rank} A=3$ and nullity $A=2$.

## 线性代数作业代写linear algebra代考|Given that the reduced row–echelon form of

[X1 X2 X3 X4 X5]=[−2X3+X5 −3X3−2X5 X3 −3X5 X5]=X3[−2 −3 1 0 0]+X5[1 −2 0 −3 1]=X3X1+X5

# 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

## 线性代数作业代写linear algebra代考|Basis of a subspace

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|belonging

We now come to the important concept of basis of a vector subspace.
DEFINITION 3.4.1 Vectors $X_{1}, \ldots, X_{m}$ belonging to a subspace $S$ are said to form a basis of $S$ if
(a) Every vector in $S$ is a linear combination of $X_{1}, \ldots, X_{m}$;
(b) $X_{1}, \ldots, X_{m}$ are linearly independent.
Note that (a) is equivalent to the statement that $S=\left\langle X_{1}, \ldots, X_{m}\right\rangle$ as we automatically have $\left\langle X_{1}, \ldots, X_{m}\right\rangle \subseteq S$. Also, in view of Remark 3.3.1 above, (a) and (b) are equivalent to the statement that every vector in $S$ is uniquely expressible as a linear combination of $X_{1}, \ldots, X_{m}$.

## 线性代数作业代写linear algebra代考|subspace

THEOREM 3.4.2 Any two bases for a subspace $S$ must contain the same number of elements.

Proof. For if $X_{1}, \ldots, X_{r}$ and $Y_{1}, \ldots, Y_{s}$ are bases for $S$, then $Y_{1}, \ldots, Y_{s}$ form a linearly independent family in $S=\left\langle X_{1}, \ldots, X_{r}\right\rangle$ and hence $s \leq r$ by Theorem 3.3.2. Similarly $r \leq s$ and hence $r=s$.

DEFINITION 3.4.2 This number is called the dimension of $S$ and is written $\operatorname{dim} S$. Naturally we define $\operatorname{dim}{0}=0$.

It follows from Theorem 3.3.1 that for any subspace $S$ of $F^{n}$, we must have $\operatorname{dim} S \leq n .$

EXAMPLE 3.4.3 If $E_{1}, \ldots, E_{n}$ denote the $n$-dimensional unit vectors in $F^{n}$, then $\operatorname{dim}\left\langle E_{1}, \ldots, E_{i}\right\rangle=i$ for $1 \leq i \leq n$.
The following result gives a useful way of exhibiting a basis.

## 线性代数作业代写linear algebra代考|belonging

(a) 中的每个向量小号是一个线性组合X1,…,X米;
(二)X1,…,X米是线性独立的。

# 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

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

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|independence

We now recall the definition of linear dependence and independence of a family of vectors in $F^{n}$ given in Chapter $2 .$

DEFINITION 3.3.1 Vectors $X_{1}, \ldots, X_{m}$ in $F^{n}$ are said to be linearly dependent if there exist scalars $x_{1}, \ldots, x_{m}$, not all zero, such that
$$x_{1} X_{1}+\cdots+x_{m} X_{m}=0$$
In other words, $X_{1}, \ldots, X_{m}$ are linearly dependent if some $X_{i}$ is expressible as a linear combination of the remaining vectors.
$X_{1}, \ldots, X_{m}$ are called linearly independent if they are not linearly dependent. Hence $X_{1}, \ldots, X_{m}$ are linearly independent if and only if the equation
$$x_{1} X_{1}+\cdots+x_{m} X_{m}=0$$
has only the trivial solution $x_{1}=0, \ldots, x_{m}=0$.
EXAMPLE 3.3.1 The following three vectors in $\mathbb{R}^{3}$
$$X_{1}=\left[\begin{array}{l} 1 \ 2 \ 3 \end{array}\right], \quad X_{2}=\left[\begin{array}{r} -1 \ 1 \ 2 \end{array}\right], \quad X_{3}=\left[\begin{array}{r} -1 \ 7 \ 12 \end{array}\right]$$
are linearly dependent, as $2 X_{1}+3 X_{2}+(-1) X_{3}=0$.

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

REMARK 3.3.1 If $X_{1}, \ldots, X_{m}$ are linearly independent and
$$x_{1} X_{1}+\cdots+x_{m} X_{m}=y_{1} X_{1}+\cdots+y_{m} X_{m},$$
then $x_{1}=y_{1}, \ldots, x_{m}=y_{m}$. For the equation can be rewritten as
$$\left(x_{1}-y_{1}\right) X_{1}+\cdots+\left(x_{m}-y_{m}\right) X_{m}=0$$
and so $x_{1}-y_{1}=0, \ldots, x_{m}-y_{m}=0$.
THEOREM 3.3.1 A family of $m$ vectors in $F^{n}$ will be linearly dependent if $m>n$. Equivalently, any linearly independent family of $m$ vectors in $F^{n}$ must satisfy $m \leq n$.
Proof. The equation
$$x_{1} X_{1}+\cdots+x_{m} X_{m}=0$$
is equivalent to $n$ homogeneous equations in $m$ unknowns. By Theorem 1.5.1, such a system has a non-trivial solution if $m>n$.

The following theorem is an important generalization of the last result and is left as an exercise for the interested student:

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

X1X1+⋯+X米X米=0

X1,…,X米如果它们不是线性相关的，则称为线性无关。因此X1,…,X米当且仅当方程是线性独立的
X1X1+⋯+X米X米=0

X1=[1 2 3],X2=[−1 1 2],X3=[−1 7 12]

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

X1X1+⋯+X米X米=和1X1+⋯+和米X米,

(X1−和1)X1+⋯+(X米−和米)X米=0

X1X1+⋯+X米X米=0

# 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

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

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|Subspaces of F^

A subset $S$ of $F^{n}$ is called a subspace of $F^{n}$ if

1. The zero vector belongs to $S$; (that is, $0 \in S$ );
2. If $u \in S$ and $v \in S$, then $u+v \in S$; (S is said to be closed under vector addition);
3. If $u \in S$ and $t \in F$, then $t u \in S ;(S$ is said to be closed under scalar multiplication).

Let $A \in M_{m \times n}(F)$. Then the set of vectors $X \in F^{n}$ satisfying $A X=0$ is a subspace of $F^{n}$ called the null space of $A$ and is denoted here by $N(A)$. (It is sometimes called the solution space of $A$.)
Proof. (1) $A 0=0$, so $0 \in N(A)$; (2) If $X, Y \in N(A)$, then $A X=0$ and $A Y=0$, so $A(X+Y)=A X+A Y=0+0=0$ and so $X+Y \in N(A)$; (3) If $X \in N(A)$ and $t \in F$, then $A(t X)=t(A X)=t 0=0$, so $t X \in N(A)$.
For example, if $A=\left[\begin{array}{ll}1 & 0 \ 0 & 1\end{array}\right]$, then $N(A)={0}$, the set consisting of just the zero vector. If $A=\left[\begin{array}{ll}1 & 2 \ 2 & 4\end{array}\right]$, then $N(A)$ is the set of all scalar multiples of $[-2,1]^{t}$.

## 线性代数作业代写linear algebra代考|consisting of all linear combinations

Let $X_{1}, \ldots, X_{m} \in F^{n}$. Then the set consisting of all linear combinations $x_{1} X_{1}+\cdots+x_{m} X_{m}$, where $x_{1}, \ldots, x_{m} \in F$, is a subspace of $F^{n}$. This subspace is called the subspace spanned or generated by $X_{1}, \ldots, X_{m}$ and is denoted here by $\left\langle X_{1}, \ldots, X_{m}\right\rangle$. We also call $X_{1}, \ldots, X_{m}$ a spanning family for $S=\left\langle X_{1}, \ldots, X_{m}\right\rangle$.

Proof. (1) $0=0 X_{1}+\cdots+0 X_{m}$, so $0 \in\left\langle X_{1}, \ldots, X_{m}\right\rangle$; (2) If $X, Y \in$ $\left\langle X_{1}, \ldots, X_{m}\right\rangle$, then $X=x_{1} X_{1}+\cdots+x_{m} X_{m}$ and $Y=y_{1} X_{1}+\cdots+y_{m} X_{m}$, so
\begin{aligned} X+Y &=\left(x_{1} X_{1}+\cdots+x_{m} X_{m}\right)+\left(y_{1} X_{1}+\cdots+y_{m} X_{m}\right) \ &=\left(x_{1}+y_{1}\right) X_{1}+\cdots+\left(x_{m}+y_{m}\right) X_{m} \in\left\langle X_{1}, \ldots, X_{m}\right\rangle \end{aligned}
(3) If $X \in\left\langle X_{1}, \ldots, X_{m}\right\rangle$ and $t \in F$, then
\begin{aligned} X &=x_{1} X_{1}+\cdots+x_{m} X_{m} \ t X &=t\left(x_{1} X_{1}+\cdots+x_{m} X_{m}\right) \ &=\left(t x_{1}\right) X_{1}+\cdots+\left(t x_{m}\right) X_{m} \in\left\langle X_{1}, \ldots, X_{m}\right\rangle \end{aligned}
For example, if $A \in M_{m \times n}(F)$, the subspace generated by the columns of $A$ is an important subspace of $F^{m}$ and is called the column space of $A$. The column space of $A$ is denoted here by $C(A)$. Also the subspace generated by the rows of $A$ is a subspace of $F^{n}$ and is called the row space of $A$ and is denoted by $R(A)$.

## 线性代数作业代写linear algebra代考|Subspaces of F^

1. 零向量属于小号; （那是，0∈小号);
2. 如果你∈小号和v∈小号， 然后你+v∈小号; （据说 S 在向量加法下是闭合的）；
3. 如果你∈小号和吨∈F， 然后吨你∈小号;(小号据说在标量乘法下是封闭的）。

## 线性代数作业代写linear algebra代考|consisting of all linear combinations

X+和=(X1X1+⋯+X米X米)+(和1X1+⋯+和米X米) =(X1+和1)X1+⋯+(X米+和米)X米∈⟨X1,…,X米⟩
(3) 如果X∈⟨X1,…,X米⟩和吨∈F， 然后
X=X1X1+⋯+X米X米 吨X=吨(X1X1+⋯+X米X米) =(吨X1)X1+⋯+(吨X米)X米∈⟨X1,…,X米⟩

# 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

## 美本或者加拿大本科，如果需要期末考试之前突击线性代数，怎样可以效率最大化？

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