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## 数学代写|线性代数代写Linear algebra代考|Properties of the range of a linear transformation

Note that the range of a linear transform $T: V \rightarrow W$ is composed of vectors in the arrival vector space $W$. (The vectors we arrive at in $W$ after applying $T$ to vectors in $V$.) Next we show that the range of $T$ is a subspace of the arrival vector space $W$.

Proposition (5.10). Let $V$ and $W$ be vector spaces and $T: V \rightarrow W$ be a linear transformation. The range or image of the transformation $T$ is a subspace of the arrival vector space $W$.
What does this mean?
The shaded part in $W$ in Fig. 5.16 is a subspace of $W$.
How do we prove this proposition?
Again, by using Proposition (3.7) of chapter 3:
A non-empty subset $S$ with vectors $\mathbf{u}$ and $\mathbf{v}$ is a subspace $\Leftrightarrow k \mathbf{u}+c \mathbf{v}$ is also in $S$.
Proof.
For a linear transform we have $T(\mathbf{O})=\mathbf{O}$, therefore $\mathbf{O}$ is in the set of $\operatorname{range}(T)$, which means the range is non-empty.
How do we prove that range $(T)$ is a subspace of the arrival space $W$ ?
By showing any linear combination is also in the range. Let $\mathbf{u}$ and $\mathbf{w}$ be vectors in $\operatorname{range}(T)$ then we need to show that
$$k \mathbf{u}+c \mathbf{w} \text { is also in range }(T) \quad[k \text { and } c \text { are scalars }]$$

Required to prove that the set $\operatorname{range}(T)$ is closed, which means that $k \mathbf{u}+c \mathbf{w}$ cannot escape from range $(T)$.

Since $\mathbf{u}$ and $\mathbf{w}$ are in $\operatorname{range}(T)$, there must exist input vectors $\mathbf{x}$ in $V$ and $\mathbf{y}$ in $V$ such that
$$T(x)=\mathbf{u} \text { and } T(\mathbf{y})=\mathbf{w}$$

## 数学代写|线性代数代写Linear algebra代考|Rank of a linear transformation (mapping)

The rank of a linear transformation tells us how much information has been transformed over and is measured as the dimension of the range. The rank also tells us whether information has been lost by the linear transform.

Definition (5.11). Let $T: V \rightarrow W$ be a linear transform (map) and range $(T)$ be the range. Then the dimension of range $(T)$ is called the rank of $T$ denoted $\operatorname{rank}(T)$ (Fig. 5.21).

Next we state one of the most important results of linear algebra. The proof of this theorem is given towards the end of the section because it is long and requires you to recall some definitions given in previous chapters.

Dimension theorem (also called the rank-nullity theorem) (5.12). Let $T: V \rightarrow W$ be a linear transformation from an $n$-dimensional vector space $V$ to a vector space $W$. Then
$$\operatorname{rank}(T)+\operatorname{nullity}(T)=n$$
What does this formula mean? It means
$$\operatorname{dim}(\operatorname{range}(T))+\operatorname{dim}(\operatorname{ker}(T))=n$$
where dim represents dimension. This is same as the dimension theorem (3.34) of chapter 3 . This suggests that adding the dimension of the range and kernel gives the dimension of the start vector space $V$. As stated earlier, this result says that all the information is contained in these two sets – kernel and range.

Why is the rank of a linear transformation important?
The rank gives us how much information has been carried over by the transform. If the rank of the linear transform $T: V \rightarrow W$ is equal to the dimension of the start vector space $V$ then all the information has been moved over and we can go back; that is, the linear transform has an inverse. We have:

1. If $\operatorname{rank}(T)=\operatorname{dim}(V)$ then all the information has been carried over by $T$.
2. If $\operatorname{rank}(T)<\operatorname{dim}(V)$ then some information has been lost by $T$.
3. If $\operatorname{rank}(T)=0<\operatorname{dim}(V)$ then virtually all the information has been lost by $T$.

## 数学代写|线性代数代写Linear algebra代考|Properties of the range of a linear transformation

$$k \mathbf{u}+c \mathbf{w} \text { is also in range }(T) \quad[k \text { and } c \text { are scalars }]$$

$$T(x)=\mathbf{u} \text { and } T(\mathbf{y})=\mathbf{w}$$

## 数学代写|线性代数代写Linear algebra代考|Rank of a linear transformation (mapping)

$$\operatorname{rank}(T)+\operatorname{nullity}(T)=n$$

$$\operatorname{dim}(\operatorname{range}(T))+\operatorname{dim}(\operatorname{ker}(T))=n$$

## 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。

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## 数学代写|线性代数代写Linear algebra代考|Orthogonal matrices

One of the most tedious problems in linear algebra is to find the inverse of a matrix. For a 3 by 3 or larger matrix, it becomes a monotonous task to determine the inverse by hand calculations. However, there is one set of matrices called orthogonal matrices where the inverse can be obtained by transposition of the matrix. It is straightforward to find the transpose of a matrix.

Orthogonal matrices arise naturally when working with orthonormal bases. Working with orthonormal bases is very handy because it allows you to use a formula like Pythagoras or it allows you to work in the field of Fourier series.

Orthogonal matrices are important in subjects such as numerical analysis because these matrices have good numerical stability.

Definition (4.18). A square matrix $\mathbf{Q}=\left(\begin{array}{lllll}\mathbf{v}_1 & \mathbf{v}_2 & \mathbf{v}_3 & \cdots & \mathbf{v}_n\end{array}\right)$, whose columns $\mathbf{v}_1, \mathbf{v}_2$, $\mathbf{v}_3, \ldots, \mathbf{v}_n$ are orthonormal (perpendicular unit) vectors, is called an orthogonal matrix.
An example of an orthogonal matrix is the identity matrix.

## 数学代写|线性代数代写Linear algebra代考|What do you notice about your final result in the above example?

We end up with the identity matrix $\mathbf{I}$, that is $\mathbf{Q}^T \mathbf{Q}=\mathbf{I}$. This is no coincidence. There is a general result which says that if matrix $\mathbf{Q}$ is orthogonal then $\mathbf{Q}^T \mathbf{Q}=\mathbf{I}$. We can also go the other way, that is if $\mathbf{Q}^T \mathbf{Q}=\mathbf{I}$ then matrix $\mathbf{Q}$ is orthogonal.

Proposition (4.19). Let $\mathbf{Q}=\left(\begin{array}{lllll}\mathbf{v}_1 & \mathbf{v}_2 & \mathbf{v}_3 & \cdots & \mathbf{v}_n\end{array}\right)$ be a square matrix. Then $\mathbf{Q}$ is an orthogonal matrix $\Leftrightarrow \mathbf{Q}^T \mathbf{Q}=\mathbf{I}$.

How do we prove this result?
We have $\Leftrightarrow$ in the statement, so we need to prove it both ways, $\Rightarrow$ and $\Leftarrow$.
Proof.
$(\Rightarrow)$. We assume that $\mathbf{Q}=\left(\begin{array}{lllll}\mathbf{v}_1 & \mathbf{v}_2 & \mathbf{v}_3 & \cdots & \mathbf{v}_n\end{array}\right)$ is an orthogonal matrix, which means that $\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_n$ is an orthonormal (perpendicular unit) set of vectors in $\mathbb{R}^n$. Required to prove $\mathbf{Q}^T \mathbf{Q}=\mathbf{I}$. We carry out the matrix multiplication:
\begin{aligned} & \mathbf{Q}^T \mathbf{Q}=\left(\begin{array}{lllll} \mathbf{v}_1 & \mathbf{v}_2 & \mathbf{v}_3 & \cdots & \mathbf{v}_n \end{array}\right)^T\left(\begin{array}{lllll} \mathbf{v}_1 & \mathbf{v}_2 & \mathbf{v}_3 & \cdots & \mathbf{v}_n \end{array}\right) \ & =\left(\begin{array}{c} \mathbf{v}_1^T \ \mathbf{v}_2^T \ \vdots \ \mathbf{v}_n^T \end{array}\right)\left(\begin{array}{lllll} \mathbf{v}_1 & \mathbf{v}_2 & \mathbf{v}_3 & \cdots & \mathbf{v}_n \end{array}\right) \quad\left[\begin{array}{l} \text { transposing to convert columns } \ \text { to rows } \end{array}\right] \ & =\left(\begin{array}{cccc} \mathbf{v}_1^T \mathbf{v}_1 & \mathbf{v}_1^T \mathbf{v}_2 & \cdots & \mathbf{v}_1^T \mathbf{v}_n \ \mathbf{v}_2^T \mathbf{v}_1 & \mathbf{v}_2^T \mathbf{v}_2 & \cdots & \mathbf{v}_2^T \mathbf{v}_n \ \vdots & \vdots & & \vdots \ \mathbf{v}_n^T \mathbf{v}_1 & \mathbf{v}_n^T \mathbf{v}_2 & \cdots & \mathbf{v}_n^T \mathbf{v}_n \end{array}\right) \quad\left[\begin{array}{l} \text { carrying out matrix } \ \text { multiplication – row by } \ \text { column } \end{array}\right] \ & \end{aligned}
Remember, our destination is to show that the final matrix in the above is the identity.

## 数学代写|线性代数代写Linear algebra代考|What do you notice about your final result in the above example?

$(\Rightarrow)$。我们假设$\mathbf{Q}=\left(\begin{array}{lllll}\mathbf{v}_1 & \mathbf{v}_2 & \mathbf{v}_3 & \cdots & \mathbf{v}_n\end{array}\right)$是一个正交矩阵，这意味着$\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_n$是$\mathbb{R}^n$中的一个正交(垂直单位)向量集。需要证明$\mathbf{Q}^T \mathbf{Q}=\mathbf{I}$。我们进行矩阵乘法运算:
\begin{aligned} & \mathbf{Q}^T \mathbf{Q}=\left(\begin{array}{lllll} \mathbf{v}_1 & \mathbf{v}_2 & \mathbf{v}_3 & \cdots & \mathbf{v}_n \end{array}\right)^T\left(\begin{array}{lllll} \mathbf{v}_1 & \mathbf{v}_2 & \mathbf{v}_3 & \cdots & \mathbf{v}_n \end{array}\right) \ & =\left(\begin{array}{c} \mathbf{v}_1^T \ \mathbf{v}_2^T \ \vdots \ \mathbf{v}_n^T \end{array}\right)\left(\begin{array}{lllll} \mathbf{v}_1 & \mathbf{v}_2 & \mathbf{v}_3 & \cdots & \mathbf{v}_n \end{array}\right) \quad\left[\begin{array}{l} \text { transposing to convert columns } \ \text { to rows } \end{array}\right] \ & =\left(\begin{array}{cccc} \mathbf{v}_1^T \mathbf{v}_1 & \mathbf{v}_1^T \mathbf{v}_2 & \cdots & \mathbf{v}_1^T \mathbf{v}_n \ \mathbf{v}_2^T \mathbf{v}_1 & \mathbf{v}_2^T \mathbf{v}_2 & \cdots & \mathbf{v}_2^T \mathbf{v}_n \ \vdots & \vdots & & \vdots \ \mathbf{v}_n^T \mathbf{v}_1 & \mathbf{v}_n^T \mathbf{v}_2 & \cdots & \mathbf{v}_n^T \mathbf{v}_n \end{array}\right) \quad\left[\begin{array}{l} \text { carrying out matrix } \ \text { multiplication – row by } \ \text { column } \end{array}\right] \ & \end{aligned}

## 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。

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## 数学代写|线性代数代写Linear algebra代考|Orthogonal vectors

Can you remember what orthogonal vectors meant in the Euclidean space $\mathbb{R}^n$ ?
Two vectors $\mathbf{u}$ and $\mathbf{v}$ in $\mathbb{R}^n$ are said to be orthogonal or perpendicular $\Leftrightarrow$
$$\mathbf{u} \cdot \mathbf{v}=0$$
Hence for the general vector space $V$ with an inner product we have:
Definition (4.8). Two vectors $\mathbf{u}$ and $\mathbf{v}$ in the vector space $V$ are said to be orthogonal $\Leftrightarrow\langle\mathbf{u}, \mathbf{v}\rangle=0$
This is a fundamental and very useful result in linear algebra.
If vectors $\mathbf{u}$ and $\mathbf{v}$ are orthogonal, we say that $\mathbf{u}$ is orthogonal to $\mathbf{v}$, or vice versa that is $\mathbf{v}$ is orthogonal to $\mathbf{u}$ because $\langle\mathbf{u}, \mathbf{v}\rangle=\langle\mathbf{v}, \mathbf{u}\rangle=0$.
Consider the vectors $\mathbf{u}, \mathbf{v}, \mathbf{w}$ and $\mathbf{x}$ in $\mathbb{R}^2$ (Fig. 4.6):

You may recall from chapter 2 that vectors acting in the same direction have a positive dot product and vectors in opposite directions have a negative dot product. If they are perpendicular ( $\mathbf{u}$ and $\mathbf{v})$ then the dot product is zero.

Applications such as signal processing, communication and radar systems rely on inner product space. In the exercises, you are asked to show that this expression $S$ lies between 0 and $1 . S$ measures the degree to which the two signals are alike. A value of $S$ close to 1 means that the signals are similar in nature. A value of $S$ close to 0 means that the signals are very different but not necessarily orthogonal.

## 数学代写|线性代数代写Linear algebra代考|Orthonormal set

A set of vectors which are orthogonal (perpendicular) to each other is called an orthogonal set.

A set of vectors in which all the vectors have a norm or length of 1 is called a normalized set.

A set of perpendicular unit vectors is called an orthonormal set. This is a set of vectors which are both orthogonal and normalized.

For example, the set of vectors $\mathbf{e}_1=\left(\begin{array}{ll}1 & 0\end{array}\right)^T$ and $\mathbf{e}_2=\left(\begin{array}{ll}0 & 1\end{array}\right)^T$ in $\mathbb{R}^2$ are orthonormal (perpendicular unit) vectors with the inner product as the dot product:
$$\left\langle\mathbf{e}_1, \mathbf{e}_2\right\rangle=\left(\begin{array}{l} 1 \ 0 \end{array}\right) \cdot\left(\begin{array}{l} 0 \ 1 \end{array}\right)=(1 \times 0)+(0 \times 1)=0$$
The vectors $\mathbf{e}_1$ and $\mathbf{e}_2$ are orthogonal (perpendicular). The norms or lengths of these vectors:
$$\left|\mathbf{e}_1\right|=\left|\mathbf{e}_2\right|=1$$
Thus the vectors $\mathbf{e}_1$ and $\mathbf{e}_2$ are orthonormal (perpendicular unit vectors) because they are both orthogonal and normalized vectors.
Examples of orthonormal sets are shown in Fig. 4.8 for $\mathbb{R}^2$ and $\mathbb{R}^3$ :

Remember that these are the standard basis vectors and they are also an orthonormal (perpendicular unit) basis for $\mathbb{R}^2$ and $\mathbb{R}^3$. These perpendicular unit vectors make a convenient basis as we will discuss in the next section.

## 数学代写|线性代数代写Linear algebra代考|Orthogonal vectors

$\mathbb{R}^n$中的两个向量$\mathbf{u}$和$\mathbf{v}$是正交的或垂直的$\Leftrightarrow$
$$\mathbf{u} \cdot \mathbf{v}=0$$

## 数学代写|线性代数代写Linear algebra代考|Orthonormal set

$$\left\langle\mathbf{e}_1, \mathbf{e}_2\right\rangle=\left(\begin{array}{l} 1 \ 0 \end{array}\right) \cdot\left(\begin{array}{l} 0 \ 1 \end{array}\right)=(1 \times 0)+(0 \times 1)=0$$

$$\left|\mathbf{e}_1\right|=\left|\mathbf{e}_2\right|=1$$

## 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。

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

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## 数学代写|线性代数代写Linear algebra代考|Properties of rank and nullity

In the above examples $3.33,3.34$ and 3.35 we had
\begin{aligned} & \operatorname{nullity}(\mathbf{A})+\operatorname{rank}(\mathbf{A})=1+2=3 \ & \operatorname{nullity}(\mathbf{B})+\operatorname{rank}(\mathbf{B})=2+2=4 \ & \operatorname{nullity}(\mathbf{C})+\operatorname{rank}(\mathbf{C})=1+1=2 \end{aligned}
Can you see any relationship between the nullity, rank and the number of unknowns in vector $\mathbf{x}$ ?
$$\text { Nullity }+ \text { Rank }=\text { Number of unknowns }$$
In general, if we have a $m$ by $n$ matrix $\mathbf{A}$ :
$$\left(\begin{array}{ccc} a_{11} & \cdots & a_{1 n} \ \vdots & \ddots & \vdots \ a_{m 1} & \cdots & a_{m n} \end{array}\right)\left(\begin{array}{c} x_1 \ \vdots \ x_n \end{array}\right)=\left(\begin{array}{c} 0 \ \vdots \ 0 \end{array}\right) \quad[\mathbf{A x}=\mathbf{O}]$$
Then
$$\operatorname{nullity}(\mathbf{A})+\operatorname{rank}(\mathbf{A})=n$$
Note that $n$ is the number of columns of matrix $\mathbf{A}$ which is the total number of unknowns in the homogeneous system $\mathbf{A x}=\mathbf{O}$. This result normally has the grand title of ‘The Dimension Theorem of Matrices’.

## 数学代写|线性代数代写Linear algebra代考|What does this proposition mean?

The solution of $\mathbf{A x}=\mathbf{b}$ consists of two parts:
(homogeneous solution) + (particular solution)
Remember, the homogeneous solution $\mathbf{x}_H$ (vector) belongs to the null space of matrix $\mathbf{A}$.
Proof.
Let $\mathbf{x}$ be the solution of $\mathbf{A x}=\mathbf{b}$ then
\begin{aligned} \mathbf{A}\left(\mathbf{x}-\mathbf{x}_P\right) & =\mathbf{A} \mathbf{x}-\mathbf{A} \mathbf{x}_P \ & =\mathbf{b}-\mathbf{b}=\mathbf{O} \end{aligned}
Since we have $\mathbf{A}\left(\mathbf{x}-\mathbf{x}_P\right)=\mathbf{O}$, therefore $\mathbf{x}-\mathbf{x}_P$ is the homogeneous solution, that is $\mathbf{x}_H=\mathbf{x}-\mathbf{x}_P$. Hence we have our result $\mathbf{x}=\mathbf{x}_P+\mathbf{x}_H$.
We need to show all the solutions are of this format $\mathbf{x}_P+\mathbf{x}_H$.
Let $\mathbf{x}^{\prime}$ be a solution of $\mathbf{A x}=\mathbf{O}$, then
$$\mathbf{A}\left(\mathbf{x}+\mathbf{x}^{\prime}\right)=\mathbf{A x}+\mathbf{A} \mathbf{x}^{\prime}=\mathbf{A x}+\mathbf{O}=\mathbf{b}+\mathbf{O}=\mathbf{b}$$
Hence $\mathbf{x}+\mathbf{x}^{\prime}$ is a solution of $\mathbf{A x}=\mathbf{b}$.
We conclude that all the solutions are of this form $\mathbf{x}=\mathbf{x}_P+\mathbf{x}_H$.

## 数学代写|线性代数代写Linear algebra代考|Properties of rank and nullity

\begin{aligned} & \operatorname{nullity}(\mathbf{A})+\operatorname{rank}(\mathbf{A})=1+2=3 \ & \operatorname{nullity}(\mathbf{B})+\operatorname{rank}(\mathbf{B})=2+2=4 \ & \operatorname{nullity}(\mathbf{C})+\operatorname{rank}(\mathbf{C})=1+1=2 \end{aligned}

$$\text { Nullity }+ \text { Rank }=\text { Number of unknowns }$$

$$\left(\begin{array}{ccc} a_{11} & \cdots & a_{1 n} \ \vdots & \ddots & \vdots \ a_{m 1} & \cdots & a_{m n} \end{array}\right)\left(\begin{array}{c} x_1 \ \vdots \ x_n \end{array}\right)=\left(\begin{array}{c} 0 \ \vdots \ 0 \end{array}\right) \quad[\mathbf{A x}=\mathbf{O}]$$

$$\operatorname{nullity}(\mathbf{A})+\operatorname{rank}(\mathbf{A})=n$$

## 数学代写|线性代数代写Linear algebra代考|What does this proposition mean?

$\mathbf{A x}=\mathbf{b}$的解决方案由两部分组成:
(齐次溶液)+(特解)

\begin{aligned} \mathbf{A}\left(\mathbf{x}-\mathbf{x}_P\right) & =\mathbf{A} \mathbf{x}-\mathbf{A} \mathbf{x}_P \ & =\mathbf{b}-\mathbf{b}=\mathbf{O} \end{aligned}

$$\mathbf{A}\left(\mathbf{x}+\mathbf{x}^{\prime}\right)=\mathbf{A x}+\mathbf{A} \mathbf{x}^{\prime}=\mathbf{A x}+\mathbf{O}=\mathbf{b}+\mathbf{O}=\mathbf{b}$$

## 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。

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## avatest™帮您通过考试

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## 数学代写|线性代数代写Linear algebra代考|What can you predict about the row space of row equivalent matrices?

The space they occupy is equal.
For example, the row space $S$ of the above matrix $\mathbf{A}$ is given by the vectors $\mathbf{v}$ in $S$ such that:
$$\mathbf{v}=k_1 \mathbf{a}_1+k_2 \mathbf{a}_2=k_1\left(\begin{array}{l} 1 \ 2 \ 3 \end{array}\right)+k_2\left(\begin{array}{c} 4 \ 8 \ 12 \end{array}\right)=k_1\left(\begin{array}{l} 1 \ 2 \ 3 \end{array}\right)+3 k_2\left(\begin{array}{l} 1 \ 2 \ 3 \end{array}\right)=c\left(\begin{array}{l} 1 \ 2 \ 3 \end{array}\right)=c \mathbf{r}_1$$
where $c=k_1+3 k_2$
Hence all the vectors $\mathbf{v}$ are in the row space $S$ of matrix $\mathbf{R}$. In fact, you can span the row space of matrix A with just one vector $\left(\begin{array}{lll}1 & 2 & 3\end{array}\right)^T$ which is the non-zero row in matrix $\mathbf{R}$. Actually the row space $S$ created by matrix $\mathbf{A}$ is the same as the row space created by matrix R. We have
Row space of $\mathbf{A}=$ Row space of $\mathbf{R}$
Proposition (3.25). If matrices $\mathbf{A}$ and $\mathbf{R}$ are row equivalent then their row spaces are equal.
How do we prove this result? By showing that

The row space of $\mathbf{A}$ is in row space of $\mathbf{R}$.

The row space of $\mathbf{R}$ is in row space of $\mathbf{A}$.
If both these conditions are satisfied then the row spaces of matrices $\mathbf{A}$ and $\mathbf{R}$ must be equal.

## 数学代写|线性代数代写Linear algebra代考|Basis of a spanned subspace of $\mathbb{R}^n$

Let $\mathbf{u}=\left(\begin{array}{lll}1 & 2 & 3\end{array}\right)^T$ and $\mathbf{v}=\left(\begin{array}{lll}4 & 5 & 6\end{array}\right)^T$ be vectors in $\mathbb{R}^3$. Let $S$ be the space spanned by these vectors which is illustrated in Fig. 3.19.

We are generally interested in finding a simple set of axes to describe this plane, or more formally, a basis for this space $S$.

How can we find such a basis?
Writing the vectors as rows of a matrix $\mathbf{A}=\left(\begin{array}{l}\mathbf{u} \ \mathbf{v}\end{array}\right)$ then the row space of $\mathbf{A}$ is the vector space spanned by vectors $\mathbf{u}$ and $\mathbf{v}$. By using row operations on $\mathbf{A}$, we can find a basis for the row space of a matrix $\mathbf{A}$ as we did in subsection 3.5.3 above.

The procedure of finding a basis for a subspace of $\mathbb{R}^n$ which is spanned by the vectors $\left{\mathbf{r}_1, \mathbf{r}_2, \mathbf{r}_3, \ldots, \mathbf{r}_m\right}$ is given by:

1. Form the matrix $\mathbf{A}=\left(\begin{array}{c}\mathbf{r}_1 \ \vdots \ \mathbf{r}_m\end{array}\right)$. The row space of $\mathbf{A}$ is the space spanned by $\mathbf{r}_1, \mathbf{r}_2, \ldots, \mathbf{r}_m$
2. Convert this matrix $\mathbf{A}$ into (reduced) row echelon form, $\mathbf{R}$ say.
3. The non-zero rows of matrix $\mathbf{R}$ form a basis for the vector space $\operatorname{span}\left{\mathbf{r}_1, \mathbf{r}_2, \ldots, \mathbf{r}_m\right}$.

## 数学代写|线性代数代写Linear algebra代考|What can you predict about the row space of row equivalent matrices?

$$\mathbf{v}=k_1 \mathbf{a}_1+k_2 \mathbf{a}_2=k_1\left(\begin{array}{l} 1 \ 2 \ 3 \end{array}\right)+k_2\left(\begin{array}{c} 4 \ 8 \ 12 \end{array}\right)=k_1\left(\begin{array}{l} 1 \ 2 \ 3 \end{array}\right)+3 k_2\left(\begin{array}{l} 1 \ 2 \ 3 \end{array}\right)=c\left(\begin{array}{l} 1 \ 2 \ 3 \end{array}\right)=c \mathbf{r}_1$$

$\mathbf{A}=$的行空间$\mathbf{R}$的行空间

$\mathbf{A}$的行空间在$\mathbf{R}$的行空间中。

$\mathbf{R}$的行空间在$\mathbf{A}$的行空间中。

## 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。

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## 数学代写|线性代数代写Linear algebra代考|Finite dimensional vector space

What does the term finite dimensional vector space mean?
It is a vector space $V$ which has a finite number of vectors in its basis.
Definition (3.18). In general, if a finite number of vectors form a basis for a vector space $V$ then we say $V$ is finite dimensional. Otherwise, the vector space $V$ is known as infinite dimensional.
If the vector space $V$ consists only of the zero vector then it is also finite dimensional.
Can you think of any finite dimensional vector spaces?
The Euclidean spaces $-\mathbb{R}^2, \mathbb{R}^3, \mathbb{R}^4, \ldots, \mathbb{R}^n$.

Are there any other examples of finite dimensional vector spaces?
The set $P_2$ of polynomials of degree 2 or less for example, or the set of all 2 by 2 matrices $M_{22}$. (These were covered in the previous section.)

Definition (3.19). In general, if $n$ vectors $\left{\mathbf{v}1, \mathbf{v}_2, \mathbf{v}_3, \ldots, \mathbf{v}_n\right}$ form a basis for a vector space $V$ then we say that $V$ is $n$-dimensional. What is $\operatorname{dim}\left(M{22}\right)$ equal to?
The standard basis for $M_{22}$ (matrices of size 2 by 2) from the Exercises 3.3 question 5 is
$$\mathbf{A}=\left(\begin{array}{ll} 1 & 0 \ 0 & 0 \end{array}\right), \mathbf{B}=\left(\begin{array}{ll} 0 & 1 \ 0 & 0 \end{array}\right), \mathbf{C}=\left(\begin{array}{ll} 0 & 0 \ 1 & 0 \end{array}\right) \text { and } \mathbf{D}=\left(\begin{array}{ll} 0 & 0 \ 0 & 1 \end{array}\right)$$
Therefore $\operatorname{dim}\left(M_{22}\right)=4$ because we have four matrices in ${\mathbf{A}, \mathbf{B}, \mathbf{C}, \mathbf{D}}$ which form a basis for $M_{22}$.
What is $\operatorname{dim}\left(P_2\right)$ equal to?
Remember that the standard basis for $P_2$ (the set of all polynomials of degree 2 or less) is the set $\left{1, t, t^2\right}$, which means $\operatorname{dim}\left(P_2\right)=3$ since the basis consists of three vectors.
Table 3.1 shows some vector spaces and their dimensions.

## 数学代写|线性代数代写Linear algebra代考|Properties of finite dimensional vector spaces

In this section we show some important properties of bases and dimension. This is a demanding section because the proofs of propositions are lengthy.
Lemma (3.21). Let $V$ be a finite $n$-dimensional vector space. We have the following:
(a) Let $\left{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3, \ldots, \mathbf{v}_n\right}$ be a set of linearly independent vectors. Then $\left{\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_m\right}$ where $m>n(m$ is greater than $n)$ is linearly dependent.
(b) If the $n$ vectors $\left{\mathbf{u}_1, \mathbf{u}_2, \mathbf{u}_3, \ldots, \mathbf{u}_n\right}$ span $V$ then $\left{\mathbf{u}_1, \mathbf{u}_2, \mathbf{u}_3, \ldots, \mathbf{u}_m\right}$ where $m<n$ does not $\operatorname{span} V$.
What does part (a) mean?
In $n$-dimensional vector space, if you add additional vectors to $n$ linearly independent vectors then the set becomes linearly dependent.
How do we prove this result?
Proof of $(a)$.
The number of basis vectors in $V$ is $n$. Suppose that $\left{\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_m\right}$ are linearly independent. Then we must have $m \leq n$ ( $m$ is less than or equal to $n)$.
Why?
Because by question 14 of Exercises 3.3 we know that the number of vectors in a linearly independent set must be less than or equal to $n, m \leq n$ (the number of basis vectors).

However, we are given that $m>n$ ( $m$ is greater than $n)$, therefore our supposition that $\left{\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_m\right}$ is a linearly independent set of vectors must be wrong, so this set is linearly dependent.

## 数学代写|线性代数代写Linear algebra代考|Finite dimensional vector space

$$\mathbf{A}=\left(\begin{array}{ll} 1 & 0 \ 0 & 0 \end{array}\right), \mathbf{B}=\left(\begin{array}{ll} 0 & 1 \ 0 & 0 \end{array}\right), \mathbf{C}=\left(\begin{array}{ll} 0 & 0 \ 1 & 0 \end{array}\right) \text { and } \mathbf{D}=\left(\begin{array}{ll} 0 & 0 \ 0 & 1 \end{array}\right)$$

$\operatorname{dim}\left(P_2\right)$等于多少?

## 数学代写|线性代数代写Linear algebra代考|Properties of finite dimensional vector spaces

(a)设$\left{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3, \ldots, \mathbf{v}_n\right}$为线性无关向量的集合。然后$\left{\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_m\right}$，其中$m>n(m$大于$n)$是线性相关的。
(b)如果$n$向量$\left{\mathbf{u}_1, \mathbf{u}_2, \mathbf{u}_3, \ldots, \mathbf{u}_n\right}$跨越$V$，则$\left{\mathbf{u}_1, \mathbf{u}_2, \mathbf{u}_3, \ldots, \mathbf{u}_m\right}$，其中$m<n$不$\operatorname{span} V$。
(a)部分是什么意思?

$V$中基向量的个数为$n$。假设$\left{\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_m\right}$是线性无关的。那么我们就得到$m \leq n$ ($m$小于等于$n)$)

## 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。

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

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## 数学代写|线性代数代写Linear algebra代考|Examples of vector subspaces

Let $V$ be a vector space and $S$ be a non-empty subset of $V$. If the set $S$ satisfies all 10 axioms of a vector space with respect to the same vector addition and scalar multiplication as $V$ then $S$ is also a vector space. We say $S$ is a subspace of $V$.

Definition (3.4). A non-empty subset $S$ of a vector space $V$ is called a subspace of $V$ if it also a vector space with respect to the same vector addition and scalar multiplication as $V$.
We illustrate this in Fig. 3.3.

Note the difference between subspace and subset. A subset is merely a specific set of elements chosen from $V$. A subset must also satisfy the 10 axioms of vector space to be called a subspace.

## 数学代写|线性代数代写Linear algebra代考|How do we prove this proposition?

$(\Rightarrow)$. We first assume that $S$ is a subspace of the vector space $V$, and from this we deduce conditions (a) and (b) [show that we have closure under vector addition and scalar multiplication]. $(\Leftarrow)$. Then we assume conditions (a) and (b) are satisfied, and from this we deduce that the set $S$ is a subspace of the vector space $V$.
Proof.
Let $\mathbf{u}$ and $\mathbf{v}$ be vectors in the set $S$.
$(\Rightarrow)$. Let $S$ be a subspace of the vector space $V$. By the above definition (3.4) we have closure under vector addition and scalar multiplication because the set $S$ is itself a vector space. [Remember, axioms 1 and 6 state that we have closure under vector addition and scalar multiplication]. Hence conditions (a) and (b) hold.
$(\Leftarrow)$. Assume conditions (a) and (b) are satisfied, that is we have closure under vector addition and scalar multiplication.
Required to prove that all 10 axioms of the last section are satisfied.
We have closure, therefore axioms 1 and 6 are satisfied. Axioms 2, 3, 7, 8, 9 and 10 are satisfied because these axioms are true for all vectors in the vector space $V$ and vectors $\mathbf{u}$ and $\mathbf{v}$ are vectors in the vector space $V$.
For the set $S$ we need to prove axioms 4 and 5 which are:

There is a zero vector $\mathbf{O}$ in $V$ which satisfies
$$\mathbf{u}+\mathbf{O}=\mathbf{u} \quad \text { for every vector } \mathbf{u} \text { in } V$$

For every vector $\mathbf{u}$ there is a vector $-\mathbf{u}$ which satisfies the following:
$$\mathbf{u}+(-\mathbf{u})=\mathbf{O}$$
We have to show (Fig. 3.6) that the zero vector, $\mathbf{O}$, and $-\mathbf{u}$ are also in $S$ for any $\mathbf{u}$ in $S$.

## 数学代写|线性代数代写Linear algebra代考|Examples of vector subspaces

Let $V$ be a vector space and $S$ be a non-empty subset of $V$. If the set $S$ satisfies all 10 axioms of a vector space with respect to the same vector addition and scalar multiplication as $V$ then $S$ is also a vector space. We say $S$ is a subspace of $V$.

Definition (3.4). A non-empty subset $S$ of a vector space $V$ is called a subspace of $V$ if it also a vector space with respect to the same vector addition and scalar multiplication as $V$.
We illustrate this in Fig. 3.3.

Note the difference between subspace and subset. A subset is merely a specific set of elements chosen from $V$. A subset must also satisfy the 10 axioms of vector space to be called a subspace.

## 数学代写|线性代数代写Linear algebra代考|How do we prove this proposition?

$(\Rightarrow)$. We first assume that $S$ is a subspace of the vector space $V$, and from this we deduce conditions (a) and (b) [show that we have closure under vector addition and scalar multiplication]. $(\Leftarrow)$. Then we assume conditions (a) and (b) are satisfied, and from this we deduce that the set $S$ is a subspace of the vector space $V$.
Proof.
Let $\mathbf{u}$ and $\mathbf{v}$ be vectors in the set $S$.
$(\Rightarrow)$. Let $S$ be a subspace of the vector space $V$. By the above definition (3.4) we have closure under vector addition and scalar multiplication because the set $S$ is itself a vector space. [Remember, axioms 1 and 6 state that we have closure under vector addition and scalar multiplication]. Hence conditions (a) and (b) hold.
$(\Leftarrow)$. Assume conditions (a) and (b) are satisfied, that is we have closure under vector addition and scalar multiplication.
Required to prove that all 10 axioms of the last section are satisfied.
We have closure, therefore axioms 1 and 6 are satisfied. Axioms 2, 3, 7, 8, 9 and 10 are satisfied because these axioms are true for all vectors in the vector space $V$ and vectors $\mathbf{u}$ and $\mathbf{v}$ are vectors in the vector space $V$.
For the set $S$ we need to prove axioms 4 and 5 which are:

There is a zero vector $\mathbf{O}$ in $V$ which satisfies
$$\mathbf{u}+\mathbf{O}=\mathbf{u} \quad \text { for every vector } \mathbf{u} \text { in } V$$

For every vector $\mathbf{u}$ there is a vector $-\mathbf{u}$ which satisfies the following:
$$\mathbf{u}+(-\mathbf{u})=\mathbf{O}$$
We have to show (Fig. 3.6) that the zero vector, $\mathbf{O}$, and $-\mathbf{u}$ are also in $S$ for any $\mathbf{u}$ in $S$.

## 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。

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## 数学代写|线性代数代写Linear algebra代考|What does this proposition mean?

There is only one way of writing any vector as a linear combination of the basis vectors.

We have already proven this result for the standard basis in Proposition (2.18) of the last section.
Proof.
Let $\mathbf{u}$ be an arbitrary vector in $\mathbb{R}^n$. We are given that the vectors $\left{\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_n\right}$ form a basis, so they span $\mathbb{R}^n$ which means that we can write the vector $\mathbf{u}$ in $\mathbb{R}^n$ as a linear combination of $\left{\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_n\right}$. There exist scalars $k_1, k_2, k_3, \ldots$ and $k_n$ which satisfy
$$\mathbf{u}=k_1 \mathbf{v}_1+k_2 \mathbf{v}_2+k_3 \mathbf{v}_3+\cdots+k_n \mathbf{v}_n$$
Suppose we can write this vector $\mathbf{u}$ as another linear combination of the basis vectors
$$\mathbf{u}=c_1 \mathbf{v}_1+c_2 \mathbf{v}_2+c_3 \mathbf{v}_3+\cdots+c_n \mathbf{v}_n$$
where the $c$ ‘s are scalars.

What do we need to prove?
We need to prove that the two sets of scalars are equal: $k_1=c_1, k_2=c_2, \ldots$ and $k_n=c_n$. Equating the two linear combinations because both are equal to $\mathbf{u}$ gives
\begin{aligned} & k_1 \mathbf{v}_1+k_2 \mathbf{v}_2+\cdots+k_n \mathbf{v}_n=c_1 \mathbf{v}_1+c_2 \mathbf{v}_2+\cdots+c_n \mathbf{v}_n=\mathbf{u} \ & k_1 \mathbf{v}_1+k_2 \mathbf{v}_2+\cdots+k_n \mathbf{v}_n-c_1 \mathbf{v}_1-c_2 \mathbf{v}_2-\cdots-c_n \mathbf{v}_n=\mathbf{u}-\mathbf{u}=\mathbf{O} \ & \left(k_1-c_1\right) \mathbf{v}_1+\left(k_2-c_2\right) \mathbf{v}_2+\cdots+\left(k_n-c_n\right) \mathbf{v}_n=\mathbf{O} \quad \text { [factorizing] } \end{aligned}
The basis vectors $\left{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3, \ldots, \mathbf{v}_n\right}$ are linearly independent, therefore all the scalars are equal to zero.

## 数学代写|线性代数代写Linear algebra代考|Why?

Because this is the definition of linear independence given in the last section (2.19):
Vectors $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3, \ldots$ and $\mathbf{v}_n$ in $\mathbb{R}^n$ are linearly independent $\Leftrightarrow$
$$m_1 \mathbf{v}_1+m_2 \mathbf{v}_2+m_3 \mathbf{v}_3+\cdots+m_n \mathbf{v}_n=\mathbf{O} \text { gives } m_1=m_2=m_3=\cdots=m_n=0$$
Applying this to the above derivation:
$$\left(k_1-c_1\right) \mathbf{v}_1+\left(k_2-c_2\right) \mathbf{v}_2+\cdots+\left(k_n-c_n\right) \mathbf{v}_n=\mathbf{O}$$
we have
$$\begin{gathered} k_1-c_1=0, k_2-c_2=0, k_3-c_3=0, \ldots \text { and } k_n-c_n=0 \ k_1=c_1, k_2=c_2, k_3=c_3, \ldots \text { and } k_n=c_n \end{gathered}$$
Hence any arbitrary vector $\mathbf{u}$ can be written uniquely as a linear combination of the basis vectors $\left{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3, \ldots, \mathbf{v}_n\right}$.

Next we prove a lemma. A lemma is a proposition or theorem, the proof of which is used as a stepping stone towards proving something of greater interest. However, there are many lemmas in mathematics which have become important results in themselves, such as Zorn’s lemma, Euclid’s lemma and Gauss’s lemma.

Lemma (2.30). Let $T=\left{\mathbf{w}_1, \mathbf{w}_2, \mathbf{w}_3, \ldots, \mathbf{w}_m\right}$ be a set of $m$ vectors that are linearly independent in $\mathbb{R}^n$ then $m \leq n$.

## 数学代写|线性代数代写Linear algebra代考|What does this proposition mean?

$$\mathbf{u}=k_1 \mathbf{v}_1+k_2 \mathbf{v}_2+k_3 \mathbf{v}_3+\cdots+k_n \mathbf{v}_n$$

$$\mathbf{u}=c_1 \mathbf{v}_1+c_2 \mathbf{v}_2+c_3 \mathbf{v}_3+\cdots+c_n \mathbf{v}_n$$

\begin{aligned} & k_1 \mathbf{v}_1+k_2 \mathbf{v}_2+\cdots+k_n \mathbf{v}_n=c_1 \mathbf{v}_1+c_2 \mathbf{v}_2+\cdots+c_n \mathbf{v}_n=\mathbf{u} \ & k_1 \mathbf{v}_1+k_2 \mathbf{v}_2+\cdots+k_n \mathbf{v}_n-c_1 \mathbf{v}_1-c_2 \mathbf{v}_2-\cdots-c_n \mathbf{v}_n=\mathbf{u}-\mathbf{u}=\mathbf{O} \ & \left(k_1-c_1\right) \mathbf{v}_1+\left(k_2-c_2\right) \mathbf{v}_2+\cdots+\left(k_n-c_n\right) \mathbf{v}_n=\mathbf{O} \quad \text { [factorizing] } \end{aligned}

## 数学代写|线性代数代写Linear algebra代考|Why?

$\mathbb{R}^n$中的向量$\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3, \ldots$和$\mathbf{v}_n$是线性无关的$\Leftrightarrow$
$$m_1 \mathbf{v}_1+m_2 \mathbf{v}_2+m_3 \mathbf{v}_3+\cdots+m_n \mathbf{v}_n=\mathbf{O} \text { gives } m_1=m_2=m_3=\cdots=m_n=0$$

$$\left(k_1-c_1\right) \mathbf{v}_1+\left(k_2-c_2\right) \mathbf{v}_2+\cdots+\left(k_n-c_n\right) \mathbf{v}_n=\mathbf{O}$$

$$\begin{gathered} k_1-c_1=0, k_2-c_2=0, k_3-c_3=0, \ldots \text { and } k_n-c_n=0 \ k_1=c_1, k_2=c_2, k_3=c_3, \ldots \text { and } k_n=c_n \end{gathered}$$

## 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。

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## 数学代写|线性代数代写Linear algebra代考|Unit vectors

A vector of length 1 is called a unit vector. In Exercises 2.1, we showed that for any non-zero vector $\mathbf{u}$ in $\mathbb{R}^n$ we have $\left|\frac{1}{|\mathbf{u}|} \mathbf{u}\right|=1$.

What does this mean?
It means that we can always find a unit vector in the direction of any non-zero vector $\mathbf{u}$ by dividing the given vector by its length $|\mathbf{u}|$ (Fig. 2.22).

For example, a vector in a particular direction of length 5 can be divided by 5 to give a vector in the same direction but length 1 (unit vector).

The process of finding a unit vector in the direction of the given vector $\mathbf{u}$ is called normalizing. The unit vector in the direction of the vector $\mathbf{u}$ is normally denoted by $\hat{\mathbf{u}}$ (pronounced as ‘ $u$ hat’) meaning it is a vector of length 1 , that is
$$\hat{\mathbf{u}}=\frac{1}{|\mathbf{u}|} \mathbf{u}$$
Later on in this chapter we will see that normalizing vectors simplifies calculations. Examples of unit vectors are shown in Fig. 2.23.

The vectors shown in Fig. $2.23, \mathbf{e}_1=\left(\begin{array}{l}1 \ 0\end{array}\right)$ and $\mathbf{e}_2=\left(\begin{array}{l}0 \ 1\end{array}\right)$, are unit vectors in $\mathbb{R}^2$, and $\mathbf{e}_1=\left(\begin{array}{lll}1 & 0 & 0\end{array}\right)^T, \mathbf{e}_2=\left(\begin{array}{lll}0 & 1 & 0\end{array}\right)^T$ and $\mathbf{e}_3=\left(\begin{array}{lll}0 & 0 & 1\end{array}\right)^T$ are unit vectors in $\mathbb{R}^3$. These are normally called the standard unit vectors.
For any $n$ space, $\mathbb{R}^n$, the standard unit vectors are defined by
$$\mathbf{e}_1=\left(\begin{array}{c} 1 \ 0 \ \vdots \ 0 \end{array}\right), \mathbf{e}_2=\left(\begin{array}{c} 0 \ 1 \ 0 \ \vdots \ 0 \end{array}\right), \ldots \mathbf{e}_k=\left(\begin{array}{c} 0 \ \vdots \ 1 \ 0 \ \vdots \end{array}\right), \cdots, \mathbf{e}_n=\left(\begin{array}{c} 0 \ \vdots \ 0 \ 0 \ 1 \end{array}\right)$$
That is, we have 1 in the $k$ th position of the vector $\mathbf{e}_k$ and zeros everywhere else.

Actually these are examples of perpendicular unit vectors called orthonormal vectors, which means that they are normalized and they are orthogonal. Hence orthonormal vectors have two properties:

1. All the vectors are orthogonal to each other (perpendicular to each other).
2. All vectors are normalized, that is they have a norm or length of 1 (unit vectors).
Orthonormal (perpendicular unit) vectors are important in linear algebra.

## 数学代写|线性代数代写Linear algebra代考|Application of vectors

An application of vectors is the support vector machine, which is a computer algorithm. The algorithm produces the best hyperplane which separates data groups (or vectors). Hyperplanes are general planes in $\mathbb{R}^n$. In support vector machines, we are interested in finding the shortest distance between the hyperplane and the vectors.

A hyperplane is a general plane in $n$-space. In two-space it is a line, as shown in Fig. 2.24.

The shortest distance from a vector $\mathbf{u}$ to any point on the hyperplane $\mathbf{v} \cdot \mathbf{x}+c=0$ where $\mathbf{x}=(x y \cdots)^T$ in $n$-space can be shown to equal, $\frac{|\mathbf{u} \cdot \mathbf{v}+c|}{|\mathbf{v}|}$.

## 数学代写|线性代数代写Linear algebra代考|Unit vectors

$$\hat{\mathbf{u}}=\frac{1}{|\mathbf{u}|} \mathbf{u}$$

$$\mathbf{e}_1=\left(\begin{array}{c} 1 \ 0 \ \vdots \ 0 \end{array}\right), \mathbf{e}_2=\left(\begin{array}{c} 0 \ 1 \ 0 \ \vdots \ 0 \end{array}\right), \ldots \mathbf{e}_k=\left(\begin{array}{c} 0 \ \vdots \ 1 \ 0 \ \vdots \end{array}\right), \cdots, \mathbf{e}_n=\left(\begin{array}{c} 0 \ \vdots \ 0 \ 0 \ 1 \end{array}\right)$$

## 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。

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

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## 数学代写|线性代数代写Linear algebra代考|Angle between two vectors

Work is defined as the product of the force applied in a particular direction and the distance it moves in that direction.

A real-world example of this is to imagine a rope tied to an object, perhaps a barge. You need to move the barge by pulling the rope (Fig. 2.12).

If you could stand directly behind the barge and push, then all your force would be in the direction of movement.
In this case, $\mathbf{F}$ and $\mathbf{d}$ are parallel and the angle between them is $0^{\circ}$ so we have
$$\text { Work done }=|\mathbf{F}| \cos \left(0^{\circ}\right) \times|\mathbf{d}|=|\mathbf{F}| \times|\mathbf{d}| \quad \text { [because } \cos \left(0^{\circ}\right)=1 \text { ] }$$
This is the least possible force used to push the object because we are pushing in the same direction as we would like the object to move.

If you push the barge in a direction perpendicular (orthogonal) to the canal, it would not move forward at all, so you would do no work because
$$\text { Work done }=|\mathbf{F}| \cos \left(90^{\circ}\right) \times|\mathbf{d}|=0 \quad\left[\text { because } \cos \left(90^{\circ}\right)=0\right]$$
The actual amount of work done in moving the barge along the canal is a value somewhere between these two possibilities, and is given by the angle the rope makes with the direction of the canal.

## 数学代写|线性代数代写Linear algebra代考|Inequalities

Next, we prove some inequalities in relation to the dot product and norm (length) of vectors.
Cauchy-Schwarz inequality (2.14). Let $\mathbf{u}$ and $\mathbf{v}$ be vectors in $\mathbb{R}^n$ then
$$|\mathbf{u} \cdot \mathbf{v}| \leq|\mathbf{u}||\mathbf{v}|$$

For the above Example 2.6 we had
$$|\mathbf{u} \cdot \mathbf{v}| \underset{\text { By part (i) }}{\equiv} 9 \leq 21.02 \underset{\text { By part (ii) }}{\equiv}|\mathbf{u}||\mathbf{v}|$$
Proof.
How can we prove this inequality for any vectors in $\mathbb{R}^n$ ? If the vectors $\mathbf{u}$ and $\mathbf{v}$ are non-zero then we can use the above formula:
$$\mathbf{u} \cdot \mathbf{v}=|\mathbf{u}||\mathbf{v}| \cos (\theta)$$
Taking the modulus of both sides we have
$$|\mathbf{u} \cdot \mathbf{v}|=||\mathbf{u}||\mathbf{v}| \cos (\theta)|$$
The lengths are positive or zero, $|\mathbf{u}| \geq 0$ and $|\mathbf{v}| \geq 0$, therefore the modulus of these is just $|\mathbf{u}|$ and $|\mathbf{v}|$ respectively.
Why?
Because if $x \geq 0$ then $|x|=x$. Hence we have $|\mathbf{u} \cdot \mathbf{v}|=|\mathbf{u}||\mathbf{v}||\cos (\theta)|$.

## 数学代写|线性代数代写Linear algebra代考|Angle between two vectors

$$\text { Work done }=|\mathbf{F}| \cos \left(0^{\circ}\right) \times|\mathbf{d}|=|\mathbf{F}| \times|\mathbf{d}| \quad \text { [because } \cos \left(0^{\circ}\right)=1 \text { ] }$$

$$\text { Work done }=|\mathbf{F}| \cos \left(90^{\circ}\right) \times|\mathbf{d}|=0 \quad\left[\text { because } \cos \left(90^{\circ}\right)=0\right]$$

## 数学代写|线性代数代写Linear algebra代考|Inequalities

Cauchy-Schwarz不等式(2.14)。设$\mathbf{u}$和$\mathbf{v}$为$\mathbb{R}^n$中的向量
$$|\mathbf{u} \cdot \mathbf{v}| \leq|\mathbf{u}||\mathbf{v}|$$

$$|\mathbf{u} \cdot \mathbf{v}| \underset{\text { By part (i) }}{\equiv} 9 \leq 21.02 \underset{\text { By part (ii) }}{\equiv}|\mathbf{u}||\mathbf{v}|$$

$$\mathbf{u} \cdot \mathbf{v}=|\mathbf{u}||\mathbf{v}| \cos (\theta)$$

$$|\mathbf{u} \cdot \mathbf{v}|=||\mathbf{u}||\mathbf{v}| \cos (\theta)|$$

## 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。