# 数学代写|线性代数代写linear algebra代考|Rank and invertible matrices

#### Doug I. Jones

Lorem ipsum dolor sit amet, cons the all tetur adiscing elit

couryes™为您提供可以保分的包课服务

couryes-lab™ 为您的留学生涯保驾护航 在代写线性代数linear algebra方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写线性代数linear algebra代写方面经验极为丰富，各种代写线性代数linear algebra相关的作业也就用不着说。

## 数学代写|线性代数代写linear algebra代考|Rank and invertible matrices

In this subsection we discuss the relationships between rank and invertible matrices.
Proposition (3.30). Let $\mathbf{A}$ be an $n$ by $n$ matrix. The matrix $\mathbf{A}$ is invertible $\Leftrightarrow \operatorname{rank}(\mathbf{A})=n$.
What does this mean?
Matrix is invertible $\Leftrightarrow$ it has no redundant rows. We say that the matrix $\mathbf{A}$ has full rank.
Proof.
$(\Rightarrow)$. We assume matrix $\mathbf{A}$ is invertible. By (1.35):
Theorem (1.35). Let $\mathbf{A}$ be a $n$ by $n$ matrix, then the following are equivalent:
(a) The matrix $\mathbf{A}$ is invertible.
(b) The reduced row echelon form of the matrix $\mathbf{A}$ is the identity matrix $\mathbf{I}$.
The reduced row echelon form of the matrix $\mathbf{A}$ is the identity $n$ by $n$ matrix $\mathbf{I}$. Thus there are $n$ non-zero rows of $\mathbf{I}$, therefore $\operatorname{rank}(\mathbf{A})=n$.
$(\Leftarrow)$. In this case, we assume that $\operatorname{rank}(\mathbf{A})=n$ and we need to prove that the matrix $\mathbf{A}$ is invertible. Since $\operatorname{rank}(\mathbf{A})=n$, the reduced row echelon form of $\mathbf{A}$ has no zero rows. By (1.39);
Proposition (1.39). $\mathbf{R}$ has at least one row of zeros $\Leftrightarrow \mathbf{A}$ is non-invertible (singular).
This means that matrix A must be invertible.
Hence if a square matrix is not of full rank then it is non-invertible.

## 数学代写|线性代数代写linear algebra代考|How can we write the general solution to Ax = O?

$$\left(\begin{array}{ccc} a_{11} & \cdots & a_{1 n} \ \vdots & \ddots & \vdots \ a_{m 1} & \cdots & a_{m n} \end{array}\right)\left(\begin{array}{c} x_1 \ \vdots \ x_n \end{array}\right)=\left(\begin{array}{c} 0 \ \vdots \ 0 \end{array}\right)$$
Multiplying these matrices we have:
\begin{aligned} & a_{11} x_1+\cdots+a_{1 n} x_n=0 \ & \vdots \quad \vdots \quad \vdots \quad \vdots \ & a_{m 1} x_1+\cdots+a_{m n} x_n=0 \ & \end{aligned}
We can write this in terms of the column vectors $\mathbf{c}1, \mathbf{c}_2, \ldots$ and $\mathbf{c}_n$ of matrix $\mathbf{A}$ : $$\left(\begin{array}{c} \mathbf{c}_1 \ a{11} \ \vdots \ a_{m 1} \end{array}\right) x_1+\cdots+\left(\begin{array}{c} \mathbf{c}n \ a{1 n} \ \vdots \ a_{m n} \end{array}\right) x_n=\left(\begin{array}{c} 0 \ \vdots \ 0 \end{array}\right)$$
Note that the left hand side is a linear combination of the column vectors $\mathbf{c}_1, \mathbf{c}_2, \ldots$ and $\mathbf{c}_n$ of matrix A. Recall that this linear combination is the column space of the matrix $\mathbf{A}$. The null space consists of vectors $\mathbf{x}=\left(\begin{array}{llll}x_1 & x_2 & \cdots & x_n\end{array}\right)^T$ such that they satisfy the linear combination $\left(^*\right)$.
The null space is a non-empty set.

How do we know it is non-empty?
Because the homogeneous system $\mathbf{A x}=\mathbf{O}$ always has the trivial solution $\mathbf{x}=\mathbf{O}$ $\left(x_1=\cdots=x_n=0\right)$ so we know the null space of matrix $\mathbf{A}$ is not empty.

Proposition (3.32). If $\mathbf{A}$ is a matrix with $n$ columns then the null space $N(\mathbf{A})$ is a subspace of $\mathbb{R}^n$.

# 线性代数代考

## 数学代写|线性代数代写linear algebra代考|Rank and invertible matrices

$(\Rightarrow)$。假设矩阵$\mathbf{A}$是可逆的。By (1.35):

(a)矩阵$\mathbf{A}$可逆。
(b)矩阵$\mathbf{A}$的行简化阶梯形是单位矩阵$\mathbf{I}$。

$(\Leftarrow)$。在这种情况下，我们假设$\operatorname{rank}(\mathbf{A})=n$我们需要证明矩阵$\mathbf{A}$是可逆的。因为$\operatorname{rank}(\mathbf{A})=n$, $\mathbf{A}$的行简化阶梯形没有零行。By (1.39);

## 数学代写|线性代数代写linear algebra代考|How can we write the general solution to Ax = O?

$$\left(\begin{array}{ccc} a_{11} & \cdots & a_{1 n} \ \vdots & \ddots & \vdots \ a_{m 1} & \cdots & a_{m n} \end{array}\right)\left(\begin{array}{c} x_1 \ \vdots \ x_n \end{array}\right)=\left(\begin{array}{c} 0 \ \vdots \ 0 \end{array}\right)$$

\begin{aligned} & a_{11} x_1+\cdots+a_{1 n} x_n=0 \ & \vdots \quad \vdots \quad \vdots \quad \vdots \ & a_{m 1} x_1+\cdots+a_{m n} x_n=0 \ & \end{aligned}

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## 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。

Days
Hours
Minutes
Seconds

# 15% OFF

## On All Tickets

Don’t hesitate and buy tickets today – All tickets are at a special price until 15.08.2021. Hope to see you there :)