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statistics-lab™ 为您的留学生涯保驾护航 在代写计算线性代数Computational 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。我们的专家在代写计算线性代数Computational Linear Algebra代写方面经验极为丰富，各种代写计算线性代数Computational Linear Algebra相关的作业也就用不着说。

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 数学代写|计算线性代数代写Computational Linear Algebra代考|Linear Dependence and Linear Independence

Definition 4.11. Let $\lambda_{1}, \lambda_{2}, \ldots, \lambda_{n}$ be $n$ scalars $\in \mathbb{R}$ and $\overrightarrow{v_{1}}, \overrightarrow{v_{2}}, \ldots, \overrightarrow{v_{n}}$ be $n$ vectors $\in \mathbb{V}{3}$. The linear combination of the $n$ vectors by means of the $n$ scalars is the vector $$\vec{w}=\lambda{1} \overrightarrow{v_{1}}+\lambda_{2} \overrightarrow{v_{2}}+\ldots, \lambda_{n} \vec{v}{n} .$$ Definition 4.12. Let $\overrightarrow{v{1}}, \overrightarrow{v_{2}}, \ldots, \overrightarrow{v_{n}}$ be $n$ vectors $\in \mathbb{V}{3}$. These vectors are said linearly dependent if the null vector can be expressed as their linear combination by means of and $n$-tuple of non-null coefficients: \begin{aligned} &\exists \lambda{1}, \lambda_{2}, \ldots, \lambda_{n} \in \mathbb{R}{\ni} \cdot \ &\vec{o}=\lambda{1} \overrightarrow{v_{1}}+\lambda_{2} \overrightarrow{v_{2}}+\ldots+\lambda_{n} \overrightarrow{v_{n}} \end{aligned}
with $\lambda_{1}, \lambda_{2}, \ldots, \lambda_{n} \neq 0,0, \ldots, 0$.
Definition 4.13. Let $\overrightarrow{v_{1}}, \overrightarrow{v_{2}}, \ldots, \overrightarrow{v_{n}}$ be $n$ vectors $\in \mathbb{V}{3}$. These vectors are said linearly independent if the null vector can be expressed as their linear combination only by means of null coefficients: \begin{aligned} &\nexists \lambda{1}, \lambda_{2}, \ldots, \lambda_{n} \in \mathbb{R}{ }{\ni}{ }^{\circ} \ &\vec{o}=\lambda{1} \overrightarrow{v_{1}}+\lambda_{2} \overrightarrow{v_{2}}+\ldots+\lambda_{n} \overrightarrow{v_{n}} \end{aligned}
with $\lambda_{1}, \lambda_{2}, \ldots, \lambda_{n} \neq 0,0, \ldots, 0$.
Example 4.7. Let us consider three vectors $\overrightarrow{v_{1}}, \overrightarrow{v_{2}}$, and $\overrightarrow{v_{3}} \in \mathbb{V}{3}$. If at least one tuple $\lambda{1}, \lambda_{2}, \lambda_{3} \in \mathbb{R}$ and $\neq 0,0,0$ such that $\vec{o}=\lambda_{1} \overrightarrow{v_{1}}+\lambda_{2} \overrightarrow{v_{2}}+\lambda_{3} \overrightarrow{v_{3}}$ can be found, the vectors are linearly dependent. For example if the tuple $-4,5,0$ is such that $\vec{o}=$ $-4 \overrightarrow{v_{1}}+5 \overrightarrow{v_{2}}+0 \overrightarrow{v_{3}}$ then the tree vectors are linearly dependent.

## 数学代写|计算线性代数代写Computational Linear Algebra代考|Matrices of Vectors

Proposition 4.4. Let $\vec{u}, \vec{v} \in \mathbb{V}{3}$ where \begin{aligned} &\vec{u}=\left(u{1}, u_{2}, u_{3}\right) \ &\vec{v}=\left(v_{1}, v_{2}, v_{3}\right) \end{aligned}
and $\mathbf{A}$ be a $2 \times 3$ matrix whose elements are the components of $\vec{u}$ and $\vec{v}$
$$\mathbf{A}=\left(\begin{array}{lll} u_{1} & u_{2} & u_{3} \ v_{1} & v_{2} & v_{3} \end{array}\right) .$$
These two vectors are parallel (and thus linearly dependent) if and only if the rank of the matrix $\mathbf{A}$ associated with the corresponding components is $<2: \rho_{\mathbf{A}}<2$.
Proof. If $\vec{u}$ and $\vec{v}$ are parallel they could be expressed as $\vec{u}=\lambda \vec{v}$ with $\lambda \in \mathbb{R}$. Thus,
$$\begin{array}{r} \vec{u}=\lambda \vec{v} \Rightarrow \ \Rightarrow u_{1} \overrightarrow{e_{1}}+u_{2} \overrightarrow{e_{2}}+u_{3} \overrightarrow{e_{3}}=\lambda\left(v_{1} \overrightarrow{e_{1}}+v_{2} \overrightarrow{e_{2}}+v_{3} \overrightarrow{e_{3}}\right) . \end{array}$$
Since two vectors are the equal if and only if they have the same components,
$$\begin{gathered} u_{1}=\lambda v_{1} \ u_{2}=\lambda v_{2} \ u_{3}=\lambda v_{3} . \end{gathered}$$
Since the two rows are proportional, there is no non-singular order 2 submatrix. Thus $\rho_{\mathbf{A}}<2$.

If $\rho_{\mathbf{A}}<2$, every two submatrix has null determinant. This can happen in the following cases.

• A row is composed of zeros. This means that one vector is the null vector $\vec{o}$, e.g.
$$\mathbf{A}=\left(\begin{array}{ccc} 0 & 0 & 0 \ v_{1} & v_{2} & v_{3} \end{array}\right) \text {. }$$
Since every vector is parallel to $\vec{o}$, the vectors are parallel.

## 数学代写|计算线性代数代写Computational Linear Algebra代考|Complex Numbers

As mentioned in Chap. 1, for a given set and an operator applied to its elements, if the result of the operation is still an element of the set regardless of the input of the operator, then the set is said closed with respect to that operator. For example it is easy to verify that $\mathbb{R}$ is closed with respect to the sum as the sum of two real numbers is certainly a real number. On the other hand, $\mathbb{R}$ is not closed with respect to the square root operation. More specifically, if a square root of a negative number has to be calculated the result is not determined and is not a real number. In order to represent these numbers Gerolamo Cardano in the sixteenth century introduced the concept of Imaginary numbers, see [11], by defining the imaginary unit $j$ as the square root of $-1: j=\sqrt{-1}$. This means that the square roots of negative numbers can be represented.
Example 5.1. $\sqrt{-9}=j 3$.
Imaginary numbers compose a set of numbers represented by the symbol I. The basic arithmetic operations can be applied to imaginary numbers.

• sum: $j a+j b=j(a+b)$
• difference: $j a-j b=j(a-b)$
• product: $j a j b=-a b$
• division: $\frac{j a}{j b}=\frac{a}{b}$
Example 5.2. Let us consider the imaginary numbers $j 2$ and $j 5$. It follows that
\begin{aligned} &j 2+j 5=j 7 \ &j 2-j 5=-j 3 \ &j 2 j 5=-10 \ &\frac{j 2}{j 5}=\frac{2}{5} . \end{aligned}

## 数学代写|计算线性代数代写Computational Linear Algebra代考|Linear Dependence and Linear Independence

$$\vec{w}=\lambda \overrightarrow{v_{1}}+\lambda_{2} \overrightarrow{v_{2}}+\ldots, \lambda_{n} \vec{v} n$$

$$\exists \lambda 1, \lambda_{2}, \ldots, \lambda_{n} \in \mathbb{R} \ni \cdot \quad \vec{o}=\lambda 1 \overrightarrow{v_{1}}+\lambda_{2} \overrightarrow{v_{2}}+\ldots+\lambda_{n} \overrightarrow{v_{n}}$$

$$\nexists \lambda 1, \lambda_{2}, \ldots, \lambda_{n} \in \mathbb{R} \ni^{\circ} \quad \vec{o}=\lambda 1 \overrightarrow{v_{1}}+\lambda_{2} \overrightarrow{v_{2}}+\ldots+\lambda_{n} \overrightarrow{v_{n}}$$

## 数学代写|计算线性代数代写Computational Linear Algebra代考|Matrices of Vectors

$$\vec{u}=\left(u 1, u_{2}, u_{3}\right) \quad \vec{v}=\left(v_{1}, v_{2}, v_{3}\right)$$

$$\vec{u}=\lambda \vec{v} \Rightarrow \Rightarrow u_{1} \overrightarrow{e_{1}}+u_{2} \overrightarrow{e_{2}}+u_{3} \overrightarrow{e_{3}}=\lambda\left(v_{1} \overrightarrow{e_{1}}+v_{2} \overrightarrow{e_{2}}+v_{3} \overrightarrow{e_{3}}\right)$$

$$u_{1}=\lambda v_{1} u_{2}=\lambda v_{2} u_{3}=\lambda v_{3}$$

• 一行由零组成。这意味着一个向量是空向量 $\vec{o}$ ，例如
$$\mathbf{A}=\left(\begin{array}{llllll} 0 & 0 & 0 & v_{1} & v_{2} & v_{3} \end{array}\right) .$$
因为每个向量都平行于 $\vec{o}$ ，向量是平行的。

## 数学代写|计算线性代数代写Computational Linear Algebra代考|Complex Numbers

• 和: $j a+j b=j(a+b)$
• 区别: $j a-j b=j(a-b)$
• 产品: $j a j b=-a b$
• 分配: $\frac{j a}{j b}=\frac{a}{b}$
例 5.2。让我们考虑虚数 $j 2$ 和 $j 5$. 它遵循
$$j 2+j 5=j 7 \quad j 2-j 5=-j 3 j 2 j 5=-10 \quad \frac{j 2}{j 5}=\frac{2}{5}$$

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assignmentutor™您的专属作业导师
assignmentutor™您的专属作业导师