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• 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|线性代数代写linear algebra代考|LINEAR INDEPENDENCE

Now we ask: How do we establish what the “redundant vectors” are in the description of the subspace generated by a set of vectors? If we want be efficient in the description of a subspace, we must be able to describe it as the subspace generated by the smallest possible number of vectors. The answer to this question comes from the concept of linear independence. This is by far the most difficult concept to understand, and it is the cornerstone of the whole theory. Shortly, the story is: if a set of generators of a subspace is a set of linearly independent vectors, then we are sure that it is the most efficient way to describe the subspace generated by those vectors, namely that we are using the smallest number of vectors. Let us see the definition and then with a series of small steps, we will arrive to prove the assertion above.

Definition 3.2.1 Let $V$ be a vector space. The vectors $\mathbf{v}_1, \ldots, \mathbf{v}_n \in V$ are linearly independent if for every linear combination
$$\lambda_1 \mathbf{v}_1+\cdots+\lambda_n \mathbf{v}_n=0,$$
we have $\lambda_1=\cdots=\lambda_n=0$. In other words, the only linear combination of the vectors $\mathbf{v}_1, \ldots, \mathbf{v}_n$ giving the zero vector is the one with all zero scalars. We will say also that the set of vectors $\left{\mathbf{v}_1, \ldots, \mathbf{v}_n\right}$ is linearly independent. ${ }^1$

The vectors $\mathbf{v}_1, \ldots, \mathbf{v}_n$ are linearly dependent, if they are not independent. In other words, the vectors of the set $\left{\mathbf{v}_1, \ldots, \mathbf{v}_n\right}$ are linearly dependent if there exist scalars $\lambda_1, \ldots, \lambda_n$, not all zero, such that $\lambda_1 \mathbf{v}_1+\cdots+\lambda_n \mathbf{v}_n=\mathbf{0}$.

Let us review the previous examples. The set of vectors ${(1,0),(0,1)}$ in $\mathbb{R}^2$ is a set of vectors which are linearly independent, as their only linear combination that gives the zero vector is obtained with all zero scalars:
$$\alpha(1,0)+\beta(0,1)=(\alpha, \beta)=(0,0),$$
only if $\alpha=\beta=0$.
On the other hand, the vectors of the set ${(1,0),(0,1),(1,1)}$ are linearly dependent, because there is a linear combination of the given vectors with scalars, not all zero, which is equal to the zero vector.
$$1 \cdot(1,0)+1 \cdot(0,1)-1 \cdot(1,1)=(0,0)$$
Let us see a more elaborate example.

## 数学代写|线性代数代写linear algebra代考|EXERCISES WITH SOLUTIONS

3.3.1 Determine for which values of $k$ the vectors $x^2+2 x+k, 5 x^2+2 k x+k^2, k x^2+x+3$ generate $\mathbb{R}_2[x]$
Solution. Consider the generic vector $a x^2+b x+c \in \mathbb{R}_2[x]$, and see when $a x^2+b x+c \in$ $\left\langle x^2+2 x+k, 5 x^2+2 k x+k^2, k x^2+x+3\right\rangle$. This happens if we have
$$a x^2+b x+c=\lambda_1\left(x^2+2 x+k\right)+\lambda_2\left(5 x^2+2 k x+k^2\right)+\lambda_3\left(k x^2+x+3\right),$$

for some $\lambda_1, \lambda_2, \lambda_3 \in \mathbb{R}$, that is if
$$a x^2+b x+c=\left(\lambda_1+5 \lambda_2+k \lambda_3\right) x^2+\left(2 \lambda_1+2 k \lambda_2+\lambda_3\right) x+\left(k \lambda_1+k^2 \lambda_2+3 \lambda_3\right) .$$
So $\lambda_1, \lambda_2, \lambda_3$ must be solutions of the system:
$$\left{\begin{array}{l} \lambda_1+5 \lambda_2+k \lambda_3=a \ 2 \lambda_1+2 k \lambda_2+\lambda_3=b \ k \lambda_1+k^2 \lambda_2+3 \lambda_3=c \end{array}\right.$$
The complete matrix associated with the system is:
$$(A \mid \underline{b})=\left(\begin{array}{rrr|r} 1 & 5 & k & a \ 2 & 2 k & 1 & b \ k & k^2 & 3 & c \end{array}\right),$$
which reduced to row echelon form with Gaussian algorithm becomes:
$$\left(A^{\prime} \mid \underline{b^{\prime}}\right)=\left(\begin{array}{rrr|r} 1 & 5 & k & a \ 0 & 2 k-10 & 1-2 k & b-2 a \ 0 & 0 & 3-\frac{1}{2} k & c-\frac{b}{2} k \end{array}\right) .$$

# 线性代数代考

## 数学代写|线性代数代写linear algebra代考|LINEAR INDEPENDENCE

$$\lambda_1 \mathbf{v}_1+\cdots+\lambda_n \mathbf{v}_n=0,$$

$$\alpha(1,0)+\beta(0,1)=(\alpha, \beta)=(0,0),$$

$$1 \cdot(1,0)+1 \cdot(0,1)-1 \cdot(1,1)=(0,0)$$

## 数学代写|线性代数代写linear algebra代考|EXERCISES WITH SOLUTIONS

3.3.1 确定哪些值 $k$ 向量 $x^2+2 x+k, 5 x^2+2 k x+k^2, k x^2+x+3$ 产生 $\mathbb{R}_2[x]$

$$a x^2+b x+c=\lambda_1\left(x^2+2 x+k\right)+\lambda_2\left(5 x^2+2 k x+k^2\right)+\lambda_3\left(k x^2+x+3\right),$$

$$a x^2+b x+c=\left(\lambda_1+5 \lambda_2+k \lambda_3\right) x^2+\left(2 \lambda_1+2 k \lambda_2+\lambda_3\right) x+\left(k \lambda_1+k^2 \lambda_2+3 \lambda_3\right) .$$

$\$ \$$Neft }$$
\lambda_1+5 \lambda_2+k \lambda_3=a 2 \lambda_1+2 k \lambda_2+\lambda_3=b k \lambda_1+k^2 \lambda_2+3 \lambda_3=c
$$正确的。 Thecompletematrixassociatedwiththesystemis : (A \mid \underline {b})=\backslash \operatorname{left}($$
15 k|a 2| 2 k|1| b k\left|k^2\right| 3 \mid c
$$正确的）， whichreducedtorowechelon formwithGaussianalgorithmbecomes : Veft(A^{prime } \backslash mid \underline \left{\mathrm{b}^{\wedge}{{\right. prime }} \backslash right )=\backslash \mathrm{Veft}^{\prime}($$
15 \quad 5|a 0| 2 k-10|1-2 k| b-2 a 0|0| 3-\frac{1}{2} k \mid c-\frac{b}{2} k
$$正确的）。 \ \$$

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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。

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