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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• Advanced Probability Theory 高等概率论
• Advanced Mathematical Statistics 高等数理统计学
• (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|SUBSPACES

How can we recognize and describe a vector space? How can we single out a subset of a vector space with the same characteristics? To answer to these questions is necessary to introduce the definition of subspace.

Definition 2.4.1 Let $W$ be a subset of the space vector $V$. We say that $W$ is a subspace of $V$ if it satisfies the following properties:

1) $W$ is different from the empty set;
2) $W$ is closed with respect to the sum, that is, for every $\mathbf{u}, \mathbf{v} \in W$ we have that $\mathbf{u}+\mathbf{v} \in W$
3) $W$ is closed with respect to the product by scalars, that is, for every $\mathbf{u} \in W$ and every $\lambda \in \mathbb{R}$ we have that $\lambda \mathbf{u} \in W$.

We note that, since $W$ is not empty, and $\lambda \mathbf{u} \in W$ for each $\lambda \in \mathbb{R}$, then $\mathbf{0}V \in W$, because we can take any vector of $W$ and multiply it by $\lambda=0$. In fact, property (1) can effectively be replaced by the property: 1) $0_V \in W$, and we obtain an equivalent definition. It is important to note that a subspace $W$ of $V$ is a vector space with the operations of $V$ restricted to $W$. In fact, property 2) of Definition $2.4 .1$ ensures that the restriction to $W$ of the sum defined in $V$ gives as a result a vector of $W$ : $$+\left._V\right|{W \times W}: W \times W \rightarrow W .$$
Similarly, property 3) of Definition 2.4.1 ensures that the restriction to $\mathbb{R} \times W$ of the product by scalars defined on $\mathbb{R} \times V$ gives as a result a vector of $W$. Then properties 1 through 8 of Definition $2.3 .1$ continue to hold because they are true in $V$.

In particular, therefore, every vector space $V$ has always at least two subspaces: $V$ itself and the zero subspace, consisting of only the zero vector $0_V$. Because of Observation 2.3.8, if $V$ itself is not trivial, every nontrivial subspace of $V$ contains infinitely many elements.

We now want to clarify the concept of subspace with some examples and counterexamples.

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

2.5.1 Determine if the set $X=\left{(r, s, r-s) \in \mathbb{R}^3\right}$ is a subspace of $\mathbb{R}^3$.
Solution. First of all, we observe that $X$ is not the empty set because $(0,0,0) \in X$ (just take $r=s=0$ ).

Let us now consider two generic elements of $X$ : $\left(r_1, s_1, r_1-s_1\right)$ and $\left(r_2, s_2, r_2-s_2\right)$. Their sum is: $\left(r_1, s_1, r_1-s_1\right)+\left(r_2, s_2, r_2-s_2\right)=\left(r_1+r_2, s_1+s_2, r_1-s_1+r_2-s_2\right)=$ $\left(r_1+r_2, s_1+s_2, r_1+r_2-\left(s_1+s_2\right)\right)$, and it still belongs to $X$ as it is of the type $(r, s, r-s)$, with $r=r_1+r_2$ and $s=s_1+s_2$.

Consider $\left(r_1, s_1, r_1-s_1\right) \in X$ and $\lambda \in \mathbb{R}$. Then $\lambda\left(r_1, s_1, r_1-s_1\right)=\left(\lambda r_1, \lambda s_1, \lambda\left(r_1-\right.\right.$ $\left.\left.s_1\right)\right)=\left(\lambda r_1, \lambda s_1, \lambda r_1-\lambda s_1\right)$ still belongs to $X$ as it is of the type $(r, s, r-s)$, with $r=\lambda r_1$ and $s=\lambda s_1$. So $X$ is a subspace of $\mathbb{R}^3$.
2.5.2 Determine whether the set $W=\left{(x, y, z) \in \mathbb{R}^3 \mid 2 x+z^2=0\right} \subseteq \mathbb{R}^3$ is a subspace of $\mathbb{R}^3$.
Solution. $W$ is not the empty set because $(0,0,0) \in W$.
Consider now two generic elements of $W,\left(x_1, y_1, z_1\right)$ and $\left(x_2, y_2, z_2\right)$, with $2 x_1+$ $z_1^2=0$ and $2 x_2+z_2^2=0$. We have that $\left(x_1, y_1, z_1\right)+\left(x_2, y_2, z_2\right)=\left(x_1+x_2, y_1+\right.$ $\left.y_2, z_1+z_2\right)$, and this sum belongs to $W$ if and only if $2\left(x_1+x_2\right)+\left(z_1+z_2\right)^2=0$. But $2\left(x_1+x_2\right)+\left(z_1+z_2\right)^2=2 x_1+2 x_2+z_1^2+z_2^2+2 z_1 z_2=\left(2 x_1+z_1^2\right)+\left(2 x_2 Z_2+{ }^2\right)+2 z_1 z_2=$ $0+0+2 z_1 z_2=2 z_1 z_2$, and it is not true that $2 z_1 z_2$ is always equal to zero.

For example, the elements $(-2,1,2)$ and $(-8,3,4)$ belong to $W$ but $(-2,1,2)+$ $(-8,3,4)=(-10,4,6) \notin W$, because $2 \cdot(-10)+(6)^2 \neq 0$. (Note that these $W$ elements were not chosen randomly but so to satisfy the request $\left.2 z_1 z_2 \neq 0\right)$.
So $W$ is not a subspace of $\mathbb{R}^3$.
2.5.3 Determine a non-empty subset of $\mathbb{R}^3$ closed with respect to the sum but not with respect to the product by scalars.
Solution. The set $X={(x, y, z) \mid x, y, z \in \mathbb{R}, x \geq 0}$ has this property. In fact, $X$ is not empty because, for example, $(0,0,0) \in X$. Let us check if $X$ is closed with respect to the sum. Let $\left(x_1, y_1, z_1\right),\left(x_2, y_2, z_2\right) \in X$, with $x_1, x_2 \geq 0$. Then $\left(x_1, y_1, z_1\right)+$ $\left(x_2, y_2, z_2\right)=\left(x_1+x_2, y_1+y_2, z_1+z_2\right) \in X$ because $x_1+x_2 \geq 0$ (the sum of two non-negative real numbers is a non-negative real number). Now let $\left(x_1, y_1, z_1\right) \in X$ and $\lambda \in \mathbb{R}$. We have that $\lambda\left(x_1, y_1, z_1\right)=\left(\lambda x_1, \lambda y_1, \lambda z_1\right)$ belongs to $X$ if and only if $\lambda x_1 \geq 0$. But if we choose $\lambda$ negative and $x_1>0$, for example $\lambda=-1$ and $\left(x_1, y_1, z_1\right)=(3,-2,1)$, this condition it is not verified. So $X$ is not closed with respect to the product by scalars.

# 线性代数代考

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

1) $W$ 不同于空集：
2) $W$ 关于总和是封闭的，也就是说，对于每个 $\mathbf{u}, \mathbf{v} \in W$ 我们有 $\mathbf{u}+\mathbf{v} \in W$
3) $W$ 是关于乘积由标量封闭的，也就是说，对于每个 $\mathbf{u} \in W$ 和每一个 $\lambda \in \mathbb{R}$ 我们有 $\lambda \mathbf{u} \in W$.

$${ }_{+}{ }_v \mid W \times W: W \times W \rightarrow W .$$

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

2.5.1判断集合是否 \left 的分隔符缺失或无法识别 是一个子空间 $\mathbb{R}^3$.

$\left(r_1, s_1, r_1-s_1\right)+\left(r_2, s_2, r_2-s_2\right)=\left(r_1+r_2, s_1+s_2, r_1-s_1+r_2-s_2\right)=\left(r_1+r_2, s_1+s_2, r_1+r_2-\left(s_1+s_2\right)\right)$ ，它仍然属于 $X$ 因为它的类型 $(r, s, r-s) ，$ 和 $r=r_1+r_2$ 和 $s=s_1+s_2$.

$2.5 .2$ 判断是否设置 $\backslash 1 \mathrm{eft}$ 的分隔符缺失或无法识别

$2.5 .3$ 确定一个非空子集 $\mathbb{R}^3$ 关于总和但不是关于标量乘积的封闭。

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## MATLAB代写

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assignmentutor™您的专属作业导师
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