### 数学代写|线性代数代写linear algebra代考|MATH1051

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• (Generalized) Linear Models 广义线性模型
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• 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|Subsets and Subspaces

Let $\left(V,+,{ }^{\circ}\right)$ be a vector space. In this section we discuss conditions on a subset of a vector space that will guarantee the subset is also a vector space. Recall that a subset of $V$ is a set that contains some of the elements of $V$. We define subset more precisely here.

Let $V$ and $W$ be sets. We say that $W$ is a subset of $V$ if every element of $W$ is an element of $V$ and we write $W \subset V$ or $W \subseteq V$. In the case where $W \neq V$ (there are elements of $V$ that are not in $W$ ), we say that $W$ is a proper subset of $V$ and we write $W \subseteq V$.

In a vector space context, we always assume the same operations on $W$ as we have defined on $V$.
Let $W$ be a subset of $V$. We are interested in subsets that also satisfy the vector space properties (recall Definition 2.3.5).

Let $(V,+, \cdot)$ be a vector space over a field $\mathbb{F}$. If $W \subseteq V$, then we say that $W$ is a subspace of $(V,+, \cdot)$ whenever $(W,+, \cdot)$ is also a vector space.

Now consider which vector space properties of $(V,+, \cdot)$ must also be true of the subset $W$. Which properties are not necessarily true? The commutative, associative, and distributive properties still hold because the operations are the same, the scalars come from the same scalar field, and elements of $W$ come from the set $V$. Therefore, since these properties are true in $V$, they are true in $W$. We say that these properties are inherited from $V$ since $V$ is like a parent set to $W$. Also, since, we do not change the scalar set when considering a subset, the scalar 1 is still an element of the scalar set. This tells us that we can determine whether a subset of a vector space is, itself, a vector space, by checking those properties that depend on how the subset differs from the parent vector space. The properties we need to check are the following
(P1) $W$ is closed under addition.
(P2) $W$ is closed under scalar multiplication.
(P8) $W$ contains the additive identity, denoted $0 .$
(P9) $W$ contains additive inverses.
With careful consideration, we see that, because $V$ contains additive inverses, then if (P1), (P2), and (P8) are true for $W$, it follows that $W$ must also contain additive inverses (see Exercise 14). Hence, as the following theorem states, we need only test for properties (P1), (P2), and (P8) in order to determine whether a subset is a subspace.

## 数学代写|线性代数代写linear algebra代考|Examples of Subspaces

Every vector space $(V,+, \cdot)$ has at least the following two subspaces.
Theorem $2.5 .7$
Let $(V,+, \cdot)$ be a vector space. Then $V$ is itself a subspace of $(V,+, \cdot)$.
Proof. Since every set is a subset of itself, the result follows from Definition $2.5 .2 .$
Theorem $2.5 .8$
Let $(V,+, \cdot)$ be a vector space. Then the set ${0}$ is a subspace of $(V,+, \cdot)$.
The proof is Exercise $19 .$
Example 2.5.9 Recall Example 2.4.9 from the last section. Let $V \subset \mathbb{R}^{3}$ be the set of all solutions to the equation $x_{1}+3 x_{2}-x_{3}=0$. Then $V$ is a subspace of $\mathbb{R}^{3}$, with the standard operations.

More generally, as we saw in the last section, the set of solutions to any homogeneous linear equation with $n$ variables is a subspace of $\left(\mathbb{R}^{n},+, \cdot\right)$.

Example 2.5.10 Consider the coordinate axes as a subset of the vector space $\mathbb{R}^{2}$. That is, let $T \subset \mathbb{R}^{2}$ be defined by
$$T=\left{x=\left(x_{1}, x_{2}\right) \in \mathbb{R}^{2} \mid x_{1}=0 \text { or } x_{2}=0\right}$$
$T$ is not a subspace of $\left(\mathbb{R}^{2},+, \cdot\right)$, because although 0 is in $T$, making $T \neq \emptyset, T$ does not have the property that for all $x, y \in T$ and for all $\alpha, \beta \in \mathbb{R}, \alpha x+\beta y \in T$. To verify this, we need only produce one example of vectors $x, y \in T$ and scalars $\alpha, \beta \in \mathbb{R}$ so that $\alpha x+\beta y$ is not in $T$. Notice that $x=(0,1)$, $y=(1,0)$ are elements of $T$ and $\alpha=\beta=1$ are in $\mathbb{R}$. Since $1 \cdot x+1 \cdot y=(1,1)$ which is not in $T$, $T$ does not satisfy the subspace property.

Example 2.5.11 Consider $W=\left{(a, b, c) \in \mathbb{R}^{3} \mid c=0\right} . W$ is a subspace of $\mathbb{R}^{3}$, with the standard operations of addition and scalar multiplication. See Exercise $9 .$

## 数学代写|线性代数代写linear algebra代考|Building New Subspaces

In this section, we investigate the question, “If we start with two subspaces of the same vector space, to some observations that will simplify some previous examples and give us new tools for proving that subsets are subspaces. We first consider intersections and unions.
Definition $2.5 .17$
Let $S$ and $T$ be sets.

• The intersection of $S$ and $T$, written $S \cap T$, is the set containing all elements that are in both $S$ and $T$.
• The union of $S$ and $T$, written $S \cup T$, is the set containing all elements that are in either $S$ or $T$ (or both).
The intersection of two subspaces is a also a subspace.

Proof. Let $W_{1}$ and $W_{2}$ be subspaces of $(V,+, \cdot)$. We will show that the intersection of $W_{1}$ and $W_{2}$ is nonempty and closed under scalar multiplication and vector addition. To show that $W_{1} \cap W_{2}$ is nonempty, we notice that since both $W_{1}$ and $W_{2}$ contain the zero vector, so does $W_{1} \cap W_{2}$.

Now, let $u$ and $v$ be elements of $W_{1} \cap W_{2}$ and let $\alpha$ be a scalar. Since $W_{1}$ and $W_{2}$ are closed under addition and scalar multiplication, we know that $\alpha \cdot u+v$ is also in both $W_{1}$ and $W_{2}$. That is, $\alpha \cdot u+v$ is in $W_{1} \cap W_{2}$, so by Corollary $2.5 .6 W_{1} \cap W_{2}$ is closed under addition and scalar multiplication.
Thus, by Corollary $2.5 .4, W_{1} \cap W_{2}$ is a subspace of $(V,+, \cdot)$.
An important example involves solutions to homogeneous equations, which we first considered in Example 2.4.12.

Example 2.5.19 The solution set of a single homogeneous equation in $n$ variables is a subspace of $\mathbb{R}^{n}$ (see Example 2.4.12). By Theorem 2.5.18, the intersection of the solution sets of any $k$ homogeneous equations in $n$ variables is also subspace of $\mathbb{R}^{n}$.

In other words, if a system of linear equations consists only of homogeneous equations, then the set of solutions forms a subspace of $\mathbb{R}^{n}$. This is such an important result that we promote it from example to theorem.

## 数学代写|线性代数代写linear algebra代考|Subsets and Subspaces

（P1）在在添加下关闭。
(P2)在在标量乘法下是闭合的。
(P8)在包含加法身份，表示为0.
(P9)在包含加法逆元。

## 数学代写|线性代数代写linear algebra代考|Examples of Subspaces

T=\left{x=\left(x_{1}, x_{2}\right) \in \mathbb{R}^{2} \mid x_{1}=0 \text { 或 } x_{2} =0\右}T=\left{x=\left(x_{1}, x_{2}\right) \in \mathbb{R}^{2} \mid x_{1}=0 \text { 或 } x_{2} =0\右}

## 数学代写|线性代数代写linear algebra代考|Building New Subspaces

• 的交叉点小号和吨, 写小号∩吨, 是包含两者中所有元素的集合小号和吨.
• 工会小号和吨, 写小号∪吨, 是包含所有元素的集合小号或者吨（或两者）。
两个子空间的交集也是一个子空间。

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