## 数学代写|线性代数代写linear algebra代考|MATHS1011

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

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 $W \subsetneq V$.

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

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.

Example 2.5.12 Consider the set of images
$V={I \mid I$ is of the form shown in Figure $2.19$ and $a, b, c \in \mathbb{R}}$.
$V$ is a subspace of the space of images with the same geometric configuration.
Proof. Let $V$ be defined as above. We will show that $V$ satisfies the hypotheses of Theorem 2.5.3. We can see that the set of images $V$ is a subset of the vector space of all images with the same geometric configuration.

Next, notice that $V$ is nonempty since the image in $V$ with $a=b=c=0$ is the zero image. Now, we need to show that the set is closed under scalar multiplication and addition. Let $\alpha \in \mathbb{R}$ be a scalar and let $I_1$ and $I_2$ be images in $V$. For the image $I_1$, let $a=a_1, b=b_1$, and $c=c_1$, and for the image $I_2$, let $a=a_2, b=b_2$, and $c=c_2$ (Figure 2.20).

We can see that $\alpha I_1+I_2$ is also in $V$ with $a=\alpha a_1+a_2, b=\alpha b_1+b_2$, and $c=\alpha c_1+c_2$. For example, the pixel intensity in the pixel on the bottom left of $\alpha I_1+I_2$ is
$$\alpha\left(a_1-c_1\right)+a_2-c_2=\alpha a_1+a_2-\left(\alpha c_1-c_2\right)=a-c .$$

# 线性代数代考

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

(P1) W在添加下关闭。
(P2) $W$ 在标量乘法下是封闭的。
(P8) $W$ 包含附加标识，表示为 0 。
(P9) $W$ 包含加法逆元。

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

$T$ 不是的子空间 $\left(\mathbb{R}^2,+, \cdot\right)$ ，因为虽然 0 在 $T$ ，制作 $T \neq \emptyset, T$ 没有所 有的财产 $x, y \in T$ 对于所有人 $\alpha, \beta \in \mathbb{R}, \alpha x+\beta y \in T$. 为了验证这 一点，我们只需要产生一个向量的例子 $x, y \in T$ 和标量 $\alpha, \beta \in \mathbb{R}$ 以便 $\alpha x+\beta y$ 不在 $T$. 请注意 $x=(0,1), y=(1,0)$ 是元溸 $T$ 和 $\alpha=\beta=1$ 在 $\mathbb{R}$. 自从 $1 \cdot x+1 \cdot y=(1,1)$ 哪个不在 $T, T$ 不满足子 空间性质。

$V=I \mid I \$$isofthe formshowninFigure \ 2.19 \ a n d \ a, b, c \in \mathbb{R} V 是具有相同几何配置的图像空间的子空间。 证明。让 V 定义如上。我们将表明 V 满足定理 2.5.3 的假设。我们可以 看到这组图像 V 是具有相同几何配置的所有图像的向量空间的子集。 接下来，请注意 V 是非空的，因为图像在 V 和 a=b=c=0 是零图像。现在，我们需要证明该集合在标量乘法和加法下是封闭的。让 \alpha \in \mathbb{R} 是一个标量，让 I_1 和 I_2 成为图像 V. 对于图像 I_1 ，让 a=a_1, b=b_1 ，和 c=c_1 ，对于图像 I_2 ， 让 a=a_2, b=b_2 ， 和 c=c_2 (图 2.20)。 我们可以看到 \alpha I_1+I_2 也在 V 和 a=\alpha a_1+a_2, b=\alpha b_1+b_2 ， 和 c=\alpha c_1+c_2. 例如，左下角像溸中的像债强度 \alpha I_1+I_2 是$$ \alpha\left(a_1-c_1\right)+a_2-c_2=\alpha a_1+a_2-\left(\alpha c_1-c_2\right)=a-c$\$

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## MATLAB代写

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