# 数学代写|线性代数代写linear algebra代考|Sum and Intersection of Subspaces

#### Doug I. Jones

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## 数学代写|线性代数代写linear algebra代考|Sum and Intersection of Subspaces

In this section we construct new spaces out of old ones.

FIGURE 3.4: The solution set $\mathcal{S}$ is obtained adding the particular solution $\mathbf{x}_p=(1,0,3)$ to the solution set of the homogeneous system.

Definition 31 Let $U$ and $W$ be subspaces of a vector space $V$ over $\mathbb{K}$. The sum of the subspaces $U$ and $W$, denoted $U+W$, is defined by
$$U+W={\boldsymbol{x}+\boldsymbol{y}: \boldsymbol{x} \in U \wedge \boldsymbol{y} \in W} .$$
Proposition 3.13 Let $U$ and $W$ be subspaces of a vector space $V$. Then $U+W$ and $U \cap W$ are vector subspaces of $V$.

Proof Notice that $U+W$ and $U \cap W$ are non-empty sets. Let $\alpha$ be a scalar and let $\boldsymbol{z} \in U+W$. Then there exist $\boldsymbol{x} \in U$ and $\boldsymbol{y} \in W$ such that $\boldsymbol{z}=\boldsymbol{x}+\boldsymbol{y}$ and
$$\alpha \boldsymbol{z}=\alpha(\boldsymbol{x}+\boldsymbol{y})=\alpha \boldsymbol{x}+\alpha \boldsymbol{y} .$$
Since $\alpha \boldsymbol{x} \in U$ and $\alpha \boldsymbol{y} \in W$, it follows that $\alpha \boldsymbol{z} \in U+W$. Hence, $U+W$ is closed under scalar multiplication.
To see that $U+W$ is closed for vector addition, let $\boldsymbol{z}_1$ and $\boldsymbol{z}_2$ be vectors in $U+W$. Then there exist $\boldsymbol{x}_1, \boldsymbol{x}_2 \in U$ and $\boldsymbol{y}_1, \boldsymbol{y}_2 \in W$ such that $\boldsymbol{z}_1=\boldsymbol{x}_1+\boldsymbol{y}_1$ and $z_2=\boldsymbol{x}_2+\boldsymbol{y}_2$. Hence
\begin{aligned} \boldsymbol{z}_1+\boldsymbol{z}_2 & =\boldsymbol{x}_1+\boldsymbol{y}_1+\boldsymbol{x}_2+\boldsymbol{y}_2 \ & =\left(\boldsymbol{x}_1+\boldsymbol{x}_2\right)+\left(\boldsymbol{y}_1+\boldsymbol{y}_2\right), \end{aligned}
which shows that $U+W$ is closed for the operation + .

## 数学代写|线性代数代写linear algebra代考|Change of Basis

Sometimes a problem becomes more tractable if one chooses an appropriate basis for the space in question. In the problem tackled in Example 5.12 of Chapter 5 , we will see that it is more convenient to use a particular basis of $\mathbb{R}^2$ and then go back to the standard basis. Is there, however, an easy way to relate the coordinate vectors in both bases? The answer is yes. We will justify our answer in this section.

Let $V$ be a vector space over $\mathbb{K}$ of dimension $k$, let $\mathcal{B}1$ and $\mathcal{B}_2$ be bases of $V$ and let $\boldsymbol{u} \in V$ be a vector. One might ponder whether there exists a relation between the coordinate vectors $\boldsymbol{u}{\mathcal{B}1}, \boldsymbol{u}{\mathcal{B}2} \in \mathbb{K}^k$ of $\boldsymbol{u}$ relative to $\mathcal{B}_1$ and $\mathcal{B}_2$, respectively. We shall see that there exists a matrix linking these two coordinate vectors making it easy to ‘pass’ from one basis to the other. Consider the bases $\mathcal{B}_1=\left(\boldsymbol{b}{\boldsymbol{1}}, \boldsymbol{b}{\boldsymbol{2}}, \cdots, \boldsymbol{b}{\boldsymbol{k}}\right)$ and $\mathcal{B}2$ of $V$. We have that $$\boldsymbol{u}=\alpha_1 \boldsymbol{b}_1+\alpha_2 \boldsymbol{b}_2+\cdots+\alpha_k \boldsymbol{b}{\boldsymbol{k}}$$
Hence, by Proposition 3.5,
\begin{aligned} (\boldsymbol{u}){\mathcal{B}_2} & =\left(\alpha_1 \boldsymbol{b}{\boldsymbol{1}}+\alpha_2 \boldsymbol{b}2+\cdots+\alpha_k \boldsymbol{b}{\boldsymbol{k}}\right){\mathcal{B}_2} \ & =\left(\alpha_1 \boldsymbol{b}{\mathbf{1}}\right){\mathcal{B}_2}+\left(\alpha_2 \boldsymbol{b}{\mathbf{2}}\right){\mathcal{B}_2}+\cdots+\left(\alpha_k \boldsymbol{b}{\boldsymbol{k}}\right){\mathcal{B}_2} \ & =\alpha_1\left(\boldsymbol{b}{\mathbf{1}}\right){\mathcal{B}_2}+\alpha_2\left(\boldsymbol{b}_2\right){\mathcal{B}2}+\cdots+\alpha_k\left(\boldsymbol{b}{\boldsymbol{k}}\right){\mathcal{B}_2} . \end{aligned} This equality can also be written in a matrix form as $$\mathbf{u}{\mathcal{B}2}=\left[\begin{array}{lllllll} \left(\mathbf{b}_1\right){\mathcal{B}2} & \mid & \left(\mathbf{b}_2\right){\mathcal{B}2} & \mid & \ldots & \mid & \left(\mathbf{b}{\mathbf{k}}\right){\mathcal{B}_2} \end{array}\right]\left[\begin{array}{c} \alpha_1 \ \alpha_2 \ \vdots \ \alpha_k \end{array}\right] .$$ That is, The matrix whose columns are the coordinate vectors of the vectors of basis $\mathcal{B}_1$ relative to the basis $\mathcal{B}_2$ is called the change of basis matrix from basis $\mathcal{B}_1$ to basis $\mathcal{B}_2$, and is denoted by $M{\mathcal{B}_2 \leftarrow \mathcal{B}_1}$.

# 线性代数代考

## 数学代写|线性代数代写linear algebra代考|Sum and Intersection of Subspaces

$$U+W=\boldsymbol{x}+\boldsymbol{y}: \boldsymbol{x} \in U \wedge \boldsymbol{y} \in W$$

$$\alpha \boldsymbol{z}=\alpha(\boldsymbol{x}+\boldsymbol{y})=\alpha \boldsymbol{x}+\alpha \boldsymbol{y}$$

$$\boldsymbol{z}_1+\boldsymbol{z}_2=\boldsymbol{x}_1+\boldsymbol{y}_1+\boldsymbol{x}_2+\boldsymbol{y}_2=\left(\boldsymbol{x}_1+\boldsymbol{x}_2\right)$$

## 数学代写|线性代数代写linear algebra代考|Change of Basis

$$\boldsymbol{u}=\alpha_1 \boldsymbol{b}_1+\alpha_2 \boldsymbol{b}_2+\cdots+\alpha_k \boldsymbol{b} \boldsymbol{k}$$

$(\boldsymbol{u}) \mathcal{B}_2=\left(\alpha_1 \boldsymbol{b} \mathbf{1}+\alpha_2 \boldsymbol{b} 2+\cdots+\alpha_k \boldsymbol{b} \boldsymbol{k}\right) \mathcal{B}_2=\left(\alpha_1\right.$

$$\mathbf{u} \mathcal{B} 2=\left[\left(\mathbf{b}_1\right) \mathcal{B} 2 \quad\left|\quad\left(\mathbf{b}_2\right) \mathcal{B} 2 \quad\right| \ldots \mid \quad(\mathbf{b k}) \mathcal{B}_2\right]$$

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