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• Statistical Inference 统计推断
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• (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|Bases and Dimension

A crucial notion is that of a basis of a vector space. Intuitively, a basis can be thought of as a system of coordinate axes with respect to which the vectors are described, much like what happens in $\mathbb{R}^2$ or $\mathbb{R}^3$, for example. Roughly speaking, the number of axes in the system is the dimension of the space.

Definition 28 A basis $\mathcal{B}$ of a vector space $V$ is a (any) linearly independent set that spañs $V$.
Example 3.7 Here are some examples of basis of $\mathbb{R}^2$.
a) The set $\mathcal{E}_2={(1,0),(0,1)}$ is a basis of $\mathbb{R}^2$, said the standard basis of $\mathbb{R}^2$.
b) The set $\mathcal{B}={(1,1),(-1,0)}$ is another basis of $\mathbb{R}^2$.
c) Another basis of $\mathbb{R}^2$ is $\mathcal{B}_1=\ldots$ (find one).

FIGURE 3.3: The coordinate vectors of $\boldsymbol{u}=(2,3)$ relative to the standard basis and relative to $\mathcal{B}=\left(\boldsymbol{b}_1, \boldsymbol{b}_2\right)$ with $\boldsymbol{b}_1=(1,1), \boldsymbol{b}_2=(-1,0)$.

Exercise 3.5 Express the vector $(2,3)$ as a linear combination of the vector of the basis $\mathcal{B}$ in b) above.
Solution. One wants to find $\alpha_1, \alpha_2 \in \mathbb{R}$ such that
$$\alpha_1\left[\begin{array}{l} 1 \ 1 \end{array}\right]+\alpha_2\left[\begin{array}{c} -1 \ 0 \end{array}\right]=\left[\begin{array}{l} 2 \ 3 \end{array}\right]$$
This leads to solving the system whose augmented matrix is
$$\left[\begin{array}{cc|c} 1 & -1 & 2 \ 1 & 0 & 3 \end{array}\right],$$
yielding $\alpha_1=3$ e $\alpha_2=1$ (see Figure 3.3).
Thé subset $\mathcal{E}_n={(1,0, \ldots, 0),(0,1,0, \ldots, 0), \ldots,(0, \ldots, 0,1)}$ of $\mathbb{R}^n$ is the standard basis of $\mathbb{R}^n$.

## 数学代写|线性代数代写linear algebra代考|Matrix spaces and spaces of polynomials

Up to this point and although our definitons are general, in practical terms we have been focusing on $\mathbb{K}^n$. It is now the time to see what we get when addressing these concepts in more general vector spaces.

The ordered standard basis of the space $\mathrm{M}{k, n}(\mathbb{K})$ of the $k \times n$ real matrices is the ordered set consisting of the real $k \times n$ matrices having all entries but one equal to zero which takes value 1 ; the ordering is such that the non-zero entry in the first matrix is entry-11 and it ‘circulates’ along the lines from left to right. For example, in the case of $\mathrm{M}_2(\mathbb{K})$, the standard basis is $$\mathcal{B}_c=\left(\left[\begin{array}{ll} 1 & 0 \ 0 & 0 \end{array}\right],\left[\begin{array}{ll} 0 & 1 \ 0 & 0 \end{array}\right],\left[\begin{array}{ll} 0 & 0 \ 1 & 0 \end{array}\right],\left[\begin{array}{ll} 0 & 0 \ 0 & 1 \end{array}\right]\right) .$$ Given a matrix $A=\left[a{i j}\right]$, we have
$$A=\left[\begin{array}{ll} a_{11} & a_{12} \ a_{21} & a_{22} \end{array}\right]=a_{11}\left[\begin{array}{ll} 1 & 0 \ 0 & 0 \end{array}\right]+a_{12}\left[\begin{array}{ll} 0 & 1 \ 0 & 0 \end{array}\right]+a_{21}\left[\begin{array}{ll} 0 & 0 \ 1 & 0 \end{array}\right]+a_{22}\left[\begin{array}{ll} 0 & 0 \ 0 & 1 \end{array}\right] .$$
Hence it is clear that $A$ is a linear combination of the vectors, i.e., of the matrices, in $\mathcal{B}c$. It is also easy to see that $\mathcal{B}_c$ is a linearly independent set. In fact, $$\alpha_1\left[\begin{array}{ll} 1 & 0 \ 0 & 0 \end{array}\right]+\alpha_2\left[\begin{array}{ll} 0 & 1 \ 0 & 0 \end{array}\right]+\alpha_3\left[\begin{array}{ll} 0 & 0 \ 1 & 0 \end{array}\right]+\alpha_4\left[\begin{array}{ll} 0 & 0 \ 0 & 1 \end{array}\right]=\left[\begin{array}{ll} 0 & 0 \ 0 & 0 \end{array}\right],$$ yields $\alpha_1=\alpha_2=\alpha_3=\alpha_4=0$, which shows that $\mathcal{B}_c$ is linearly independent. We see that $\mathcal{B}_c$ is a basis of $\mathrm{M}_2(\mathbb{K})$ and that, for a matrix $A$ as above, the coordinate vector $A{\mathcal{B}c}$ of $A$ relative to the basis $\mathcal{B}_c$ is $\left(a{11}, a_{12}, a_{21}, a_{22}\right)$ which lies in $\mathbb{K}^4$. We have then that any matrix
$$A=\left[\begin{array}{ll} a_{11} & a_{12} \ a_{21} & a_{22} \end{array}\right]$$
has an image in $\mathbb{K}^4$ according to
\begin{aligned} T: \mathrm{M}2(\mathbb{K}) & \rightarrow \mathbb{K}^4 \ A & \mapsto A{\mathcal{B}}=\left(a_{11}, a_{12}, a_{21}, a_{22}\right) . \end{aligned}
Observe that Proposition $3.5$ guarantees that $T$ is bijective.

# 线性代数代考

## 数学代写|线性代数代写linear algebra代考|Bases and Dimension

a) 集合 $\mathcal{E}_2=(1,0),(0,1)$ 是一个基础䄳 ${ }^2$ ，表示标准基础 $\mathbb{R}^2$.
b) 集合 $\mathcal{B}=(1,1),(-1,0)$ 是另一个基础 $\mathbb{R}^2$.
c) 另一个基础 $\mathbb{R}^2$ 是 $\mathcal{B}_1=\ldots$ (找一个)。

$$\alpha_1\left[\begin{array}{ll} 1 & 1 \end{array}\right]+\alpha_2\left[\begin{array}{ll} -1 & 0 \end{array}\right]=\left[\begin{array}{ll} 2 & 3 \end{array}\right]$$

## 数学代写|线性代数代写linear algebra代考|Matrix spaces and spaces of polynomials

$$T: \mathrm{M} 2(\mathbb{K}) \rightarrow \mathbb{K}^4 A \quad \mapsto A \mathcal{B}=\left(a_{11}, a_{12}, a_{21}, a_{22}\right) .$$

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