### 数学代写|线性代数代写linear algebra代考|MTH2106

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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|The Dimension of a Subspace

It can be shown that if a subspace $H$ has a basis of $p$ vectors, then every basis of $H$ must consist of exactly $p$ vectors. (See Exercises 27 and 28 .) Thus the following definition makes sense.
The dimension of a nonzero subspace $H$, denoted by $\operatorname{dim} H$, is the number of vectors in any basis for $H$. The dimension of the zero subspace ${0}$ is defined to be zero. ${ }^2$
The space $\mathbb{R}^n$ has dimension $n$. Every basis for $\mathbb{R}^n$ consists of $n$ vectors. A plane through 0 in $\mathbb{R}^3$ is two-dimensional, and a line through $\mathbf{0}$ is one-dimensional.

EXAMPLE 2 Recall that the null space of the matrix $A$ in Example 6 in Section $2.8$ had a basis of 3 vectors. So the dimension of $\operatorname{Nul} A$ in this case is 3 . Observe how each basis vector corresponds to a free variable in the equation $A \mathbf{x}=\mathbf{0}$. Our construction always produces a basis in this way. So, to find the dimension of $\mathrm{Nul} A$, simply identify and count the number of free variables in $A \mathbf{x}=\mathbf{0}$.
The rank of a matrix $A$, denoted by rank $A$, is the dimension of the column space of $A$.
Since the pivot columns of $A$ form a basis for $\operatorname{Col} A$, the rank of $A$ is just the number of pivot columns in $A$.

The row reduction in Example 3 reveals that there are two free variables in $A \mathbf{x}=\mathbf{0}$, because two of the five columns of $A$ are not pivot columns. (The nonpivot columns correspond to the free variables in $A \mathbf{x}=\mathbf{0}$.) Since the number of pivot columns plus the number of nonpivot columns is exactly the number of columns, the dimensions of Col $A$ and $\mathrm{Nul} A$ have the following useful connection. (See the Rank Theorem in Section $4.6$ for additional details.)
The Rank Theorem
If a matrix $A$ has $n$ columns, then $\operatorname{rank} A+\operatorname{dim} \operatorname{Nul} A=n$.
The following theorem is important for applications and will be needed in Chapters 5 and 6. The theorem (proved in Section 4.5) is certainly plausible, if you think of a $p$-dimensional subspace as isomorphic to $\mathbb{R}^p$. The Invertible Matrix Theorem shows that $p$ vectors in $\mathbb{R}^p$ are linearly independent if and only if they also span $\mathbb{R}^p$.

## 数学代写|线性代数代写linear algebra代考|Column Space and Null Space of a Matrix

Subspaces of $\mathbb{R}^n$ usually occur in applications and theory in one of two ways. In both cases, the subspace can be related to a matrix.
The column space of a matrix $A$ is the set $\operatorname{Col} A$ of all linear combinations of the columns of $A$.
If $A=\left[\begin{array}{lll}\mathbf{a}_1 & \cdots & \mathbf{a}_n\end{array}\right]$, with the columns in $\mathbb{R}^m$, then $\operatorname{Col} A$ is the same as Span $\left{\mathbf{a}_1, \ldots, \mathbf{a}_n\right}$. Example 4 shows that the column space of an $\boldsymbol{m} \times \boldsymbol{n}$ matrix is a subspace of $\mathbb{R}^m$. Note that $\operatorname{Col} A$ equals $\mathbb{R}^m$ only when the columns of $A$ span $\mathbb{R}^m$. Otherwise, $\operatorname{Col} A$ is only part of $\mathbb{R}^m$.

EXAMPLE 4 Let $A=\left[\begin{array}{rrr}1 & -3 & -4 \ -4 & 6 & -2 \ -3 & 7 & 6\end{array}\right]$ and $\mathbf{b}=\left[\begin{array}{r}3 \ 3 \ -4\end{array}\right]$. Determine whether $\mathbf{b}$ is in the column space of $A$.

SOLUTION The vector $\mathbf{b}$ is a linear combination of the columns of $A$ if and only if $\mathbf{b}$ can be written as $A \mathbf{x}$ for some $\mathbf{x}$, that is, if and only if the equation $A \mathbf{x}=\mathbf{b}$ has a solution. Row reducing the augmented matrix $\left[A \begin{array}{ll}A & \mathbf{b}\end{array}\right]$,
$$\left[\begin{array}{rrrr} 1 & -3 & -4 & 3 \ -4 & 6 & -2 & 3 \ -3 & 7 & 6 & -4 \end{array}\right] \sim\left[\begin{array}{rrrr} 1 & -3 & -4 & 3 \ 0 & -6 & -18 & 15 \ 0 & -2 & -6 & 5 \end{array}\right] \sim\left[\begin{array}{rrrr} 1 & -3 & -4 & 3 \ 0 & -6 & -18 & 15 \ 0 & 0 & 0 & 0 \end{array}\right]$$
we conclude that $A \mathbf{x}=\mathbf{b}$ is consistent and $\mathbf{b}$ is in $\operatorname{Col} A$.

The solution of Example 4 shows that when a system of linear equations is written in the form $A \mathbf{x}=\mathbf{b}$, the column space of $A$ is the set of all $\mathbf{b}$ for which the system has a solution.
The null space of a matrix $A$ is the set $\mathrm{Nul} A$ of all solutions of the homogeneous equation $A \mathbf{x}=\mathbf{0}$
When $A$ has $n$ columns, the solutions of $A \mathbf{x}=\mathbf{0}$ belong to $\mathbb{R}^n$, and the null space of $A$ is a subset of $\mathbb{R}^n$. In fact, $\mathrm{Nul} A$ has the properties of a subspace of $\mathbb{R}^n$.
The null space of an $m \times n$ matrix $A$ is a subspace of $\mathbb{R}^n$. Equivalently, the set of all solutions of a system $A \mathbf{x}=\mathbf{0}$ of $m$ homogeneous linear equations in $n$ unknowns is a subspace of $\mathbb{R}^n$.
PROOF The zero vector is in $\operatorname{Nul} A$ (because $A 0=0$ ). To show that $\mathrm{Nul} A$ satisfies the other two properties required for a subspace, take any $\mathbf{u}$ and $\mathbf{v}$ in $\mathrm{Nul} A$. That is, suppose $A \mathbf{u}=\mathbf{0}$ and $A \mathbf{v}=\mathbf{0}$. Then, by a property of matrix multiplication,
$$A(\mathbf{u}+\mathbf{v})=A \mathbf{u}+A \mathbf{v}=\mathbf{0}+\mathbf{0}=\mathbf{0}$$
Thus $\mathbf{u}+\mathbf{v}$ satisfies $A \mathbf{x}=\mathbf{0}$, and so $\mathbf{u}+\mathbf{v}$ is in $\operatorname{Nul} A$. Also, for any scalar $c, A(c \mathbf{u})=$ $c(A \mathbf{u})=c(0)=\mathbf{0}$, which shows that $c \mathbf{u}$ is in $\mathrm{Nul} A$.

To test whether a given vector $\mathbf{v}$ is in $\operatorname{Nul} A$, just compute $A \mathbf{v}$ to see whether $A \mathbf{v}$ is the zero vector. Because $\mathrm{Nul} A$ is described by a condition that must be checked for each vector, we say that the null space is defined implicitly. In contrast, the column space is defined explicitly, because vectors in Col A can be constructed (by linear combinations) from the columns of $A$. To create an explicit description of $\mathrm{Nul} A$, solve the equation $A \mathbf{x}=\mathbf{0}$ and write the solution in parametric vector form. (See Example 6 , below.) ${ }^2$

# 线性代数代考

## 数学代写|线性代数代写linear algebra代考|Column Space and Null Space of a Matrix

Veft{ $\left.\backslash m a t h b f{a} _1, \backslash d o t s, \backslash m a t h b f{a} _n \backslash r i g h t\right}$. 示例 4 显示了一个列空间 $\boldsymbol{m} \times \boldsymbol{n}$ 矩阵是一个子空间 $\mathbb{R}^m$. 注意 $\operatorname{Col} A$ 等于 $\mathbb{R}^m$ 只有当列 $A$ 跨度 $\mathbb{R}^m$. 否则， $\operatorname{Col} A$ 只是一部分 $\mathbb{R}^m$. $A$.

$$A(\mathbf{u}+\mathbf{v})=A \mathbf{u}+A \mathbf{v}=\mathbf{0}+\mathbf{0}=\mathbf{0}$$

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## MATLAB代写

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