# 数学代写|线性代数代写linear algebra代考|MAST10022

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## 数学代写|线性代数代写linear algebra代考|An Explicit Description of Nul A

There is no obvious relation between vectors in $\mathrm{Nul} A$ and the entries in $A$. We say that $\operatorname{Nul} A$ is defined implicitly, because it is defined by a condition that must be checked. No explicit list or description of the elements in $\mathrm{Nul} A$ is given. However, solving the equation $A \mathbf{x}=\mathbf{0}$ amounts to producing an explicit description of $\mathrm{Nul} A$. The next example reviews the procedure from Section $1.5$.
EXAMPLE 3 Find a spanning set for the null space of the matrix
$$A=\left[\begin{array}{rrrrr} -3 & 6 & -1 & 1 & -7 \ 1 & -2 & 2 & 3 & -1 \ 2 & -4 & 5 & 8 & -4 \end{array}\right]$$
SOLUTION The first step is to find the general solution of $A \mathbf{x}=\mathbf{0}$ in terms of free variables. Row reduce the augmented matrix $\left[\begin{array}{cc}A & \mathbf{0}\end{array}\right]$ to reduced echelon form in order to write the basic variables in terms of the free variables:
$$\left[\begin{array}{rrrrrr} 1 & -2 & 0 & -1 & 3 & 0 \ 0 & 0 & 1 & 2 & -2 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 \end{array}\right]$$
$$\begin{array}{r} x_1-2 x_2-x_4+3 x_5=0 \ x_3+2 x_4-2 x_5=0 \ 0=0 \end{array}$$

The general solution is $x_1=2 x_2+x_4-3 x_5, x_3=-2 x_4+2 x_5$, with $x_2, x_4$, and $x_5$ free. Next, decompose the vector giving the general solution into a linear combination of vectors where the weights are the free variables. That is,
\begin{aligned} {\left[\begin{array}{l} x_1 \ x_2 \ x_3 \ x_4 \ x_5 \end{array}\right] } & =\left[\begin{array}{c} 2 x_2+x_4-3 x_5 \ x_2 \ -2 x_4+2 x_5 \ x_4 \ x_5 \end{array}\right]=x_2\left[\begin{array}{l} 2 \ 1 \ 0 \ 0 \ 0 \end{array}\right]+x_4\left[\begin{array}{r} 1 \ 0 \ -2 \ 1 \ 0 \end{array}\right]+x_5\left[\begin{array}{r} -3 \ 0 \ 2 \ 0 \ 1 \end{array}\right] \ & =x_2 \mathbf{u}+x_4 \mathbf{v}+x_5 \mathbf{w} \end{aligned}
Every linear combination of $\mathbf{u}, \mathbf{v}$, and $\mathbf{w}$ is an element of $\mathrm{Nul} A$ and vice versa. Thus ${\mathbf{u}, \mathbf{v}, \mathbf{w}}$ is a spanning set for $\operatorname{Nul} A$.

Two points should be made about the solution of Example 3 that apply to all problems of this type where $\operatorname{Nul} A$ contains nonzero vectors. We will use these facts later.

1. The spanning set produced by the method in Example 3 is automatically linearly independent because the free variables are the weights on the spanning vectors. For instance, look at the $2 \mathrm{nd}, 4$ th, and 5 th entries in the solution vector in (3) and note that $x_2 \mathbf{u}+x_4 \mathbf{v}+x_5 \mathbf{w}$ can be $\mathbf{0}$ only if the weights $x_2, x_4$, and $x_5$ are all zero.
2. When Nul $A$ contains nonzero vectors, the number of vectors in the spanning set for Nul $A$ equals the number of free variables in the equation $A \mathbf{x}=\mathbf{0}$.

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

Another important subspace associated with a matrix is its column space. Unlike the null space, the column space is defined explicitly via linear combinations.
The column space of an $m \times n$ matrix $A$, written as $\operatorname{Col} A$, is the set of all linear combinations of the columns of $A$. If $A=\left[\begin{array}{lll}\mathbf{a}_1 & \cdots & \mathbf{a}_n\end{array}\right]$, then
$$\operatorname{Col} A=\operatorname{Span}\left{\mathbf{a}_1, \ldots, \mathbf{a}_n\right}$$
Since Span $\left{\mathbf{a}_1, \ldots, \mathbf{a}_n\right}$ is a subspace, by Theorem 1 , the next theorem follows from the definition of $\operatorname{Col} A$ and the fact that the columns of $A$ are in $\mathbb{R}^m$.
The column space of an $m \times n$ matrix $A$ is a subspace of $\mathbb{R}^m$.
Note that a typical vector in $\operatorname{Col} A$ can be written as $A \mathbf{x}$ for some $\mathbf{x}$ because the notation $A \mathbf{x}$ stands for a linear combination of the columns of $A$. That is,
$$\operatorname{Col} A=\left{\mathbf{b}: \mathbf{b}=A \mathbf{x} \text { for some } \mathbf{x} \text { in } \mathbb{R}^n\right}$$
The notation $A \mathbf{x}$ for vectors in $\operatorname{Col} A$ also shows that $\operatorname{Col} A$ is the range of the linear transformation $\mathbf{x} \mapsto A \mathbf{x}$. We will return to this point of view at the end of the section.

# 线性代数代考

## 数学代写|线性代数代写linear algebra代考|An Explicit Description of Nul A

$$A=\left[\begin{array}{llllllllll} -3 & 6 & -1 & 1 & -7 & 1 & -2 & 2 & 3 & -1 \end{array}\right.$$

$$\left[\begin{array}{llllllllllllll} 1 & -2 & 0 & -1 & 3 & 0 & 0 & 0 & 1 & 2 & -2 & 0 & 0 & 0 \end{array}\right.$$
$$x_1-2 x_2-x_4+3 x_5=0 x_3+2 x_4-2 x_5=00=0$$

$$x_1=2 x_2+x_4-3 x_5, x_3=-2 x_4+2 x_5 \text { ， 和 }$$
$x_2, x_4$ ，和 $x_5$ 自由。接下来，将给出通解的向量分解为 向量的线性组合，其中权重是自由变量。那是，
$$\left[\begin{array}{lllll} x_1 & x_2 & x_3 & x_4 & x_5 \end{array}\right]=\left[\begin{array}{ll} 2 x_2+x_4-3 x_5 x_2-2 x_4+2 x_5 \end{array}\right.$$

1. 示例 3 中的方法生成的生成集自动线性无关，因 为自由变量是生成向量上的权重。例如，看看 2nd, 4(3) 中的解向量中的第 th 和第 5 个条目， 并注意 $x_2 \mathbf{u}+x_4 \mathbf{v}+x_5 \mathbf{w}$ 可0仅当权重 $x_2, x_4$ ，和 $x_5$ 都是零。
2. 当为空 $A$ 包含非零向量，Nul 的生成集中的向量 数 $A$ 等于方程中自由变量的数量 $A \mathbf{x}=\mathbf{0}$.

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

loperatorname{Col} A=loperatorname{Span $} \backslash l e f t{\backslash m a t$

loperatorname ${$ Col $} A=\backslash \operatorname{left}{\backslash \operatorname{mathbf}{b}: \backslash \operatorname{Imathbf}{b}=A \backslash$

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