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# 数学代考|线性代数代写Linear algebra代考|MA2210 NULL SPACES, COLUMN SPACES, AND LINEAR TRANSFORMATIONS

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## 数学代考|线性代数代写LINEAR ALGEBRA代考|NULL SPACES, COLUMN SPACES, AND LINEAR TRANSFORMATIONS

In applications of linear algebra, subspaces of $\mathbb{R}^{n}$ usually arise in one of two ways: (1) as the set of all solutions to a system of homogeneous linear equations or (2) as the set of all linear combinations of certain specified vectors. In this section, we compare and contrast these two descriptions of subspaces, allowing us to practice using the concept of a subspace. Actually, as you will soon discover, we have been working with subspaces ever since Section 1.3. The main new feature here is the terminology. The section concludes with a discussion of the kernel and range of a linear transformation.
The Null Space of a Matrix
Consider the following system of homogeneous equations:
$$\begin{array}{r} x_{1}-3 x_{2}-2 x_{3}=0 \ -5 x_{1}+9 x_{2}+x_{3}=0 \end{array}$$
In matrix form, this system is written as $A \mathbf{x}=\mathbf{0}$, where
$$A=\left[\begin{array}{rrr} 1 & -3 & -2 \ -5 & 9 & 1 \end{array}\right]$$
Recall that the set of all $\mathbf{x}$ that satisfy (1) is called the solution set of the system (1). Often it is convenient to relate this set directly to the matrix $A$ and the equation $A \mathbf{x}=\mathbf{0}$. We call the set of $\mathbf{x}$ that satisfy $A \mathbf{x}=\mathbf{0}$ the null space of the matrix $A$.

## 数学代考|线性代数代写LINEAR ALGEBRA代考|An Explicit Description of

There is no obvious relation between vectors in $\mathrm{Nul} A$ and the entries in $A$. We say that $\mathrm{Nul} A$ 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 $\left[\begin{array}{ll}A & \mathbf{0}\end{array}\right]$ 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{aligned} x_{1}-2 x_{2}-x_{4}+3 x_{5} &=0 \ x_{3}+2 x_{4}-2 x_{5} &=0 \ 0 &=0 \end{aligned}

## 数学代考|线性代数代写LINEAR ALGEBRA代考|NULL SPACES, COLUMN SPACES, AND LINEAR TRANSFORMATIONS

$$x_{1}-3 x_{2}-2 x_{3}=0-5 x_{1}+9 x_{2}+x_{3}=0$$

$$A=\left[\begin{array}{llllll} 1 & -3 & -2 & -5 & 9 & 1 \end{array}\right]$$

## 数学代考|线性代数代写LINEAR ALGEBRA代考|An Explicit Description of

$$A=\left[\begin{array}{lllllllllllll} -3 & 6 & -1 & 1 & -71 & -2 & 2 & 3 & -12 & -4 & 5 & 8 & -4 \end{array}\right]$$
SOLUTION 第一步是找到通解 $A \mathrm{x}=\mathbf{0}$ 就自由变量而言。行减少增广矩阵 $\left[\begin{array}{ll}A & 0\end{array}\right]$ 简化梯形形式，以便根据自由

$\left[\begin{array}{llllllllllllllllll}1 & -2 & 0 & -1 & 3 & 0 & 0 & 0 & 1 & 2 & -2 & 0 & 0 & 0 & 0 & 0 & 0 & 0\end{array}\right]$,
$x_{1}-2 x_{2}-x_{4}+3 x_{5}=0 x_{3}+2 x_{4}-2 x_{5} \quad=00=0$

## MATLAB代写

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