## 数学代写|线性代数代写linear algebra代考|MATHS1011

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

Theorem 5 in Section $1.4$ shows that if $A$ is $m \times n$, then the transformation $\mathbf{x} \mapsto A \mathbf{x}$ has the properties
$$A(\mathbf{u}+\mathbf{v})=A \mathbf{u}+A \mathbf{v} \quad \text { and } \quad A(c \mathbf{u})=c A \mathbf{u}$$
for all $\mathbf{u}, \mathbf{v}$ in $\mathbb{R}^n$ and all scalars $c$. These properties, written in function notation, identify the most important class of transformations in linear algebra.
A transformation (or mapping) $T$ is linear if:
(i) $T(\mathbf{u}+\mathbf{v})=T(\mathbf{u})+T(\mathbf{v})$ for all $\mathbf{u}, \mathbf{v}$ in the domain of $T$;
(ii) $T(c \mathbf{u})=c T(\mathbf{u})$ for all scalars $c$ and all $\mathbf{u}$ in the domain of $T$.
Every matrix transformation is a linear transformation. Important examples of linear transformations that are not matrix transformations will be discussed in Chapters 4 and 5.

Linear transformations preserve the operations of vector addition and scalar multiplication. Property (i) says that the result $T(\mathbf{u}+\mathbf{v})$ of first adding $\mathbf{u}$ and $\mathbf{v}$ in $\mathbb{R}^n$ and then applying $T$ is the same as first applying $T$ to $\mathbf{u}$ and to $\mathbf{v}$ and then adding $T(\mathbf{u})$ and $T(\mathbf{v})$ in $\mathbb{R}^m$. These two properties lead easily to the following useful facts.
If $T$ is a linear transformation, then
$$T(0)=\mathbf{0}$$
and
$$T(c \mathbf{u}+d \mathbf{v})=c T(\mathbf{u})+d T(\mathbf{v})$$
for all vectors $\mathbf{u}, \mathbf{v}$ in the domain of $T$ and all scalars $c, d$.
Property (3) follows from condition (ii) in the definition, because $T(0)=T(0 \mathbf{u})=$ $0 T(\mathbf{u})=\mathbf{0}$. Property (4) requires both (i) and (ii):
$$T(c \mathbf{u}+d \mathbf{v})=T(c \mathbf{u})+T(d \mathbf{v})=c T(\mathbf{u})+d T(\mathbf{v})$$
Observe that if a transformation satisfies (4) for all $\mathbf{u}, \mathbf{v}$ and $c$, $d$, it must be linear. (Set $c=d=1$ for preservation of addition, and set $d=0$ for preservation of scalar multiplication.) Repeated application of (4) produces a useful generalization:
$$T\left(c_1 \mathbf{v}_1+\cdots+c_p \mathbf{v}_p\right)=c_1 T\left(\mathbf{v}_1\right)+\cdots+c_p T\left(\mathbf{v}_p\right)$$

## 数学代写|线性代数代写linear algebra代考|THE MATRIX OF A LINEAR TRANSFORMATION

Whenever a linear transformation $T$ arises geometrically or is described in words, we usually want a “formula” for $T(\mathbf{x})$. The discussion that follows shows that every linear transformation from $\mathbb{R}^n$ to $\mathbb{R}^m$ is actually a matrix transformation $\mathbf{x} \mapsto A \mathbf{x}$ and that important properties of $T$ are intimately related to familiar properties of $A$. The key to finding $A$ is to observe that $T$ is completely determined by what it does to the columns of the $n \times n$ identity matrix $I_n$.
EXAMPLE 1 The columns of $I_2=\left[\begin{array}{ll}1 & 0 \ 0 & 1\end{array}\right]$ are $\mathbf{e}_1=\left[\begin{array}{l}1 \ 0\end{array}\right]$ and $\mathbf{e}_2=\left[\begin{array}{l}0 \ 1\end{array}\right]$ Suppose $T$ is a linear transformation from $\mathbb{R}^2$ into $\mathbb{R}^3$ such that
$$T\left(\mathbf{e}_1\right)=\left[\begin{array}{r} 5 \ -7 \ 2 \end{array}\right] \quad \text { and } \quad T\left(\mathbf{e}_2\right)=\left[\begin{array}{r} -3 \ 8 \ 0 \end{array}\right]$$
With no additional information, find a formula for the image of an arbitrary $\mathbf{x}$ in $\mathbb{R}^2$.

SOLUTION Write
$$\mathbf{x}=\left[\begin{array}{l} x_1 \ x_2 \end{array}\right]=x_1\left[\begin{array}{l} 1 \ 0 \end{array}\right]+x_2\left[\begin{array}{l} 0 \ 1 \end{array}\right]=x_1 \mathbf{e}_1+x_2 \mathbf{e}_2$$
Since $T$ is a linear transformation,
\begin{aligned} T(\mathbf{x}) & =x_1 T\left(\mathbf{e}_1\right)+x_2 T\left(\mathbf{e}_2\right) \ & =x_1\left[\begin{array}{r} 5 \ -7 \ 2 \end{array}\right]+x_2\left[\begin{array}{r} -3 \ 8 \ 0 \end{array}\right]=\left[\begin{array}{c} 5 x_1-3 x_2 \ -7 x_1+8 x_2 \ 2 x_1+0 \end{array}\right] \end{aligned}
The step from equation (1) to equation (2) explains why knowledge of $T\left(\mathbf{e}_1\right)$ and $T\left(\mathbf{e}_2\right)$ is sufficient to determine $T(\mathbf{x})$ for any $\mathbf{x}$. Moreover, since (2) expresses $T(\mathbf{x})$ as a linear combination of vectors, we can put these vectors into the columns of a matrix $A$ and write (2) as
$$T(\mathbf{x})=\left[\begin{array}{ll} T\left(\mathbf{e}_1\right) & T\left(\mathbf{e}_2\right) \end{array}\right]\left[\begin{array}{l} x_1 \ x_2 \end{array}\right]=A \mathbf{x}$$

# 线性代数代考

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

$A(\mathbf{u}+\mathbf{v})=A \mathbf{u}+A \mathbf{v} \quad$ and $\quad A(c \mathbf{u})=c A \mathbf{u}$

(二) $T(c \mathbf{u})=c T(\mathbf{u})$ 对于所有标量 $c$ 和所有 $\mathbf{u}$ 在的领域 $T$.

$$T(0)=\mathbf{0}$$

$$T(c \mathbf{u}+d \mathbf{v})=c T(\mathbf{u})+d T(\mathbf{v})$$

$T(0)=T(0 \mathbf{u})=0 T(\mathbf{u})=\mathbf{0}$. 属性 (4) 需要 (i) 和
(ii):
$$T(c \mathbf{u}+d \mathbf{v})=T(c \mathbf{u})+T(d \mathbf{v})=c T(\mathbf{u})+d T(\mathbf{v})$$

$$T\left(c_1 \mathbf{v}_1+\cdots+c_p \mathbf{v}_p\right)=c_1 T\left(\mathbf{v}_1\right)+\cdots+c_p T\left(\mathbf{v}_p\right)$$

## 数学代写|线性代数代写linear algebra代考|THE MATRIX OF A LINEAR TRANSFORMATION

$$\mathbf{x}=\left[\begin{array}{ll} x_1 & x_2 \end{array}\right]=x_1\left[\begin{array}{ll} 1 & 0 \end{array}\right]+x_2\left[\begin{array}{ll} 0 & 1 \end{array}\right]=x_1 \mathbf{e}_1+x_2 \mathbf{e}_2$$

$$T(\mathbf{x})=x_1 T\left(\mathbf{e}_1\right)+x_2 T\left(\mathbf{e}_2\right) \quad=x_1[5-72]$$

$$T(\mathbf{x})=\left[\begin{array}{ll} T\left(\mathbf{e}_1\right) & T\left(\mathbf{e}_2\right) \end{array}\right]\left[\begin{array}{ll} x_1 & x_2 \end{array}\right]=A \mathbf{x}$$

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## MATLAB代写

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