## 数学代写|线性代数代写linear algebra代考|THE COMPOSITION OF LINEAR TRANSFORMATIONS

In this section, we deal with the composition of linear transformations. If $F$ and $G$ are linear transformations respectively associated with two matrices $A$ and $B$ (with respect to the canonical bases), we have that the composition $F \circ G$ is a linear transformation and we want to know which matrix is associated with it.

We first recall the definition of composition of functions. If $g: A \rightarrow B$ and $f: B \rightarrow C$ are functions, and the domain of $f$ coincides with the codomain of $g$, we define the composition as $f \circ g: A \rightarrow C$, which corresponds to applying first $g$, then f. Formally:
$$\begin{array}{rlcc} f \circ g: A & \longrightarrow & C \ a & \mapsto & f(g(a)) . \end{array}$$
We observe that the composition is associative, that is, if we have three functions $h: A \rightarrow B, g: B \rightarrow C, f: C \rightarrow D$, we have that $f \circ(g \circ h)=(f \circ g) \circ h$. In fact, for every $a \in A$ :
$$(f \circ(g \circ h))(a)=f((g \circ h)(a))=f(g(h(a)))=(f \circ g)(h(a))=((f \circ g) \circ h)(a) .$$
The composition operation is not commutative, i.e. general $f \circ g \neq g \circ f$ and it might even happen that $g \circ f$ is not defined.
Example 5.3.1 Let

We compute $f \circ g: \mathbb{R} \rightarrow \mathbb{R}$ and $g \circ f: \mathbb{R} \rightarrow \mathbb{R}$
$$f \circ g: x \mapsto g(x)=x+2 \mapsto f(x+2)=(x+2)^2-1=x^2+2 x+3,$$
$$g \circ f: y \mapsto f(y)=y^2-1 \mapsto g\left(y^2-1\right)=\left(y^2-1\right)+2=y^2+1 \text {. }$$
In this case, $f \circ g \neq g \circ f$.
Let us now see an important example in linear algebra.
Example 5.3.2 Consider the two linear transformations $L_A: \mathbb{R}^3 \longrightarrow \mathbb{R}^2, L_B$ : $\mathbb{R}^2 \longrightarrow \mathbb{R}^2$ associated with the matrices:
$$A=\left(\begin{array}{ccc} -1 & 1 & 2 \ 3 & 1 & 0 \end{array}\right), \quad B=\left(\begin{array}{ll} 2 & 1 \ 1 & 3 \end{array}\right) \text {, }$$
with respect to the canonical bases in $\mathbb{R}^2$ and $\mathbb{R}^3$. We see immediately that $L_B \circ L_A$ is defined, while $L_A \circ L_B$ it is not defined. This is because $L_A$ must have as argument a vector in $\mathbb{R}^3$, while for every $\mathbf{v} \in \mathbb{R}^2$ we have that $L_B(\mathbf{v}) \in \mathbb{R}^2$, then $L_A\left(L_B(\mathbf{v})\right)$ does not make sense.

Let us now see that, whenever we can take the composition of two linear transformations, we still obtain a linear transformation.

## 数学代写|线性代数代写linear algebra代考|KERNEL AND IMAGE

We now want to get into the theory of linear transformation and introduce the concepts of kernel and image, which are respectively subspaces of the domain and codomain of a linear transformation.

Definition 5.4.1 Let $V$ and $W$ be two vector spaces and $L: V \rightarrow W$ be a linear transformation. We call kernel of $L$, the set of vectors in $V$ whose image is the zero vector of $W$. This set is denoted by $\operatorname{Ker} L$.

We call image of $L$, the set of vectors in $W$ which are images of some vectors of $V$, that is,
$$\operatorname{Im}(L)={\mathbf{w} \in W \quad \mid \mathbf{w}=L(\mathbf{v}) \text { for some } \mathbf{v} \in V} .$$
Let us see some examples.
Example 5.4.2 1. Consider the derivation $D: \mathbb{R}[x] \longrightarrow \mathbb{R}[x]$, defined by $D(p(x))=p^{\prime}(x)$. As we have seen $D$ is a linear transformation. We want to know which are the polynomial $p(x)$ whose image is zero, i.e. such that $D(p(x))=0$. From calculus, we know they are all the constant polynomials. So $\operatorname{Ker}(D)={c \mid c \in \mathbb{R}}$. Let us look at the image of $D$. We ask which polynomials are derivatives of other polynomials. From calculus we know they are all the polynomials (we are in fact integrating), so $\operatorname{Im}(D)=\mathbb{R}[x]$.

1. Consider now the linear transformation $L: \mathbb{R}^3 \longrightarrow \mathbb{R}^2$ defined by: $L\left(\mathbf{e}_1\right)=2 \mathbf{e}_1-\mathbf{e}_2$, $L\left(\mathbf{e}_2\right)=\mathbf{e}_1, L\left(\mathbf{e}_3\right)=\mathbf{e}_1+2 \mathbf{e}_2$.

As seen in Theorem $5.2 .2$, we know that $L(x, y, z)=(2 x+y+z,-x+2 z)$. We want to determine kernel and image of $L$. We have that $\operatorname{Ker}(L)$ is the set of vectors whose image is the zero vector, that is,
\begin{aligned} \operatorname{Ker}(L) &={(x, y, z) \mid(2 x+y+z,-x+2 z)=(0,0)} \ &={(x, y, z) \mid 2 x+y+z=0,-x+2 z=0} \ &=\left{(x, y, z) \mid z=\frac{1}{2} x, y=-2 x-z=-\frac{5}{2} x\right}=\left{\left(x,-\frac{5}{2} x, x\right) \mid x \in \mathbb{R}\right} \ &=\langle(2,-5,1)\rangle \end{aligned}

# 抽象代数代考

## 数学代写|线性代数代写linear algebra代考|THE COMPOSITION OF LINEAR TRANSFORMATIONS

$$f \circ g: A \longrightarrow C a \mapsto f(g(a)) \text {. }$$

$$(f \circ(g \circ h))(a)=f((g \circ h)(a))=f(g(h(a)))=(f \circ g)(h(a))=((f \circ g) \circ h)(a) .$$

$$f \circ g: x \mapsto g(x)=x+2 \mapsto f(x+2)=(x+2)^2-1=x^2+2 x+3,$$
$$g \circ f: y \mapsto f(y)=y^2-1 \mapsto g\left(y^2-1\right)=\left(y^2-1\right)+2=y^2+1 .$$

$$A=\left(\begin{array}{llllll} -1 & 1 & 2 & 3 & 1 & 0 \end{array}\right), \quad B=\left(\begin{array}{llll} 2 & 1 & 1 & 3 \end{array}\right),$$

## 数学代写|线性代数代写linear algebra代考|KERNEL AND IMAGE

$$\operatorname{Im}(L)=\mathbf{w} \in W \quad \mid \mathbf{w}=L(\mathbf{v}) \text { for some } \mathbf{v} \in V .$$

1. 现在考虑线性变换 $L: \mathbb{R}^3 \longrightarrow \mathbb{R}^2$ 被定义为: $L\left(\mathbf{e}_1\right)=2 \mathbf{e}_1-\mathbf{e}_2, L\left(\mathbf{e}_2\right)=\mathbf{e}_1, L\left(\mathbf{e}_3\right)=\mathbf{e}_1+2 \mathbf{e}_2$.
如定理所示 $5.2 .2$ ， 我们知道 $L(x, y, z)=(2 x+y+z,-x+2 z)$. 我们要确定内核和图像 $L$. 我们有那个 $\operatorname{Ker}(L)$ 是其图像为雴向量的向量集合，即

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