[선형대수학] Left Product Map

$\underline{Def}$ (Left Product Map)

Let $A \in M_{m \times n}(\mathbb{F})$. We define a map $L_{A} \colon \mathbb{F}^{n} \rightarrow \mathbb{F}^{m}$ by $$L_{A}(x)=Ax \;\; \forall a \in \mathbb{F}^{n},$$

called the left-product map by $A$.

 

$\underline{Ex}$

$$A = \begin{bmatrix} 1 & 2 & 1 \\ 0 & 1 & 2 \end{bmatrix} \in M_{2 \times 3}(\mathbb{R})$$

$$L_{A} \colon \mathbb{R}^{3} \rightarrow \mathbb{R}^{2};\, L_{A}(x)=L_{A} \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix} =\begin{bmatrix} 1 & 2 & 1 \\ 0 & 1 & 2 \end{bmatrix} \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix} = \begin{bmatrix} x_{1}+2x_{2}+x_{3} \\ x_{2}+2x_{3} \end{bmatrix}$$

 

$\underline{Thm}$ (Basic Properties of Left-product maps)

Let $A, B \in M_{m \times n}(\mathbb{F})$, and let $\beta, \gamma$ be the standard ordered bases for $\mathbb{F}^{m},\mathbb{F}^{n}$, respectively. Then the following hold:

$(1)$ $L_{A} \colon \mathbb{F}^{m} \rightarrow \mathbb{F}^{n}$ is linear.

$(2)$ $[L_{A}]_{\beta}^{\gamma}=A$

$(3)$ $L_{A}=L_{B} \Leftrightarrow A=B$

$(4)$ $L_{A+B}=L_{A}+L_{B},\, L_{aA}=aL_{A}\,(a \in \mathbb{F})$

$(5)$ If $T \colon \mathbb{F}^{m} \rightarrow \mathbb{F}^{n}$ is linear, $\exists ! C \in M_{m \times n}(\mathbb{F})\,s.t.\, T=L_{C},\, C=[T]_{\beta}^{\gamma}$

$(6)$ If $E \in M_{n \times p}(\mathbb{F})$, then $L_{AE}=L_{A}L_{E}$

$(7)$ If $n=m$, then $L_{I_{n}}=I_{\mathbb{F}^{n}}$

 

$\underline{Proof}$

$(1)$

$A(ax+y)=aAx+Ay\;\; \forall x,y \in \mathbb{F}^{m},\, \forall a \in \mathbb{F}$

 

$(2)$

$L_{A}(e_{j})=Ae_{j}$ is the j-th column of $A$.

The j-th column of $[L_{A}]_{\beta}^{\gamma}$ is $[L_{A}(e_{j})]_{\gamma}$ by definition.

$$\begin{equation} \begin{split} [L_{A}(e_{j})]_{\gamma} & = [Ae_{j}]_{\gamma} \\ & = Ae_{j} \\ & = the\,j-th\,column\,of\,A \end{split}\end{equation}$$

So, $[L_{A}]_{\beta}^{\gamma}=A.$

 

$(3)$

$L_{A}=L_{B} \Rightarrow [L_{A}]_{\beta}^{\gamma}=[L_{B}]_{\beta}^{\gamma}$ by (2).

$(\Leftarrow)$ By definition.

 

$(4)$

By definition & thm.

 

$(5)$

Let $C=[T]_{\beta}^{\gamma}.$ For $x=\begin{bmatrix} x_{1} \\ \vdots \\ x_{n} \end{bmatrix} \in \mathbb{F}^{n}$,

$$T(x)=[T(x)]_{\gamma}=[T]_{\beta}^{\gamma}[x]_{\beta}=Cx=L_{C}(x)$$

 

$(6)$

By definitinon.

 

$(7)$

If $m=n$, then $\beta=\gamma$. Then $[L_{I_{n}}]_{\beta}=I_{n}=[I_{\mathbb{F}^{n}]_{\beta}$.

 

$\underline{Thm}$

Let $A \in M_{m \times n},\, B \in M_{n \times p},\, M_{p \times r}$.

Then $A(BC)=(AB)C$. (associativity)

 

$\underline{Proof}$

$$\begin{equation} \begin{split} A(BC) & = L_{A}L_{BC} \\ & = L_{A}(L_{B}L_{C}) \\ & = (L_{A}L_{B})L_{C} \\ & = \cdots \\ & = L_{(AB)C} \end{split} \end{equation}$$

by above thm (7).

So, by above thm (3), $A(BC)=(AB)C$.