[제대로된 선형대수] Week 1-05 Subspace of Linear Operator

영상 링크: https://www.youtube.com/watch?v=kQIBdeb1ppI

해당 글은 위 영상 링크의 내용을 토대로 작성한 것입니다.
좋은 컨텐츠를 제작하고 공유해주신 훈러닝님 감사합니다.

 

들어가기에 앞서

우리는 아래 빨간 공간 $R(L)$ 의 $dim$ 이 $X$ 전체의 $dim$ 에서 $Ker\left ( L \right )$ 의 $dim$ 을 뺀 것과 같다는 것을 보일 것이다.

* $dim$: 어느 한 vsp 를 span 하는 lin indep. 한 벡터의 개수

 

 

Def 81. $X, U$ vsp / F, $L: X \to U$ linear

① $Ker\left ( L \right ) = \left \{ x \in X | L\left ( x \right ) = 0 \right \}$ ② $R\left ( L \right ) = \left \{ u \in U | u = L\left ( x \right ), \exists x \in X \right \}$.

 

Thrm 92. $X, U$ finite vsp / F, $T: X \to U$ linear

$T$ injective $\rightleftharpoons Ker\left ( T \right ) = \left \{ 0 \right \}$ (i.e. trivial 함 / 너무 자명함)

 

pf)

($\Rightarrow$) $T$ injective $\rightleftharpoons T\left ( u \right ) = T\left ( v \right ) \Rightarrow u = v$ (by Def 69)

Sps. $u \in Ker\left ( T \right )$

$T\left ( u \right ) = 0 = T\left ( 0 \right )$ (by $T\left ( 0 \right ) = 0$ if $T$ linear)

$\Rightarrow u = 0$

 

($\Leftarrow$) $Ker\left ( T \right ) = \left \{ 0 \right \}$, WTS $T\left ( u \right ) = T\left ( v \right ) \Rightarrow u = v$

Sps. $T\left ( u \right ) = T\left ( v \right )$

$T\left ( u - v \right ) = 0 \in Ker\left ( T \right )$

$\Rightarrow u = v$ $\square$

 

Why $T\left ( 0 \right ) = 0$ if $T$ linear ?

    $T: X \to U$

    $u, -u \in X$

    $T\left ( u - u \right ) = T\left ( u \right ) - T\left ( u \right )$ (by Def 55)

    $T\left ( u \right ) - T\left ( u \right ) = 0$

    $\therefore T\left ( 0 \right ) = 0$

 

Thrm 93. $X, U$: finite vsp / F, $T: X \to U$ linear

$T$ injective $\Rightarrow dim\left ( X \right ) \leq dim\left ( U \right )$

(대우) $dim\left ( X \right ) > dim\left ( U \right ) \Rightarrow T$ not injective

 

pf)

$X = sp\left \{ x_1, x_2, \cdots, x_n \right \} \Rightarrow \left \{ T\left ( x_1 \right ), T\left ( x_2 \right ), \cdots, T\left ( x_n \right ) \right \}$ $\left ( T\left ( x_i \right ) \in U \right )$

$\left ( dim\left ( X \right ) = n \right )$

우린 $\left \{ T\left ( x_1 \right ), T\left ( x_2 \right ), \cdots, T\left ( x_n \right ) \right \}$ 가 lin indep. 하다는 걸 보여주면 적어도 $dim\left (U \right )$ 이 $dim\left (X \right )$ 정도는 될 것이다, 혹은 $dim\left ( U \right )$ 이 $dim \left (X \right )$ 보다 크거나 같다라고 말할 수 있다.

 

WTS $\sum_{i=1}^{n} \alpha_i T\left ( x_i \right) = 0 \to \forall \alpha_i = 0$

$\Rightarrow \sum_{i=1}^{n} T \left ( \alpha_i x_i \right ) = 0$ (by Def 55)

$\Rightarrow T\left ( \sum_{i=1}^{n} \alpha_i x_i \right ) = 0$ (by Def 55)

$\Rightarrow \sum_{i=1}^{n} \alpha_i x_i \in Ker\left ( T \right )$ (by Def 81)

 

But $Ker\left ( T \right ) = \left \{ 0 \right \}$ (by Thrm 92)

$\Rightarrow \sum_{i=1}^{n} \alpha_i x_i =0 \to \forall \alpha_i = 0$ (Def 20) $\square$

 

Def 95. $dim \left ( Ker \left ( T \right ) \right ) = nullity \left ( T \right )$

 

Def 96. $T$ singular $\rightleftharpoons$ $Ker \left ( T \right )$ not trivial

 

Def 97. $dim\left ( R \left ( T \right ) \right ) = rank\left ( T \right )$

 

Thrm 99. $X \in \mathbb{R}^n, U \in \mathbb{R}^m$: finite vsp / F, $T: X \to U$ linear

$T$ surjective $\Rightarrow dim\left ( X \right ) \geq dim\left ( U \right )$

 

pf)

$T$ surjective $\rightleftharpoons rank\left ( T \right ) = dim\left ( U \right )$ ($dim$(치역) = $dim$(공역))

                $\rightleftharpoons U = sp\left \{ T\left ( x_1 \right ), \cdots, T\left ( x_m \right ) \right \}$ (m개의 $T\left ( x_i \right )$ 가 $U$ 의 basis 다.)

 

WTS $\left \{ x_1, \cdots, x_m \right \}$ lin indep. $\Rightarrow m \leq dim\left ( X \right )$

만약 $\left \{ x_1, \cdots, x_m \right \}$ 가 lin indep. 하면 $X$ 는 최소 m개의 basis vector 가 있을 것이다.

이는 $dim\left ( X \right )$ 가 m 이상이라는 것과 같다.

 

$\left \{ T\left ( x_1 \right ), \cdots, T\left ( x_m \right ) \right \}$ lin indep. $\rightleftharpoons \sum \alpha_i T\left ( x_i \right ) = 0 \Rightarrow \forall \alpha_i = 0$ (by Def 20)

 

Sps. $\sum_{i=1}^m \alpha_i x_i = 0$

$\Rightarrow T\left ( \sum \alpha_i x_i \right ) = T\left ( 0 \right )$

$\Rightarrow T\left ( \sum \alpha_i x_i \right ) = 0$ (by $T\left ( 0 \right ) = 0$ if $T$ is linear)

$\Rightarrow \sum \alpha_i T\left ( x_i \right ) = 0$ (by Def 55)

$\Rightarrow \forall \alpha_i = 0$ (by $\left \{ T\left ( x_1 \right ), \cdots, T\left ( x_m \right ) \right \}$ lin indep.)

$\therefore \left \{ x_1, \cdots, x_m \right \}$ lin indep. $\to m \leq dim\left ( X \right )$ $\square$

 

Def 101. $rank\left ( T \right ) = dim\left ( X \right ) \rightleftharpoons T$ is full rank.

 

Thrm 102. $T$ bijective $\Rightarrow dim\left ( X \right ) = dim\left ( U \right )$

 

Corl 103. $X, U$ isomorphic $\rightleftharpoons dim\left ( X \right ) = dim\left ( U \right )$

 

Video watch on 2023. 03. 25.
Last update: 2023. 04. 09.
Written by Taejun Lim

 

복습

1. 2023/05/20