[선형대수학] 사상이 선형이고 가역인 경우 벡터 공간, 좌표 벡터의 관계

$\underline{Lemma}$

Let $T \colon V \rightarrow W$ be linear & invertible.

Then $dim(V) < \infty \Leftrightarrow dim(W) < \infty.$

In this case, $dim(V)=dim(W).$

 

$\underline{Proof}$

$(\Rightarrow)$

Let $\beta= \{ x_{1}, \cdots, x_{n} \}$ be a basis for $V$.

By thm, $T(\beta)$ spans $R(T)=W$. By thm, $dim(W) < \infty$.

$(\Leftarrow)$

Since $T^{-1} \colon W \rightarrow V$ is linear & invertible, if $dim(W) < \infty$, by the same argument $dim(V) < \infty$.

Next, suppose $dim(V) < \infty$ (and so $dim(W) < \infty$) Since $T$ is 1-1 & onto, $nullity(T)=0$ & $rank(T)=dim(W)$. By Dimenstion Thm, $dim(V)=dim(W)$.

 

$\underline{Thm}$

Let $V,W$ be finite dimensional vector spaces with ordered basis $\beta, \gamma$, respectively. Let $T \colon V \rightarrow W$ be linear. Then

$$T\,is\,invertible \Leftrightarrow [T]_{\beta}^{\gamma} \, is \, invertible.$$

In this case, $([T]_{\beta}^{\gamma})^{-1}=[T^{-1}]_{\gamma}^{\beta}$.

 

$\underline{Proof}$

$(\Rightarrow)$

Suppose $T$ is invertible. By the above lemma, $dim(V)=dim(W)=n$. So, $[T]_{\beta}^{\gamma} \in M_{n \times n}$.

Since $TT^{-1}=I_{W}$ & $T^{-1}T=I_{V}$, we have

$$I_{n}=[I_{W}]_{\gamma}=[TT^{-1}]_{\gamma}=[T]_{\beta}^{\gamma}[T^{-1}]_{\gamma}^{\beta},$$

$$I_{n}=[I_{V}]_{\beta}=[T^{-1}T]_{\beta}=[T^{-1}]_{\beta}^{\gamma}[T]_{\beta}^{\gamma}.$$

Thus, $[T]_{\beta}^{\gamma}$ is invertible & $([T]_{\beta}^{\gamma})^{-1}=[T^{-1}]_{\gamma}^{\beta}$.

$(\Leftarrow)$

Suppose $[T]_{\beta}^{\gamma}=A$ is invertible. Then $\exists B \in M_{n \times n}$ s.t. $$AB=BA=I_{n}$$

Write $\beta=\{v_{1},\cdots,v_{n} \},\, \gamma=\{w_{1},\cdots,w_{n} \}$.

Then $\exists ! U \in \mathcal{L}(W,V)$ s.t. $$\displaystyle U(w_{j})=\sum_{i=1}^{n} B_{ij}v_{i} \;\; for \; 1 \leq j \leq n$$

So, $[U]_{\gamma}^{\beta}=B$. Note $$[UT]_{\beta}=[U]_{\gamma}^{\beta}[T]_{\beta}^{\gamma}=BA=I_{n}=[I_{V}]_{\beta}$$

Thus, $UT=I_{V}$. Similarlym we get $TU=I_{W}$.

 

$\underline{Coro.1}$

Let $V$ be a finite-dimensional vector space with an ordered basis $\beta$, and let $T \colon V \rightarrow V$ be linear. Then

$$T\, is\, invertible \Leftrightarrow [T]_{\beta} \, is \, invertible.$$

In this case, $([T]_{\beta})^{-1}=[T^{-1}]_{\beta}.$

 

$\underline{Coro.2}$

Let $A \in M_{n \times n}$. Then $$L_{A}\, is\, invertible \Leftrightarrow A \, is \, invertible.$$

where $A=[L_{A}]_{\beta}$ ($\beta$ is standard ordered basis for $\mathbb{F}^{n})

In this case, $$[L_{A^{-1}}]_{\beta}=A^{-1}=([L_{A}]_{\beta})^{-1}=[L_{A}^{-1}]_{\beta},$$

and so $L_{A}^{-1}=L_{A^{-1}}$.