[선형대수학] 선형 변환의 행렬 표현 (Matrix Representation of a Linear Transform)

$\underline{Def}$

Let $V,W$ be finite-dimensional vector spaces with ordered basis $\beta=\{v_{1},\cdots,v_{n}\},\,\gamma=\{w_{1},\cdots,w_{m}\}$, respectively.

Let $T \colon V \rightarrow W$ be linear. For each $v_{j}(1\leq j\leq n)$, since $T(v_{j}) \in W$, $\exists ! a_{1j},\cdots,a_{mj} \in \mathbb{F}$ such that $\displaystyle T(v_{j})=\sum_{i=1}^{m} a_{ij}w_{i}$.

We write $[T]_{\beta}^{\gamma}=(a_{ij})_{m \times n} \in M_{m \times n}(\mathbb{F})$, called the matrix representation of $T$ by $\beta$ & $\gamma$.

In the case that $V=W$ & $\beta=\gamma$, we write $[T]_{\beta}$ for $[T]_{\beta}^{\beta}$.

 

$\underline{Rmk}$

$\displaystyle [T]_{\beta}^{\gamma}=\begin{bmatrix} [T(v_{1})]_{\gamma} & \cdots & [T(v_{n})]_{\gamma} \end{bmatrix}$ where $[T(v_{i})]_{\gamma}$ is column vector.

If $U \colon V \rightarrow W$ is also linear & $[U]_{\beta}^{\gamma}=[T]_{\beta}^{\gamma}$, then $U(v_{j})=T(v_{j})\, (1\leq j \leq n)$, and so $U=T$ by thm.

 

$\underline{Ex}$

$T \colon \mathbb{R}^{2} \rightarrow \mathbb{R}^{3}$ linear.

$T(a_{1}, a_{2})=(a_{1}+3a_{2},0,2a_{1}-4a_{2})$

$\beta=\{ e_{1}, e_{2} \},\, \gamma =\{ e_{1}, e_{2}, e_{3} \}$: standard ordered basis for $\mathbb{R}^{2},\mathbb{R}^{3}$, respectively.

$T(e_{1})=T(1,0)=(1,0,2)=1e_{1}+0e_{2}+2e_{3}$

$T(e_{2})=T(0,1)=(3,0,-4)=3e_{1}+0e_{2}-4e_{3}$

So, $\displaystyle [T]_{\beta}^{\gamma}=\begin{bmatrix} 1 & 3 \\ 0 & 0 \\ 2 & -4 \end{bmatrix}$

 

$\beta^{'}=\{ e_{2},e_{1} \},\, \gamma^{'}=\{ e_{3},e_{2},e_{1} \}$

Then $\displaystyle [T]_{\beta}^{\gamma^{'}}=\begin{bmatrix} 2 & -4 \\ 0 & 0 \\ 1 & 3 \end{bmatrix},[T]_{\beta^{'}}^{\gamma}=\begin{bmatrix} 3 & 1 \\ 0 & 0 \\ -4 & 2 \end{bmatrix},[T]_{\beta^{'}}^{\gamma^{'}}=\begin{bmatrix} -4 & 2 \\ 0 & 0 \\ 3 & 1 \end{bmatrix}$