[선형대수학] 선형 사상의 값에 대응되는 행렬 표현 (Matrix Representation Corresponding to The Value of Linear Map)

$\underline{Thm}$

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

Let $$AB=\begin{bmatrix} u_{1} & u_{2} & \cdots & u_{p} \end{bmatrix},\, B=\begin{bmatrix} v_{1} & v_{2} & \cdots & v_{p} \end{bmatrix}$$ 

Then

$$u_{j}=Av_{j}, v_{j}=Be_{j}\;\;\;(B=BI_{p})$$

 

$\underline{Thm}$

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

$$[T(u)]_{\gamma}=[T]_{\beta}^{\gamma}[u]_{\beta}\;\;\; \forall u \in V$$

 

$\underline{Proof}$

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

Let $[u]_{\beta}=\begin{bmatrix} a_{1} \\ \vdots \\ a_{n} \end{bmatrix}$; that is, $u=a_{1}v_{1}+\cdots+a_{n}v_{n}.$

$[T]_{\beta}^{\gamma}=(b_{ij})_{m \times n}$; that is, $\displaystyle T(v_{j})=\sum_{i=1}^{n} b_{ij}w_{i}\;\;(1 \leq j \leq n).$

Then

$$\displaystyle \begin{equation} \begin{split} T(u) & =T(\sum_{j=1}^{n} a_{j}v_{j}) \\ & =\sum_{j=1}^{n}a_{j} \sum_{i=1}^{m}b_{ij}w_{i} \\ & = \sum_{i=1}^{m}(\sum_{j=1}^{n} b_{ij}a_{j}) w_{i} \end{split} \end{equation}$$

Hence,

$$\displaystyle \begin{equation} \begin{split} [T(u)]_{\gamma} & = \begin{bmatrix} \sum_{j=1}^{n} b_{1j}a_{j} \\ \vdots \\ \sum_{j=1}^{n} b_{mj}a_{j} \end{bmatrix} \\ & = (b_{ij})_{m \times n} \begin{bmatrix} a_{1} \\ \vdots \\ a_{n} \end{bmatrix} \\ & = [T]_{\beta}^{\gamma}[u]_{\beta} \end{split} \end{equation}$$

 

$\underline{Ex}$

Let $T \colon P_{3}(\mathbb{R}) \rightarrow P_{2}(\mathbb{R})$ be the linear map given by $T(f(x))=f^{'}(x)$.

$\beta = \{ 1,x,x^{2}x^{3} \},\, \gamma=\{1,x,x^{2} \}.$

Note

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

Let $P(x)=2-4x+x^{2}+3x^{3}$; then

$$[P(x)]_{\beta}=\begin{bmatrix} 2 \\ -4 \\ 1 \\ 3 \end{bmatrix},\, T(P(x))= -4+2x+9x^{2}$$

so, $[T(P(x))]_{\gamma}=\begin{bmatrix} -4 \\ 2 \\ 9 \end{bmatrix}$.

On the other hand,

$$\begin{equation} \begin{split} [T]_{\beta}^{\gamma}[P(x)]_{\beta} & = \begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 3 \end{bmatrix} \begin{bmatrix} 2 \\ -4 \\ 1 \\ 3 \end{bmatrix} \\ & = \begin{bmatrix} -4 \\ 2 \\ 0 \end{bmatrix} \end{split} \end{equation}$$