[선형대수학] 선형 사상의 합성과 행렬 곱셈 (Compostion of Linear Maps & Matrix Multiplication)

$\underline{Thm}$

Let $V,W,Z$ be vector spaces over $\mathbb{F}$, and let $T \colon V \rightarrow W$ & $U \colon W \rightarrow Z$ be linear. Then $UT=U \circ T \colon V \rightarrow Z$ is linear.

 

$\underline{Thm}$

Let $V$ be a vector space. Let $T, U_{1}, U_{2} \in \mathcal{L}(V)=\mathcal{L}(V,V)$. Then

$(a)$ $T(U_{1}+U_{2})=TU_{1}+TU_{2}, (U_{1}+U_{2})T=U_{1}T+U_{2}T$

$(b)$ $T(U_{1}U_{2})=(TU_{1})U_{2}$

$(c)$ $IT=TI=T$ $(I \colon V \rightarrow$ is the identity map)

$(c)$ $a(U_{1}U_{2})=(aU_{1})U_{2}=U_{1}(aU_{2})\, \forall a \in \mathbb{F}$

(Caution, $U_{1}U_{2} \neq U_{2}U_{1}$ in general.)

 

$\underline{Compostion\,vs\,Matrix\,Product}$

Let $T \colon V \rightarrow W, U \colon W \rightarrow Z$ be linear. where $\alpha=\{v_{1},\cdots,v_{p} \}, \beta=\{w_{1},\cdots,w_{n} \}, \gamma=\{ z_{1},\cdots,z_{m} \}$ are ordered basis for $V,W,Z$, respectively.

Let $A=(a_{ik})_{m \times n}=[U]_{\beta}^{\gamma}, B=(b_{kj})_{n \times p}=[T]_{\alpha}^{\beta}, C=(c_{ij})_{m \times p}=[UT]_{\alpha}^{\gamma}$.

Then

$$\displaystyle T(v_{j})=\sum_{k=1}^{n} b_{kj}w_{k} \; (1 \leq j \leq p)$$

$$\displaystyle U(w_{k})=\sum_{i=1}^{m} a_{ik}z_{i} \; (1 \leq k \leq n)$$

$$\displaystyle (UT)(v_{j})=\sum_{i=1}^{m} c_{ij}z_{i} \; (1 \leq j \leq p)$$

On the other hand,

$$\displaystyle \begin{equation} \begin{split} (UT)(v_{j}) & = U(T(v_{j})) \\ & =  U(\sum_{k=1}^{n} b_{kj}w_{k}) \\ & = \sum_{k=1}^{n} b_{kj} \sum_{i=1}^{m} a_{ik}z_{i} \\ & = \sum_{i=1}^{m} (\sum_{k=1}^{n} a_{ik}b_{kj})z_{i} \end{split}\end{equation}$$

Thus,

$$\displaystyle c_{ij}=\sum_{k=1}^{n} a_{ik}b_{kj} \; (1 \leq i \leq m, 1 \leq j \leq p)$$.

 

$\underline{Def}$

Let $A \in M_{m \times n}(\mathbb{F}), B \in M_{n \times p}(\mathbb{F})$.

The product $AB$ of $A$ & $B$ is defined to be the matrix $AB \in M_{m \times p}(\mathbb{F})$ given by

$$\displaystyle (AB)_{ij}=\sum_{k=1}^{n} A_{ik}B_{kj} \; (1 \leq i \leq m, 1 \leq j \leq p).$$

 

With this definition, we have proved the following above.

 

$\underline{Thm}$

Let $V,W,Z$ be finite dimensional vector spaces with ordered bases $\alpha, \beta, \gamma$, respectively.

Let $T \colon V \rightarrow W,\, U \colon W \rightarrow Z$ be linear.

Then $[UT]_{\alpha}^{\gamma}=[U]_{\beta}^{\gamma}[T]_{\alpha}^{\beta}$.

 

$\underline{Prop}$

Let $A \in M_{m \times n}=M_{m \times n}(\mathbb{F}),\, B \in M_{n \times p}$.

Then $(AB)^{t}=B^{t}A^{t}$

 

$\underline{Proof}$

$$\displaystyle \begin{equation} \begin{split} (AB)_{ij}^{t} & = (AB)_{ij} \\ & = \sum_{k=1}^{n} A_{jk}B_{ki} \\ & = \sum_{k=1}^{n} (B_{t})_{ik}(A^{t})_{ki} \\ & = (B^{t}A^{t})_{ij}\;\;\; for\,\; 1 \leq i \leq p,\, 1 \leq j \leq m \end{split} \end{equation}$$