Matrix Multiplication

    이 포스트에서 $V, W, Z$는 모두 유한차원 $F$-벡터공간으로 취급한다. 함수의 합성은 보통 $g \circ f$로 표기하는데, linear transformation의 경우 $gf$로 표기하도록 하자.

Theorem 1

Theorem 1. Let $T, U_1, U_2 \in \mathcal{L}(V, W)$, and let $U \in \mathcal{L}(W, Z)$. Then 
(a) $UT \in \mathcal{L}(V, Z)$.
(b) If $UT$ is injective, then so is $T$.
(c) If $UT$ is surjective, then so is $U$.
(d) IF $T$ and $U$ are bijective, then so is $UT$. 

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Introduction

    선형 변환은 하나의 행렬로 표현할 수 있었고, 선형 변환의 합과 상수배도 행렬로 표현했을 때 $[T + U]_{\beta}^{\gamma} = [T]_{\beta}^{\gamma} + [U]_{\beta}^{\gamma}$, $[aT]_{\beta}^{\gamma} = a[T]_{\beta}^{\gamma}$으로 잘 맞아떨어졌다. 동일하게 선형 변환의 합성의 행렬 표현을 구해보자.

Let $T \in \mathcal{L}(V, W), U \in \mathcal{L}(W, Z)$, and let $A := [U]_{\beta}^{\gamma}, B := [T]_{\alpha}^{\beta}$, where $\alpha = \{v_1, ..., v_n\}, \beta = \{w_1, ..., w_m\}$, and $\gamma = \{z_1, ..., z_p\}$ are ordered bases for $V, W$, and $Z$, respectively. 
Consider the matrix $[UT]_{\alpha}^{\gamma}$. For $1 \leq j \leq n$, we have
$$\begin{align*}(UT)(v_j) &= U(T(v_j)) = U(\sum_{k=1}^{m} B_{kj}w_k) = \sum_{k=1}^{m}B_{kj}U(w_k) \\ &= \sum_{k=1}^{m}B_{kj}(\sum_{i=1}^{p} A_{ik}z_i) = \sum_{i=1}^{p}(\sum_{k=1}^{m} A_{ik}B_{kj})z_i \\ &= \sum_{i=1}^{p}C_{ij}z_i,\end{align*}$$ where 
$$C_{ij} := \sum_{k=1}^{m} A_{ik}B_{kj}.$$

    즉 두 행렬의 곱 $AB$를 위와 같이 $C$로 정의하는 것이 자연스럽다.

Matrix multiplication

Definition 1. Let $A \in M_{m \times n}(F), B \in M_{n \times p}(F)$. We define the product of $A$ and $B$, denoted $AB$, to be the $n \times p$ matrix such that 
$$(AB)_{ij} = \sum_{k=1}^{n}A_{ik}B_{kj} \,\, \text{for} \,\, 1 \leq i \leq m, 1 \leq j \leq p.$$

    두 행렬의 곱셈은 각 행렬이 표현하는 선형 변환의 합성의 행렬 표현과 같으므로, 일반적으로 함수의 합성은 교환 법칙이 성립하지 않듯 행렬의 곱셈도 성립하지 않는다. 즉 일반적으로 $AB \neq BA$이다.

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Theorem 2 ($[UT]_{\alpha}^{\gamma} = [U]_{\beta}^{\gamma} [T]_{\alpha}^{\beta}$)

Theorem 2. Let $T \in \mathcal{L}(V, W), U \in \mathcal{L}(W, Z)$. Then $[UT]_{\alpha}^{\gamma} = [U]_{\beta}^{\gamma} [T]_{\alpha}^{\beta}$, where $\alpha, \beta, \gamma$ are ordered bases for $V, W, Z$, respectively.

Proof. Using the above introduction's notation, $[UT]_{\alpha}^{\gamma} = C = AB = [U]_{\beta}^{\gamma}[T]_{\alpha}^{\beta}$. $\blacksquare$

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Theorem 3

Theorem 3. Let $A \in M_{m \times n}(F), B, C \in M_{n \times p}(F)$, and $D, E \in M_{q \times m}(F)$. Then
(a) $A(B + C) = AB + AC$ and $(D + E)A = DA + EA.$
(b) $a(AB) = (aA)B = A(aB), \, \forall a \in F.$
(c) $I_mA = A = AI_n.$
(d) $[I_V]_{\beta} = I_n$ where $\beta$ is an ordered basis for $n$-dimensional $V$.

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Theorem 4

Theorem 4. Let $A,B$, and $C$ be matrices such that $A(BC)$ is defined. Then $(AB)C$ is also defined and $A(BC) = (AB)C$.

Proof. To define $A(BC)$, suppose that $A, B$, and $C$ are an $m \times n$, $n \times p$, $p \times q$ matrices, respectively. Then $A(BC)$ is an $m \times q$ matrix, and so is $(AB)C$. 
We have $L_{A(BC)} = L_A(L_BL_C) = (L_AL_B)L_C = L_{(AB)C} \Longrightarrow A(BC) = (AB)C.$ $\blacksquare$

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Reference is here: https://product.kyobobook.co.kr/detail/S000003155051

Linear Algebra | Stephen Friedberg - 교보문고