66. 阵的积, 分块地写

我们知道, 若 $B$ 是 $s \times m$ 阵, 且 $a_{1}$ , $a_{2}$ , $\dots$ , $a_{n}$ 是 $m \times 1$ 阵, 则 $B [a_{1}, a_{2}, \dots, a_{n}] = [B a_{1}, B a_{2}, \dots, B a_{n}] .$ 形象地, 这是分块地写阵的积. 证明 $det (B A) = det (B) det (A)$ ( $A$ , $B$ 是同级的方阵) 时, 我们用了此事, 使推理更简单.

我要介绍另一个分块地写阵的积的方式.

定理 66.1. 设 $A_{1, 1}$ , $A_{1, 2}$ , $A_{2, 1}$ , $A_{2, 2}$ , $B_{1, 1}$ , $B_{1, 2}$ , $B_{2, 1}$ , $B_{2, 2}$ 是阵. 设 $A_{1, 1}$ , $A_{1, 2}$ 都有 $r_{1}$ 行; 设 $A_{2, 1}$ , $A_{2, 2}$ 都有 $r_{2}$ 行; 设 $A_{1, 1}$ , $A_{2, 1}$ 都有 $u_{1}$ 列; 设 $A_{1, 2}$ , $A_{2, 2}$ 都有 $u_{2}$ 列. 设 $B_{1, 1}$ , $B_{1, 2}$ 都有 $u_{1}$ 行; 设 $B_{2, 1}$ , $B_{2, 2}$ 都有 $u_{2}$ 行; 设 $B_{1, 1}$ , $B_{2, 1}$ 都有 $c_{1}$ 列; 设 $B_{1, 2}$ , $B_{2, 2}$ 都有 $c_{2}$ 列. 作 $(r_{1} + r_{2}) \times (u_{1} + u_{2})$ 阵 $A = [A_{1, 1} A_{2, 1} A_{1, 2} A_{2, 2}];$ 具体地, $[A]_{i, j} = ⎩ ⎨ ⎧ [A_{1, 1}]_{i, j}, [A_{1, 2}]_{i, j - u_{1}}, [A_{2, 1}]_{i - r_{1}, j}, [A_{2, 2}]_{i - r_{1}, j - u_{1}}, i ⩽ r_{1}, j ⩽ u_{1}; i ⩽ r_{1}, j > u_{1}; i > r_{1}, j ⩽ u_{1}; i > r_{1}, j > u_{1} .$ 作 $(u_{1} + u_{2}) \times (c_{1} + c_{2})$ 阵 $B = [B_{1, 1} B_{2, 1} B_{1, 2} B_{2, 2}];$ 具体地, $[B]_{i, j} = ⎩ ⎨ ⎧ [B_{1, 1}]_{i, j}, [B_{1, 2}]_{i, j - c_{1}}, [B_{2, 1}]_{i - u_{1}, j}, [B_{2, 2}]_{i - u_{1}, j - c_{1}}, i ⩽ u_{1}, j ⩽ c_{1}; i ⩽ u_{1}, j > c_{1}; i > u_{1}, j ⩽ c_{1}; i > u_{1}, j > c_{1} .$ 因为 $A$ 的列数等于 $B$ 的行数, 故 $A B$ 有意义.

记 $C_{i, j} = A_{i, 1} B_{1, j} + A_{i, 2} B_{2, j}$ , $i$ , $j = 1$ , $2$ . 不难看出, $C_{i, j}$ 是 $r_{i} \times c_{j}$ 阵. 作 $(r_{1} + r_{2}) \times (c_{1} + c_{2})$ 阵 $C = [C_{1, 1} C_{2, 1} C_{1, 2} C_{2, 2}];$ 具体地, $[C]_{i, j} = ⎩ ⎨ ⎧ [C_{1, 1}]_{i, j}, [C_{1, 2}]_{i, j - c_{1}}, [C_{2, 1}]_{i - r_{1}, j}, [C_{2, 2}]_{i - r_{1}, j - c_{1}}, i ⩽ r_{1}, j ⩽ c_{1}; i ⩽ r_{1}, j > c_{1}; i > r_{1}, j ⩽ c_{1}; i > r_{1}, j > c_{1} .$ 则 $A B = C$ .

形象地, $[A_{1, 1} A_{2, 1} A_{1, 2} A_{2, 2}] [B_{1, 1} B_{2, 1} B_{1, 2} B_{2, 2}] = [A_{1, 1} B_{1, 1} + A_{1, 2} B_{2, 1} A_{2, 1} B_{1, 1} + A_{2, 2} B_{2, 1} A_{1, 1} B_{1, 2} + A_{1, 2} B_{2, 2} A_{2, 1} B_{1, 2} + A_{2, 2} B_{2, 2}] .$ 于是, 形式地, 分块地写阵的积与直接用定义写阵的积没有区别.

证. 注意到 $[A B]_{i, j} = = k = 1 \sum u_{1} + u_{2} [A]_{i, k} [B]_{k, j} k = 1 \sum u_{1} [A]_{i, k} [B]_{k, j} + k = u_{1} + 1 \sum u_{1} + u_{2} [A]_{i, k} [B]_{k, j} .$

若 $j ⩽ c_{1}$ , 则 $[A B]_{i, j} = = k = 1 \sum u_{1} [A]_{i, k} [B]_{k, j} + k = u_{1} + 1 \sum u_{1} + u_{2} [A]_{i, k} [B]_{k, j} k = 1 \sum u_{1} [A]_{i, k} [B_{1, 1}]_{k, j} + k = u_{1} + 1 \sum u_{1} + u_{2} [A]_{i, k} [B_{2, 1}]_{k - u_{1}, j} .$ 当 $i ⩽ r_{1}$ 时, $[A B]_{i, j} = = = = = = k = 1 \sum u_{1} [A]_{i, k} [B_{1, 1}]_{k, j} + k = u_{1} + 1 \sum u_{1} + u_{2} [A]_{i, k} [B_{2, 1}]_{k - u_{1}, j} k = 1 \sum u_{1} [A_{1, 1}]_{i, k} [B_{1, 1}]_{k, j} + k = u_{1} + 1 \sum u_{1} + u_{2} [A_{1, 2}]_{i, k - u_{1}} [B_{2, 1}]_{k - u_{1}, j} [A_{1, 1} B_{1, 1}]_{i, j} + [A_{1, 2} B_{2, 1}]_{i, j} [A_{1, 1} B_{1, 1} + A_{1, 2} B_{2, 1}]_{i, j} [C_{1, 1}]_{i, j} [C]_{i, j} .$ 当 $i > r_{1}$ 时, $[A B]_{i, j} = = = = = = k = 1 \sum u_{1} [A]_{i, k} [B_{1, 1}]_{k, j} + k = u_{1} + 1 \sum u_{1} + u_{2} [A]_{i, k} [B_{2, 1}]_{k - u_{1}, j} k = 1 \sum u_{1} [A_{2, 1}]_{i - r_{1}, k} [B_{1, 1}]_{k, j} + k = u_{1} + 1 \sum u_{1} + u_{2} [A_{2, 2}]_{i - r_{1}, k - u_{1}} [B_{2, 1}]_{k - u_{1}, j} [A_{2, 1} B_{1, 1}]_{i - r_{1}, j} + [A_{2, 2} B_{2, 1}]_{i - r_{1}, j} [A_{2, 1} B_{1, 1} + A_{2, 2} B_{2, 1}]_{i - r_{1}, j} [C_{2, 1}]_{i - r_{1}, j} [C]_{i, j} .$

若

j > c_{1}

, 则

[A B]_{i, j} = = k = 1 \sum u_{1} [A]_{i, k} [B]_{k, j} + k = u_{1} + 1 \sum u_{1} + u_{2} [A]_{i, k} [B]_{k, j} k = 1 \sum u_{1} [A]_{i, k} [B_{1, 2}]_{k, j - c_{1}} + k = u_{1} + 1 \sum u_{1} + u_{2} [A]_{i, k} [B_{2, 2}]_{k - u_{1}, j - c_{1}} .

当

i ⩽ r_{1}

时,

[A B]_{i, j} = = = = = = k = 1 \sum u_{1} [A]_{i, k} [B_{1, 2}]_{k, j - c_{1}} + k = u_{1} + 1 \sum u_{1} + u_{2} [A]_{i, k} [B_{2, 2}]_{k - u_{1}, j - c_{1}} k = 1 \sum u_{1} [A_{1, 1}]_{i, k} [B_{1, 2}]_{k, j - c_{1}} + k = u_{1} + 1 \sum u_{1} + u_{2} [A_{1, 2}]_{i, k - u_{1}} [B_{2, 2}]_{k - u_{1}, j - c_{1}} [A_{1, 1} B_{1, 2}]_{i, j - c_{1}} + [A_{1, 2} B_{2, 2}]_{i, j - c_{1}} [A_{1, 1} B_{1, 2} + A_{1, 2} B_{2, 2}]_{i, j - c_{1}} [C_{1, 2}]_{i, j - c_{1}} [C]_{i, j} .

当

i > r_{1}

时,

[A B]_{i, j} = = = = = = k = 1 \sum u_{1} [A]_{i, k} [B_{1, 2}]_{k, j - c_{1}} + k = u_{1} + 1 \sum u_{1} + u_{2} [A]_{i, k} [B_{2, 2}]_{k - u_{1}, j - c_{1}} k = 1 \sum u_{1} [A_{2, 1}]_{i - r_{1}, k} [B_{1, 2}]_{k, j - c_{1}} + k = u_{1} + 1 \sum u_{1} + u_{2} [A_{2, 2}]_{i - r_{1}, k - u_{1}} [B_{2, 2}]_{k - u_{1}, j - c_{1}} [A_{2, 1} B_{1, 2}]_{i - r_{1}, j - c_{1}} + [A_{2, 2} B_{2, 2}]_{i - r_{1}, j - c_{1}} [A_{2, 1} B_{1, 2} + A_{2, 2} B_{2, 2}]_{i - r_{1}, j - c_{1}} [C_{2, 2}]_{i - r_{1}, j - c_{1}} [C]_{i, j} .

证毕.

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视图

66. 阵的积, 分块地写