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3A p3

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Signed-off-by: szdytom <szdytom@qq.com>
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方而静 2025-07-27 22:27:17 +08:00
parent 0450974138
commit 0142424012
Signed by: szTom
GPG Key ID: 072D999D60C6473C
2 changed files with 27 additions and 1 deletions
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1
math.typ
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@ -2,6 +2,7 @@
#let ii = "i"
#let span = $op("span")$
#let Poly = math.cal("P")
#let LinearMap = math.cal("L")
#let complexification(vv) = $vv_upright(C)$
#let rhs = "R.H.S."
#let lhs = "L.H.S."

27
sections/3A.typ
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@ -1,5 +1,5 @@
#import "../styles.typ": exercise_sol, note, tab
#import "../math.typ": Poly
#import "../math.typ": Poly, LinearMap
#exercise_sol(type: "proof")[
设 $b, c in RR$。定义
@ -71,3 +71,28 @@
#tab 解得 $b = c = 0$。这就是说 $T$ 是线性映射，当且仅当 $b = c = 0$。
]
#exercise_sol(type: "proof")[
设 $T in LinearMap(FF^n, FF^m)$。证明：对于 $j in {1, dots, m}$ 和 $k in {1, dots, n}$，存在标量 $A_(j, k) in FF$，使得对于任意 $(x_1, dots, x_n) in FF^n$，
$ T(x_1, dots, x_n) = (A_(1, 1)x_1 + dots.c + A_(1, n) x_n, dots, A_(m, 1) x_1 + dots.c + A_(m, n) x_n) $
#note[此习题表明，线性映射 $T$ 具有在原书例3.3的倒数第二个例子中预示的形式。]
][
对于 $j in {1, dots, m}$ 和 $k in {1, dots, n}$，令 $w_j in FF^m$ 和 $v_k in FF^n$ 分别为第 $j$ 个和第 $k$ 个分量为 $1$，其余分量为 $0$ 的向量。
#tab 容易发现，$w_1, dots, w_m$ 是 $FF^m$ 的基，$v_1, dots, v_n$ 是 $FF^n$ 的基。于是，对于任意 $k in {1, dots, n}$，可以找到 $A_(1, 1), dots, A_(m, 1) in FF$，使得
$ T v_k = A_(1, k) w_1 + dots.c + A_(m, k) w_m $
#tab 另一方面，根据 $v_k$ 的定义，我们知道 $(x_1, dots, x_n) = x_1 v_1 + dots.c + x_n v_n$。同时，考虑到 $T$ 是线性映射，我们有
$ T(x_1, dots, x_n) &= T(sum_(k = 1)^n x_k v_k) \
&= sum_(k = 1)^n T(x_k v_k) \
&= sum_(k = 1)^n x_k T v_k \
&= sum_(k = 1)^n x_k sum_(j = 1)^m A_(j, k) w_j \
&= sum_(j = 1)^m w_j sum_(k = 1)^n A_(j, k) x_k \
&= (A_(1, 1)x_1 + dots.c + A_(1, n) x_n, dots, A_(m, 1) x_1 + dots.c + A_(m, n) x_n) $
#tab 这立即给出了我们想要的结果。
]