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3C p4

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Signed-off-by: szdytom <szdytom@qq.com>
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方而静 2025-09-08 20:06:28 +08:00
parent faa17267a6
commit ee001d70c4
Signed by: szTom
GPG Key ID: 072D999D60C6473C
2 changed files with 21 additions and 1 deletions
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21
sections/3C.typ
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@ -1,5 +1,5 @@
#import "../styles.typ": exercise_sol, note, tab
#import "../math.typ": LinearMap, range, Matrix, null, span
#import "../math.typ": LinearMap, range, Matrix, null, span, Poly
#exercise_sol(type: "proof", label: "tricky")[
设 $T in LinearMap(V, W)$。证明：对于 $V$ 和 $W$ 的任意基，$T$ 所对应的矩阵至少有 $dim range T$ 个非零元素。
@ -61,3 +61,22 @@
#tab 故 $B_(k, j) = lambda A_(k, j)$，即 $B = lambda A$。
]
#exercise_sol(type: "answer")[
设 $D in LinearMap(Poly_3(RR), Poly_2(RR))$ 是微分映射，定义为 $p |-> p'$。求 $Poly_3(RR)$ 的一个基和 $Poly_2(RR)$ 的一个基，使得 $Matrix(D)$ 为
$ mat(1, 0, 0, 0; 0, 1, 0, 0; 0, 0, 1, 0) $
#note[和原书3.33比较一下。下一题拓展了本题。]
][
取 $Poly_3(RR)$ 的基为 $x^3, x^2, x, 1$，$Poly_2(RR)$ 的基为 $3 x^2, 2 x, 1$。则
$ D x^3 &= 3 x^2 &= 1 dot.c (3 x^2) + 0 dot.c (2 x) + 0 dot.c 1 \
D x^2 &= 2 x &= 0 dot.c (3 x^2) + 1 dot.c (2 x) + 0 dot.c 1 \
D x &= 1 &= 0 dot.c (3 x^2) + 0 dot.c (2 x) + 1 dot.c 1 \
D 1 &= 1 &= 0 dot.c (3 x^2) + 0 dot.c (2 x) + 0 dot.c 1 $
#tab 故
$ Matrix(D) = mat(1, 0, 0, 0; 0, 1, 0, 0; 0, 0, 1, 0) $
]

1
styles.typ
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@ -99,6 +99,7 @@
set footnote(numbering: "注1")
show math.equation: set text(font: mathfont)
set math.vec(delim: ("[", "]"))
set math.mat(delim: ("[", "]"))
hide[#title <book-title>]