数学 - 線形代数学 - ベクトル空間 – 基底(1次従属、1次独立、座標, 指数関数、対数関数、三角関数、実関数)

学習環境

ラング線形代数学(上)(S.ラング (著)、芹沢正三 (翻訳)、ちくま学芸文庫)の2章(ベクトル空間)、3(基底)、練習問題4、5、6、7、8.を取り組んでみる。

$\begin{array}{l} a_{1} (a, b) + a_{2} (c, d) = 0 \\ a_{1} a + a_{2} c = 0 \\ a_{1} b + a_{2} d = 0 \\ a_{1} a b + a_{2} b c = 0 \\ a_{1} a b + a_{2} a d = 0 \\ a_{2} (a d - b c) = 0 \\ a d - b c = 0 \\ a_{2} = 1 \\ a_{1} a = - c \\ a \neq 0 \\ a_{1} = - \frac{c}{a} \\ a = 0 \\ c = 0 \\ a_{1} = 1 \\ a d - b c \neq 0 \\ a_{2} = 0 \\ a_{1} a = 0 \\ a_{1} b = 0 \\ a_{1} \neq 0 \Rightarrow a = 0 \land b = 0 \Rightarrow a d - b c = 0 \\ a_{1} = 0 \end{array}$
1. $\begin{array}{l} a_{1} + a_{2} t = 0 \\ a_{1} = - a_{2} t \\ a_{1} = a_{2} = 0 \end{array}$
2. $\begin{array}{l} a_{1} t + a_{2} t^{2} = 0 \\ t (a_{1} + a_{2} t) = 0 \\ a_{1} + a_{2} t = 0 \\ a_{1} = - a_{2} t \\ a_{1} = a_{2} = 0 \end{array}$
3. $\begin{array}{l} a_{1} t + a_{2} t^{4} = 0 \\ t (a_{1} + a_{2} t^{3}) = 0 \\ a_{1} + a_{2} t^{3} = 0 \\ a_{1} = a_{2} = 0 \end{array}$
4. $\begin{array}{l} a_{1} e^{t} + a_{2} t = 0 \\ a_{1} e^{t} = - a_{2} t \\ a_{1} = a_{2} = 0 \end{array}$
$(3, 5)$
$\begin{array}{l} D f (t) = 3 \cos t - 5 \sin t \\ (- 5, 3) \end{array}$

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