2020年1月6日月曜日

数学 - 線形代数学 - ベクトル空間 - 基底 - 実関数のなすベクトル空間、直線、累乗、指数関数、三角関数(正弦と余弦、倍角)、微分、導関数、関数の対、一次独立

ラング線形代数学(上) (ちくま学現文庫)(S.ラング (著)、芹沢 正三 (翻訳)、筑摩書房)の2章(ベクトル空間)、3(基底)、練習問題5の解答を求めてみる。

1. $\begin{array}{l}{x}_{1}+{x}_{2}t=0\\ \frac{d}{\mathrm{dt}}\left({x}_{1}+{x}_{2}t\right)=0\\ {x}_{2}=0\\ {x}_{1}=0\end{array}$

よって1次独立。

2. $\begin{array}{l}{x}_{1}t+{x}_{2}{t}^{2}=0\\ t\left({x}_{1}+{x}_{2}t\right)=0\\ {x}_{1}+{x}_{2}t=0\\ {x}_{1}={x}_{2}=0\end{array}$

3. $\begin{array}{l}{x}_{1}t+{x}_{2}{t}^{4}=0\\ t\left({x}_{1}+{x}_{2}{t}^{3}\right)=0\\ {x}_{1}+{x}_{2}{t}^{3}=0\\ 3{x}_{2}{t}^{2}=0\\ {x}_{1}=0\\ {x}_{2}=0\end{array}$

4. $\begin{array}{l}{x}_{1}{e}^{t}+{x}_{2}t=0\\ {x}_{1}{e}^{t}+{x}_{2}=0\\ {x}_{1}{e}^{t}=0\\ {x}_{1}={x}_{2}=0\end{array}$

5. $\begin{array}{l}{x}_{1}t{e}^{t}+{x}_{2}{e}^{2t}=0\\ {e}^{t}\left({x}_{1}t+{x}_{2}{e}^{t}\right)=0\\ {x}_{1}t+{x}_{2}{e}^{t}=0\\ {x}_{1}={x}_{2}=0\end{array}$

6. $\begin{array}{l}{x}_{1}\mathrm{sin}t+{x}_{2}\mathrm{cos}t=0\\ {x}_{1}\mathrm{cos}t-{x}_{2}\mathrm{sin}t=0\\ \left\{\begin{array}{l}{x}_{1}{x}_{2}\mathrm{sin}t+{x}_{2}^{2}\mathrm{cos}t=0\\ {x}_{1}^{2}\mathrm{cos}t-{x}_{1}{x}_{2}\mathrm{sin}t=0\end{array}\\ \left({x}_{1}^{2}+{x}_{2}^{2}\right)\mathrm{cos}t=0\\ {x}_{1}^{2}+{x}_{2}^{2}=0\\ {x}_{1}={x}_{2}=0\end{array}$

7. $\begin{array}{l}{x}_{1}t+{x}_{2}\mathrm{sin}t=0\\ {x}_{1}+{x}_{2}\mathrm{cos}t=0\\ -{x}_{2}\mathrm{sin}t=0\\ {x}_{2}=0\\ {x}_{1}=0\end{array}$

8. $\begin{array}{l}{x}_{1}\mathrm{sin}t+{x}_{2}\mathrm{sin}2t=0\\ {x}_{1}\mathrm{sin}t+2{x}_{2}\mathrm{sin}t\mathrm{cos}t=0\\ \mathrm{sin}t\left({x}_{1}+2{x}_{2}\mathrm{cos}t\right)=0\\ {x}_{1}+2{x}_{2}\mathrm{cos}t=0\\ -2{x}_{2}\mathrm{sin}t=0\\ {x}_{2}=0\\ {x}_{1}=0\end{array}$

9. $\begin{array}{l}{x}_{1}\mathrm{cos}t+{x}_{2}\mathrm{cos}3t=0\\ {x}_{1}\mathrm{cos}t+{x}_{2}\left(\mathrm{cos}t\mathrm{cos}2t-\mathrm{sin}t\mathrm{sin}2t\right)=0\\ \left(\mathrm{cos}t\right)\left({x}_{1}+{x}_{2}\mathrm{cos}2t\right)-{x}_{2}\left(\mathrm{sin}t\right)2\mathrm{sin}t\mathrm{cos}t=0\\ \mathrm{cos}t\left({x}_{1}+{x}_{2}\mathrm{cos}2t-2{x}_{2}{\mathrm{sin}}^{2}t\right)=0\\ {x}_{1}+{x}_{2}\mathrm{cos}2t-2{x}_{2}{\mathrm{sin}}^{2}t=0\\ {x}_{1}+{x}_{2}\left({\mathrm{cos}}^{2}t-{\mathrm{sin}}^{2}t\right)-2{x}_{2}{\mathrm{sin}}^{2}t=0\\ {x}_{1}+{x}_{2}\left(1-2{\mathrm{sin}}^{2}t\right)-2{x}_{2}{\mathrm{sin}}^{2}t=0\\ {x}_{1}+{x}_{2}-4{x}_{2}{\mathrm{sin}}^{2}t=0\\ {x}_{1}+{x}_{2}-4{x}_{2}\frac{1-\mathrm{cos}2t}{2}=0\\ {x}_{1}+{x}_{2}-2{x}_{2}+2{x}_{2}\mathrm{cos}2t=0\\ -4{x}_{2}\mathrm{sin}2t=0\\ {x}_{2}=0\\ {x}_{1}=0\end{array}$