行列の指数関数三角関数(正弦と余弦)、マクローリン展開、指数関数の積、和の指数関数

学習環境

手を動かしてまなぶ線形代数 (藤岡敦(著)、裳華房)の第4章(行列の指数関数)、12(行列の指数関数)、基本問題の問12.5の解答を求めてみる。

\begin{array}{l} \exp A \\ = \exp [\begin{matrix} a & b \\ - b & a \end{matrix}] \\ = \exp ([\begin{matrix} a & 0 \\ 0 & a \end{matrix}] + [\begin{matrix} 0 & b \\ - b & 0 \end{matrix}]) \\ = \exp [\begin{matrix} a & 0 \\ 0 & a \end{matrix}] \exp [\begin{matrix} 0 & b \\ - b & 0 \end{matrix}] \\ = (E + \sum_{n = 1}^{\infty} \frac{1}{n!} {[\begin{matrix} a & 0 \\ 0 & a \end{matrix}]}^{n}) (E + \sum_{n = 1}^{\infty} \frac{1}{n!} {[\begin{matrix} 0 & b \\ - b & 0 \end{matrix}]}^{n}) \\ = (E + \sum_{n = 1}^{\infty} \frac{1}{n!} [\begin{matrix} a^{n} & 0 \\ 0 & a^{n} \end{matrix}]) (E + \sum_{n = 1}^{\infty} \frac{1}{n!} {(b [\begin{matrix} 0 & 1 \\ - 1 & 0 \end{matrix}])}^{n}) \\ = (\sum_{n = 0}^{\infty} \frac{1}{n!} [\begin{matrix} a^{n} & 0 \\ 0 & a^{n} \end{matrix}]) (E + \sum_{n = 1}^{\infty} \frac{b^{n}}{n!} {[\begin{matrix} 0 & 1 \\ - 1 & 0 \end{matrix}]}^{n}) \\ = [\begin{matrix} \sum_{n = 0}^{\infty} \frac{a^{n}}{n!} & 0 \\ 0 & \sum_{n = 0}^{\infty} \frac{a^{n}}{n!} \end{matrix}] (E + \sum_{n = 1}^{\infty} \frac{b^{n}}{n!} {[\begin{matrix} 0 & 1 \\ - 1 & 0 \end{matrix}]}^{n}) \\ = [\begin{matrix} e^{a} & 0 \\ 0 & e^{a} \end{matrix}] (E + \sum_{n = 1}^{\infty} \frac{b^{n}}{n!} {[\begin{matrix} 0 & 1 \\ - 1 & 0 \end{matrix}]}^{n}) \end{array}

また、

\begin{array}{l} {[\begin{matrix} 0 & 1 \\ - 1 & 0 \end{matrix}]}^{2} = [\begin{matrix} - 1 & 0 \\ 0 & - 1 \end{matrix}] \\ {[\begin{matrix} 0 & 1 \\ - 1 & 0 \end{matrix}]}^{3} = [\begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix}] \\ {[\begin{matrix} 0 & 1 \\ - 1 & 0 \end{matrix}]}^{4} = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] \\ {[\begin{matrix} 0 & 1 \\ - 1 & 0 \end{matrix}]}^{5} = [\begin{matrix} 0 & 1 \\ - 1 & 0 \end{matrix}] \end{array}

なので、

\begin{array}{l} E + \sum_{n = 1}^{\infty} \frac{b^{n}}{n!} {[\begin{matrix} 0 & 1 \\ - 1 & 0 \end{matrix}]}^{n} \\ = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] + [\begin{matrix} \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{(2 n)!} b^{2 n} & \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{(2 n + 1)!} b^{2 n + 1} \\ - \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{(2 n + 1)!} b^{2 n + 1} & \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{(2 n)!} b^{2 n} \end{matrix}] \\ = [\begin{matrix} \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{(2 n)!} b^{2 n} & \sin b \\ - \sin b & \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{2 n!} b^{2 n} \end{matrix}] \\ = [\begin{matrix} \cos b & \sin b \\ - \sin b & \cos b \end{matrix}] \end{array}

よって、

\begin{array}{l} \exp A \\ = [\begin{matrix} e^{a} & 0 \\ 0 & e^{a} \end{matrix}] [\begin{matrix} \cos b & \sin b \\ - \sin b & \cos b \end{matrix}] \\ = [\begin{matrix} e^{a} \cos b & e^{a} \sin b \\ - e^{a} \sin b & e^{a} \cos b \end{matrix}] \end{array}

\begin{array}{l} \exp A \\ = \exp [\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}] \\ = E + \sum_{n = 1}^{\infty} \frac{1}{n!} {[\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}]}^{n} \\ = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] + [\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}] \\ = [\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix}] \end{array}

また、

\begin{array}{l} \exp B \\ = \exp [\begin{matrix} 0 & 0 \\ - 1 & 0 \end{matrix}] \\ = E + \sum_{n = 1}^{\infty} \frac{1}{n!} {[\begin{matrix} 0 & 0 \\ - 1 & 0 \end{matrix}]}^{n} \\ = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] + [\begin{matrix} 0 & 0 \\ - 1 & 0 \end{matrix}] \\ = [\begin{matrix} 1 & 0 \\ - 1 & 1 \end{matrix}] \end{array}

よって、

\begin{array}{l} (\exp A) (\exp B) \\ = [\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix}] [\begin{matrix} 1 & 0 \\ - 1 & 1 \end{matrix}] \\ = [\begin{matrix} 0 & 1 \\ - 1 & 1 \end{matrix}] \end{array}

また、

\begin{array}{l} \exp (A + B) \\ = \exp [\begin{matrix} 0 & 1 \\ - 1 & 0 \end{matrix}] \end{array}

1より、

\exp (A + B) = [\begin{matrix} \cos 1 & \sin 1 \\ - \sin 1 & \cos 1 \end{matrix}]

\begin{array}{l} {[\begin{matrix} a & b \\ c & - a \end{matrix}]}^{2} = [\begin{matrix} a^{2} + b c & 0 \\ 0 & a^{2} + b c \end{matrix}] = - m^{2} π^{2} [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] \\ {[\begin{matrix} a & b \\ c & - a \end{matrix}]}^{3} = - m^{2} π^{2} [\begin{matrix} a & b \\ c & - a \end{matrix}] \\ {[\begin{matrix} a & b \\ c & - a \end{matrix}]}^{4} = m^{4} π^{4} [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] \\ {[\begin{matrix} a & b \\ c & - a \end{matrix}]}^{5} = m^{4} π^{4} [\begin{matrix} a & b \\ c & - a \end{matrix}] \\ {[\begin{matrix} a & b \\ c & - a \end{matrix}]}^{6} = - m^{6} π^{6} [\begin{matrix} a & b \\ c & - a \end{matrix}] \end{array}

よって、

\begin{array}{l} \exp [\begin{matrix} a & b \\ c & - a \end{matrix}] \\ = E + \sum_{n = 1}^{\infty} \frac{1}{n!} {[\begin{matrix} a & b \\ c & - a \end{matrix}]}^{n} \\ = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] + \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{(2 n)!} {(m π)}^{2 n} [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] + \frac{1}{m π} \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{(2 n + 1)!} {(m π)}^{(2 n + 1)} [\begin{matrix} a & b \\ c & - a \end{matrix}] \\ = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{(2 n)!} {(m π)}^{2 n} [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] + \frac{1}{m π} \sin (m π) [\begin{matrix} a & b \\ c & - a \end{matrix}] \\ = (\cos (m π)) E \end{array}

ゆえに、 mが偶数のとき、

(\cos (m π)) E = E

mが奇数のとき、

(\cos (m π)) E = - E