数学のブログ

行列の指数関数 三角関数(正弦と余弦)、マクローリン展開、非零行列、3次、交代行列

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)、チャレンジ問題、数物系の問12.6の解答を求めてみる。

A = [ 0 a b - a 0 c - b - c 0 ] A 2 = [ 0 a b - a 0 c - b - c 0 ] [ 0 a b - a 0 c - b - c 0 ] = [ - ( a 2 + b 2 ) - b c a c - b c - ( a 2 + c 2 ) - a b a c - a b - ( b 2 + c 2 ) ] A 3 = [ - ( a 2 + b 2 ) - b c a c - b c - ( a 2 + c 2 ) - a b a c - a b - ( b 2 + c 2 ) ] [ 0 a b - a 0 c - b - c 0 ] = [ 0 - a ( a 2 + b 2 ) - a c 2 - b ( a 2 + b 2 ) - b c 2 a ( a 2 + c 2 ) + a b 2 0 - b 2 c - c ( a 2 + c 2 ) a 2 b + b ( b 2 + c 2 ) a 2 c + c ( b 2 + c 2 ) 0 ] = [ 0 - a r 2 - b r 2 a r 2 0 - c r 2 b r 2 c r 2 0 ] = ( - 1 ) r 2 [ 0 a b - a 0 c - b - c 0 ] = ( - 1 ) r 2 A A 4 = ( - 1 ) r 2 A 2

また、

A 2 n + 1 = A 2 ( n - 1 ) + 1 + 2 = A 2 ( n - 1 ) + 1 A 2 = ( - 1 ) n - 1 r 2 ( n - 1 ) A · A 2 = ( - 1 ) n - 1 r 2 ( n - 1 ) A 3 = ( - 1 ) n - 1 r 2 ( n - 1 ) ( - 1 ) r 2 A = ( - 1 ) n r 2 n A A 2 n + 2 = A 2 ( n - 1 ) + 2 + 2 = ( - 1 ) n - 1 r 2 ( n - 1 ) A 2 A 2 = ( - 1 ) n - 1 r 2 ( n - 1 ) A 4 = ( - 1 ) n - 1 r 2 ( n - 1 ) ( - 1 ) r 2 A 2 = ( - 1 ) n r 2 n A 2

よって、 帰納法によりすべての正の自然数nに対して成り立つ。

ゆえに、

exp A = E + n = 1 1 n ! A n = E + n = 0 1 ( 2 n + 1 ) ! A 2 n + 1 + n = 1 1 ( 2 n ) ! A 2 n = E + n = 0 1 ( 2 n + 1 ) ! ( - 1 ) n r 2 n A + n = 1 1 ( 2 n ) ! A 2 ( n - 1 ) + 2 = E + 1 r n = 0 ( - 1 ) n ( 2 n + 1 ) ! r 2 n + 1 A + n = 1 1 ( 2 n ) ! ( - 1 ) ( n - 1 ) r 2 ( n - 1 ) A 2 = E + sin r r A - 1 r 2 n = 1 ( - 1 ) n ( 2 n ) ! r 2 n A 2 = E + sin r r A - 1 r 2 ( - 1 + n = 0 ( - 1 ) n ( 2 n ) ! r 2 n ) A 2 = E + sin r r A + 1 - cos r r 2 A 2