数学のブログ

行列の指数関数 正方行列、成分、指数関数、累乗、帰納法

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12.2の解答を求めてみる。

A = [ λ 1 0 λ ] A 2 = [ λ 2 2 λ 0 λ 2 ] A 3 = [ λ 2 2 λ 0 λ 2 ] [ λ 1 0 λ ] = [ λ 3 3 λ 2 0 λ 3 ] A 4 = [ λ 3 3 λ 2 0 λ 3 ] [ λ 1 0 λ ] = [ λ 4 4 λ 3 0 λ 4 ] A 5 = [ λ 4 4 λ 3 0 λ 4 ] [ λ 1 0 λ ] = [ λ 5 5 λ 4 0 λ 5 ]
A n = A n - 1 A = [ λ n - 1 ( n - 1 ) λ n - 2 0 λ n - 1 ] [ λ 1 0 λ ] = [ λ n n λ n - 1 0 λ n ]

よって帰納法により成り立つ。

exp [ λ 1 0 λ ] = E + k = 1 1 k ! [ λ k k λ k - 1 0 λ k ] = [ k = 0 λ k k ! 0 0 k = 0 λ k k ! ] + k = 1 λ k - 1 ( k - 1 ) ! [ 0 1 0 0 ] = [ e λ 0 0 e λ ] + k = 0 λ k k ! [ 0 1 0 0 ] = [ e λ 0 0 e λ ] + [ 0 e λ 0 0 ] = [ e λ e λ 0 e λ ]