実数を成分にもつ行列 \(A = \left( \begin{array}{cc} a & b \\ c & d \end{array} \right)\) と実数 \(r , s\) が 下の条件 (i) , (ii) , (iii) をみたすとする.

1. (i)　\(s \gt 1\)

2. (ii)　\(A \left( \begin{array}{c} r \\ 1 \end{array} \right) = s \left( \begin{array}{c} r \\ 1 \end{array} \right)\)

3. (iii)　\(A^n \left( \begin{array}{c} 1 \\ 0 \end{array} \right) = \left( \begin{array}{c} x _ n \\ y _ n \end{array} \right)\) （ \(n= 1, 2, \cdots\) ）とするとき, \(\displaystyle\lim _ {n \rightarrow \infty} x _ n = \displaystyle\lim _ {n \rightarrow \infty} y _ n = 0\)

このとき以下の問に答えよ.

1. (1)　\(B = \left( \begin{array}{cc} 1 & r \\ 0 & 1 \end{array} \right)^{-1} A \left( \begin{array}{cc} 1 & r \\ 0 & 1 \end{array} \right)\) を \(a , c , r , s\) を用いて表せ.

2. (2)　\(B^n \left( \begin{array}{c} 1 \\ 0 \end{array} \right) = \left( \begin{array}{c} z _ n \\ w _ n \end{array} \right)\) （ \(n= 1, 2, \cdots\) ）とするとき, \(\displaystyle\lim _ {n \rightarrow \infty} z _ n = \displaystyle\lim _ {n \rightarrow \infty} w _ n = 0\) を示せ.

3. (3)　\(c=0\) かつ \(| a | \lt 1\) を示せ.

【 解 答 】

(1)

\(A \left( \begin{array}{c} 1 \\ 0 \end{array} \right) = \left( \begin{array}{c} a \\ c \end{array} \right)\) と, 条件(ii)より \[ A \left( \begin{array}{cc} 1 & r \\ 0 & 1 \end{array} \right) = \left( \begin{array}{cc} a & c \\ sr & r \end{array} \right) \] これと, \(\left( \begin{array}{cc} 1 & r \\ 0 & 1 \end{array} \right)^{-1} = \left( \begin{array}{cc} 1 & 0 \\ r & 1 \end{array} \right)\) を用いれば \[\begin{align} B & = \left( \begin{array}{cc} 1 & 0 \\ r & 1 \end{array} \right) \left( \begin{array}{cc} a & c \\ sr & r \end{array} \right) \\ & = \underline{\left( \begin{array}{cc} a-rc & c \\ 0 & s \end{array} \right)} \end{align}\]

(2)

\(B^n = \left( \begin{array}{cc} 1 & 0 \\ r & 1 \end{array} \right) A^n \left( \begin{array}{cc} 1 & r \\ 0 & 1 \end{array} \right)\) なので \[\begin{align} B^n \left( \begin{array}{c} 1 \\ 0 \end{array} \right) & = \left( \begin{array}{cc} 1 & 0 \\ r & 1 \end{array} \right) A^n \left( \begin{array}{c} 1 \\ 0 \end{array} \right) \\ & = \left( \begin{array}{cc} 1 & 0 \\ r & 1 \end{array} \right) \left( \begin{array}{c} x _ n \\ y _ n \end{array} \right) \\ & = \left( \begin{array}{c} x _ n -r y _ n \\ y _ n \end{array} \right) = \left( \begin{array}{c} z _ n \\ w _ n \end{array} \right) \end{align}\] これと, 条件 (iii) より \[ \underline{\displaystyle\lim _ {n \rightarrow \infty} z _ n = \displaystyle\lim _ {n \rightarrow \infty} w _ n = 0} \]

(3)

\(B^n = \left( \begin{array}{cc} a _ n & 0 \\ c _ n & d _ n \end{array} \right)\) とおくと \[ \left( \begin{array}{c} z _ n \\ w _ n \end{array} \right) = B^n \left( \begin{array}{c} 1 \\ 0 \end{array} \right) = \left( \begin{array}{c} a _ n \\ c _ n \end{array} \right) \] したがって, (2) の結果より \[\begin{align} \displaystyle\lim _ {n \rightarrow \infty} a _ n & = 0 \quad ... [1] , \\ \displaystyle\lim _ {n \rightarrow \infty} c _ n & = 0 \quad ... [2] \end{align}\] また \[\begin{align} B^{n+1} & = \left( \begin{array}{cc} a-rc & c \\ 0 & s \end{array} \right) \left( \begin{array}{cc} a _ n & 0 \\ c _ n & d _ n \end{array} \right) \\ & = \left( \begin{array}{cc} (a-rc) a _ n & 0 \\ (a-rc)c _ n +cd _ n & sd _ n \end{array} \right) = \left( \begin{array}{cc} a _ {n+1} & 0 \\ c _ {n+1} & d _ {n+1} \end{array} \right) \\ \text{∴} \quad & \left\{ \begin{array}{ll} a _ {n+1} =(a-rc)a _ n & ...[3] \\ c _ {n+1} =(a-rc)c _ n +cd _ n & ...[4] \\ d _ {n+1} =sd _ n & ...[5] \end{array} \right. \end{align}\] \(a _ 1 = a-rc\) , \(d _ 1 =s\) なので, [3] [5]より \[ a _ n = (a-rc)^n , \quad d _ n = s^n \] 条件 (i) より, \(\displaystyle\lim _ {n \rightarrow \infty} d _ n = \infty\) であるが, [2] [4] より \[ \underline{c=0} \] このとき, \(a _ n =a^n\) であり, [1] より \[ \underline{| a | \lt 1} \]