実数を成分とする正方行列 \[ A = \left( \begin{array}{cc} a & b \\ c & d \end{array} \right) , \ B = \left( \begin{array}{cc} 1 & 1 \\ -1 & 2 \end{array} \right) , \ E = \left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right) \] について, 以下の問いに答えよ.
(1) \(AB = BA\) を満たす \(A\) は, \(x , y\) を用いて \(A = x B +y E\) と表せることを示せ.
(2) \(A^3 = E\) のとき \[ ( t^2 -\Delta ) A = ( t \Delta +1 ) E \] を示せ. ただし, \(t = a+d\) , \(\Delta = ad -bc\) とする.
(3) \(AB = BA\) かつ \(A^3 = E\) を満たす \(A\) をすべて求めよ.
解答
(1)
\[\begin{align} AB & = \left( \begin{array}{cc} a & b \\ c & d \end{array} \right) \left( \begin{array}{cc} 1 & 1 \\ -1 & 2 \end{array} \right) = \left( \begin{array}{cc} a-b & a+2b \\ c-d & c+2d \end{array} \right) , \\ BA & = \left( \begin{array}{cc} 1 & 1 \\ -1 & 2 \end{array} \right) \left( \begin{array}{cc} a & b \\ c & d \end{array} \right) = \left( \begin{array}{cc} a+c & b+d \\ -a+2c & -b+2d \end{array} \right) \end{align}\] \(AB = BA\) なので, 各成分を比較すると \[ \left\{ \begin{array}{ll} a-b = a+c & ... [1] \\ a+2b = b+d & ... [2] \\ c-d = a+2c & ... [3] \\ c+2d = -b+2d & ... [4] \end{array} \right. \] [1] [4] より \[ c = -b \] [2] [3] より \[ d = a+b \] よって \[ A = \left( \begin{array}{cc} a & b \\ -b & a+b \end{array} \right) = b A +(a-b) E \]
(2)
ケーリー・ハミルトンの定理より \[ A^2 = t A -\Delta E \] これを用いれば \[\begin{align} A^3 & = \left( t A -\Delta E \right) A \\ & = t \left( t A -\Delta E \right) -\Delta A \\ & = \left( t^2 -\Delta \right) A -\Delta t E = E \end{align}\] よって \[ \left( t^2 -\Delta \right) A = \left( \Delta t +1 \right) E \]
(3)
\[\begin{align} t & = a +(a+b) = 2a +b , \\ \Delta & = a (a+b) +b^2 = a^2 +ab +b^2 \end{align}\] (2) の結果から, 場合分けして考える.
1* \(t^2 -\Delta = 0\) のとき \[\begin{align} t^2 -\Delta & = (2a+b)^2 -( a^2 +ab +b^2 ) \\ & = 3a (a+b) = 0 \\ \text{∴} \quad & a = 0 , \ b = -a \end{align}\]
\(a = 0\) のとき \[\begin{align} t \Delta +1 & = b \cdot b^2 +1 = 0 \\ \text{∴} \quad b & = -1 \end{align}\] このとき \[ A = \left( \begin{array}{cc} 0 & -1 \\ 1 & -1 \end{array} \right) \]
\(b = -a\) のとき \[\begin{align} t \Delta +1 & = a \cdot a^2 +1 = 0 \\ \text{∴} \quad a & = -1 \end{align}\] このとき \[ A = \left( \begin{array}{cc} -1 & 1 \\ -1 & 0 \end{array} \right) \]
2* \(t^2 -\Delta \neq 0\) のとき
\(A = k E\) （ \(k\) は実数）と表せるので \[\begin{align} A^3 & = k^3 E = E \\ \text{∴} \quad k & = 1 \end{align}\] このとき \[ A = E \]
以上より, 求める行列 \(A\) は \[ A = \underline{\left( \begin{array}{cc} 0 & -1 \\ 1 & -1 \end{array} \right) , \left( \begin{array}{cc} -1 & 1 \\ -1 & 0 \end{array} \right) , \left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right)} \]