第17章 17.20

(1)

$AB = \begin{pmatrix} 1 & 0 & 1 \\ 0 & 2 & 0 \\ 4 & 0 & 1 \end{pmatrix} \begin{pmatrix} 3 & 0 & 6 \\ 0 & -3 & 0 \\ 24 & 0 & 3 \end{pmatrix} = \begin{pmatrix} 27 & 0 & 9 \\ 0 & -6 & 0 \\ 36 & 0 & 27 \end{pmatrix}$

$BA = \begin{pmatrix} 3 & 0 & 6 \\ 0 & -3 & 0 \\ 24 & 0 & 3 \end{pmatrix} \begin{pmatrix} 1 & 0 & 1 \\ 0 & 2 & 0 \\ 4 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 27 & 0 & 9 \\ 0 & -6 & 0 \\ 36 & 0 & 27 \end{pmatrix}$

よって$AB = BA$

(2)

$|A - \lambda E| = \begin{vmatrix} 1-\lambda & 0 & 1 \\ 0 & 2-\lambda & 0 \\ 4 & 0 & 1-\lambda \end{vmatrix} = (1-\lambda)(2-\lambda)(1-\lambda) - 4(2-\lambda) = (2-\lambda)(1-2\lambda+\lambda^2-4)$

$= (2-\lambda)(\lambda^2-2\lambda-3) = (2-\lambda)(\lambda-3)(\lambda+1) = 0 \quad \lambda = -1,2,3$

$\lambda = -1$のとき

$\begin{pmatrix} 2 & 0 & 1 \\ 0 & 3 & 0 \\ 4 & 0 & 2 \end{pmatrix} - \begin{pmatrix} 2 & 0 & 1 \\ 0 & 1 & 0 \\ 2 & 0 & 1 \end{pmatrix} \to \begin{pmatrix} 2 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix} \quad p_1 = C_1 \begin{pmatrix} -\frac{1}{2} \\ 0 \\ 1 \end{pmatrix} = C_1 \begin{pmatrix} 1 \\ 0 \\ -2 \end{pmatrix}$

$\lambda = 2$のとき

$\begin{pmatrix} -1 & 0 & 1 \\ 0 & 0 & 0 \\ 4 & 0 & -1 \end{pmatrix} - \begin{pmatrix} 1 & 0 & -1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} \to \begin{pmatrix} 1 & 0 & -1 \\ 0 & 0 & 3 \\ 0 & 0 & 0 \end{pmatrix} \to \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix} \quad p_2 = C_2 \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}$

$\lambda = 3$のとき

$\begin{pmatrix} -2 & 0 & 1 \\ 0 & -1 & 0 \\ 4 & 0 & -2 \end{pmatrix} \to \begin{pmatrix} 2 & 0 & -1 \\ 0 & 1 & 0 \\ 2 & 0 & -1 \end{pmatrix} \to \begin{pmatrix} 2 & 0 & -1 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix} \quad p_3 = C_3 \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix}$

(3)

$|B - \lambda E| = \begin{vmatrix} 3-\lambda & 0 & 6 \\ 0 & -3-\lambda & 0 \\ 24 & 0 & 3-\lambda \end{vmatrix} = (3-\lambda)(-3-\lambda)(3-\lambda) - 144(-3-\lambda)$

$= (-3-\lambda)(9-6\lambda+\lambda^2-144) = (-3-\lambda)(\lambda^2-6\lambda-135) = (-3-\lambda)(\lambda+9)(\lambda-15) = 0$

$\lambda = -3, -9, 15$

これから全ての固有ベクトルにおいて $Bp_i = p_i\lambda$ （$\lambda$は$-3, -9, 15$のうちのどれか）となればよい

$Bp_1 = \begin{pmatrix} 3 & 0 & 6 \\ 0 & -3 & 0 \\ 24 & 0 & 3 \end{pmatrix} \begin{pmatrix} 1 \\ 0 \\ -2 \end{pmatrix} = \begin{pmatrix} -9 \\ 0 \\ 18 \end{pmatrix} = -9 \begin{pmatrix} 1 \\ 0 \\ -2 \end{pmatrix}$

$Bp_2 = \begin{pmatrix} 3 & 0 & 6 \\ 0 & -3 & 0 \\ 24 & 0 & 3 \end{pmatrix} \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ -3 \\ 0 \end{pmatrix} = -3 \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}$

$Bp_3 = \begin{pmatrix} 3 & 0 & 6 \\ 0 & -3 & 0 \\ 24 & 0 & 3 \end{pmatrix} \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix} = \begin{pmatrix} 15 \\ 0 \\ 30 \end{pmatrix} = 15 \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix}$

よって$A$の固有ベクトルは$B$の固有ベクトルでもある

(4)

$P^{-1}AP = \begin{pmatrix} -1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{pmatrix}, \quad P^{-1}BP = \begin{pmatrix} -9 & 0 & 0 \\ 0 & -3 & 0 \\ 0 & 0 & 15 \end{pmatrix}, \quad P = \begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ -2 & 0 & 2 \end{pmatrix}$