高専の数学解説

第17章 17.11

(1)

A=(101011110)A=\begin{pmatrix}1 & 0 & 1\\0 & 1 & 1\\1 & 1 & 0\end{pmatrix} とおくと

∣A−λE∣=∣1−λ0101−λ111−λ∣=−λ(1−λ)2−(1−λ)−(1−λ)=λ3+2λ2+λ−2|A-\lambda E|=\begin{vmatrix}1-\lambda & 0 & 1\\0 & 1-\lambda & 1\\1 & 1 & -\lambda\end{vmatrix}=-\lambda(1-\lambda)^2-(1-\lambda)-(1-\lambda)=\lambda^3+2\lambda^2+\lambda-2

=(λ−1)(λ+1)(λ−2)=0=(\lambda-1)(\lambda+1)(\lambda-2)=0 λ=1,2,−1\lambda=1,2,-1

λ=1\lambda=1 のとき

(00100111−1)→(11−1001001)→(11−1001000)→(110001000)\begin{pmatrix}0 & 0 & 1\\0 & 0 & 1\\1 & 1 & -1\end{pmatrix}\to\begin{pmatrix}1 & 1 & -1\\0 & 0 & 1\\0 & 0 & 1\end{pmatrix}\to\begin{pmatrix}1 & 1 & -1\\0 & 0 & 1\\0 & 0 & 0\end{pmatrix}\to\begin{pmatrix}1 & 1 & 0\\0 & 0 & 1\\0 & 0 & 0\end{pmatrix}

p1=C1(−110)→2C1(−12120)→C1′(−12120)p_1=C_1\begin{pmatrix}-1\\1\\0\end{pmatrix}\to\sqrt{2}C_1\begin{pmatrix}-\frac{1}{\sqrt{2}}\\\frac{1}{\sqrt{2}}\\0\end{pmatrix}\to C_1'\begin{pmatrix}-\frac{1}{\sqrt{2}}\\\frac{1}{\sqrt{2}}\\0\end{pmatrix}

λ=2\lambda=2 のとき

(−1010−1111−2)→(10−10−1111−2)→(10−10−1101−1)→(10−101−1000)\begin{pmatrix}-1 & 0 & 1\\0 & -1 & 1\\1 & 1 & -2\end{pmatrix}\to\begin{pmatrix}1 & 0 & -1\\0 & -1 & 1\\1 & 1 & -2\end{pmatrix}\to\begin{pmatrix}1 & 0 & -1\\0 & -1 & 1\\0 & 1 & -1\end{pmatrix}\to\begin{pmatrix}1 & 0 & -1\\0 & 1 & -1\\0 & 0 & 0\end{pmatrix}

p2=C2(111)→C2(131313)p_2=C_2\begin{pmatrix}1\\1\\1\end{pmatrix}\to C_2\begin{pmatrix}\frac{1}{\sqrt{3}}\\\frac{1}{\sqrt{3}}\\\frac{1}{\sqrt{3}}\end{pmatrix}

λ=−1\lambda=-1 のとき

(201021111)→(111021201)→(1110210−2−1)→(111021000)→(1110112000)→(10120112000)\begin{pmatrix}2 & 0 & 1\\0 & 2 & 1\\1 & 1 & 1\end{pmatrix}\to\begin{pmatrix}1 & 1 & 1\\0 & 2 & 1\\2 & 0 & 1\end{pmatrix}\to\begin{pmatrix}1 & 1 & 1\\0 & 2 & 1\\0 & -2 & -1\end{pmatrix}\to\begin{pmatrix}1 & 1 & 1\\0 & 2 & 1\\0 & 0 & 0\end{pmatrix}\to\begin{pmatrix}1 & 1 & 1\\0 & 1 & \frac{1}{2}\\0 & 0 & 0\end{pmatrix}\to\begin{pmatrix}1 & 0 & \frac{1}{2}\\0 & 1 & \frac{1}{2}\\0 & 0 & 0\end{pmatrix}

p3=C3(−1/2−1/21)→C3′(−16−1626)p_3 = C_3 \begin{pmatrix} -1/2 \\ -1/2 \\ 1 \end{pmatrix} \to C_3' \begin{pmatrix} -\frac{1}{\sqrt{6}} \\ -\frac{1}{\sqrt{6}} \\ \frac{2}{\sqrt{6}} \end{pmatrix}

これから直行行列Pは (1213−16−1213−1601326)\begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{6}} \\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{6}} \\ 0 & \frac{1}{\sqrt{3}} & \frac{2}{\sqrt{6}} \end{pmatrix} 対角化 P−1AP=(10002000−1)P^{-1}AP = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & -1 \end{pmatrix}

(2)

A=(311120102)A = \begin{pmatrix} 3 & 1 & 1 \\ 1 & 2 & 0 \\ 1 & 0 & 2 \end{pmatrix} とすると

∣A−λE∣=∣3−λ1112−λ0102−λ∣=(3−λ)(2−λ)2−(2−λ)−(2−λ)=(2−λ)(6−5λ+λ2−2)|A - \lambda E| = \begin{vmatrix} 3-\lambda & 1 & 1 \\ 1 & 2-\lambda & 0 \\ 1 & 0 & 2-\lambda \end{vmatrix} = (3-\lambda)(2-\lambda)^2 - (2-\lambda) - (2-\lambda) = (2-\lambda)(6 - 5\lambda + \lambda^2 - 2)

=(2−λ)(λ2−5λ+4)=(2−λ)(λ−4)(λ−1)= (2-\lambda)(\lambda^2 - 5\lambda + 4) = (2-\lambda)(\lambda-4)(\lambda-1) λ=1,2,4\lambda = 1,2,4

λ = 1 のとき

(211110101)→(110211101)→(1100−110−11)→(11001−1000)→(10101−1000)\begin{pmatrix} 2 & 1 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{pmatrix} \to \begin{pmatrix} 1 & 1 & 0 \\ 2 & 1 & 1 \\ 1 & 0 & 1 \end{pmatrix} \to \begin{pmatrix} 1 & 1 & 0 \\ 0 & -1 & 1 \\ 0 & -1 & 1 \end{pmatrix} \to \begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{pmatrix} \to \begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{pmatrix}

p1=C1(−111)→C1′(−131313)p_1 = C_1 \begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix} \to C_1' \begin{pmatrix} -\frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{3}} \end{pmatrix}

λ=2\lambda = 2 のとき

(111100100)→(111100000)→(100111000)→(100011000)\begin{pmatrix} 1 & 1 & 1 \\ 1 & 0 & 0 \\ 1 & 0 & 0 \end{pmatrix} \to \begin{pmatrix} 1 & 1 & 1 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} \to \begin{pmatrix} 1 & 0 & 0 \\ 1 & 1 & 1 \\ 0 & 0 & 0 \end{pmatrix} \to \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{pmatrix}

p2=C2(0−11)→C2′(012−12)p_2 = C_2 \begin{pmatrix} 0 \\ -1 \\ 1 \end{pmatrix} \to C_2' \begin{pmatrix} 0 \\ \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} \end{pmatrix}

λ=4\lambda = 4 のとき

(−1111−2010−2)→(1−1−11−2010−2)→(1−1−10−1101−1)→(1−1−101−1000)→(10−201−1000)\begin{pmatrix} -1 & 1 & 1 \\ 1 & -2 & 0 \\ 1 & 0 & -2 \end{pmatrix} \to \begin{pmatrix} 1 & -1 & -1 \\ 1 & -2 & 0 \\ 1 & 0 & -2 \end{pmatrix} \to \begin{pmatrix} 1 & -1 & -1 \\ 0 & -1 & 1 \\ 0 & 1 & -1 \end{pmatrix} \to \begin{pmatrix} 1 & -1 & -1 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{pmatrix} \to \begin{pmatrix} 1 & 0 & -2 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{pmatrix}

p3=C3(211)→C3′(261616)p_3 = C_3 \begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix} \to C_3' \begin{pmatrix} \frac{2}{\sqrt{6}} \\ \frac{1}{\sqrt{6}} \\ \frac{1}{\sqrt{6}} \end{pmatrix}

これから直行行列 P は

P=(13026131216−13−1216)P = \begin{pmatrix} \frac{1}{\sqrt{3}} & 0 & \frac{2}{\sqrt{6}} \\ \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} \\ -\frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} \end{pmatrix}

対角化 P−1AP=(100020004)P^{-1}AP = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 4 \end{pmatrix}

(3)

A=(1−2−2−21−2−2−21)とするもとA = \begin{pmatrix} 1 & -2 & -2 \\ -2 & 1 & -2 \\ -2 & -2 & 1 \end{pmatrix} \text{とするもと}

∣A−λE∣=∣1−λ−2−2−21−λ−2−2−21−λ∣=(1−λ)3−8−8−12(1−λ)=−λ3+3λ2−3λ+1−16−12+12λ|A - \lambda E| = \begin{vmatrix} 1-\lambda & -2 & -2 \\ -2 & 1-\lambda & -2 \\ -2 & -2 & 1-\lambda \end{vmatrix} = (1-\lambda)^3 - 8 - 8 - 12(1-\lambda) = -\lambda^3 + 3\lambda^2 - 3\lambda + 1 - 16 - 12 + 12\lambda

=−λ3+3λ2+9λ−27=−(λ+3)(λ−3)2= -\lambda^3 + 3\lambda^2 + 9\lambda - 27 = -(\lambda + 3)(\lambda - 3)^2

λ=−3,3 （３は重解）\lambda = -3, 3 \text{ （３\text{は重解}）}

λ=−3\lambda = -3 のとき

(4−2−2−24−2−2−24)→(2−1−1−24−2−2−24)→(2−1−103−30−33)→(2−1−101−1000)→(20−201−1000)→(10−101−1000)\begin{pmatrix} 4 & -2 & -2 \\ -2 & 4 & -2 \\ -2 & -2 & 4 \end{pmatrix} \to \begin{pmatrix} 2 & -1 & -1 \\ -2 & 4 & -2 \\ -2 & -2 & 4 \end{pmatrix} \to \begin{pmatrix} 2 & -1 & -1 \\ 0 & 3 & -3 \\ 0 & -3 & 3 \end{pmatrix} \to \begin{pmatrix} 2 & -1 & -1 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{pmatrix} \to \begin{pmatrix} 2 & 0 & -2 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{pmatrix} \to \begin{pmatrix} 1 & 0 & -1 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{pmatrix}

p1=C(111)→C1′(131313)p_1 = C\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} \to C_1'\begin{pmatrix} \frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{3}} \end{pmatrix}

λ=3\lambda = 3 のとき

(−2−2−2−2−2−2−2−2−2)→(111000000)\begin{pmatrix} -2 & -2 & -2 \\ -2 & -2 & -2 \\ -2 & -2 & -2 \end{pmatrix} \to \begin{pmatrix} 1 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}

p2=C2(−110)+C3(−101)→C2′(−12120)+C3′(−12012)p_2 = C_2\begin{pmatrix} -1 \\ 1 \\ 0 \end{pmatrix} + C_3\begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix} \to C_2'\begin{pmatrix} -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \\ 0 \end{pmatrix} + C_3'\begin{pmatrix} -\frac{1}{\sqrt{2}} \\ 0 \\ \frac{1}{\sqrt{2}} \end{pmatrix}

これから直行行列 P は

P=(13−12−121312013012)P = \begin{pmatrix} \frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{2}} & 0 \\ \frac{1}{\sqrt{3}} & 0 & \frac{1}{\sqrt{2}} \end{pmatrix}

対角化 P−1AP=P^{-1}AP =

(−300030003)\begin{pmatrix} -3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{pmatrix}

解説: r31bn1z