Question #314303

Find all Eigen values and corresponding Eigen vectors for the matrix A=

0 0 3

2 5 0

2 3 0



1
Expert's answer
2022-03-20T06:44:11-0400

A=(003250230)AλI=0λ0325λ0230λ==λ2(5λ)+0+186(5λ)00==λ3+5λ2+6λ12=0λ35λ26λ+12=012:±1,±2,±3,±4,±6,±12λ=±1:(±1)35(±1)26(±1)+120λ=±2:(±2)35(±2)26(±2)+120λ=±3:(±3)35(±3)26(±3)+120λ=±4:(±4)35(±4)26(±4)+120λ=±6:(±6)35(±6)26(±6)+120λ=±12:(±12)35(±12)26(±12)+120A=\begin{pmatrix} 0 & 0&3 \\ 2 & 5&0\\ 2&3&0 \end{pmatrix}\\ |A-\lambda I|=\begin{vmatrix} 0-\lambda & 0&3 \\ 2 & 5-\lambda&0\\ 2&3&0-\lambda \end{vmatrix}=\\ =\lambda^2(5-\lambda)+0+18-6(5-\lambda)-0-0=\\ =-\lambda^3+5\lambda^2+6\lambda-12=0\\ \lambda^3-5\lambda^2-6\lambda+12=0\\ 12:\pm1, \pm2,\pm3,\pm4,\pm6,\pm12\\ \lambda=\pm1: (\pm1)^3-5(\pm1)^2-6(\pm1)+12\neq0\\ \lambda=\pm2:(\pm2)^3-5(\pm2)^2-6(\pm2)+12\neq0\\ \lambda=\pm3: (\pm3)^3-5(\pm3)^2-6(\pm3)+12\neq0\\ \lambda=\pm4: (\pm4)^3-5(\pm4)^2-6(\pm4)+12\neq0\\ \lambda=\pm6: (\pm6)^3-5(\pm6)^2-6(\pm6)+12\neq0\\ \lambda=\pm12: (\pm12)^3-5(\pm12)^2-6(\pm12)+12\neq0





λ12λ21λ36\lambda_1\approx -2\\ \lambda_2\approx1\\ \lambda_3\approx6

All Eigen values

(AλI)x=0,x01.λ1=2x1=(x1,x2,x3)0(Aλ1I)x1=0(203270232)(x1x2x3)=(000)2x1+3x3=02x1+7x2=02x1+3x2+2x3=0x3=23x1x2=27x1x1=1,x2=27,x3=23x1=(1,27,23)(A-\lambda I)\vec{x}=\vec{0}, \vec{x}\neq\vec{0}\\ 1. \lambda_1=-2\\ \vec{x_1}=(x_1,x_2,x_3)\neq\vec{0}\\ (A-\lambda_1I)\vec{x_1}=\vec{0}\\ \begin{pmatrix} 2 &0&3 \\ 2 & 7&0\\ 2&3&2 \end{pmatrix}\begin{pmatrix} x_1 \\ x_2\\ x_3 \end{pmatrix}=\begin{pmatrix} 0\\0\\0 \end{pmatrix}\\ 2x_1+3x_3=0\\ 2x_1+7x_2=0\\ 2x_1+3x_2+2x_3=0\\ x_3=-\frac{2}{3}x_1\\ x_2=-\frac{2}{7}x_1\\ x_1=1, x_2=-\frac{2}{7}, x_3=-\frac{2}{3}\\ \vec{x_1}=(1,-\frac{2}{7}, -\frac{2}{3})

2.λ2=1x2=(x1,x2,x3)0(Aλ2I)x2=0(103240231)(x1x2x3)=(000)x1+3x3=02x1+4x2=02x1+3x2x3=0x3=13x1x2=12x1x1=1,x2=12,x3=13x2=(1,12,13)2. \lambda_2=1\\ \vec{x_2}=(x_1,x_2,x_3)\neq\vec{0}\\ (A-\lambda_2I)\vec{x_2}=\vec{0}\\ \begin{pmatrix} -1 &0&3 \\ 2 & 4&0\\ 2&3&-1 \end{pmatrix}\begin{pmatrix} x_1 \\ x_2\\ x_3 \end{pmatrix}=\begin{pmatrix} 0\\0\\0 \end{pmatrix}\\ -x_1+3x_3=0\\ 2x_1+4x_2=0\\ 2x_1+3x_2-x_3=0\\ x_3=\frac{1}{3}x_1\\ x_2=-\frac{1}{2}x_1\\ x_1=1, x_2=-\frac{1}{2}, x_3=\frac{1}{3}\\ \vec{x_2}=(1,-\frac{1}{2}, \frac{1}{3})

3.λ3=6x3=(x1,x2,x3)0(Aλ3I)x3=0(603210236)(x1x2x3)=(000)6x1+3x3=02x1x2=02x1+3x26x3=0x3=2x1x2=2x1x1=1,x2=2,x3=2x3=(1,2,2)3. \lambda_3=6\\ \vec{x_3}=(x_1,x_2,x_3)\neq\vec{0}\\ (A-\lambda_3I)\vec{x_3}=\vec{0}\\ \begin{pmatrix} -6 &0&3 \\ 2 & -1&0\\ 2&3&-6 \end{pmatrix}\begin{pmatrix} x_1 \\ x_2\\ x_3 \end{pmatrix}=\begin{pmatrix} 0\\0\\0 \end{pmatrix}\\ -6x_1+3x_3=0\\ 2x_1-x_2=0\\ 2x_1+3x_2-6x_3=0\\ x_3=2x_1\\ x_2=2x_1\\ x_1=1, x_2=2, x_3=2\\ \vec{x_3}=(1,2,2)

x1=(1,27,23),x2=(1,12,13),x3=(1,2,2)\vec{x_1}=(1,-\frac{2}{7}, -\frac{2}{3}),\vec{x_2}=(1,-\frac{1}{2}, \frac{1}{3}), \vec{x_3}=(1,2,2)

All corresponding Eigen vectors


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