Question #260988

Consider the linear eigenproblem, 𝐴𝑥=𝜆𝑥, for the matrix

D=[1 1 2

2 1 1

1 1 3 ]


1. Solve for the largest eigenvalue by the direct method using the secant method. Let 𝜆^((0))=5 and 𝜆^((1))=4

2.Solve for the eigenvalues by the QR method


1
Expert's answer
2021-11-09T00:12:51-0500

1.

f(λ0)=f(5)=151121511135=47+52=25f(\lambda_0)=f(5)=\begin{vmatrix} 1-5 & 1&1 \\ 2 & 1-5&1\\ 1&1&3-5 \end{vmatrix}=-4\cdot7+5-2=-25


f(λ1)=f(4)=141121411134=32+3+5=2f(\lambda_1)=f(4)=\begin{vmatrix} 1-4 & 1&1 \\ 2 & 1-4&1\\ 1&1&3-4 \end{vmatrix}=-3\cdot2+3+5=2


slope = f(λ1)f(λ0)λ1λ0=2+2545=27\frac{f(\lambda_1)-f(\lambda_0)}{\lambda_1-\lambda_0}=\frac{2+25}{4-5}=-27


the largest eigenvalue:


λ2=λ1f(λ1)slope=4+2/27=4.074\lambda_2=\lambda_1-\frac{f(\lambda_1)}{slope}=4+2/27=4.074


2.

d1=(121)d_1=\begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix} , d2=(111)d_2=\begin{pmatrix} 1 \\ 1\\ 1 \end{pmatrix} , d3=(213)d_3=\begin{pmatrix} 2 \\ 1 \\ 3 \end{pmatrix}


d1=1+22+1=2.449||d_1||=\sqrt{1+2^2+1}=2.449

q1=d1/d1q_1=d_1/||d_1||

q1T=(0.4080.8160.408)q_1^T=\begin{pmatrix} 0.408 & 0.816&0.408 \\ \end{pmatrix}


d2=d2(q1Td2)q1d'_2=d_2-(q_1^Td_2)q_1


d2=(111)(0.4080.8160.408)(111)(0.4080.8160.408)=(0.8340.3340.834)d'_2=\begin{pmatrix} 1 \\ 1\\ 1 \end{pmatrix}-\begin{pmatrix} 0.408 & 0.816&0.408 \\ \end{pmatrix}\begin{pmatrix} 1 \\ 1\\ 1 \end{pmatrix}\begin{pmatrix} 0.408 \\ 0.816\\ 0.408 \end{pmatrix}=\begin{pmatrix} 0.834 \\ 0.334\\ 0.834 \end{pmatrix}


d2=1.226||d'_2||=1.226


q2T=d2/d2=(0.6800.2720.680)q_2^T=d'_2/||d'_2||=\begin{pmatrix} 0.680 & 0.272&0.680 \\ \end{pmatrix}


d3=d3(q1Td3)q1(q2Td3)q2=(213)(0.4080.8160.408)(213)(0.4080.8160.408)(0.6800.2720.680)(213)(0.6800.2720.680)=(20.3330.92410.6660.07430.4991.387)=(0.7430.2601.114)d'_3=d_3-(q_1^Td_3)q_1-(q_2^Td_3)q_2=\begin{pmatrix} 2 \\ 1\\ 3 \end{pmatrix}-\begin{pmatrix} 0.408 & 0.816&0.408 \\ \end{pmatrix}\begin{pmatrix} 2 \\ 1\\ 3 \end{pmatrix}\begin{pmatrix} 0.408 \\ 0.816\\ 0.408 \end{pmatrix}--\begin{pmatrix} 0.680 & 0.272&0.680 \\ \end{pmatrix}\begin{pmatrix} 2 \\ 1\\ 3 \end{pmatrix}\begin{pmatrix} 0.680 \\ 0.272\\ 0.680 \end{pmatrix}=\begin{pmatrix} 2- 0.333-0.924\\ 1-0.666-0.074\\ 3-0.499-1.387 \end{pmatrix}=\begin{pmatrix} 0.743\\ 0.260\\ 1.114 \end{pmatrix}


d3=1.364||d'_3||=1.364

q3=d3/d3q_3=d'_3/||d'_3||

q3T=(0.5450.1910.817)q_3^T=\begin{pmatrix} 0.545 & 0.191&0.817 \\ \end{pmatrix}


Q(0)=(q1q2q3)=(0.4080.6800.5450.8160.2720.1910.4080.6800.817)Q^{(0)}=\begin{pmatrix} q_1 & q_2&q_3 \\ \end{pmatrix}=\begin{pmatrix} 0.408 & 0.680&0.545 \\ 0.816 & 0.272&0.191\\ 0.408&0.680&0.817 \end{pmatrix}


for R(0)R^{(0)} :

r11=d1=2.449,r22=d2=1.226,r33=d3=1.364r_{11}=||d'_1||=2.449,r_{22}=||d'_2||=1.226,r_{33}=||d'_3||=1.364

r12=q1Td2=1.632,r13=q1Td3=2.856,r23=q2Td3=3.672r_{12}=q_1^Td_2=1.632,r_{13}=q_1^Td_3=2.856,r_{23}=q_2^Td_3=3.672


R(0)=(2.4491.6322.85601.2263.672001.364)R^{(0)}=\begin{pmatrix} 2.449 & 1.632&2.856 \\ 0 & 1.226&3.672\\ 0&0&1.364 \end{pmatrix}


D(1)=R(0)Q(0)=(2.4491.6322.85601.2263.672001.364)(0.4080.6800.5450.8160.2720.1910.4080.6800.817)=D^{(1)}=R^{(0)}Q^{(0)}=\begin{pmatrix} 2.449 & 1.632&2.856 \\ 0 & 1.226&3.672\\ 0&0&1.364 \end{pmatrix}\begin{pmatrix} 0.408 & 0.680&0.545 \\ 0.816 & 0.272&0.191\\ 0.408&0.680&0.817 \end{pmatrix}=


=(3.4964.0514.0002.4992.8303.2340.5570.9281.114)=\begin{pmatrix} 3.496 & 4.051&4.000 \\ 2.499 & 2.830&3.234\\ 0.557&0.928&1.114 \end{pmatrix}


The diagonal elements in matrix D(1)D^{(1)} are the first approximation to the eigenvalues of matrix DD.

So,

λ1=3.496,λ2=2.830,λ3=1.114\lambda_1=3.496,\lambda_2=2.830,\lambda_3=1.114


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