Question #259092

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

D=[1 1 2

2 1 1

1 1 3 ]

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


1
Expert's answer
2021-11-09T01:05:09-0500

Part one


D=(112211113)D= \begin{pmatrix} 1&1 &2 \\ 2&1 & 1\\ 1&1&3 \end{pmatrix}


Characteristic equation detDλI=0det||D-\lambda\Iota||=0


Det(1λ1221λ1113λ)Det \begin{pmatrix} 1-\lambda&1&2 \\ 2&1-\lambda&1\\ 1&1&3-\lambda \end{pmatrix}


Characteristic polynomial is of the form;


λ3A1λ2+A2λA3=0\lambda^3-A_1\lambda^2+A_2\lambda-A_3=0

Where A1=A_1= sum of main diagonal element

=1+1+3=5=1+1+3=5


A2=A_2= sum of minors of the main diagonal element

A2=1113+1213+1121=2+11=2A_2= \begin{vmatrix} 1 & 1 \\ 1 & 3 \end{vmatrix}+\begin{vmatrix} 1 & 2 \\ 1 & 3 \end{vmatrix}+\begin{vmatrix} 1& 1 \\ 2 & 1 \end{vmatrix}=2+1-1=2


A3=det(D)=1111312113+22111A_3= det(D)=\quad\quad\quad\quad1\begin{vmatrix} 1 & 1\\ 1 & 3 \end{vmatrix}-1\begin{vmatrix} 2& 1\\ 1 & 3 \end{vmatrix}+2\begin{vmatrix} 2& 1\\ 1& 1 \end{vmatrix}


=1(2)1(5)+2(1)=1(2)-1(5)+2(1)

=1=-1


Characteristic polynomial is


λ35λ2+2λ+1=0\lambda^3-5\lambda^2+2\lambda+1=0



Let λ35λ2+2λ+1=0\lambda^3-5\lambda^2+2\lambda+1=0

f(λ)=λ35λ2+2λ+1f(\lambda)=\lambda^3-5\lambda^2+2\lambda+1



1st iteration


λ0=5\lambda_0=5 and λ1=4\lambda_1=4

f(λ0)=f(s)=11f(\lambda_0)=f(s)=11

f(λ1)=f(4)=7f(\lambda_1)=f(4)=-7


λ2=511\therefore\>\lambda_2=5-11 45711=4.3889\frac{4-5}{-7-11}=4.3889

f(λ2)=1.9937f(\lambda_2)=-1.9937


2nd iteration


λ1=4\lambda_1=4 and λ2=4.3889\lambda_2=4.3889

f(λ1)=f(4)=7f(\lambda_1)=f(4)=-7

f(λ2)=f(4.3889)=1.9937f(\lambda_2)=f(4.3889)= -1.9937


λ3=4(7)\lambda_3=4-(-7) 4.388941.9937(7)=4.5438\frac{4.3889-4}{-1.9937-(-7)}=4.5438


3rd iteration


λ2=4.3889\lambda_2=4.3889 and λ3=4.5438\lambda_3=4.5438

f(λ2)=1.9937f(\lambda_2)=-1.9937

f(λ3)=f(4.5438)=0.6688f(\lambda_3)=f(4.5438)=0.6688

λ4=4.3889(1.9937)\lambda_4=4.3889-(-1.9937)


4.54384.38890.66881.9937\frac{4.5438-4.3889}{0.6688-\>-1.9937}

=4.5049=4.5049

f(λ4)=f(4.5049)=0.0378f(\lambda_4)=f(4.5049)=-0.0378



4th iteration


λ3=4.5438\lambda_3=4.5438 and λ4=4.5049\lambda_4=4.5049


f(λ3)=0.6688f(\lambda_3)=0.6688

f(λ4)=0.0378f(\lambda_4)=-0.0378


λ5=4.54380.6688\lambda_5=4.5438-0.6688

4.50494.54380.03780.6688\frac{4.5049-4.5438}{-0.0378-0.6688}

=4.5070=4.5070


f(λ5)=0.0003f(\lambda_5)=-0.0003

Approximate root =4.507=4.507


Part two

QQ -RR method


Let a1=[121]a_1= \begin{bmatrix} 1&2&1 \end{bmatrix} a2=[111]a_2=\begin{bmatrix} 1&1& 1 \end{bmatrix}

and

a3=[213]a_3=\begin{bmatrix} 2&1 & 3 \end{bmatrix}


Let orthogonal set =(b1b2b3)=\begin{pmatrix} b_1& b_2&b_3 \end{pmatrix}


Let orthonormal set =(q1q2q3)=\begin{pmatrix} q_1&q_2& q_3 \end{pmatrix}


Let b1=[121],b_1= \begin{bmatrix} 1&2 & 1 \end{bmatrix},


b1=12+22+12=2.45||b_1||=\sqrt{1^2+2^2+1^2}=2.45


q1=12.45[121]=[0.41,0.82,0.41]q_1=\frac{1}{2.45}\begin{bmatrix} 1&2 & 1 \end{bmatrix}=\begin{bmatrix} 0.41,&0.82,& 0.41 \end{bmatrix}


b2=[111]b_2=\begin{bmatrix} 1&1& 1 \end{bmatrix}-

<(1,1,1),(0.41,0.82,0.41)>(0.41,0.82,0.41)<(1,1,1),(0.41,0.82,0.41)>(0.41,0.82,0.41)


=(0.33,0.33,0.33)=\begin{pmatrix} 0.33,&-0.33,&0.33 \end{pmatrix}


b2=0.332+0.332+0.332=||b_2||= \sqrt{ 0.33^2+0.33^2+0.33^2}=

0.580.58


q2=10.58[0.33,0.330.33]=q_2= \frac{1}{0.58}\begin{bmatrix} 0.33,&-0.33& 0.33 \end{bmatrix}=

(0.58,0.58,0.58)\begin{pmatrix} 0.58,&-0.58,&0.58 \end{pmatrix}


b3=(2,1,3)b_3=\begin{pmatrix} 2,&1,&3 \end{pmatrix} -

<(2,1,3),(0.41,0.82,0.41)>-<(2,1,3),(0.41,0.82,0.41)>

(0.41,0.82,0.41)(0.41,0.82,0.41)


<(2,1,3),(0.58,0.58,0.58)>- <(2,1,3),(0.58,-0.58,0.58)>


b3=(0.5,0,0.5)b3=b_3=(-0.5,0,0.5)\quad ||b_3||=


0.52+02+0.52=0.71\sqrt{0.5^2+0^2+0.5^2}=0.71


q3=10.71(0.5,0,0.5)=q_3=\frac{1}{0.71}(-0.5,0,0.5)=

(0.71,0,0.71)(-0.71,0,0.71)


Q=[q1q2q3]=\therefore Q= \begin{bmatrix} q_1&q_2& q_3 \end{bmatrix}=

[0.410.580.710.820.5800.410.580.71]\begin{bmatrix} 0.41&0.58 & -0.71\\ 0.82&-0.58 & 0\\ 0.41&0.58&0.71 \end{bmatrix}


R=QTA=(0.410.820.410.580.580.580.7100.71)(112211113)R=Q^TA=\begin{pmatrix} 0.41&0.82 & 0.41\\ 0.58&-0.58 & 0.58\\ -0.71&0&0.71 \end{pmatrix}\begin{pmatrix} 1&1 & 2\\ 2&1& 1\\ 1&1&3 \end{pmatrix}


=(2.451.632.8600.582.31000.71)=\begin{pmatrix} 2.45&1.63& 2.86\\ 0&0.58&2.31\\ 0&0&0.71 \end{pmatrix}


Denoting the above RR as R0R_0 and the QQ as Q0Q_0

Let A1=R0Q0A_1=R_0Q_0

Let A1=Q1R1A_1=Q_1R_1 be QRQR factorization of A1A_1 and similarly create A2=R1Q1A_2=R_1Q_1


This process is continued until AmA_m is created such that Am=QmRmA_m=Q_mR_m and Am1=RmQmA_{m1}=R_mQ_m are equal.

At convergence the diagonal entries below the main diagonal are sufficiently small.


The diagonal entries of AmA_m will be eigenvalues of matrix AA


The expected values will be approximate to 0.28514,0.77812and4.507010.28514,0.77812 \>and \> 4 .50701



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