Answer to Question #261367 in Linear Algebra for kofi

Question

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

Answer 1

1.

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

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

slope = $\frac{f(\lambda_1)-f(\lambda_0)}{\lambda_1-\lambda_0}=\frac{2+25}{4-5}=-27$

the largest eigenvalue:

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

2.

$d_1=\begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}$ , $d_2=\begin{pmatrix} 1 \\ 1\\ 1 \end{pmatrix}$ , $d_3=\begin{pmatrix} 2 \\ 1 \\ 3 \end{pmatrix}$

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

$q_1=d_1/||d_1||$

$q_1^T=\begin{pmatrix} 0.408 & 0.816&0.408 \\ \end{pmatrix}$

$d'_2=d_2-(q_1^Td_2)q_1$

$d'_2=\begin{pmatrix} 1 \\ 1\\ 1 \end{pmatrix}-\begin{pmatrix} 0.408 & 0.816&0.408 \\ \end{pmatrix}\begin{pmatrix} 1 \\ 1\\ 1 \end{pmatrix}\begin{pmatrix} 0.408 \\ 0.816\\ 0.408 \end{pmatrix}=\begin{pmatrix} 0.834 \\ 0.334\\ 0.834 \end{pmatrix}$

$||d'_2||=1.226$

$q_2^T=d'_2/||d'_2||=\begin{pmatrix} 0.680 & 0.272&0.680 \\ \end{pmatrix}$

$d'_3=d_3-(q_1^Td_3)q_1-(q_2^Td_3)q_2=\begin{pmatrix} 2 \\ 1\\ 3 \end{pmatrix}-\begin{pmatrix} 0.408 & 0.816&0.408 \\ \end{pmatrix}\begin{pmatrix} 2 \\ 1\\ 3 \end{pmatrix}\begin{pmatrix} 0.408 \\ 0.816\\ 0.408 \end{pmatrix}--\begin{pmatrix} 0.680 & 0.272&0.680 \\ \end{pmatrix}\begin{pmatrix} 2 \\ 1\\ 3 \end{pmatrix}\begin{pmatrix} 0.680 \\ 0.272\\ 0.680 \end{pmatrix}=\begin{pmatrix} 2- 0.333-0.924\\ 1-0.666-0.074\\ 3-0.499-1.387 \end{pmatrix}=\begin{pmatrix} 0.743\\ 0.260\\ 1.114 \end{pmatrix}$

$||d'_3||=1.364$

$q_3=d'_3/||d'_3||$

$q_3^T=\begin{pmatrix} 0.545 & 0.191&0.817 \\ \end{pmatrix}$

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

for $R^{(0)}$ :

$r_{11}=||d'_1||=2.449,r_{22}=||d'_2||=1.226,r_{33}=||d'_3||=1.364$

$r_{12}=q_1^Td_2=1.632,r_{13}=q_1^Td_3=2.856,r_{23}=q_2^Td_3=3.672$

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

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

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

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

So,

$\lambda_1=3.496,\lambda_2=2.830,\lambda_3=1.114$