Question 9.3.3: Apply the Inverse Power method with x^(0) = (1, 1, 1)^t to t...
Apply the Inverse Power method with x ^{(0)}=(1,1,1)^{t} to the matrix
A=\left[\begin{array}{rrr} -4 & 14 & 0 \\ -5 & 13 & 0 \\ -1 & 0 & 2 \end{array}\right] \quad \text { with } \quad q=\frac{ x ^{(0) t} A x ^{(0)}}{ x ^{(0) t} x ^{(0)}}=\frac{19}{3} ,
and use Aitken’s Δ² method to accelerate the convergence.
The Power method was applied to this matrix in Example 1 using the initial vector
x ^{(0)}=(1,1,1)^{t} . It gave the approximate eigenvalue \mu^{(12)}=6.000837 and eigenvector \left( x ^{(12)}\right)^{t}=(1,0.714316,-0.249895)^{t} .
For the Inverse Power method we consider
A-q I=\left[\begin{array}{rrr} -\frac{31}{3} & 14 & 0 \\ -5 & \frac{20}{3} & 0 \\ -1 & 0 & -\frac{13}{3} \end{array}\right]
With x ^{(0)}=(1,1,1)^{t} , the method first finds y ^{(1)} \text { by solving }(A-q I) y ^{(1)}= x ^{(0)} . This gives
y ^{(1)}=\left(-\frac{33}{5},-\frac{24}{5}, \frac{84}{65}\right)^{t}=(-6.6,-4.8,1.292 \overline{307692})^{t}
So
\left\| y ^{(1)}\right\|_{\infty}=6.6, \quad x ^{(1)}=\frac{1}{-6.6} y ^{(1)}=(1,0.7272727,-0.1958042)^{t},
and
\mu^{(1)}=-\frac{1}{6.6}+\frac{19}{3}=6.1818182
Subsequent results are listed in Table 9.4, and the right column lists the results of Aitken’s Δ² method applied to the \mu^{(m)} . These are clearly superior results to those obtained with the Power method.
Table 9.4
\begin{array}{llll}\hline m & { x ^{(m) t}} & \mu^{(m)} & \hat{\mu}^{(m)} \\\hline 0 & (1,1,1) & & \\1 & (1,0.7272727,-0.1958042) & 6.1818182 & 6.000098 \\2 & (1,0.7155172,-0.2450520) & 6.0172414 & 6.000001 \\3 & (1,0.7144082,-0.2495224) & 6.0017153 & 6.000000 \\4 & (1,0.7142980,-0.2499534) & 6.0001714 & 6.000000 \\5 & (1,0.7142869,-0.2499954) & 6.0000171 & \\6 & (1,0.7142858,-0.2499996) & 6.0000017 & \\\hline\end{array}