Question 9.3.4: The matrix A = [ 4 −1 1 −1 3 −2 1 −2 3 ] has the dominant ei...

The procedure for obtaining a second eigenvalue \lambda_{2} proceeds as follows:

\begin{aligned}x &=\frac{1}{6}\left[\begin{array}{r}4 \\-1 \\1\end{array}\right]=\left(\frac{2}{3},-\frac{1}{6}, \frac{1}{6}\right)^{t}, \\v ^{(1)} x ^{t} &=\left[\begin{array}{r}1 \\-1 \\1\end{array}\right]\left[\begin{array}{lll}\frac{2}{3}, & -\frac{1}{6}, & \frac{1}{6}\end{array}\right]=\left[\begin{array}{rrr}\frac{2}{3} & -\frac{1}{6} & \frac{1}{6} \\-\frac{2}{3} & \frac{1}{6} & -\frac{1}{6} \\\frac{2}{3} & -\frac{1}{6} & \frac{1}{6}\end{array}\right]\end{aligned}

and

B=A-\lambda_{1} v ^{(1)} x ^{t}=\left[\begin{array}{rrr}4 & -1 & 1 \\-1 & 3 & -2 \\1 & -2 & 3\end{array}\right]-6\left[\begin{array}{rrr}\frac{2}{3} & -\frac{1}{6} & \frac{1}{6} \\-\frac{2}{3} & \frac{1}{6} & -\frac{1}{6} \\\frac{2}{3} & -\frac{1}{6} & \frac{1}{6}\end{array}\right]=\left[\begin{array}{rrr}0 & 0 & 0 \\3 & 2 & -1 \\-3 & -1 & 2\end{array}\right].

Deleting the first row and column gives

B^{\prime}=\left[\begin{array}{rr} 2 & -1 \\ -1 & 2 \end{array}\right] ,

which has eigenvalues \lambda_{2}=3 \text { and } \lambda_{3}=1 \text {. For } \lambda_{2}=3 \text {, the eigenvector } w ^{(2)^{\prime}} can be obtained by solving the linear system

\left(B^{\prime}-3 I\right) w ^{(2)^{\prime}}= 0 , \quad \text { resulting in } w ^{(2)^{\prime}}=(1,-1)^{t} .

Adding a zero for the first component gives w ^{(2)}=(0,1,-1)^{t} and, from Eq. (9.6), we have the eigenvector v ^{(2)} of A corresponding to x_{2}=3 :

v ^{(i)}=\left(\lambda_{i}-\lambda_{1}\right) w ^{(i)}+\lambda_{1}\left( x ^{t} w ^{(i)}\right) v ^{(1)}                                 (9.6)

\begin{aligned} v ^{(2)} &=\left(\lambda_{2}-\lambda_{1}\right) w ^{(2)}+\lambda_{1}\left( x ^{t} w ^{(2)}\right) v ^{(1)} \\ &=(3-6)(0,1,-1)^{t}+6\left[\left(\frac{2}{3},-\frac{1}{6}, \frac{1}{6}\right)(0,1,-1)^{t}\right](1,-1,1)^{t}=(-2,-1,1)^{t} . \end{aligned}