2.4.1 Largest Eigenvalue and Related Eigenvector

Regard the $N$ eigenvectors $\vec{a}_{i}$ of $\mbox{${\bf A}$}$ as the base vectors in $N$-space. Then any $N$-vector may be written

$$\vec{x}_{0}= \sum_{i=1}^{N} c_{i} \, \vec{a}_{i}$$

with appropriate coefficients $c_{i}$. Let $\vec{a}_{m}$ be the eigenvector corresponding to the largest (by absolute value) eigenvalue $\lambda_{m}$. Multiply $\vec{x}_{0}$ several times by $\mbox{${\bf A}$}$, each time normalizing the result:



After a few iterations

$$\vec{x}_{k} \propto \sum_{i=1}^{N} c_{i} \, \lambda_{i}^{k} \, \vec{a}_{i} \approx c_{m} \, \lambda_{m}^{k} \, \vec{a}_{m}$$ (2.5)

To obtain the eigenvalue $\lambda_{m}$, use



where ${x_{\beta}'}^{(k)}$ is any cartesian component of the unnormalized vector $\vec{x}_{k}\,'$.

EXAMPLE: Let

$$\mbox{${\bf A}$}=\mbox{$\left( \begin{array}{cc}3&1\\ \vspace{-9pt}\\ 2&4\end{array} \right)$}$$

and use

$$\vec{x}_{0}=\mbox{$\left( \begin{array}{r} \sqrt{2}/2 \\ \vspace{-9 pt}\\ \sqrt{2}/2 \end{array} \right)$}$$

choose as the starting vector. The iterated and normalized vectors are

$$\vec{x}_{1}=\mbox{$\left( \begin{array}{r} 0.555 \\ \vspace{... ....464 \\ \vspace{-9 pt}\\ 0.886 \end{array} \right)$}\; ; \dots$$

From $\vec{x}_{3}$ and the unnormalized

$$\vec{x}_{4}\,'=\mbox{$\left( \begin{array}{r} 2.279 \\ \vspace{-9 pt}\\ 4.471 \end{array} \right)$}$$

we find $\lambda_{m}=4.907$. The exact solution is

$$\vec{a}_{m}=\mbox{$\left( \begin{array}{r} 0.45 \\ \vspace{-... ...90 \end{array} \right)$} \;\; {\rm and} \;\; \lambda_{m}=5\, .$$