2.4.2 Arbitrary Eigenvalue/-vector: Inverse Iteration

Task: Find the eigenvalue $\lambda_{n}$ nearest to some given number $\lambda$.

Method: Start with arbitrary vector $\vec{x}_{0}$, and proceed:



After a few iterations the vector

$$\vec{x}_{k} \propto \sum_{i=1}^{N} c_{i} \, \left[ \lambda_{i} - \lambda \right]^{-k} \, \vec{a}_{i}$$

will approach

$$\vec{x}_{k} \rightarrow c_{n} \, \left[ \lambda_{n} - \lambda \right]^{-k} \, \vec{a}_{n}$$

To evaluate $\lambda_{n}$, use





EXAMPLE: Use $\mbox{${\bf A}$}$ as before, but search in the vicinity of $\lambda=1$:

$$\left[ \mbox{${\bf A}$} - \lambda \, \mbox{${\bf I}$} \right... ...gin{array}{cc}3&-1\\ \vspace{-9pt}\\ -2&2\end{array} \right)$}$$

Starting out from

$$\vec{x}_{0}=\mbox{$\left( \begin{array}{r} 1 \\ \vspace{-9 pt}\\ 0 \end{array} \right)$}$$

we get

$$\vec{x}_{1}=\mbox{$\left( \begin{array}{r} 0.832 \\ \vspace{... ... 0.709 \\ \vspace{-9 pt}\\ -0.705 \end{array} \right)$}\,\dots$$

From

$$\vec{x}_{5}\,'=\mbox{$\left( \begin{array}{r} 0.708 \\ \vspace{-9 pt}\\ -0.707 \end{array} \right)$}$$

we have $\vec{x}_{5}\,' \cdot \vec{x}_{4}=1.0015$, yielding $\lambda_{n}=2.001$.

The exact eigenvalues are $5$ and $2$; the eigenvector corresponding to $\lambda=2$ is

$$\vec{a}=\mbox{$\left( \begin{array}{r} 0.707 \\ \vspace{-9 pt}\\ -0.707 \end{array} \right)$}$$