Question

Perform three iterations of the inverse power method to obtain the smallest eigenvalue in magnitude of the matrix [{2,5},{3,6}].
Take appropriate initial approximation to the eigenvector.

Answer 1

Answer on Question #85783 – Math – Quantitative Methods

Question

Perform three iterations of the inverse power method to obtain the smallest eigenvalue in magnitude of the matrix [\{2,5\}, \{3,6\}].

Take appropriate initial approximation to the eigenvector.

Solution

$$A = \left( \begin{array}{cc} 2 & 5 \\ 3 & 6 \end{array} \right)$$

a non-zero vector $x$ is an eigenvector of $A$ if there exists a scalar $\lambda$ such that

$$A x = \lambda x \leftrightarrow x = A^{-1}(\lambda x) \leftrightarrow A^{-1} x = \frac{1}{\lambda} x$$

the smallest eigenvalue of $A$ in magnitude is the largest eigenvalue of $A^{-1}$ and we can use power method on the $A^{-1}$ to compute its dominant eigenvalue:

$$x^{k+1} = \frac{A^{-1} x^k}{\|A^{-1} x^k\|_{\infty}}, \qquad x^k - k^{th} \text{ approximation to the dominant eigenvector}$$

to speed up the convergence of the method, we choose the eigenvector as the initial approximation:

$$x^0 = \left( \begin{array}{c} -2.12 \\ 1 \end{array} \right), \qquad A^{-1} = \left( \begin{array}{cc} -2 & 5/3 \\ 1 & -2/3 \end{array} \right)$$ $$A^{-1} x^0 = \left( \begin{array}{cc} -2 & 5/3 \\ 1 & -2/3 \end{array} \right) \left( \begin{array}{c} -2.12 \\ 1 \end{array} \right) = \left( \begin{array}{c} 5.907 \\ -2.787 \end{array} \right), \qquad x^1 = \left( \begin{array}{c} 1 \\ -0.472 \end{array} \right), \qquad |\lambda^1| = 0.169$$ $$A^{-1} x^1 = \left( \begin{array}{cc} -2 & 5/3 \\ 1 & -2/3 \end{array} \right) \left( \begin{array}{c} 1 \\ -0.472 \end{array} \right) = \left( \begin{array}{c} -2.786 \\ 1.315 \end{array} \right), \qquad x^2 = \left( \begin{array}{c} -1 \\ 0.472 \end{array} \right), \qquad |\lambda^2| = 0.359$$ $$A^{-1} x^2 = \left( \begin{array}{cc} -2 & 5/3 \\ 1 & -2/3 \end{array} \right) \left( \begin{array}{c} -1 \\ 0.472 \end{array} \right) = \left( \begin{array}{c} 2.786 \\ -1.314 \end{array} \right), \qquad x^3 = \left( \begin{array}{c} 1 \\ -0.472 \end{array} \right), \qquad |\lambda^3| = 0.359$$

the smallest eigenvalue of $A$ in magnitude is 0.359.

Answer provided by https://www.AssignmentExpert.com