Question

Starting with x^(0)=[1 1 1]^T, find the dominant eigenvalue and corresponding eigenvector for the matrix A= [4 -1 1, 4 -8 1, -2 1 5] using the power mwthod

Answer 1

Question #75212 - Math - Quantitative Methods

Starting with $x^{\wedge}(0) = [1 1 1]^{\wedge}T$ , find the dominant eigenvalue and corresponding eigenvector for the matrix $A = [4 - 1 1, 4 - 8 1, -2 1 5]$ using the power method

Solution: the power method is described by the recurrence relation:

$$x _ {k + 1} = \frac {A x _ {k}}{\| A x _ {k} \|} = \frac {A ^ {k + 1} x _ {0}}{\| A ^ {k + 1} x _ {0} \|}$$

if $A = \left[ \begin{array}{rrr}4 & -1 & 1\\ 4 & -8 & 1\\ -2 & 1 & 5 \end{array} \right]$ and $x_0 = \left[ \begin{array}{l}1\\ 1\\ 1 \end{array} \right]$ then $Ax_0 = \left[ \begin{array}{rrr}4 & -1 & 1\\ 4 & -8 & 1\\ -2 & 1 & 5 \end{array} \right]\left[ \begin{array}{l}1\\ 1\\ 1 \end{array} \right] = \left[ \begin{array}{l}4\\ -3\\ 4 \end{array} \right]$ and $\| Ax_0\| = 4$

$$x _ {1} = \frac {1}{4} \cdot \left[ \begin{array}{c} 4 \\ - 3 \\ 4 \end{array} \right] = \left[ \begin{array}{c} 1 \\ - 0. 7 5 \\ 1 \end{array} \right]$$

at each iteration the vector $x$ is multiplied by the matrix $A$ and normalized. The subsequence $x_k$ ( $k \to \infty$ ) converges to an eigenvector associated with the dominant eigenvalue.

Using the Rayleigh quotient, we can approximate the dominant eigenvalue of A:

$$\lambda = \frac {A x \cdot x}{x \cdot x}$$

Let $\lambda_{i}$ be the approximate dominant eigenvalue at the i-th iteration. After the first iteration:

$$\lambda_ {1} = \left\{A x _ {1} = \left[ \begin{array}{c c c} 4 & - 1 & 1 \\ 4 & - 8 & 1 \\ - 2 & 1 & 5 \end{array} \right] \left[ \begin{array}{c} 1 \\ - 0. 7 5 \\ 1 \end{array} \right] = \left[ \begin{array}{c} 5. 7 5 \\ 1 1 \\ 2. 2 5 \end{array} \right] \text {a n d} x _ {1} = \left[ \begin{array}{c} 1 \\ - 0. 7 5 \\ 1 \end{array} \right] \right\} = - 0. 0 9 7 5 6 1$$

Continuing this process, we obtain the sequence of approximations shown in the table:

15 iterations are required to obtain successive approximations that converge when rounded to three significant digits.

Answer: eigenvector $\cong$ [0.09, 1, -0.06], eigenvalue $\cong$ 7.70

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