Answer to Question #110538 in Quantitative Methods for Anju Jayachandran

Question #110538
Compute the largest eigenvalue in magnitude, and the corresponding eigenvector of
the matrix
A=1 −1 1
2 0 3
1 4 −1
by performing four iterations of the power method
1
Expert's answer
2020-04-22T19:30:49-0400

A=[111203141]A=\begin{bmatrix} 1&-1&1 \\ 2&0&3\\ 1&4&-1 \end{bmatrix}

Begin with an initial nonzero approximation of

x0=[111]x_0=\begin{bmatrix} 1 \\ 1\\ 1 \end{bmatrix}

Then obtain the following approximations

Iteration

x1=Ax0=[111203141][111]=[154]x_1=Ax_0=\begin{bmatrix} 1 & -1&1 \\ 2 & 0&3\\ 1&4&-1 \end{bmatrix}\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} =\begin{bmatrix} 1 \\ 5\\ 4 \end{bmatrix}

“Scaled” Approximation

1[154]\to1\begin{bmatrix} 1 \\ 5\\ 4 \end{bmatrix}

x2=Ax1=[111203141][154]=[01417]x_2=Ax_1=\begin{bmatrix} 1 & -1&1 \\ 2 & 0&3\\ 1&4&-1 \end{bmatrix}\begin{bmatrix} 1 \\ 5\\ 4 \end{bmatrix} =\begin{bmatrix} 0 \\ 14\\ 17 \end{bmatrix}

14[011.2]\to14\begin{bmatrix} 0 \\ 1\\ 1.2 \end{bmatrix}

x3=Ax2=[111203141][01417]=[35139]3[11713]x4=Ax3=[111203141][35139]=[9123168]9[113.718.7]x_3=Ax_2=\begin{bmatrix} 1 & -1&1 \\ 2 & 0&3\\ 1&4&-1 \end{bmatrix}\begin{bmatrix} 0 \\ 14\\ 17 \end{bmatrix} =\begin{bmatrix} 3 \\ 51\\ 39 \end{bmatrix}\\ \to3\begin{bmatrix} 1 \\ 17\\ 13 \end{bmatrix}\\ x_4=Ax_3=\begin{bmatrix} 1 & -1&1 \\ 2 & 0&3\\ 1&4&-1 \end{bmatrix}\begin{bmatrix} 3 \\ 51\\ 39 \end{bmatrix} =\begin{bmatrix} -9 \\ 123\\ 168 \end{bmatrix}\\ \to-9\begin{bmatrix} 1 \\ -13.7\\ -18.7 \end{bmatrix}

Note that the approximations appear to be approaching scalar multiples of

[11419]\begin{bmatrix} 1 \\ -14\\ -19 \end{bmatrix}

With x=(1,-13.7,-18.7) as the approximation of a dominant eigenvector of A , 

use the Rayleigh quotient to obtain an approximation of the dominant eigenvalue of A . First compute the product Ax

Ax=[111203141][113.718.7]=[454.135.1]Ax=\begin{bmatrix} 1 & -1&1 \\ 2 & 0&3\\ 1&4&-1 \end{bmatrix}\begin{bmatrix} 1 \\ -13.7\\ -18.7 \end{bmatrix} =\begin{bmatrix} -4 \\ -54.1\\ -35.1 \end{bmatrix}

Then, because

Axx=41+(54.1)(13.7)+(35.1)(18.7)=1393.54xx=11+(13.7)(13.7)+(18.7)(18.7)=538.38Ax\cdot x=-4\cdot 1+(-54.1)(-13.7)+(-35.1)(-18.7)=1393.54\\ x\cdot x=1\cdot1+(-13.7)(-13.7)+(-18.7)(-18.7)=538.38

you can compute the Rayleigh quotient to be

λ=Axxxx=1393.54538.38=2.6\lambda=\frac{Ax\cdot x}{x\cdot x}=\frac{1393.54}{538.38}=2.6

which is a good approximation of the dominant eigenvalue

λ=3\lambda=3

Answer:

[11419],λ=3\begin{bmatrix} 1 \\ -14\\ -19 \end{bmatrix}, \lambda=3


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