Answer to Question #110430 in Quantitative Methods for Gayatri Yadav

Question

Answer to Question #110430 in Quantitative Methods for Gayatri Yadav

Question #110430

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.

Expert's answer

"A=\\begin{bmatrix}\n 1 & -1&1 \\\\\n 2 & 0&3\\\\\n1&4&-1\n\\end{bmatrix}"

Begin with an initial nonzero approximation of

"x_0=\\begin{bmatrix}\n 1 \\\\\n 1\\\\\n1\n\\end{bmatrix}"

Then obtain the following approximations

Iteration

"x_1=Ax_0=\\begin{bmatrix}\n 1 & -1&1 \\\\\n 2 & 0&3\\\\\n1&4&-1\n\\end{bmatrix}\\begin{bmatrix}\n 1 \\\\\n 1\\\\\n1\n\\end{bmatrix}\n=\\begin{bmatrix}\n 1 \\\\\n 5\\\\\n4\n\\end{bmatrix}"

“Scaled” Approximation

"\\to1\\begin{bmatrix}\n 1 \\\\\n 5\\\\\n4\n\\end{bmatrix}"

Iteration

"x_2=Ax_1=\\begin{bmatrix}\n 1 & -1&1 \\\\\n 2 & 0&3\\\\\n1&4&-1\n\\end{bmatrix}\\begin{bmatrix}\n 1 \\\\\n 5\\\\\n4\n\\end{bmatrix}\n=\\begin{bmatrix}\n 0 \\\\\n 14\\\\\n17\n\\end{bmatrix}"

“Scaled” Approximation

"\\to14\\begin{bmatrix}\n 0 \\\\\n 1\\\\\n1.2\n\\end{bmatrix}"

Iteration

"x_3=Ax_2=\\begin{bmatrix}\n 1 & -1&1 \\\\\n 2 & 0&3\\\\\n1&4&-1\n\\end{bmatrix}\\begin{bmatrix}\n 0 \\\\\n 14\\\\\n17\n\\end{bmatrix}\n=\\begin{bmatrix}\n3 \\\\\n51\\\\\n39\n\\end{bmatrix}"

“Scaled” Approximation

"\\to3\\begin{bmatrix}\n 1 \\\\\n 17\\\\\n13\n\\end{bmatrix}"

Iteration

"x_4=Ax_3=\\begin{bmatrix}\n 1 & -1&1 \\\\\n 2 & 0&3\\\\\n1&4&-1\n\\end{bmatrix}\\begin{bmatrix}\n 3 \\\\\n 51\\\\\n39\n\\end{bmatrix}\n=\\begin{bmatrix}\n-9 \\\\\n123\\\\\n168\n\\end{bmatrix}"

“Scaled” Approximation

"\\to-9\\begin{bmatrix}\n 1 \\\\\n -13.7\\\\\n-18.7\n\\end{bmatrix}"

Note that the approximations appear to be approaching scalar multiples of

"\\begin{bmatrix}\n 1 \\\\\n -14\\\\\n-19\n\\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=\\begin{bmatrix}\n 1 & -1&1 \\\\\n 2 & 0&3\\\\\n1&4&-1\n\\end{bmatrix}\\begin{bmatrix}\n 1 \\\\\n -13.7\\\\\n-18.7\n\\end{bmatrix}\n=\\begin{bmatrix}\n-4 \\\\\n-54.1\\\\\n-35.1\n\\end{bmatrix}"

Then, because

"Ax\\cdot x=-4\\cdot 1+(-54.1)(-13.7)+(-35.1)(-18.7)=1393.54\\\\\nx\\cdot x=1\\cdot1+(-13.7)(-13.7)+(-18.7)(-18.7)=538.38"

you can compute the Rayleigh quotient to be

"\\lambda=\\frac{Ax\\cdot x}{x\\cdot x}=\\frac{1393.54}{538.38}=2.6"

which is a good approximation of the dominant eigenvalue

"\\lambda=3"

Answer "\\begin{bmatrix}\n 1 \\\\\n -14\\\\\n-19\n\\end{bmatrix}, \\lambda=3"