Answer on Question #74012 – Math – Quantitative Methods
Question
find the dominant eigenvalue and the corresponding eigenvector for the matrix A = ( − 4 14 0 − 5 13 0 − 1 0 2 ) A = \begin{pmatrix} -4 & 14 & 0 \\ -5 & 13 & 0 \\ -1 & 0 & 2 \end{pmatrix} A = ⎝ ⎛ − 4 − 5 − 1 14 13 0 0 0 2 ⎠ ⎞ y ( 0 ) = ( 1 1 1 ) y(0) = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} y ( 0 ) = ⎝ ⎛ 1 1 1 ⎠ ⎞
Solution
We begin with an initial non-zero approximation of dominant eigenvector x 0 x_0 x 0 x 0 = ( 1 1 1 ) x_0 = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} x 0 = ⎝ ⎛ 1 1 1 ⎠ ⎞
x 1 = A x ~ 1 = ( − 4 14 0 − 5 13 0 − 1 0 2 ) ( 1 1 1 ) = ( 10 8 1 ) = 10 ( 1 0.8 0.1 ) x_1 = A \widetilde{x}_1 = \begin{pmatrix} -4 & 14 & 0 \\ -5 & 13 & 0 \\ -1 & 0 & 2 \end{pmatrix} \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 10 \\ 8 \\ 1 \end{pmatrix} = 10 \begin{pmatrix} 1 \\ 0.8 \\ 0.1 \end{pmatrix} x 1 = A x 1 = ⎝ ⎛ − 4 − 5 − 1 14 13 0 0 0 2 ⎠ ⎞ ⎝ ⎛ 1 1 1 ⎠ ⎞ = ⎝ ⎛ 10 8 1 ⎠ ⎞ = 10 ⎝ ⎛ 1 0.8 0.1 ⎠ ⎞
We take out the largest element of the resultant matrix and will be our new initial vector. Proceeding in this manner we obtain a series of approximations as follows:
x 2 = A x ~ 2 = ( − 4 14 0 − 5 13 0 − 1 0 2 ) ( 1 0.8 0.1 ) = ( 7.2 5.4 − 0.8 ) = 7.2 ( 1 0.75 − 0.11 ) x_2 = A \widetilde{x}_2 = \begin{pmatrix} -4 & 14 & 0 \\ -5 & 13 & 0 \\ -1 & 0 & 2 \end{pmatrix} \begin{pmatrix} 1 \\ 0.8 \\ 0.1 \end{pmatrix} = \begin{pmatrix} 7.2 \\ 5.4 \\ -0.8 \end{pmatrix} = 7.2 \begin{pmatrix} 1 \\ 0.75 \\ -0.11 \end{pmatrix} x 2 = A x 2 = ⎝ ⎛ − 4 − 5 − 1 14 13 0 0 0 2 ⎠ ⎞ ⎝ ⎛ 1 0.8 0.1 ⎠ ⎞ = ⎝ ⎛ 7.2 5.4 − 0.8 ⎠ ⎞ = 7.2 ⎝ ⎛ 1 0.75 − 0.11 ⎠ ⎞ x 3 = A x ~ 3 = ( − 4 14 0 − 5 13 0 − 1 0 2 ) ( 1 0.75 − 0.11 ) = ( 6.5 4.75 − 1.22 ) = 6.5 ( 1 0.730 − 0.187 ) x_3 = A \widetilde{x}_3 = \begin{pmatrix} -4 & 14 & 0 \\ -5 & 13 & 0 \\ -1 & 0 & 2 \end{pmatrix} \begin{pmatrix} 1 \\ 0.75 \\ -0.11 \end{pmatrix} = \begin{pmatrix} 6.5 \\ 4.75 \\ -1.22 \end{pmatrix} = 6.5 \begin{pmatrix} 1 \\ 0.730 \\ -0.187 \end{pmatrix} x 3 = A x 3 = ⎝ ⎛ − 4 − 5 − 1 14 13 0 0 0 2 ⎠ ⎞ ⎝ ⎛ 1 0.75 − 0.11 ⎠ ⎞ = ⎝ ⎛ 6.5 4.75 − 1.22 ⎠ ⎞ = 6.5 ⎝ ⎛ 1 0.730 − 0.187 ⎠ ⎞ x 4 = A x ~ 4 = ( − 4 14 0 − 5 13 0 − 1 0 2 ) ( 1 0.730 − 0.187 ) = ( 6.22 4.49 − 1.37 ) = 6.22 ( 1 0.7218 − 0.2202 ) x_4 = A \widetilde{x}_4 = \begin{pmatrix} -4 & 14 & 0 \\ -5 & 13 & 0 \\ -1 & 0 & 2 \end{pmatrix} \begin{pmatrix} 1 \\ 0.730 \\ -0.187 \end{pmatrix} = \begin{pmatrix} 6.22 \\ 4.49 \\ -1.37 \end{pmatrix} = 6.22 \begin{pmatrix} 1 \\ 0.7218 \\ -0.2202 \end{pmatrix} x 4 = A x 4 = ⎝ ⎛ − 4 − 5 − 1 14 13 0 0 0 2 ⎠ ⎞ ⎝ ⎛ 1 0.730 − 0.187 ⎠ ⎞ = ⎝ ⎛ 6.22 4.49 − 1.37 ⎠ ⎞ = 6.22 ⎝ ⎛ 1 0.7218 − 0.2202 ⎠ ⎞ x 5 = A x ~ 5 = ( − 4 14 0 − 5 13 0 − 1 0 2 ) ( 1 0.7218 − 0.2202 ) = ( 6.1052 4.3834 − 1.4404 ) = 6.1052 ( 1 0.7179 − 0.2359 ) x_5 = A \widetilde{x}_5 = \begin{pmatrix} -4 & 14 & 0 \\ -5 & 13 & 0 \\ -1 & 0 & 2 \end{pmatrix} \begin{pmatrix} 1 \\ 0.7218 \\ -0.2202 \end{pmatrix} = \begin{pmatrix} 6.1052 \\ 4.3834 \\ -1.4404 \end{pmatrix} = 6.1052 \begin{pmatrix} 1 \\ 0.7179 \\ -0.2359 \end{pmatrix} x 5 = A x 5 = ⎝ ⎛ − 4 − 5 − 1 14 13 0 0 0 2 ⎠ ⎞ ⎝ ⎛ 1 0.7218 − 0.2202 ⎠ ⎞ = ⎝ ⎛ 6.1052 4.3834 − 1.4404 ⎠ ⎞ = 6.1052 ⎝ ⎛ 1 0.7179 − 0.2359 ⎠ ⎞ x 6 = A x ~ 6 = ( − 4 14 0 − 5 13 0 − 1 0 2 ) ( 1 0.7179 − 0.2359 ) = ( 6.05 4.33 − 1.47 ) = 6.05 ( 1 0.715 − 0.242 ) x _ {6} = A \widetilde {x} _ {6} = \left( \begin{array}{c c c} - 4 & 14 & 0 \\ - 5 & 13 & 0 \\ - 1 & 0 & 2 \end{array} \right) \left( \begin{array}{c} 1 \\ 0.7179 \\ -0.2359 \end{array} \right) = \left( \begin{array}{c} 6.05 \\ 4.33 \\ -1.47 \end{array} \right) = 6.05 \left( \begin{array}{c} 1 \\ 0.715 \\ -0.242 \end{array} \right) x 6 = A x 6 = ⎝ ⎛ − 4 − 5 − 1 14 13 0 0 0 2 ⎠ ⎞ ⎝ ⎛ 1 0.7179 − 0.2359 ⎠ ⎞ = ⎝ ⎛ 6.05 4.33 − 1.47 ⎠ ⎞ = 6.05 ⎝ ⎛ 1 0.715 − 0.242 ⎠ ⎞
Therefore, the dominant eigen-value is approximately 6.05 and the corresponding eigenvector is
( 1 0.715 − 0.242 ) . \left( \begin{array}{c} 1 \\ 0.715 \\ -0.242 \end{array} \right). ⎝ ⎛ 1 0.715 − 0.242 ⎠ ⎞ .
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