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Exercise 5.3.7

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Using the language of the stochastic matrix, we direct write the stochastic matrix to be

A = (\begin{matrix} 1 & \frac{1}{3} & 0 & 0 \\ 0 & 0 & \frac{1}{3} & 0 \\ 0 & \frac{2}{3} & 0 & 0 \\ 0 & 0 & \frac{2}{3} & 1 \end{matrix}) .

I don’t think there is too much difference between the process of diagonalizing and finding the eigenvectors. It’s much easier to observe the sequence

e_{2}, A e_{2}, A^{2} e_{2}, \dots,

which is

(\begin{matrix} 0 \\ 1 \\ 0 \\ 0 \end{matrix}), (\begin{matrix} \frac{1}{3} \\ 0 \\ \frac{2}{3} \\ 0 \end{matrix}), (\begin{matrix} \frac{1}{3} \\ \frac{2}{9} \\ 0 \\ \frac{4}{9} \end{matrix}), (\begin{matrix} \frac{11}{27} \\ 0 \\ \frac{4}{27} \\ \frac{4}{9} \end{matrix}), (\begin{matrix} \frac{11}{27} \\ \frac{4}{81} \\ 0 \\ \frac{44}{81} \end{matrix}), \dots

And find the limit of the first entry to be

\frac{1}{3} + \frac{2}{3} \cdot \frac{1}{9} + \frac{2}{3} \cdot \frac{2}{9} \cdot \frac{1}{9} + \frac{2}{3} \cdot {(\frac{2}{9})}^{2} \cdot \frac{1}{9} + \dots

\frac{1}{3} + \frac{\frac{2}{3} \cdot \frac{1}{9}}{1 - \frac{2}{9}} = \frac{3}{7} .

So the answer is $\frac{3}{7}$ .

jephianlin

2011-06-27 00:00

Exercise 5.3.7

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