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Exercise 5.3.14

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We prove it by induction on $m$ . When $m = 1$ , the formula meet the matrix $A$ . Suppose the formula holds for the case $m = k - 1$ . Then the case $m = k$ would be

A^{k} = A^{k - 1} A = (\begin{matrix} r_{k - 1} & r_{k} & r_{k} \\ r_{k} & r_{k - 1} & r_{k} \\ r_{k} & r_{k} & r_{k - 1} \end{matrix}) A

= (\begin{matrix} r_{k} & \frac{r_{k - 1} + r_{k}}{2} & \frac{r_{k - 1} + r_{k}}{2} \\ \frac{r_{k - 1} + r_{k}}{2} & r_{k} & \frac{r_{k - 1} + r_{k}}{2} \\ \frac{r_{k - 1} + r_{k}}{2} & \frac{r_{k - 1} + r_{k}}{2} & r_{k} \end{matrix}) .

After check that

\frac{r_{k - 1} + r_{k}}{2} = \frac{1}{3} \cdot \frac{1 + \frac{{(- 1)}^{k}}{2^{k - 2}} + 1 + \frac{{(- 1)}^{k}}{2^{k - 1}}}{2} = \frac{1}{3} [1 + \frac{{(- 1)}^{k}}{2^{k}}] = r_{k + 1}

we get the desired result. To deduce the second equality, just replace $A^{m}$ by the right hand side in the formula.

jephianlin

2011-06-27 00:00

Exercise 5.3.14

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