Exercise 5.6.5

Answers

For the first $A$ , it has eigenvector of $v_{1} = [\begin{matrix} 0.5 \\ 0.5 \end{matrix}]$ corresponding to the eigenvalue of 1 (Note we need make the vector sum to 1 here). it has eigenvector of $v_{2} = [\begin{matrix} 1 \\ - 1 \end{matrix}]$ corresponding to eigenvalue of $\frac{1}{3}$ , so $A = XΛ X^{- 1}$ , where $X = [\begin{matrix} 0.5 & 1 \\ 0.5 & - 1 \end{matrix}]$ , so $A^{n} = X Λ^{n} X^{- 1} \to [\begin{matrix} 0.5 & 0.5 \\ 0.5 & 0.5 \end{matrix}]$ . This can also be obtained by $A^{n} = v_{1} 1^{T} = [\begin{matrix} 0.5 \\ 0.5 \end{matrix}] [\begin{matrix} 1 & 1 \end{matrix}] = [\begin{matrix} 0.5 & 0.5 \\ 0.5 & 0.5 \end{matrix}]$
For the second $A$ , it has eigenvalues of $λ_{1} = 1, λ_{2} = λ_{3} = 0.25$ and eigenvectors of $v_{1} = [\begin{matrix} \frac{1}{3} \\ \frac{1}{3} \\ \frac{1}{3} \end{matrix}]$ corresponding to the eigenvalue of 1. So we have $A^{n} = v_{1} 1^{T} = [\begin{matrix} \frac{1}{3} & \frac{1}{3} & \frac{1}{3} \\ \frac{1}{3} & \frac{1}{3} & \frac{1}{3} \\ \frac{1}{3} & \frac{1}{3} & \frac{1}{3} \end{matrix}]$

niuers

2020-03-20 00:00