Question 7.4.2: Consider the positive transition matrix A =[0.4 0.3 0.6 0.7]......

a. The characteristic polynomial of A is f_A(λ) = λ^2 − 1.1λ + 0.1 =(λ − 1)(λ − 0.1), so that the eigenvalues are 1 and 0.1.

Now E-1=ker\begin{bmatrix}-0.6&0.3\\0.6&-0.3\end{bmatrix}=span\begin{bmatrix}1\\2\end{bmatrix}  and  E_{0.1}=ker\begin{bmatrix}0.3&0.3\\0.6&0.6\end{bmatrix}=span\begin{bmatrix}1\\-1\end{bmatrix}. We can diagonalize A with S=\begin{bmatrix}1&1\\2&-1\end{bmatrix}  and  B=\begin{bmatrix}1&0\\0&0.1\end{bmatrix}.

Then S^{−1}AS = B and A = SBS^{−1}, so that

A^t=SB^tS^{-1}=\frac{1}{3}\begin{bmatrix}1&1\\2&-1\end{bmatrix}\begin{bmatrix}1&0\\0&(0.1)^t\end{bmatrix}\begin{bmatrix}1&1\\2&-1\end{bmatrix}

=\frac{1}{3}\begin{bmatrix}1+2(0.1)^t&1-(0.1)^t\\2-2(0.1)^t&2+(0.1)^t\end{bmatrix}.

Now  \lim\limits_{t \to \infty} A^t=\lim\limits_{t \to \infty}\left(\begin{bmatrix}1+2(0.1)^t&1-(0.1)^t\\2-2(0.1)^t&2+(0.1)^t\end{bmatrix}\right)=\frac{1}{3}\begin{bmatrix}1&1\\2&2\end{bmatrix}.

b. A^t\begin{bmatrix}0.5\\0.5\end{bmatrix}=\frac{1}{6}\begin{bmatrix}1+2(0.1)^t&1-(0.1)^t\\2-2(0.1)^t&2+(0.1)^t\end{bmatrix}\begin{bmatrix}1\\1\end{bmatrix}=\frac{1}{6}\begin{bmatrix}2+(0.1)^t\\4-(0.1)^t\end{bmatrix}  and  \lim\limits_{t \to \infty}\left(A^t\begin{bmatrix}0.5\\0.5\end{bmatrix}\right)=\lim\limits_{t \to \infty}\left(\frac{1}{6}\begin{bmatrix}2+(0.1)^t\\4-(0.1)^t\end{bmatrix}\right)=\frac{1}{3}\begin{bmatrix}1\\2\end{bmatrix}.