Calculate:
$P_{11}(n)=P(X_n=1|X_0=1)$
where the transition matrix is of the form:
$$\left[\begin{matrix}0 & 1 &0 \\ 0 & \dfrac{1}{2} & \dfrac{1}{2} \\ \dfrac{1}{2} & 0 & \dfrac{1}{2}\end{matrix}\right]$$
okay so I worked out my eigenvalues of the matrix, and got $\lambda = 1, \pm \dfrac{i}{2}$
$P_n$ should have the form $P_n=C_1(\lambda _ 1)^n + C_2\bigg(\dfrac{i}{2}\bigg)^n + C_3\bigg(\dfrac{-i}{2}\bigg)^n$
Then as we want $P_{11}(n)$ to be real we know that: $\bigg(\pm\dfrac{i}{2}\bigg)^n=\bigg(\dfrac{1}{2}\bigg)^n \bigg(\cos\bigg(\dfrac{n \pi}{2} \bigg) \pm i \sin\bigg(\dfrac{n \pi}{2} \bigg) \bigg)$
Subbing this all back in I get the equation: $$P_{11}(n)=C_1 + \bar{C_2}\bigg(\dfrac{1}{2}\bigg)^n \cos\bigg(\dfrac{n \pi}{2}\bigg) + \bar{C_3}\bigg(\dfrac{1}{2}\bigg)^n \sin\bigg(\dfrac{n \pi}{2}\bigg)$$
I am then stuck as to how to calculate the values of my coefficients. Any guidance would be great, thank you
Since the matrix $P$ has $3$ distinct eigenvalues it is diagonalizable. So, $\exists$ an invertible matrix $U$ such that $P=U\Lambda U^{-1}$ where $\Lambda$ is a diagonal matrix. Here, $U$ will be formed with columns as the eigenvectors of the of the matrix $P$. So say if the eigenpairs are $(\lambda_i,v_i),\ 1\le i\le 3$, then the matrices $U$ and $\Lambda$ will be $$U=[v_1\quad v_2\quad v_3]\\ \Lambda=diag(\lambda_1\lambda_2,\lambda_3)$$ Then, $$P=U\Lambda U^{-1}\Rightarrow P^n=U\Lambda^n U^{-1}$$. So, now, $P_{ij}(n)$ is just the element in the $(i,j)$ entry of $P^n$.