I dont quite understand this question that I am doing some practice questions for and was wondering if someone could help explain it.
The question is a s follows:
"Let $P$ be the stochastic matrix given by: $$ P = \left[ \begin{matrix} 0 & 0 & 0 & 0 & 0 & 0 \\ 1/2 & 0 & 1/2 & 0 & 0 & 0 \\ 0 & 1/2 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1/2 & 1/2 & 1/2 & 0 \\ 1/2 & 1/2 & 0 & 1/2 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1/2 & 1 \end{matrix}\right]$$
The matrix $P$ has six distinct eigenvalues $\lambda_1 = 1, \lambda_2, \ldots, \lambda_6$ with $\left| \lambda_i \right| < 1$ for $i = 2,3,4,5,6$. Let $v_1, \ldots, v_6$ be the corresponding eigenvectors. What is $v_1$ ? If $X_0 = c_1v_1 + \ldots + c_6v_6$. Write down a formula for the probability of winning after $n$ steps in terms of $n, \lambda_i, v_i$ and $c_i$."
I know that $$v_1 = \left[ \begin{matrix} 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 1 \end{matrix} \right]$$ but what I don't understand is how to write the probability of winning after $n$ steps in terms of $c_i$.
The first thought that came to my mind was the idea of diagonalization and writing $$P^nx_0 = UD^nU^T$$and then $$P^nx_0 = UD^nU^Tx_0$$but then I couldn't figure out how to include the coefficients.
Would someone mind explaining or helping me out please ?
Advice is greatly appreciated.
Provided that $X_0 = \sum_{i}c_iv_i$, we have $$ X_1 = PX_0 = \sum_{i}c_iPv_i=\sum_ic_i\lambda_iv_i $$ and $$ X_2 = PX_1 = \sum_{i}c_i\lambda_iPv_i=\sum_ic_i\lambda_i^2v_i $$ More generally, $$ X_n = P^nX_0=\sum_ic_i\lambda_i^nv_i $$ Finally, we normalize $X_n$ such that its components constitute a probability, i.e., their sum is $1$.