we discussed matrices in class and had the following task: Given
$$U=\begin{bmatrix}0 & 0 & 0 & 0 & 0 & 0\\1 & \frac{1}{15} & 0 & 0 & 0 &0\\0 & \frac{8}{15} & \frac{3}{15} & 0 & 0 & 0\\0 & \frac{6}{15} & \frac{9}{15} & \frac{6}{15} & 0 & 0\\ 0 & 0 & \frac{3}{15} & \frac{8}{15} & \frac{10}{15} & 0\\0 & 0 & 0 & \frac{1}{15} & \frac{5}{15} & 1\end{bmatrix}, \quad \overrightarrow{s}=\begin{bmatrix}1\\0\\0\\0\\0\\0\end{bmatrix},$$
solve for $x$ such that $U^x\overrightarrow{s}=\overrightarrow{s_x}$ where the last element (Row) of $\overrightarrow{s_x}$ should be equal or greater than 0.99.
We were told that the only way to get $x$ is by inserting random numbers and "search" for it. By doing this we actually found:
$$\overrightarrow{s_{15}}=\begin{bmatrix}0\\0\\0\\0\\0,0137\\0,9863\end{bmatrix},\quad \overrightarrow{s_{16}}=\begin{bmatrix}0\\0\\0\\0\\0,0091\\0,9909\end{bmatrix}.$$
So the answer is pretty much $x=16$, but is there no way to solve for $x$ instead of inserting random numbers until you find the answer?
Thanks for any answers
Hint :
Using this you get the diagonalization of $U=PDP^{-1}$, then $U^x = PDP^{-1} \dots PDP^{-1} = P D^x P^{-1}$ where $D$ is the matrix with the elements of $D$ elevated to power $x$. Using this you can solve for the last element being greater to $0.99$