How to find the limit of a matrix $P^n = UD^nU^{-1}$ where $D$ is a diagonal matrix of eigenvalues and $U$ a matrix of eigenvectors?

Question

If we have a matrix where $P = UDU^{-1}$, where $D$ is a diagonal matrix of real eigenvalues that are less than or equal to 1, and $U$ is the corresponding matrix of eigenvectors, how can we show that the limit of $P^n = UD^nU^{-1} \to u_1v_1'$, where $u_1$ is the column vector for $U$ and $v_1$ is the row vector of $U^{-1}$?

Answer 1

From your notation and tags we can surmise that you are dealing with a transition matrix $P$ for a Markov chain. A typical situation in this context is that $1$ is an eigenvalue with multiplicity $1$ and that all other eigenvalues are less than $1$ in modulus. Your result follows immediately under those conditions as explained by user7530.

Answer 2

What you say isn't true: $$P^n = UD^nU^{-1} = U\left[\begin{array}{cccc}\epsilon_1 & & & \\ & \epsilon_2 & & \\ & & \ddots & \\ & & & \epsilon_m\end{array}\right]U^{-1}$$ where $$\epsilon_i = \begin{cases}0, & |\lambda_i| < 1\\ 1, & \lambda_i = 1\end{cases}.$$

Notice that $P^n$ does not converge if any eigenvalues are $\leq -1$. And moreover, $P^n$ will have the form you seek only if exactly one eigenvalue is 1, and all the others are in $(-1,1).$