# Existence of $\lim_{n \rightarrow \infty}A^n$

I was studying Markov chain and in it, it is useful to have a transition matrix $$M$$ that $$\lim_{n \rightarrow \infty}M^n$$ exist.

So I thought about the existence of $$\lim_{n \rightarrow \infty}A^n$$ in general($$A$$ is a square matrix). and it was easy to prove:

for every diagonalizable matrix $$A= QDQ^{-1}$$, the $$\lim_{n \rightarrow \infty}A^n$$ exists if and only if elements of $$D$$ be in the range $$(-1,1]$$.

but what about not diagonalizable matrices? what is the condition for them?

EXAMPLES: $$E_1$$ is not diagonalizable and $$\lim_{n \rightarrow \infty}E_1^n$$ exists, $$E_2$$ is not diagonalizable and $$\lim_{n \rightarrow \infty}E_2^n$$ does not exists. $$E_1 = \begin{bmatrix}0.1 & 0.1 \\0 & 0.1 \end{bmatrix}, E_2 = \begin{bmatrix}1 & 1 \\0 & 1 \end{bmatrix}$$

You can basically do the same thing with the Jordan normal form. You need all eigenvalues to be either in the interior of the unit disk or $$1$$. Additionally, if $$1$$ is an eigenvalue, it cannot be defective (defective eigenvalues of $$1$$ lead to polynomial growth).