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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. $$ E_1 = \begin{bmatrix}0.1 & 0.1 \\0 & 0.1 \end{bmatrix}, E_2 = \begin{bmatrix}1 & 1 \\0 & 1 \end{bmatrix}$$

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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).

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