TITLE: Limit of matrix $A$ raised to power of $n$, as $n$ approaches infinity.
QUESTION [9 upvotes]: I understand that the limit of $n$ approaching infinity of a matrix $A^n$, can be computed, in some cases, by looking at the diagonalization of that matrix, and then looking at the limit of $n$ going to infinity of the resulting diagonal matrix, $D$, whose elements are raised to the power $n$.
What I do not understand is when we do not raise the matrix, call it $P$, consisting of the eigenvectors of $A$, and its inverse, to the power of $n$ as well?
So:
$ P^{-1}AP = D    $
$A = PDP^{-1}  $
$A^n = (PDP^{-1})^n$
$A^n = P^nD^n(P^{-1})^n$
Why do the matrices $P^n$ and $(P^{-1})^n$ not have to be taken into account when looking at the limit of $n$ going to infinity?

REPLY [26 votes]: In general, the statement
$$
(AB)^n=A^nB^n
$$
is false for square matrices. So it's not true in general that, from $A=PDP^{-1}$ it follows that $A^n=P^nD^n(P^{-1})^n$.
Rather you should note that
$$
A^2=(PDP^{-1})(PDP^{-1})=PDP^{-1}PDP^{-1}=PDDP^{-1}=PD^2P^{-1}
$$
and, by easy induction,
$$
A^n=PD^nP^{-1}
$$
for every $n$. Do you see the difference?
Now, in order to compute the limit, it is sufficient to compute the limit of $D^n$, because matrix multiplication is continuous.