mjhjmtu

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^-1AP = 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?

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
$$
A^2=(PDP^-1)(PDP^-1)=PDP^-1PDP^-1=PDDP^-1=PD^2P^-1
$$
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.

We have that

$$A = PDP^-1implies A^2 = PDP^-1 PDP^-1= PD(P^-1P)DP^-1= PD (I)DP^-1=PD^2P^-1$$

and so on we can generalize the result rigorously for any $n$ by induction.