Limit of matrix $A$ raised to power of $n$, as $n$ approaches infinity.

Question

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?

Answer 1

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.

Answer 2

We have that

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

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