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

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?

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.

• Perhaps it is worth remarking that we do still have $P$ and $P^-$ to take into account, but that is straightforward and we do not have to worry what $P^n$ might be. – PJTraill Nov 19 '18 at 14:23
• @PJTraill Isn't the “Rather” part covering it? – egreg Nov 19 '18 at 14:25
• Formally it certainly does, but given the level of the questioner I thought one might make that explicit (though an alternative would be to nudge them to chew on some thought inclining them in that direction). – PJTraill Nov 19 '18 at 14:26

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.

• thank you! the "inner" factors on $P$ and $P^{-1}$ will cancel and you'll only be left with the one $P$ and its inverse. – TunaBooties Nov 18 '18 at 22:32
• @Tyna Yes exactly and then we can generalize that for any $n$ (rigoursly by induction). – gimusi Nov 18 '18 at 22:34