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?


$ 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.

  • $\begingroup$ 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. $\endgroup$ – PJTraill Nov 19 '18 at 14:23
  • $\begingroup$ @PJTraill Isn't the “Rather” part covering it? $\endgroup$ – egreg Nov 19 '18 at 14:25
  • $\begingroup$ 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). $\endgroup$ – 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.

  • 4
    $\begingroup$ 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. $\endgroup$ – TunaBooties Nov 18 '18 at 22:32
  • 2
    $\begingroup$ @Tyna Yes exactly and then we can generalize that for any $n$ (rigoursly by induction). $\endgroup$ – gimusi Nov 18 '18 at 22:34

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