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












8














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?






linear-algebra matrices limits







share|cite|improve this question





























    8














    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?






    linear-algebra matrices limits







    share|cite|improve this question



























      8












      8








      8







      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?






      linear-algebra matrices limits







      share|cite|improve this 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?






      linear-algebra matrices limits




      linear-algebra matrices limits






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 3 '18 at 2:02









      the_fox

      2,47211431




      2,47211431










      asked Nov 18 '18 at 22:26









      TynaTyna

      876




      876






















          2 Answers
          2






          active

          oldest

          votes


















          25














          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.






          share|cite|improve this answer























          • 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





















          13














          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.






          share|cite|improve this answer



















          • 4




            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.
            – Tyna
            Nov 18 '18 at 22:32








          • 2




            @Tyna Yes exactly and then we can generalize that for any $n$ (rigoursly by induction).
            – gimusi
            Nov 18 '18 at 22:34











          Your Answer





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          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          25














          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.






          share|cite|improve this answer























          • 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


















          25














          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.






          share|cite|improve this answer























          • 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
















          25












          25








          25






          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.






          share|cite|improve this answer














          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.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 19 '18 at 14:38

























          answered Nov 18 '18 at 22:35









          egregegreg

          179k1485202




          179k1485202












          • 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




















          • 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


















          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






          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




          @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






          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













          13














          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.






          share|cite|improve this answer



















          • 4




            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.
            – Tyna
            Nov 18 '18 at 22:32








          • 2




            @Tyna Yes exactly and then we can generalize that for any $n$ (rigoursly by induction).
            – gimusi
            Nov 18 '18 at 22:34
















          13














          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.






          share|cite|improve this answer



















          • 4




            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.
            – Tyna
            Nov 18 '18 at 22:32








          • 2




            @Tyna Yes exactly and then we can generalize that for any $n$ (rigoursly by induction).
            – gimusi
            Nov 18 '18 at 22:34














          13












          13








          13






          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.






          share|cite|improve this answer














          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.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 19 '18 at 14:24

























          answered Nov 18 '18 at 22:28









          gimusigimusi

          1




          1








          • 4




            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.
            – Tyna
            Nov 18 '18 at 22:32








          • 2




            @Tyna Yes exactly and then we can generalize that for any $n$ (rigoursly by induction).
            – gimusi
            Nov 18 '18 at 22:34














          • 4




            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.
            – Tyna
            Nov 18 '18 at 22:32








          • 2




            @Tyna Yes exactly and then we can generalize that for any $n$ (rigoursly by induction).
            – gimusi
            Nov 18 '18 at 22:34








          4




          4




          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.
          – Tyna
          Nov 18 '18 at 22:32






          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.
          – Tyna
          Nov 18 '18 at 22:32






          2




          2




          @Tyna Yes exactly and then we can generalize that for any $n$ (rigoursly by induction).
          – gimusi
          Nov 18 '18 at 22:34




          @Tyna Yes exactly and then we can generalize that for any $n$ (rigoursly by induction).
          – gimusi
          Nov 18 '18 at 22:34


















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