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











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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







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    up vote
    7
    down vote

    favorite












    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


























      up vote
      7
      down vote

      favorite









      up vote
      7
      down vote

      favorite











      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






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      share|cite|improve this question













      share|cite|improve this question




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      edited 8 hours ago









      Xander Henderson

      13.7k93552




      13.7k93552










      asked yesterday









      Tyna

      675




      675






















          2 Answers
          2






          active

          oldest

          votes

















          up vote
          24
          down vote













          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
            12 hours ago












          • @PJTraill Isn't the “Rather” part covering it?
            – egreg
            12 hours ago










          • 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
            12 hours ago




















          up vote
          12
          down vote













          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
            yesterday








          • 2




            @Tyna Yes exactly and then we can generalize that for any $n$ (rigoursly by induction).
            – gimusi
            yesterday











          Your Answer





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          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          24
          down vote













          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
            12 hours ago












          • @PJTraill Isn't the “Rather” part covering it?
            – egreg
            12 hours ago










          • 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
            12 hours ago

















          up vote
          24
          down vote













          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
            12 hours ago












          • @PJTraill Isn't the “Rather” part covering it?
            – egreg
            12 hours ago










          • 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
            12 hours ago















          up vote
          24
          down vote










          up vote
          24
          down vote









          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 11 hours ago

























          answered yesterday









          egreg

          173k1383197




          173k1383197












          • 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
            12 hours ago












          • @PJTraill Isn't the “Rather” part covering it?
            – egreg
            12 hours ago










          • 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
            12 hours ago




















          • 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
            12 hours ago












          • @PJTraill Isn't the “Rather” part covering it?
            – egreg
            12 hours ago










          • 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
            12 hours ago


















          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
          12 hours ago






          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
          12 hours ago














          @PJTraill Isn't the “Rather” part covering it?
          – egreg
          12 hours ago




          @PJTraill Isn't the “Rather” part covering it?
          – egreg
          12 hours ago












          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
          12 hours ago






          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
          12 hours ago












          up vote
          12
          down vote













          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
            yesterday








          • 2




            @Tyna Yes exactly and then we can generalize that for any $n$ (rigoursly by induction).
            – gimusi
            yesterday















          up vote
          12
          down vote













          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
            yesterday








          • 2




            @Tyna Yes exactly and then we can generalize that for any $n$ (rigoursly by induction).
            – gimusi
            yesterday













          up vote
          12
          down vote










          up vote
          12
          down vote









          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 12 hours ago

























          answered yesterday









          gimusi

          85.5k74293




          85.5k74293








          • 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
            yesterday








          • 2




            @Tyna Yes exactly and then we can generalize that for any $n$ (rigoursly by induction).
            – gimusi
            yesterday














          • 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
            yesterday








          • 2




            @Tyna Yes exactly and then we can generalize that for any $n$ (rigoursly by induction).
            – gimusi
            yesterday








          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
          yesterday






          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
          yesterday






          2




          2




          @Tyna Yes exactly and then we can generalize that for any $n$ (rigoursly by induction).
          – gimusi
          yesterday




          @Tyna Yes exactly and then we can generalize that for any $n$ (rigoursly by induction).
          – gimusi
          yesterday


















           

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