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?
linear-algebra matrices limits
edited Dec 3 at 2:02
the_fox
2,43411431
asked Nov 18 at 22:26
Tyna
876
add a comment |
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
edited Dec 3 at 2:02
the_fox
2,43411431
asked Nov 18 at 22:26
Tyna
876
add a comment |
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
edited Dec 3 at 2:02
the_fox
2,43411431
asked Nov 18 at 22:26
Tyna
876
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
edited Dec 3 at 2:02
the_fox
2,43411431
asked Nov 18 at 22:26
Tyna
876
edited Dec 3 at 2:02
the_fox
2,43411431
asked Nov 18 at 22:26
Tyna
876
edited Dec 3 at 2:02
the_fox
2,43411431
edited Dec 3 at 2:02
the_fox
2,43411431
edited Dec 3 at 2:02
the_fox
2,43411431
2,43411431
asked Nov 18 at 22:26
Tyna
876
asked Nov 18 at 22:26
Tyna
876
asked Nov 18 at 22:26
Tyna
876
876
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
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.
answered Nov 18 at 22:35
egreg
178k1484201
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 at 14:23
@PJTraill Isn't the “Rather” part covering it?
– egreg
Nov 19 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 at 14:26
add a comment |
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.
answered Nov 18 at 22:28
gimusi
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 at 22:32
2
@Tyna Yes exactly and then we can generalize that for any $n$ (rigoursly by induction).
– gimusi
Nov 18 at 22:34
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
answered Nov 18 at 22:35
egreg
178k1484201
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 at 14:23
@PJTraill Isn't the “Rather” part covering it?
– egreg
Nov 19 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 at 14:26
add a comment |
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.
answered Nov 18 at 22:35
egreg
178k1484201
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 at 14:23
@PJTraill Isn't the “Rather” part covering it?
– egreg
Nov 19 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 at 14:26
add a comment |
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.
answered Nov 18 at 22:35
egreg
178k1484201
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.
answered Nov 18 at 22:35
egreg
178k1484201
edited Nov 19 at 14:38
answered Nov 18 at 22:35
egreg
178k1484201
answered Nov 18 at 22:35
egreg
178k1484201
answered Nov 18 at 22:35
egreg
178k1484201
178k1484201
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 at 14:23
@PJTraill Isn't the “Rather” part covering it?
– egreg
Nov 19 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 at 14:26
add a comment |
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 at 14:23
@PJTraill Isn't the “Rather” part covering it?
– egreg
Nov 19 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 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 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 at 14:23
@PJTraill Isn't the “Rather” part covering it?
– egreg
Nov 19 at 14:25
@PJTraill Isn't the “Rather” part covering it?
– egreg
Nov 19 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 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 at 14:26
add a comment |
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.
answered Nov 18 at 22:28
gimusi
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 at 22:32
2
@Tyna Yes exactly and then we can generalize that for any $n$ (rigoursly by induction).
– gimusi
Nov 18 at 22:34
add a comment |
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.
answered Nov 18 at 22:28
gimusi
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 at 22:32
2
@Tyna Yes exactly and then we can generalize that for any $n$ (rigoursly by induction).
– gimusi
Nov 18 at 22:34
add a comment |
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.
answered Nov 18 at 22:28
gimusi
1
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.
answered Nov 18 at 22:28
gimusi
1
edited Nov 19 at 14:24
answered Nov 18 at 22:28
gimusi
1
answered Nov 18 at 22:28
gimusi
1
answered Nov 18 at 22:28
gimusi
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 at 22:32
2
@Tyna Yes exactly and then we can generalize that for any $n$ (rigoursly by induction).
– gimusi
Nov 18 at 22:34
add a comment |
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 at 22:32
2
@Tyna Yes exactly and then we can generalize that for any $n$ (rigoursly by induction).
– gimusi
Nov 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 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 at 22:32
2
2
@Tyna Yes exactly and then we can generalize that for any $n$ (rigoursly by induction).
– gimusi
Nov 18 at 22:34
@Tyna Yes exactly and then we can generalize that for any $n$ (rigoursly by induction).
– gimusi
Nov 18 at 22:34
add a comment |
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