Mixing
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
edited 8 hours ago
Xander Henderson
13.7k93552
asked yesterday
Tyna
675
add a comment |
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
edited 8 hours ago
Xander Henderson
13.7k93552
asked yesterday
Tyna
675
add a comment |
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
edited 8 hours ago
Xander Henderson
13.7k93552
asked yesterday
Tyna
675
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 8 hours ago
Xander Henderson
13.7k93552
asked yesterday
Tyna
675
edited 8 hours ago
Xander Henderson
13.7k93552
asked yesterday
Tyna
675
edited 8 hours ago
Xander Henderson
13.7k93552
edited 8 hours ago
Xander Henderson
13.7k93552
edited 8 hours ago
Xander Henderson
13.7k93552
13.7k93552
asked yesterday
Tyna
675
asked yesterday
Tyna
675
asked yesterday
Tyna
675
675
add a comment |
add a comment |
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.
answered yesterday
egreg
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
add a comment |
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.
answered yesterday
gimusi
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
add a comment |
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.
answered yesterday
egreg
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
add a comment |
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.
answered yesterday
egreg
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
add a comment |
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.
answered yesterday
egreg
173k1383197
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 yesterday
egreg
173k1383197
edited 11 hours ago
answered yesterday
egreg
173k1383197
answered yesterday
egreg
173k1383197
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
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
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
add a comment |
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.
answered yesterday
gimusi
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
add a comment |
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.
answered yesterday
gimusi
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
add a comment |
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.
answered yesterday
gimusi
85.5k74293
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 yesterday
gimusi
85.5k74293
edited 12 hours ago
answered yesterday
gimusi
85.5k74293
answered yesterday
gimusi
85.5k74293
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
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
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
add a comment |
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