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 '18 at 2:02
the_fox
2,47211431
asked Nov 18 '18 at 22:26
TynaTyna
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 '18 at 2:02
the_fox
2,47211431
asked Nov 18 '18 at 22:26
TynaTyna
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 '18 at 2:02
the_fox
2,47211431
asked Nov 18 '18 at 22:26
TynaTyna
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 '18 at 2:02
the_fox
2,47211431
asked Nov 18 '18 at 22:26
TynaTyna
876
edited Dec 3 '18 at 2:02
the_fox
2,47211431
asked Nov 18 '18 at 22:26
TynaTyna
876
edited Dec 3 '18 at 2:02
the_fox
2,47211431
edited Dec 3 '18 at 2:02
the_fox
2,47211431
edited Dec 3 '18 at 2:02
the_fox
2,47211431
2,47211431
asked Nov 18 '18 at 22:26
TynaTyna
876
asked Nov 18 '18 at 22:26
TynaTyna
876
asked Nov 18 '18 at 22:26
TynaTyna
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 '18 at 22:35
egregegreg
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
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 '18 at 22:28
gimusigimusi
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
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3004222%2flimit-of-matrix-a-raised-to-power-of-n-as-n-approaches-infinity%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
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 '18 at 22:35
egregegreg
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
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 '18 at 22:35
egregegreg
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
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 '18 at 22:35
egregegreg
179k1485202
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 '18 at 22:35
egregegreg
179k1485202
edited Nov 19 '18 at 14:38
answered Nov 18 '18 at 22:35
egregegreg
179k1485202
answered Nov 18 '18 at 22:35
egregegreg
179k1485202
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
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 '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
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 '18 at 22:28
gimusigimusi
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
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 '18 at 22:28
gimusigimusi
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
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 '18 at 22:28
gimusigimusi
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 '18 at 22:28
gimusigimusi
1
edited Nov 19 '18 at 14:24
answered Nov 18 '18 at 22:28
gimusigimusi
1
answered Nov 18 '18 at 22:28
gimusigimusi
1
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
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 '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
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3004222%2flimit-of-matrix-a-raised-to-power-of-n-as-n-approaches-infinity%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
