Mixing
Effect of perturbation on eigenvalues
$begingroup$
Given a diagonal matrix $A$ with diagonal elements $a_1,dotsc,a_n$ what can we say about the eigenvalues of the perturbation $A+E(t)$ satisfying $E(t) = O(t^k)$ so that there exists a constant $K>0$ so that $|E(t)|leq Kt^k$.
Now the determinant of $det(A+E(t)-zeta I)$ is a sum of the form
$$p(zeta) = sum_j left(prod_{i=l_j}^{n}(a_{pi_j(i)}-zeta+O(t^k))O(t^{kcdot l_j})right)$$
where $pi_j$ denotes permutations on ${1,dotsc,n}$.
I would like to say that the roots of $p$ with respect to $zeta$,
here denoted $(b_i)_{i=1}^{n}$, should satisfy
$$|a_{i}-b_i| = O(t^k)$$
however I don't know how to prove this or if it is true?
By expanding the product $prod_{i=l_j}^{n}(a_{pi_j(i)}-zeta+O(t^k))$ into a sum of the form
$$sum prod(a_{tilde{pi}(i)}-zeta)O(t^{k tilde{l}_j})$$
i get an even bigger sum and this does not seem to help me at all so I am kind of stuck at this point.
eigenvalues-eigenvectors perturbation-theory
asked Jan 16 at 10:45
Olof RubinOlof Rubin
1,160316
$endgroup$
add a comment |
$begingroup$
Given a diagonal matrix $A$ with diagonal elements $a_1,dotsc,a_n$ what can we say about the eigenvalues of the perturbation $A+E(t)$ satisfying $E(t) = O(t^k)$ so that there exists a constant $K>0$ so that $|E(t)|leq Kt^k$.
Now the determinant of $det(A+E(t)-zeta I)$ is a sum of the form
$$p(zeta) = sum_j left(prod_{i=l_j}^{n}(a_{pi_j(i)}-zeta+O(t^k))O(t^{kcdot l_j})right)$$
where $pi_j$ denotes permutations on ${1,dotsc,n}$.
I would like to say that the roots of $p$ with respect to $zeta$,
here denoted $(b_i)_{i=1}^{n}$, should satisfy
$$|a_{i}-b_i| = O(t^k)$$
however I don't know how to prove this or if it is true?
By expanding the product $prod_{i=l_j}^{n}(a_{pi_j(i)}-zeta+O(t^k))$ into a sum of the form
$$sum prod(a_{tilde{pi}(i)}-zeta)O(t^{k tilde{l}_j})$$
i get an even bigger sum and this does not seem to help me at all so I am kind of stuck at this point.
eigenvalues-eigenvectors perturbation-theory
asked Jan 16 at 10:45
Olof RubinOlof Rubin
1,160316
$endgroup$
$begingroup$
Do you mean $O(t^k)$ as $t to 0$ ?
$endgroup$
– Keith McClary
Jan 16 at 18:02
$begingroup$
Yes that is what I meant
$endgroup$
– Olof Rubin
Jan 16 at 18:03
$begingroup$
If we take $F(t)= E(t^frac{1}{k})$, doesn't this reduce to the $k=1$ problem for $A+F(t)$ ?
$endgroup$
– Keith McClary
Jan 16 at 23:54
add a comment |
$begingroup$
Given a diagonal matrix $A$ with diagonal elements $a_1,dotsc,a_n$ what can we say about the eigenvalues of the perturbation $A+E(t)$ satisfying $E(t) = O(t^k)$ so that there exists a constant $K>0$ so that $|E(t)|leq Kt^k$.
Now the determinant of $det(A+E(t)-zeta I)$ is a sum of the form
$$p(zeta) = sum_j left(prod_{i=l_j}^{n}(a_{pi_j(i)}-zeta+O(t^k))O(t^{kcdot l_j})right)$$
where $pi_j$ denotes permutations on ${1,dotsc,n}$.
I would like to say that the roots of $p$ with respect to $zeta$,
here denoted $(b_i)_{i=1}^{n}$, should satisfy
$$|a_{i}-b_i| = O(t^k)$$
however I don't know how to prove this or if it is true?
By expanding the product $prod_{i=l_j}^{n}(a_{pi_j(i)}-zeta+O(t^k))$ into a sum of the form
$$sum prod(a_{tilde{pi}(i)}-zeta)O(t^{k tilde{l}_j})$$
i get an even bigger sum and this does not seem to help me at all so I am kind of stuck at this point.
eigenvalues-eigenvectors perturbation-theory
asked Jan 16 at 10:45
Olof RubinOlof Rubin
1,160316
$endgroup$
Given a diagonal matrix $A$ with diagonal elements $a_1,dotsc,a_n$ what can we say about the eigenvalues of the perturbation $A+E(t)$ satisfying $E(t) = O(t^k)$ so that there exists a constant $K>0$ so that $|E(t)|leq Kt^k$.
Now the determinant of $det(A+E(t)-zeta I)$ is a sum of the form
$$p(zeta) = sum_j left(prod_{i=l_j}^{n}(a_{pi_j(i)}-zeta+O(t^k))O(t^{kcdot l_j})right)$$
where $pi_j$ denotes permutations on ${1,dotsc,n}$.
I would like to say that the roots of $p$ with respect to $zeta$,
here denoted $(b_i)_{i=1}^{n}$, should satisfy
$$|a_{i}-b_i| = O(t^k)$$
however I don't know how to prove this or if it is true?
By expanding the product $prod_{i=l_j}^{n}(a_{pi_j(i)}-zeta+O(t^k))$ into a sum of the form
$$sum prod(a_{tilde{pi}(i)}-zeta)O(t^{k tilde{l}_j})$$
i get an even bigger sum and this does not seem to help me at all so I am kind of stuck at this point.
eigenvalues-eigenvectors perturbation-theory
eigenvalues-eigenvectors perturbation-theory
asked Jan 16 at 10:45
Olof RubinOlof Rubin
1,160316
asked Jan 16 at 10:45
Olof RubinOlof Rubin
1,160316
asked Jan 16 at 10:45
Olof RubinOlof Rubin
1,160316
asked Jan 16 at 10:45
Olof RubinOlof Rubin
1,160316
asked Jan 16 at 10:45
Olof RubinOlof Rubin
1,160316
1,160316
$begingroup$
Do you mean $O(t^k)$ as $t to 0$ ?
$endgroup$
– Keith McClary
Jan 16 at 18:02
$begingroup$
Yes that is what I meant
$endgroup$
– Olof Rubin
Jan 16 at 18:03
$begingroup$
If we take $F(t)= E(t^frac{1}{k})$, doesn't this reduce to the $k=1$ problem for $A+F(t)$ ?
$endgroup$
– Keith McClary
Jan 16 at 23:54
add a comment |
$begingroup$
Do you mean $O(t^k)$ as $t to 0$ ?
$endgroup$
– Keith McClary
Jan 16 at 18:02
$begingroup$
Yes that is what I meant
$endgroup$
– Olof Rubin
Jan 16 at 18:03
$begingroup$
If we take $F(t)= E(t^frac{1}{k})$, doesn't this reduce to the $k=1$ problem for $A+F(t)$ ?
$endgroup$
– Keith McClary
Jan 16 at 23:54
$begingroup$
Do you mean $O(t^k)$ as $t to 0$ ?
$endgroup$
– Keith McClary
Jan 16 at 18:02
$begingroup$
Do you mean $O(t^k)$ as $t to 0$ ?
$endgroup$
– Keith McClary
Jan 16 at 18:02
$begingroup$
Yes that is what I meant
$endgroup$
– Olof Rubin
Jan 16 at 18:03
$begingroup$
Yes that is what I meant
$endgroup$
– Olof Rubin
Jan 16 at 18:03
$begingroup$
If we take $F(t)= E(t^frac{1}{k})$, doesn't this reduce to the $k=1$ problem for $A+F(t)$ ?
$endgroup$
– Keith McClary
Jan 16 at 23:54
$begingroup$
If we take $F(t)= E(t^frac{1}{k})$, doesn't this reduce to the $k=1$ problem for $A+F(t)$ ?
$endgroup$
– Keith McClary
Jan 16 at 23:54
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Here are a couple hints. I'll come back and fill in more details once you've had a chance to look at them.
A necessary condition for $ Vert E(t) Vert = mathcal{O}(t^k) $ as $tto0$ is that each entry also goes to zero like $t^k$.
Gershgorin's circle theorem says that all eigenvalues are contained within the union of disks centered at the diagonal elements of a matrix, where each disk has radius equal to the sum of magnitude of the off diagonal elements in the corresponding row. What can you tell about the row sums and diagonal elements of $A+E(t)$?
answered Jan 17 at 2:20
tchtch
803310
$endgroup$
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%2f3075584%2feffect-of-perturbation-on-eigenvalues%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Here are a couple hints. I'll come back and fill in more details once you've had a chance to look at them.
A necessary condition for $ Vert E(t) Vert = mathcal{O}(t^k) $ as $tto0$ is that each entry also goes to zero like $t^k$.
Gershgorin's circle theorem says that all eigenvalues are contained within the union of disks centered at the diagonal elements of a matrix, where each disk has radius equal to the sum of magnitude of the off diagonal elements in the corresponding row. What can you tell about the row sums and diagonal elements of $A+E(t)$?
answered Jan 17 at 2:20
tchtch
803310
$endgroup$
add a comment |
$begingroup$
Here are a couple hints. I'll come back and fill in more details once you've had a chance to look at them.
A necessary condition for $ Vert E(t) Vert = mathcal{O}(t^k) $ as $tto0$ is that each entry also goes to zero like $t^k$.
Gershgorin's circle theorem says that all eigenvalues are contained within the union of disks centered at the diagonal elements of a matrix, where each disk has radius equal to the sum of magnitude of the off diagonal elements in the corresponding row. What can you tell about the row sums and diagonal elements of $A+E(t)$?
answered Jan 17 at 2:20
tchtch
803310
$endgroup$
add a comment |
$begingroup$
Here are a couple hints. I'll come back and fill in more details once you've had a chance to look at them.
A necessary condition for $ Vert E(t) Vert = mathcal{O}(t^k) $ as $tto0$ is that each entry also goes to zero like $t^k$.
Gershgorin's circle theorem says that all eigenvalues are contained within the union of disks centered at the diagonal elements of a matrix, where each disk has radius equal to the sum of magnitude of the off diagonal elements in the corresponding row. What can you tell about the row sums and diagonal elements of $A+E(t)$?
answered Jan 17 at 2:20
tchtch
803310
$endgroup$
Here are a couple hints. I'll come back and fill in more details once you've had a chance to look at them.
A necessary condition for $ Vert E(t) Vert = mathcal{O}(t^k) $ as $tto0$ is that each entry also goes to zero like $t^k$.
Gershgorin's circle theorem says that all eigenvalues are contained within the union of disks centered at the diagonal elements of a matrix, where each disk has radius equal to the sum of magnitude of the off diagonal elements in the corresponding row. What can you tell about the row sums and diagonal elements of $A+E(t)$?
answered Jan 17 at 2:20
tchtch
803310
answered Jan 17 at 2:20
tchtch
803310
answered Jan 17 at 2:20
tchtch
803310
answered Jan 17 at 2:20
tchtch
803310
803310
add a comment |
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%2f3075584%2feffect-of-perturbation-on-eigenvalues%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
$begingroup$
Do you mean $O(t^k)$ as $t to 0$ ?
$endgroup$
– Keith McClary
Jan 16 at 18:02
$begingroup$
Yes that is what I meant
$endgroup$
– Olof Rubin
Jan 16 at 18:03
$begingroup$
If we take $F(t)= E(t^frac{1}{k})$, doesn't this reduce to the $k=1$ problem for $A+F(t)$ ?
$endgroup$
– Keith McClary
Jan 16 at 23:54