Mixing
Eigenvalue Comparison
$begingroup$
Let $A$ be a real matrix, if some of its diagonal entries perturbed by subtracting postive real numbers, then what can we say about the eigenvalues of perturbed matrix in comparison to eigenvalues of the original?
I know that if every diagonal entry of $A$ decreased uniformly by $k(>0)$ amount then real part of every eigenvalues of perturbed will be shifted by $k$ amount left side.
I am especially intrested in any comparison between $max Relambda (A)$ and $ max Relambda (A -D),~D$ is a diagonal matrix whose at least one entry is positive.
linear-algebra matrices eigenvalues-eigenvectors matrix-analysis
asked Feb 2 at 10:30
user602672user602672
283
$endgroup$
|
show 6 more comments
$begingroup$
Let $A$ be a real matrix, if some of its diagonal entries perturbed by subtracting postive real numbers, then what can we say about the eigenvalues of perturbed matrix in comparison to eigenvalues of the original?
I know that if every diagonal entry of $A$ decreased uniformly by $k(>0)$ amount then real part of every eigenvalues of perturbed will be shifted by $k$ amount left side.
I am especially intrested in any comparison between $max Relambda (A)$ and $ max Relambda (A -D),~D$ is a diagonal matrix whose at least one entry is positive.
linear-algebra matrices eigenvalues-eigenvectors matrix-analysis
asked Feb 2 at 10:30
user602672user602672
283
$endgroup$
$begingroup$
There is a notion of $le$ for Hermitian matrices but for arbitrary matrices I don't know what you mean. And what is $lambda$, any complex number?
$endgroup$
– Keith McClary
Feb 3 at 0:41
$begingroup$
@Keith McClary arbitrary means; no restriction on the matrix type. $lambda(M)$ represents the eigenvalue of the matrix.
$endgroup$
– user602672
Feb 3 at 12:10
$begingroup$
Do you have reason to believe this is true, say by numerical simulation?
$endgroup$
– Nao
Feb 3 at 12:35
1
$begingroup$
Subtracting $kI$ just reduces the real part of the eigenvalue by $k$, so you can move $kI$ to the LHS, if that's any help.
$endgroup$
– Keith McClary
Feb 4 at 21:21
1
$begingroup$
I mean, you could reframe your question as "what happens to the eigenvalues of $A$ under the perturbation $(kI-D)$".
$endgroup$
– Keith McClary
Feb 11 at 21:12
|
show 6 more comments
$begingroup$
Let $A$ be a real matrix, if some of its diagonal entries perturbed by subtracting postive real numbers, then what can we say about the eigenvalues of perturbed matrix in comparison to eigenvalues of the original?
I know that if every diagonal entry of $A$ decreased uniformly by $k(>0)$ amount then real part of every eigenvalues of perturbed will be shifted by $k$ amount left side.
I am especially intrested in any comparison between $max Relambda (A)$ and $ max Relambda (A -D),~D$ is a diagonal matrix whose at least one entry is positive.
linear-algebra matrices eigenvalues-eigenvectors matrix-analysis
asked Feb 2 at 10:30
user602672user602672
283
$endgroup$
Let $A$ be a real matrix, if some of its diagonal entries perturbed by subtracting postive real numbers, then what can we say about the eigenvalues of perturbed matrix in comparison to eigenvalues of the original?
I know that if every diagonal entry of $A$ decreased uniformly by $k(>0)$ amount then real part of every eigenvalues of perturbed will be shifted by $k$ amount left side.
I am especially intrested in any comparison between $max Relambda (A)$ and $ max Relambda (A -D),~D$ is a diagonal matrix whose at least one entry is positive.
linear-algebra matrices eigenvalues-eigenvectors matrix-analysis
linear-algebra matrices eigenvalues-eigenvectors matrix-analysis
asked Feb 2 at 10:30
user602672user602672
283
asked Feb 2 at 10:30
user602672user602672
283
edited Feb 12 at 13:31
user602672
asked Feb 2 at 10:30
user602672user602672
283
asked Feb 2 at 10:30
user602672user602672
283
asked Feb 2 at 10:30
user602672user602672
283
283
$begingroup$
There is a notion of $le$ for Hermitian matrices but for arbitrary matrices I don't know what you mean. And what is $lambda$, any complex number?
$endgroup$
– Keith McClary
Feb 3 at 0:41
$begingroup$
@Keith McClary arbitrary means; no restriction on the matrix type. $lambda(M)$ represents the eigenvalue of the matrix.
$endgroup$
– user602672
Feb 3 at 12:10
$begingroup$
Do you have reason to believe this is true, say by numerical simulation?
$endgroup$
– Nao
Feb 3 at 12:35
1
$begingroup$
Subtracting $kI$ just reduces the real part of the eigenvalue by $k$, so you can move $kI$ to the LHS, if that's any help.
$endgroup$
– Keith McClary
Feb 4 at 21:21
1
$begingroup$
I mean, you could reframe your question as "what happens to the eigenvalues of $A$ under the perturbation $(kI-D)$".
$endgroup$
– Keith McClary
Feb 11 at 21:12
|
show 6 more comments
$begingroup$
There is a notion of $le$ for Hermitian matrices but for arbitrary matrices I don't know what you mean. And what is $lambda$, any complex number?
$endgroup$
– Keith McClary
Feb 3 at 0:41
$begingroup$
@Keith McClary arbitrary means; no restriction on the matrix type. $lambda(M)$ represents the eigenvalue of the matrix.
$endgroup$
– user602672
Feb 3 at 12:10
$begingroup$
Do you have reason to believe this is true, say by numerical simulation?
$endgroup$
– Nao
Feb 3 at 12:35
1
$begingroup$
Subtracting $kI$ just reduces the real part of the eigenvalue by $k$, so you can move $kI$ to the LHS, if that's any help.
$endgroup$
– Keith McClary
Feb 4 at 21:21
1
$begingroup$
I mean, you could reframe your question as "what happens to the eigenvalues of $A$ under the perturbation $(kI-D)$".
$endgroup$
– Keith McClary
Feb 11 at 21:12
$begingroup$
There is a notion of $le$ for Hermitian matrices but for arbitrary matrices I don't know what you mean. And what is $lambda$, any complex number?
$endgroup$
– Keith McClary
Feb 3 at 0:41
$begingroup$
There is a notion of $le$ for Hermitian matrices but for arbitrary matrices I don't know what you mean. And what is $lambda$, any complex number?
$endgroup$
– Keith McClary
Feb 3 at 0:41
$begingroup$
@Keith McClary arbitrary means; no restriction on the matrix type. $lambda(M)$ represents the eigenvalue of the matrix.
$endgroup$
– user602672
Feb 3 at 12:10
$begingroup$
@Keith McClary arbitrary means; no restriction on the matrix type. $lambda(M)$ represents the eigenvalue of the matrix.
$endgroup$
– user602672
Feb 3 at 12:10
$begingroup$
Do you have reason to believe this is true, say by numerical simulation?
$endgroup$
– Nao
Feb 3 at 12:35
$begingroup$
Do you have reason to believe this is true, say by numerical simulation?
$endgroup$
– Nao
Feb 3 at 12:35
1
1
$begingroup$
Subtracting $kI$ just reduces the real part of the eigenvalue by $k$, so you can move $kI$ to the LHS, if that's any help.
$endgroup$
– Keith McClary
Feb 4 at 21:21
$begingroup$
Subtracting $kI$ just reduces the real part of the eigenvalue by $k$, so you can move $kI$ to the LHS, if that's any help.
$endgroup$
– Keith McClary
Feb 4 at 21:21
1
1
$begingroup$
I mean, you could reframe your question as "what happens to the eigenvalues of $A$ under the perturbation $(kI-D)$".
$endgroup$
– Keith McClary
Feb 11 at 21:12
$begingroup$
I mean, you could reframe your question as "what happens to the eigenvalues of $A$ under the perturbation $(kI-D)$".
$endgroup$
– Keith McClary
Feb 11 at 21:12
|
show 6 more comments
0
active
oldest
votes
Your Answer
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%2f3097174%2feigenvalue-comparison%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
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%2f3097174%2feigenvalue-comparison%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
$begingroup$
There is a notion of $le$ for Hermitian matrices but for arbitrary matrices I don't know what you mean. And what is $lambda$, any complex number?
$endgroup$
– Keith McClary
Feb 3 at 0:41
$begingroup$
@Keith McClary arbitrary means; no restriction on the matrix type. $lambda(M)$ represents the eigenvalue of the matrix.
$endgroup$
– user602672
Feb 3 at 12:10
$begingroup$
Do you have reason to believe this is true, say by numerical simulation?
$endgroup$
– Nao
Feb 3 at 12:35
1
$begingroup$
Subtracting $kI$ just reduces the real part of the eigenvalue by $k$, so you can move $kI$ to the LHS, if that's any help.
$endgroup$
– Keith McClary
Feb 4 at 21:21
1
$begingroup$
I mean, you could reframe your question as "what happens to the eigenvalues of $A$ under the perturbation $(kI-D)$".
$endgroup$
– Keith McClary
Feb 11 at 21:12