How to understand the convergence to the dominant eigenvalue-eigenvector pair from the Power Method?
$begingroup$
For the Power Method I make an initial guess for the dominant eigenvector of matrix A. It can be expressed as a linear combination of the set of eigenvectors of A.
Why is it that the Power Method converges to the dominant eigenvector? Is it that the initial guess contains a non-zero component along the dominant eigenvector (by assumption) and since it has the largest (by magnitude) eigenvalue of A the scaling within the recurrence relation "brings" us closer and closer to the dominant eigenvector while the non-dominant eigenvectors are being "silenced" by the scaling?
linear-algebra eigenvalues-eigenvectors algorithms
$endgroup$
add a comment |
$begingroup$
For the Power Method I make an initial guess for the dominant eigenvector of matrix A. It can be expressed as a linear combination of the set of eigenvectors of A.
Why is it that the Power Method converges to the dominant eigenvector? Is it that the initial guess contains a non-zero component along the dominant eigenvector (by assumption) and since it has the largest (by magnitude) eigenvalue of A the scaling within the recurrence relation "brings" us closer and closer to the dominant eigenvector while the non-dominant eigenvectors are being "silenced" by the scaling?
linear-algebra eigenvalues-eigenvectors algorithms
$endgroup$
$begingroup$
The power method does not always converge. Some conditions are needed to guarantee convergence. One sufficient condition is that the dominant eigenvalue is simple and the initial vector does not lie in the sum of the the generalised eigenspaces for other eigenvalues. This is always satisfied if, for instance, the matrix is nonnegative and irreducible and the initial vector is nonnegative but nonzero.
$endgroup$
– user1551
Dec 21 '18 at 9:54
$begingroup$
@user1551 That's fine, however, I am just checking my understanding of convergence toward the dominant eigenpair with the community. Other than that thanks for the side note.
$endgroup$
– G. LC
Dec 21 '18 at 13:22
add a comment |
$begingroup$
For the Power Method I make an initial guess for the dominant eigenvector of matrix A. It can be expressed as a linear combination of the set of eigenvectors of A.
Why is it that the Power Method converges to the dominant eigenvector? Is it that the initial guess contains a non-zero component along the dominant eigenvector (by assumption) and since it has the largest (by magnitude) eigenvalue of A the scaling within the recurrence relation "brings" us closer and closer to the dominant eigenvector while the non-dominant eigenvectors are being "silenced" by the scaling?
linear-algebra eigenvalues-eigenvectors algorithms
$endgroup$
For the Power Method I make an initial guess for the dominant eigenvector of matrix A. It can be expressed as a linear combination of the set of eigenvectors of A.
Why is it that the Power Method converges to the dominant eigenvector? Is it that the initial guess contains a non-zero component along the dominant eigenvector (by assumption) and since it has the largest (by magnitude) eigenvalue of A the scaling within the recurrence relation "brings" us closer and closer to the dominant eigenvector while the non-dominant eigenvectors are being "silenced" by the scaling?
linear-algebra eigenvalues-eigenvectors algorithms
linear-algebra eigenvalues-eigenvectors algorithms
asked Dec 21 '18 at 3:48
G. LCG. LC
1033
1033
$begingroup$
The power method does not always converge. Some conditions are needed to guarantee convergence. One sufficient condition is that the dominant eigenvalue is simple and the initial vector does not lie in the sum of the the generalised eigenspaces for other eigenvalues. This is always satisfied if, for instance, the matrix is nonnegative and irreducible and the initial vector is nonnegative but nonzero.
$endgroup$
– user1551
Dec 21 '18 at 9:54
$begingroup$
@user1551 That's fine, however, I am just checking my understanding of convergence toward the dominant eigenpair with the community. Other than that thanks for the side note.
$endgroup$
– G. LC
Dec 21 '18 at 13:22
add a comment |
$begingroup$
The power method does not always converge. Some conditions are needed to guarantee convergence. One sufficient condition is that the dominant eigenvalue is simple and the initial vector does not lie in the sum of the the generalised eigenspaces for other eigenvalues. This is always satisfied if, for instance, the matrix is nonnegative and irreducible and the initial vector is nonnegative but nonzero.
$endgroup$
– user1551
Dec 21 '18 at 9:54
$begingroup$
@user1551 That's fine, however, I am just checking my understanding of convergence toward the dominant eigenpair with the community. Other than that thanks for the side note.
$endgroup$
– G. LC
Dec 21 '18 at 13:22
$begingroup$
The power method does not always converge. Some conditions are needed to guarantee convergence. One sufficient condition is that the dominant eigenvalue is simple and the initial vector does not lie in the sum of the the generalised eigenspaces for other eigenvalues. This is always satisfied if, for instance, the matrix is nonnegative and irreducible and the initial vector is nonnegative but nonzero.
$endgroup$
– user1551
Dec 21 '18 at 9:54
$begingroup$
The power method does not always converge. Some conditions are needed to guarantee convergence. One sufficient condition is that the dominant eigenvalue is simple and the initial vector does not lie in the sum of the the generalised eigenspaces for other eigenvalues. This is always satisfied if, for instance, the matrix is nonnegative and irreducible and the initial vector is nonnegative but nonzero.
$endgroup$
– user1551
Dec 21 '18 at 9:54
$begingroup$
@user1551 That's fine, however, I am just checking my understanding of convergence toward the dominant eigenpair with the community. Other than that thanks for the side note.
$endgroup$
– G. LC
Dec 21 '18 at 13:22
$begingroup$
@user1551 That's fine, however, I am just checking my understanding of convergence toward the dominant eigenpair with the community. Other than that thanks for the side note.
$endgroup$
– G. LC
Dec 21 '18 at 13:22
add a comment |
0
active
oldest
votes
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%2f3048154%2fhow-to-understand-the-convergence-to-the-dominant-eigenvalue-eigenvector-pair-fr%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%2f3048154%2fhow-to-understand-the-convergence-to-the-dominant-eigenvalue-eigenvector-pair-fr%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
$begingroup$
The power method does not always converge. Some conditions are needed to guarantee convergence. One sufficient condition is that the dominant eigenvalue is simple and the initial vector does not lie in the sum of the the generalised eigenspaces for other eigenvalues. This is always satisfied if, for instance, the matrix is nonnegative and irreducible and the initial vector is nonnegative but nonzero.
$endgroup$
– user1551
Dec 21 '18 at 9:54
$begingroup$
@user1551 That's fine, however, I am just checking my understanding of convergence toward the dominant eigenpair with the community. Other than that thanks for the side note.
$endgroup$
– G. LC
Dec 21 '18 at 13:22