Inverting the sum of a circulant and a diagonal matrix
$begingroup$
Is there an efficient way of (numerically) solving a linear system, where the matrix is a sum of circulant and diagonal matrices?
I need to solve a linear system of the form
(V+D)x=y
where
V is a circulant matrix, symmetric, positive definite, and invertible, but with eigenvalues spanning a wide range.
D is a diagonal matrix, the values on the diagonal varying between 0 and 1.
Either component can be easily inverted, D directly and V using Fourier technique, but their sum is not. I do get a solution via conjugate-gradient iteration, with V+I as preconditioner, but the convergence is painfully slow (without preconditioning it is horrible). I would need to solve this kind of system for thousands of times, with V always staying the same but D and y changing. The size of the matrix is of the order of 10^5.
Is there a general technique for solving this kind of linear systems, or alternatively, can someone suggest a better preconditioner?
linear-algebra numerical-linear-algebra
asked Jan 8 at 18:01
KosmoelinaKosmoelina
112
$endgroup$
|
show 1 more comment
$begingroup$
Is there an efficient way of (numerically) solving a linear system, where the matrix is a sum of circulant and diagonal matrices?
I need to solve a linear system of the form
(V+D)x=y
where
V is a circulant matrix, symmetric, positive definite, and invertible, but with eigenvalues spanning a wide range.
D is a diagonal matrix, the values on the diagonal varying between 0 and 1.
Either component can be easily inverted, D directly and V using Fourier technique, but their sum is not. I do get a solution via conjugate-gradient iteration, with V+I as preconditioner, but the convergence is painfully slow (without preconditioning it is horrible). I would need to solve this kind of system for thousands of times, with V always staying the same but D and y changing. The size of the matrix is of the order of 10^5.
Is there a general technique for solving this kind of linear systems, or alternatively, can someone suggest a better preconditioner?
linear-algebra numerical-linear-algebra
asked Jan 8 at 18:01
KosmoelinaKosmoelina
112
$endgroup$
$begingroup$
If either matrix is low-rank (or if the diagonal matrix has a lot of repeating values), the Woodbury matrix identity would be useful
$endgroup$
– Omnomnomnom
Jan 8 at 18:17
$begingroup$
Do you mean $VV^T=I$?
$endgroup$
– Mostafa Ayaz
Jan 8 at 18:31
$begingroup$
@MostafaAyaz circulant matrices are defined as is explained here
$endgroup$
– Omnomnomnom
Jan 8 at 18:37
$begingroup$
@Kosmoelina: $V$ is circulant and symmetric - so all off-diagonal elements are identical, similarly all diagonal elements?
$endgroup$
– user376343
Jan 8 at 21:58
$begingroup$
@user376343: Yes, that is true for V. The diagonal elements of D vary.
$endgroup$
– Kosmoelina
Jan 9 at 5:35
|
show 1 more comment
$begingroup$
Is there an efficient way of (numerically) solving a linear system, where the matrix is a sum of circulant and diagonal matrices?
I need to solve a linear system of the form
(V+D)x=y
where
V is a circulant matrix, symmetric, positive definite, and invertible, but with eigenvalues spanning a wide range.
D is a diagonal matrix, the values on the diagonal varying between 0 and 1.
Either component can be easily inverted, D directly and V using Fourier technique, but their sum is not. I do get a solution via conjugate-gradient iteration, with V+I as preconditioner, but the convergence is painfully slow (without preconditioning it is horrible). I would need to solve this kind of system for thousands of times, with V always staying the same but D and y changing. The size of the matrix is of the order of 10^5.
Is there a general technique for solving this kind of linear systems, or alternatively, can someone suggest a better preconditioner?
linear-algebra numerical-linear-algebra
asked Jan 8 at 18:01
KosmoelinaKosmoelina
112
$endgroup$
Is there an efficient way of (numerically) solving a linear system, where the matrix is a sum of circulant and diagonal matrices?
I need to solve a linear system of the form
(V+D)x=y
where
V is a circulant matrix, symmetric, positive definite, and invertible, but with eigenvalues spanning a wide range.
D is a diagonal matrix, the values on the diagonal varying between 0 and 1.
Either component can be easily inverted, D directly and V using Fourier technique, but their sum is not. I do get a solution via conjugate-gradient iteration, with V+I as preconditioner, but the convergence is painfully slow (without preconditioning it is horrible). I would need to solve this kind of system for thousands of times, with V always staying the same but D and y changing. The size of the matrix is of the order of 10^5.
Is there a general technique for solving this kind of linear systems, or alternatively, can someone suggest a better preconditioner?
linear-algebra numerical-linear-algebra
linear-algebra numerical-linear-algebra
asked Jan 8 at 18:01
KosmoelinaKosmoelina
112
asked Jan 8 at 18:01
KosmoelinaKosmoelina
112
asked Jan 8 at 18:01
KosmoelinaKosmoelina
112
asked Jan 8 at 18:01
KosmoelinaKosmoelina
112
asked Jan 8 at 18:01
KosmoelinaKosmoelina
112
112
$begingroup$
If either matrix is low-rank (or if the diagonal matrix has a lot of repeating values), the Woodbury matrix identity would be useful
$endgroup$
– Omnomnomnom
Jan 8 at 18:17
$begingroup$
Do you mean $VV^T=I$?
$endgroup$
– Mostafa Ayaz
Jan 8 at 18:31
$begingroup$
@MostafaAyaz circulant matrices are defined as is explained here
$endgroup$
– Omnomnomnom
Jan 8 at 18:37
$begingroup$
@Kosmoelina: $V$ is circulant and symmetric - so all off-diagonal elements are identical, similarly all diagonal elements?
$endgroup$
– user376343
Jan 8 at 21:58
$begingroup$
@user376343: Yes, that is true for V. The diagonal elements of D vary.
$endgroup$
– Kosmoelina
Jan 9 at 5:35
|
show 1 more comment
$begingroup$
If either matrix is low-rank (or if the diagonal matrix has a lot of repeating values), the Woodbury matrix identity would be useful
$endgroup$
– Omnomnomnom
Jan 8 at 18:17
$begingroup$
Do you mean $VV^T=I$?
$endgroup$
– Mostafa Ayaz
Jan 8 at 18:31
$begingroup$
@MostafaAyaz circulant matrices are defined as is explained here
$endgroup$
– Omnomnomnom
Jan 8 at 18:37
$begingroup$
@Kosmoelina: $V$ is circulant and symmetric - so all off-diagonal elements are identical, similarly all diagonal elements?
$endgroup$
– user376343
Jan 8 at 21:58
$begingroup$
@user376343: Yes, that is true for V. The diagonal elements of D vary.
$endgroup$
– Kosmoelina
Jan 9 at 5:35
$begingroup$
If either matrix is low-rank (or if the diagonal matrix has a lot of repeating values), the Woodbury matrix identity would be useful
$endgroup$
– Omnomnomnom
Jan 8 at 18:17
$begingroup$
If either matrix is low-rank (or if the diagonal matrix has a lot of repeating values), the Woodbury matrix identity would be useful
$endgroup$
– Omnomnomnom
Jan 8 at 18:17
$begingroup$
Do you mean $VV^T=I$?
$endgroup$
– Mostafa Ayaz
Jan 8 at 18:31
$begingroup$
Do you mean $VV^T=I$?
$endgroup$
– Mostafa Ayaz
Jan 8 at 18:31
$begingroup$
@MostafaAyaz circulant matrices are defined as is explained here
$endgroup$
– Omnomnomnom
Jan 8 at 18:37
$begingroup$
@MostafaAyaz circulant matrices are defined as is explained here
$endgroup$
– Omnomnomnom
Jan 8 at 18:37
$begingroup$
@Kosmoelina: $V$ is circulant and symmetric - so all off-diagonal elements are identical, similarly all diagonal elements?
$endgroup$
– user376343
Jan 8 at 21:58
$begingroup$
@Kosmoelina: $V$ is circulant and symmetric - so all off-diagonal elements are identical, similarly all diagonal elements?
$endgroup$
– user376343
Jan 8 at 21:58
$begingroup$
@user376343: Yes, that is true for V. The diagonal elements of D vary.
$endgroup$
– Kosmoelina
Jan 9 at 5:35
$begingroup$
@user376343: Yes, that is true for V. The diagonal elements of D vary.
$endgroup$
– Kosmoelina
Jan 9 at 5:35
|
show 1 more comment
0
active
oldest
votes
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%2f3066513%2finverting-the-sum-of-a-circulant-and-a-diagonal-matrix%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%2f3066513%2finverting-the-sum-of-a-circulant-and-a-diagonal-matrix%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
$begingroup$
If either matrix is low-rank (or if the diagonal matrix has a lot of repeating values), the Woodbury matrix identity would be useful
$endgroup$
– Omnomnomnom
Jan 8 at 18:17
$begingroup$
Do you mean $VV^T=I$?
$endgroup$
– Mostafa Ayaz
Jan 8 at 18:31
$begingroup$
@MostafaAyaz circulant matrices are defined as is explained here
$endgroup$
– Omnomnomnom
Jan 8 at 18:37
$begingroup$
@Kosmoelina: $V$ is circulant and symmetric - so all off-diagonal elements are identical, similarly all diagonal elements?
$endgroup$
– user376343
Jan 8 at 21:58
$begingroup$
@user376343: Yes, that is true for V. The diagonal elements of D vary.
$endgroup$
– Kosmoelina
Jan 9 at 5:35