Lowest Magnitude Eigenvalues of Large Sparse MatricesShould eigenvalues be ordered?Efficiently find all values of parameter such that any of the eigenvalues of a matrix is equal to 1Computing eigenvectors and eigenvaluesManipulating sparse array elementsCalculate the algebraic multiplicity of known eigenvalues of a large, sparse matrixNegative eigenvalues of a covariance matrixFind Eigenvalues close to a Target for a Generalised Eigenvalue problem of the type Ax=λBx with A and B very sparseEigenvalues of a symbolic 28×28 Hessian matrixfinding eigenenergy/eigenvector pairsImproving speed for calculating eigenvalues of a sequence of large matricesHow to compute eigenvalues of a large symbolic matrix?Quickly creating a sparse array
Are intrusions within a foreign embassy considered an act of war?
How to make all magic-casting innate, but still rare?
「捨ててしまう」why is there two て’s used here?
Are there any individual aliens that have gained superpowers in the Marvel universe?
Why isn't my calculation that we should be able to see the sun well beyond the observable universe valid?
"Correct me if I'm wrong"
Print the new site header
I just entered the USA without passport control at Atlanta airport
Predict the product from the reaction
Teferi's Time Twist and Gideon's Sacrifice
How can a clan of females defend themselves in the ancient world against wandering bands?
Would a 7805 5 V regulator drain a 9 V battery?
Am I legally required to provide a (GPL licensed) source code even after a project is abandoned?
Is there any possible way to get these hearts as Adult Link?
Name for a function whose effect is canceled by another function?
Justifying Affordable Bespoke Spaceships
Unable to import binding 'SforceServiceBinding' with the Summer 19 Tooling API WSDL
Unrecognized IC Package Style
'No arbitrary choices' intuition for natural transformation.
First occurrence in the Sixers sequence
How is the idea of "girlfriend material" naturally expressed in Russian?
Why are there no file insertion syscalls
Can a character learn spells from someone else's spellbook and then sell it?
In the US, can a former president run again?
Lowest Magnitude Eigenvalues of Large Sparse Matrices
Should eigenvalues be ordered?Efficiently find all values of parameter such that any of the eigenvalues of a matrix is equal to 1Computing eigenvectors and eigenvaluesManipulating sparse array elementsCalculate the algebraic multiplicity of known eigenvalues of a large, sparse matrixNegative eigenvalues of a covariance matrixFind Eigenvalues close to a Target for a Generalised Eigenvalue problem of the type Ax=λBx with A and B very sparseEigenvalues of a symbolic 28×28 Hessian matrixfinding eigenenergy/eigenvector pairsImproving speed for calculating eigenvalues of a sequence of large matricesHow to compute eigenvalues of a large symbolic matrix?Quickly creating a sparse array
$begingroup$
I am trying to find the first three lowest eigenvalues of large sparse matrices of size range $10^3 - 10^5$. The matrices depend on some parameter $x$, so I first construct the matrices and then use Eigenvalues[Mat(x), 3]. Mathematica however orders the eigenvalues by absolute value so there is no guarantee in finding the lowest eigenvalue by the method above. See figure below
I can find all the eigenvalues, order them, and then find the minimum using the instructions found here but that defeats the point of using sparse matrices or Mathematica altogether. Moreover, the code slows down even more when cycling through all values of $x$. So, can this problem be done in Mathematica?
linear-algebra sorting sparse-arrays eigenvalues
asked Jun 10 at 17:31
WikawTirsoWikawTirso
283
New contributor
WikawTirso is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I am trying to find the first three lowest eigenvalues of large sparse matrices of size range $10^3 - 10^5$. The matrices depend on some parameter $x$, so I first construct the matrices and then use Eigenvalues[Mat(x), 3]. Mathematica however orders the eigenvalues by absolute value so there is no guarantee in finding the lowest eigenvalue by the method above. See figure below
I can find all the eigenvalues, order them, and then find the minimum using the instructions found here but that defeats the point of using sparse matrices or Mathematica altogether. Moreover, the code slows down even more when cycling through all values of $x$. So, can this problem be done in Mathematica?
linear-algebra sorting sparse-arrays eigenvalues
asked Jun 10 at 17:31
WikawTirsoWikawTirso
283
New contributor
WikawTirso is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
2
$begingroup$
Why not useEigenvalues[Mat[x], -3]?
$endgroup$
– Carl Woll
Jun 10 at 17:58
$begingroup$
It is not clear what you want. Show on a simple example on a matrixm[x]of low dimension.
$endgroup$
– Alex Trounev
Jun 10 at 18:05
$begingroup$
Duplicate of mathematica.stackexchange.com/q/194807 ?
$endgroup$
– Roman
Jun 10 at 18:11
1
$begingroup$
The title is a little misleading. If you asked me for the "lowest magnitude" eigenvalues, I would assume that you wanted the ones closest to zero. Maybe "largest-magnitude negative eigenvalues" would be more accurate?
$endgroup$
– Michael Seifert
Jun 12 at 12:48
add a comment |
$begingroup$
I am trying to find the first three lowest eigenvalues of large sparse matrices of size range $10^3 - 10^5$. The matrices depend on some parameter $x$, so I first construct the matrices and then use Eigenvalues[Mat(x), 3]. Mathematica however orders the eigenvalues by absolute value so there is no guarantee in finding the lowest eigenvalue by the method above. See figure below
I can find all the eigenvalues, order them, and then find the minimum using the instructions found here but that defeats the point of using sparse matrices or Mathematica altogether. Moreover, the code slows down even more when cycling through all values of $x$. So, can this problem be done in Mathematica?
linear-algebra sorting sparse-arrays eigenvalues
asked Jun 10 at 17:31
WikawTirsoWikawTirso
283
New contributor
WikawTirso is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I am trying to find the first three lowest eigenvalues of large sparse matrices of size range $10^3 - 10^5$. The matrices depend on some parameter $x$, so I first construct the matrices and then use Eigenvalues[Mat(x), 3]. Mathematica however orders the eigenvalues by absolute value so there is no guarantee in finding the lowest eigenvalue by the method above. See figure below
I can find all the eigenvalues, order them, and then find the minimum using the instructions found here but that defeats the point of using sparse matrices or Mathematica altogether. Moreover, the code slows down even more when cycling through all values of $x$. So, can this problem be done in Mathematica?
linear-algebra sorting sparse-arrays eigenvalues
linear-algebra sorting sparse-arrays eigenvalues
asked Jun 10 at 17:31
WikawTirsoWikawTirso
283
New contributor
WikawTirso is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Jun 10 at 17:31
WikawTirsoWikawTirso
283
New contributor
WikawTirso is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Jun 12 at 12:56
WikawTirso
asked Jun 10 at 17:31
WikawTirsoWikawTirso
283
New contributor
WikawTirso is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Jun 10 at 17:31
WikawTirsoWikawTirso
283
asked Jun 10 at 17:31
WikawTirsoWikawTirso
283
283
New contributor
WikawTirso is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
WikawTirso is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
2
$begingroup$
Why not useEigenvalues[Mat[x], -3]?
$endgroup$
– Carl Woll
Jun 10 at 17:58
$begingroup$
It is not clear what you want. Show on a simple example on a matrixm[x]of low dimension.
$endgroup$
– Alex Trounev
Jun 10 at 18:05
$begingroup$
Duplicate of mathematica.stackexchange.com/q/194807 ?
$endgroup$
– Roman
Jun 10 at 18:11
1
$begingroup$
The title is a little misleading. If you asked me for the "lowest magnitude" eigenvalues, I would assume that you wanted the ones closest to zero. Maybe "largest-magnitude negative eigenvalues" would be more accurate?
$endgroup$
– Michael Seifert
Jun 12 at 12:48
add a comment |
2
$begingroup$
Why not useEigenvalues[Mat[x], -3]?
$endgroup$
– Carl Woll
Jun 10 at 17:58
$begingroup$
It is not clear what you want. Show on a simple example on a matrixm[x]of low dimension.
$endgroup$
– Alex Trounev
Jun 10 at 18:05
$begingroup$
Duplicate of mathematica.stackexchange.com/q/194807 ?
$endgroup$
– Roman
Jun 10 at 18:11
1
$begingroup$
The title is a little misleading. If you asked me for the "lowest magnitude" eigenvalues, I would assume that you wanted the ones closest to zero. Maybe "largest-magnitude negative eigenvalues" would be more accurate?
$endgroup$
– Michael Seifert
Jun 12 at 12:48
2
2
$begingroup$
Why not use
Eigenvalues[Mat[x], -3]?$endgroup$
– Carl Woll
Jun 10 at 17:58
$begingroup$
Why not use
Eigenvalues[Mat[x], -3]?$endgroup$
– Carl Woll
Jun 10 at 17:58
$begingroup$
It is not clear what you want. Show on a simple example on a matrix
m[x] of low dimension.$endgroup$
– Alex Trounev
Jun 10 at 18:05
$begingroup$
It is not clear what you want. Show on a simple example on a matrix
m[x] of low dimension.$endgroup$
– Alex Trounev
Jun 10 at 18:05
$begingroup$
Duplicate of mathematica.stackexchange.com/q/194807 ?
$endgroup$
– Roman
Jun 10 at 18:11
$begingroup$
Duplicate of mathematica.stackexchange.com/q/194807 ?
$endgroup$
– Roman
Jun 10 at 18:11
1
1
$begingroup$
The title is a little misleading. If you asked me for the "lowest magnitude" eigenvalues, I would assume that you wanted the ones closest to zero. Maybe "largest-magnitude negative eigenvalues" would be more accurate?
$endgroup$
– Michael Seifert
Jun 12 at 12:48
$begingroup$
The title is a little misleading. If you asked me for the "lowest magnitude" eigenvalues, I would assume that you wanted the ones closest to zero. Maybe "largest-magnitude negative eigenvalues" would be more accurate?
$endgroup$
– Michael Seifert
Jun 12 at 12:48
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
To calculate the lowest eigenvalues using Mathematica, I always introduce a "shift"
in the following way:
mat1 = mat - IdentityMatrix[Length[mat]]*large
and then add large to the result of Eigenvalues[mat1]. This operation leaves
the eigenvectors unchanged and I do not need to use Arnoldi specifically. Of course,
Arnoldi does a similar shift inside, however I am not sure if that is used correctly in the Mathematica implementation.
Example:
mat=1.,2.,3.,4.;
The eigenvalues are in the order given by Eigenvalues
5.37228, -0.372281.
The procedure above produces for any large enough large:
-0.372281, 5.37228
with the smallest first. Or Eigenvalues[mat1,1]+large just the lowest eigenvalue.
answered Jun 12 at 12:10
Michael WeyrauchMichael Weyrauch
34628
$endgroup$
$begingroup$
It worked! I also realize original question was needlessly ambiguous. I get the general idea of the shift but is there a way to systematically guess a value for large? Or alternatively, 'large' compared to what (|large| > 1)? Thanks.
$endgroup$
– WikawTirso
Jun 12 at 13:16
$begingroup$
Well, the idea is to introduce a shift that all eigenvalues of the "shifted" matrix are negative. So you could calculate the largest eigenvalue of the shifted matrix, and if is negative, your shift is fine. if not, shift more.
$endgroup$
– Michael Weyrauch
Jun 12 at 14:16
add a comment |
$begingroup$
Use the Arnoldi method with shift-inversion:
Eigenvalues[A, 3, Method -> "Arnoldi", "Criteria" -> "Magnitude", "Shift" -> 0]
gives you the three smallest eigenvalues by absolute value (by magnitude).
See here: Efficiently find all values of parameter such that any of the eigenvalues of a matrix is equal to 1
After comments by @HenrikSchumacher it appears that the same can be achieved with
Eigenvalues[A, -3, Method -> "Arnoldi", "Criteria" -> "Magnitude"]
and there is no need for explicit shift-inversion.
And as @CarlWoll points out, this method is the default method for sparse matrices, so even
Eigenvalues[A, -3]
achieves the same effect. For non-sparse matrices, however, specifying the method can give a large speedup.
update
It now looks like you want the eigenvalues with smallest real part, not those with smallest magnitude. This you can also achieve with the Arnoldi method:
Eigenvalues[M, 3, Method -> "Arnoldi", "Criteria" -> "RealPart"]
gives the three eigenvalues with largest real part. To get the smallest ones, do
-Eigenvalues[-M, 3, Method -> "Arnoldi", "Criteria" -> "RealPart"]
The trick is that M and -M have the same eigenvalues (up to sign) and eigenvectors.
Of course this trick also works with the other methods presented above.
answered Jun 10 at 18:06
RomanRoman
11.4k11944
$endgroup$
$begingroup$
Reading this, I realized once more that Mathematica's implementation of Arnoldi's method is really inconsistent: BothEigenvalues[A, 1, Method -> "Arnoldi", "Criteria" -> "Magnitude", "Shift" -> 0]andEigenvalues[A, -1, Method -> "Arnoldi", "Criteria" -> "Magnitude", "Shift" -> 0]return the same.
$endgroup$
– Henrik Schumacher
Jun 10 at 21:30
$begingroup$
@HenrikSchumacher I would report this to Wolfram support. This problem only happens with theShiftoption.
$endgroup$
– Roman
Jun 10 at 21:43
$begingroup$
I have reported it already ([CASE:4270162]), but it cannot harm to build up some peer pressure. =)
$endgroup$
– Henrik Schumacher
Jun 10 at 21:49
1
$begingroup$
I believe Eigenvalues with the default method already uses the "Arnoldi" method when the input matrix is sparse and only a limited set of eigenvalues is requested. So, for sparse matrices,Eigenvalues[A, -1],Eigenvalues[A, -1, Method->"Arnoldi"]andEigenvalues[A, -1, Method->"Arnoldi", "Criteria"->"Magnitude"]should all be equivalent.
$endgroup$
– Carl Woll
Jun 10 at 22:08
$begingroup$
@CarlWoll I think you're right.
$endgroup$
– Roman
Jun 10 at 22:21
add a comment |
Your Answer
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "387"
;
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: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
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
,
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
WikawTirso is a new contributor. Be nice, and check out our Code of Conduct.
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%2fmathematica.stackexchange.com%2fquestions%2f200074%2flowest-magnitude-eigenvalues-of-large-sparse-matrices%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
$begingroup$
To calculate the lowest eigenvalues using Mathematica, I always introduce a "shift"
in the following way:
mat1 = mat - IdentityMatrix[Length[mat]]*large
and then add large to the result of Eigenvalues[mat1]. This operation leaves
the eigenvectors unchanged and I do not need to use Arnoldi specifically. Of course,
Arnoldi does a similar shift inside, however I am not sure if that is used correctly in the Mathematica implementation.
Example:
mat=1.,2.,3.,4.;
The eigenvalues are in the order given by Eigenvalues
5.37228, -0.372281.
The procedure above produces for any large enough large:
-0.372281, 5.37228
with the smallest first. Or Eigenvalues[mat1,1]+large just the lowest eigenvalue.
answered Jun 12 at 12:10
Michael WeyrauchMichael Weyrauch
34628
$endgroup$
$begingroup$
It worked! I also realize original question was needlessly ambiguous. I get the general idea of the shift but is there a way to systematically guess a value for large? Or alternatively, 'large' compared to what (|large| > 1)? Thanks.
$endgroup$
– WikawTirso
Jun 12 at 13:16
$begingroup$
Well, the idea is to introduce a shift that all eigenvalues of the "shifted" matrix are negative. So you could calculate the largest eigenvalue of the shifted matrix, and if is negative, your shift is fine. if not, shift more.
$endgroup$
– Michael Weyrauch
Jun 12 at 14:16
add a comment |
$begingroup$
To calculate the lowest eigenvalues using Mathematica, I always introduce a "shift"
in the following way:
mat1 = mat - IdentityMatrix[Length[mat]]*large
and then add large to the result of Eigenvalues[mat1]. This operation leaves
the eigenvectors unchanged and I do not need to use Arnoldi specifically. Of course,
Arnoldi does a similar shift inside, however I am not sure if that is used correctly in the Mathematica implementation.
Example:
mat=1.,2.,3.,4.;
The eigenvalues are in the order given by Eigenvalues
5.37228, -0.372281.
The procedure above produces for any large enough large:
-0.372281, 5.37228
with the smallest first. Or Eigenvalues[mat1,1]+large just the lowest eigenvalue.
answered Jun 12 at 12:10
Michael WeyrauchMichael Weyrauch
34628
$endgroup$
$begingroup$
It worked! I also realize original question was needlessly ambiguous. I get the general idea of the shift but is there a way to systematically guess a value for large? Or alternatively, 'large' compared to what (|large| > 1)? Thanks.
$endgroup$
– WikawTirso
Jun 12 at 13:16
$begingroup$
Well, the idea is to introduce a shift that all eigenvalues of the "shifted" matrix are negative. So you could calculate the largest eigenvalue of the shifted matrix, and if is negative, your shift is fine. if not, shift more.
$endgroup$
– Michael Weyrauch
Jun 12 at 14:16
add a comment |
$begingroup$
To calculate the lowest eigenvalues using Mathematica, I always introduce a "shift"
in the following way:
mat1 = mat - IdentityMatrix[Length[mat]]*large
and then add large to the result of Eigenvalues[mat1]. This operation leaves
the eigenvectors unchanged and I do not need to use Arnoldi specifically. Of course,
Arnoldi does a similar shift inside, however I am not sure if that is used correctly in the Mathematica implementation.
Example:
mat=1.,2.,3.,4.;
The eigenvalues are in the order given by Eigenvalues
5.37228, -0.372281.
The procedure above produces for any large enough large:
-0.372281, 5.37228
with the smallest first. Or Eigenvalues[mat1,1]+large just the lowest eigenvalue.
answered Jun 12 at 12:10
Michael WeyrauchMichael Weyrauch
34628
$endgroup$
To calculate the lowest eigenvalues using Mathematica, I always introduce a "shift"
in the following way:
mat1 = mat - IdentityMatrix[Length[mat]]*large
and then add large to the result of Eigenvalues[mat1]. This operation leaves
the eigenvectors unchanged and I do not need to use Arnoldi specifically. Of course,
Arnoldi does a similar shift inside, however I am not sure if that is used correctly in the Mathematica implementation.
Example:
mat=1.,2.,3.,4.;
The eigenvalues are in the order given by Eigenvalues
5.37228, -0.372281.
The procedure above produces for any large enough large:
-0.372281, 5.37228
with the smallest first. Or Eigenvalues[mat1,1]+large just the lowest eigenvalue.
answered Jun 12 at 12:10
Michael WeyrauchMichael Weyrauch
34628
edited Jun 12 at 19:42
answered Jun 12 at 12:10
Michael WeyrauchMichael Weyrauch
34628
answered Jun 12 at 12:10
Michael WeyrauchMichael Weyrauch
34628
answered Jun 12 at 12:10
Michael WeyrauchMichael Weyrauch
34628
34628
$begingroup$
It worked! I also realize original question was needlessly ambiguous. I get the general idea of the shift but is there a way to systematically guess a value for large? Or alternatively, 'large' compared to what (|large| > 1)? Thanks.
$endgroup$
– WikawTirso
Jun 12 at 13:16
$begingroup$
Well, the idea is to introduce a shift that all eigenvalues of the "shifted" matrix are negative. So you could calculate the largest eigenvalue of the shifted matrix, and if is negative, your shift is fine. if not, shift more.
$endgroup$
– Michael Weyrauch
Jun 12 at 14:16
add a comment |
$begingroup$
It worked! I also realize original question was needlessly ambiguous. I get the general idea of the shift but is there a way to systematically guess a value for large? Or alternatively, 'large' compared to what (|large| > 1)? Thanks.
$endgroup$
– WikawTirso
Jun 12 at 13:16
$begingroup$
Well, the idea is to introduce a shift that all eigenvalues of the "shifted" matrix are negative. So you could calculate the largest eigenvalue of the shifted matrix, and if is negative, your shift is fine. if not, shift more.
$endgroup$
– Michael Weyrauch
Jun 12 at 14:16
$begingroup$
It worked! I also realize original question was needlessly ambiguous. I get the general idea of the shift but is there a way to systematically guess a value for large? Or alternatively, 'large' compared to what (|large| > 1)? Thanks.
$endgroup$
– WikawTirso
Jun 12 at 13:16
$begingroup$
It worked! I also realize original question was needlessly ambiguous. I get the general idea of the shift but is there a way to systematically guess a value for large? Or alternatively, 'large' compared to what (|large| > 1)? Thanks.
$endgroup$
– WikawTirso
Jun 12 at 13:16
$begingroup$
Well, the idea is to introduce a shift that all eigenvalues of the "shifted" matrix are negative. So you could calculate the largest eigenvalue of the shifted matrix, and if is negative, your shift is fine. if not, shift more.
$endgroup$
– Michael Weyrauch
Jun 12 at 14:16
$begingroup$
Well, the idea is to introduce a shift that all eigenvalues of the "shifted" matrix are negative. So you could calculate the largest eigenvalue of the shifted matrix, and if is negative, your shift is fine. if not, shift more.
$endgroup$
– Michael Weyrauch
Jun 12 at 14:16
add a comment |
$begingroup$
Use the Arnoldi method with shift-inversion:
Eigenvalues[A, 3, Method -> "Arnoldi", "Criteria" -> "Magnitude", "Shift" -> 0]
gives you the three smallest eigenvalues by absolute value (by magnitude).
See here: Efficiently find all values of parameter such that any of the eigenvalues of a matrix is equal to 1
After comments by @HenrikSchumacher it appears that the same can be achieved with
Eigenvalues[A, -3, Method -> "Arnoldi", "Criteria" -> "Magnitude"]
and there is no need for explicit shift-inversion.
And as @CarlWoll points out, this method is the default method for sparse matrices, so even
Eigenvalues[A, -3]
achieves the same effect. For non-sparse matrices, however, specifying the method can give a large speedup.
update
It now looks like you want the eigenvalues with smallest real part, not those with smallest magnitude. This you can also achieve with the Arnoldi method:
Eigenvalues[M, 3, Method -> "Arnoldi", "Criteria" -> "RealPart"]
gives the three eigenvalues with largest real part. To get the smallest ones, do
-Eigenvalues[-M, 3, Method -> "Arnoldi", "Criteria" -> "RealPart"]
The trick is that M and -M have the same eigenvalues (up to sign) and eigenvectors.
Of course this trick also works with the other methods presented above.
answered Jun 10 at 18:06
RomanRoman
11.4k11944
$endgroup$
$begingroup$
Reading this, I realized once more that Mathematica's implementation of Arnoldi's method is really inconsistent: BothEigenvalues[A, 1, Method -> "Arnoldi", "Criteria" -> "Magnitude", "Shift" -> 0]andEigenvalues[A, -1, Method -> "Arnoldi", "Criteria" -> "Magnitude", "Shift" -> 0]return the same.
$endgroup$
– Henrik Schumacher
Jun 10 at 21:30
$begingroup$
@HenrikSchumacher I would report this to Wolfram support. This problem only happens with theShiftoption.
$endgroup$
– Roman
Jun 10 at 21:43
$begingroup$
I have reported it already ([CASE:4270162]), but it cannot harm to build up some peer pressure. =)
$endgroup$
– Henrik Schumacher
Jun 10 at 21:49
1
$begingroup$
I believe Eigenvalues with the default method already uses the "Arnoldi" method when the input matrix is sparse and only a limited set of eigenvalues is requested. So, for sparse matrices,Eigenvalues[A, -1],Eigenvalues[A, -1, Method->"Arnoldi"]andEigenvalues[A, -1, Method->"Arnoldi", "Criteria"->"Magnitude"]should all be equivalent.
$endgroup$
– Carl Woll
Jun 10 at 22:08
$begingroup$
@CarlWoll I think you're right.
$endgroup$
– Roman
Jun 10 at 22:21
add a comment |
$begingroup$
Use the Arnoldi method with shift-inversion:
Eigenvalues[A, 3, Method -> "Arnoldi", "Criteria" -> "Magnitude", "Shift" -> 0]
gives you the three smallest eigenvalues by absolute value (by magnitude).
See here: Efficiently find all values of parameter such that any of the eigenvalues of a matrix is equal to 1
After comments by @HenrikSchumacher it appears that the same can be achieved with
Eigenvalues[A, -3, Method -> "Arnoldi", "Criteria" -> "Magnitude"]
and there is no need for explicit shift-inversion.
And as @CarlWoll points out, this method is the default method for sparse matrices, so even
Eigenvalues[A, -3]
achieves the same effect. For non-sparse matrices, however, specifying the method can give a large speedup.
update
It now looks like you want the eigenvalues with smallest real part, not those with smallest magnitude. This you can also achieve with the Arnoldi method:
Eigenvalues[M, 3, Method -> "Arnoldi", "Criteria" -> "RealPart"]
gives the three eigenvalues with largest real part. To get the smallest ones, do
-Eigenvalues[-M, 3, Method -> "Arnoldi", "Criteria" -> "RealPart"]
The trick is that M and -M have the same eigenvalues (up to sign) and eigenvectors.
Of course this trick also works with the other methods presented above.
answered Jun 10 at 18:06
RomanRoman
11.4k11944
$endgroup$
$begingroup$
Reading this, I realized once more that Mathematica's implementation of Arnoldi's method is really inconsistent: BothEigenvalues[A, 1, Method -> "Arnoldi", "Criteria" -> "Magnitude", "Shift" -> 0]andEigenvalues[A, -1, Method -> "Arnoldi", "Criteria" -> "Magnitude", "Shift" -> 0]return the same.
$endgroup$
– Henrik Schumacher
Jun 10 at 21:30
$begingroup$
@HenrikSchumacher I would report this to Wolfram support. This problem only happens with theShiftoption.
$endgroup$
– Roman
Jun 10 at 21:43
$begingroup$
I have reported it already ([CASE:4270162]), but it cannot harm to build up some peer pressure. =)
$endgroup$
– Henrik Schumacher
Jun 10 at 21:49
1
$begingroup$
I believe Eigenvalues with the default method already uses the "Arnoldi" method when the input matrix is sparse and only a limited set of eigenvalues is requested. So, for sparse matrices,Eigenvalues[A, -1],Eigenvalues[A, -1, Method->"Arnoldi"]andEigenvalues[A, -1, Method->"Arnoldi", "Criteria"->"Magnitude"]should all be equivalent.
$endgroup$
– Carl Woll
Jun 10 at 22:08
$begingroup$
@CarlWoll I think you're right.
$endgroup$
– Roman
Jun 10 at 22:21
add a comment |
$begingroup$
Use the Arnoldi method with shift-inversion:
Eigenvalues[A, 3, Method -> "Arnoldi", "Criteria" -> "Magnitude", "Shift" -> 0]
gives you the three smallest eigenvalues by absolute value (by magnitude).
See here: Efficiently find all values of parameter such that any of the eigenvalues of a matrix is equal to 1
After comments by @HenrikSchumacher it appears that the same can be achieved with
Eigenvalues[A, -3, Method -> "Arnoldi", "Criteria" -> "Magnitude"]
and there is no need for explicit shift-inversion.
And as @CarlWoll points out, this method is the default method for sparse matrices, so even
Eigenvalues[A, -3]
achieves the same effect. For non-sparse matrices, however, specifying the method can give a large speedup.
update
It now looks like you want the eigenvalues with smallest real part, not those with smallest magnitude. This you can also achieve with the Arnoldi method:
Eigenvalues[M, 3, Method -> "Arnoldi", "Criteria" -> "RealPart"]
gives the three eigenvalues with largest real part. To get the smallest ones, do
-Eigenvalues[-M, 3, Method -> "Arnoldi", "Criteria" -> "RealPart"]
The trick is that M and -M have the same eigenvalues (up to sign) and eigenvectors.
Of course this trick also works with the other methods presented above.
answered Jun 10 at 18:06
RomanRoman
11.4k11944
$endgroup$
Use the Arnoldi method with shift-inversion:
Eigenvalues[A, 3, Method -> "Arnoldi", "Criteria" -> "Magnitude", "Shift" -> 0]
gives you the three smallest eigenvalues by absolute value (by magnitude).
See here: Efficiently find all values of parameter such that any of the eigenvalues of a matrix is equal to 1
After comments by @HenrikSchumacher it appears that the same can be achieved with
Eigenvalues[A, -3, Method -> "Arnoldi", "Criteria" -> "Magnitude"]
and there is no need for explicit shift-inversion.
And as @CarlWoll points out, this method is the default method for sparse matrices, so even
Eigenvalues[A, -3]
achieves the same effect. For non-sparse matrices, however, specifying the method can give a large speedup.
update
It now looks like you want the eigenvalues with smallest real part, not those with smallest magnitude. This you can also achieve with the Arnoldi method:
Eigenvalues[M, 3, Method -> "Arnoldi", "Criteria" -> "RealPart"]
gives the three eigenvalues with largest real part. To get the smallest ones, do
-Eigenvalues[-M, 3, Method -> "Arnoldi", "Criteria" -> "RealPart"]
The trick is that M and -M have the same eigenvalues (up to sign) and eigenvectors.
Of course this trick also works with the other methods presented above.
answered Jun 10 at 18:06
RomanRoman
11.4k11944
edited Jun 12 at 12:31
answered Jun 10 at 18:06
RomanRoman
11.4k11944
answered Jun 10 at 18:06
RomanRoman
11.4k11944
answered Jun 10 at 18:06
RomanRoman
11.4k11944
11.4k11944
$begingroup$
Reading this, I realized once more that Mathematica's implementation of Arnoldi's method is really inconsistent: BothEigenvalues[A, 1, Method -> "Arnoldi", "Criteria" -> "Magnitude", "Shift" -> 0]andEigenvalues[A, -1, Method -> "Arnoldi", "Criteria" -> "Magnitude", "Shift" -> 0]return the same.
$endgroup$
– Henrik Schumacher
Jun 10 at 21:30
$begingroup$
@HenrikSchumacher I would report this to Wolfram support. This problem only happens with theShiftoption.
$endgroup$
– Roman
Jun 10 at 21:43
$begingroup$
I have reported it already ([CASE:4270162]), but it cannot harm to build up some peer pressure. =)
$endgroup$
– Henrik Schumacher
Jun 10 at 21:49
1
$begingroup$
I believe Eigenvalues with the default method already uses the "Arnoldi" method when the input matrix is sparse and only a limited set of eigenvalues is requested. So, for sparse matrices,Eigenvalues[A, -1],Eigenvalues[A, -1, Method->"Arnoldi"]andEigenvalues[A, -1, Method->"Arnoldi", "Criteria"->"Magnitude"]should all be equivalent.
$endgroup$
– Carl Woll
Jun 10 at 22:08
$begingroup$
@CarlWoll I think you're right.
$endgroup$
– Roman
Jun 10 at 22:21
add a comment |
$begingroup$
Reading this, I realized once more that Mathematica's implementation of Arnoldi's method is really inconsistent: BothEigenvalues[A, 1, Method -> "Arnoldi", "Criteria" -> "Magnitude", "Shift" -> 0]andEigenvalues[A, -1, Method -> "Arnoldi", "Criteria" -> "Magnitude", "Shift" -> 0]return the same.
$endgroup$
– Henrik Schumacher
Jun 10 at 21:30
$begingroup$
@HenrikSchumacher I would report this to Wolfram support. This problem only happens with theShiftoption.
$endgroup$
– Roman
Jun 10 at 21:43
$begingroup$
I have reported it already ([CASE:4270162]), but it cannot harm to build up some peer pressure. =)
$endgroup$
– Henrik Schumacher
Jun 10 at 21:49
1
$begingroup$
I believe Eigenvalues with the default method already uses the "Arnoldi" method when the input matrix is sparse and only a limited set of eigenvalues is requested. So, for sparse matrices,Eigenvalues[A, -1],Eigenvalues[A, -1, Method->"Arnoldi"]andEigenvalues[A, -1, Method->"Arnoldi", "Criteria"->"Magnitude"]should all be equivalent.
$endgroup$
– Carl Woll
Jun 10 at 22:08
$begingroup$
@CarlWoll I think you're right.
$endgroup$
– Roman
Jun 10 at 22:21
$begingroup$
Reading this, I realized once more that Mathematica's implementation of Arnoldi's method is really inconsistent: Both
Eigenvalues[A, 1, Method -> "Arnoldi", "Criteria" -> "Magnitude", "Shift" -> 0] and Eigenvalues[A, -1, Method -> "Arnoldi", "Criteria" -> "Magnitude", "Shift" -> 0] return the same.$endgroup$
– Henrik Schumacher
Jun 10 at 21:30
$begingroup$
Reading this, I realized once more that Mathematica's implementation of Arnoldi's method is really inconsistent: Both
Eigenvalues[A, 1, Method -> "Arnoldi", "Criteria" -> "Magnitude", "Shift" -> 0] and Eigenvalues[A, -1, Method -> "Arnoldi", "Criteria" -> "Magnitude", "Shift" -> 0] return the same.$endgroup$
– Henrik Schumacher
Jun 10 at 21:30
$begingroup$
@HenrikSchumacher I would report this to Wolfram support. This problem only happens with the
Shift option.$endgroup$
– Roman
Jun 10 at 21:43
$begingroup$
@HenrikSchumacher I would report this to Wolfram support. This problem only happens with the
Shift option.$endgroup$
– Roman
Jun 10 at 21:43
$begingroup$
I have reported it already ([CASE:4270162]), but it cannot harm to build up some peer pressure. =)
$endgroup$
– Henrik Schumacher
Jun 10 at 21:49
$begingroup$
I have reported it already ([CASE:4270162]), but it cannot harm to build up some peer pressure. =)
$endgroup$
– Henrik Schumacher
Jun 10 at 21:49
1
1
$begingroup$
I believe Eigenvalues with the default method already uses the "Arnoldi" method when the input matrix is sparse and only a limited set of eigenvalues is requested. So, for sparse matrices,
Eigenvalues[A, -1], Eigenvalues[A, -1, Method->"Arnoldi"] and Eigenvalues[A, -1, Method->"Arnoldi", "Criteria"->"Magnitude"] should all be equivalent.$endgroup$
– Carl Woll
Jun 10 at 22:08
$begingroup$
I believe Eigenvalues with the default method already uses the "Arnoldi" method when the input matrix is sparse and only a limited set of eigenvalues is requested. So, for sparse matrices,
Eigenvalues[A, -1], Eigenvalues[A, -1, Method->"Arnoldi"] and Eigenvalues[A, -1, Method->"Arnoldi", "Criteria"->"Magnitude"] should all be equivalent.$endgroup$
– Carl Woll
Jun 10 at 22:08
$begingroup$
@CarlWoll I think you're right.
$endgroup$
– Roman
Jun 10 at 22:21
$begingroup$
@CarlWoll I think you're right.
$endgroup$
– Roman
Jun 10 at 22:21
add a comment |
WikawTirso is a new contributor. Be nice, and check out our Code of Conduct.
WikawTirso is a new contributor. Be nice, and check out our Code of Conduct.
WikawTirso is a new contributor. Be nice, and check out our Code of Conduct.
WikawTirso is a new contributor. Be nice, and check out our Code of Conduct.
Thanks for contributing an answer to Mathematica 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%2fmathematica.stackexchange.com%2fquestions%2f200074%2flowest-magnitude-eigenvalues-of-large-sparse-matrices%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
2
$begingroup$
Why not use
Eigenvalues[Mat[x], -3]?$endgroup$
– Carl Woll
Jun 10 at 17:58
$begingroup$
It is not clear what you want. Show on a simple example on a matrix
m[x]of low dimension.$endgroup$
– Alex Trounev
Jun 10 at 18:05
$begingroup$
Duplicate of mathematica.stackexchange.com/q/194807 ?
$endgroup$
– Roman
Jun 10 at 18:11
1
$begingroup$
The title is a little misleading. If you asked me for the "lowest magnitude" eigenvalues, I would assume that you wanted the ones closest to zero. Maybe "largest-magnitude negative eigenvalues" would be more accurate?
$endgroup$
– Michael Seifert
Jun 12 at 12:48