Eigenvalue problem - Right hand matrix is singular
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I am constructing an eigenvalue problem of the form
$$[R]{c} = lambda [F]{c}$$
The matrices are populated by the results of some integrals
$$
I_{i,j} = int f(x,y,i,j) dxdy quad for quad i=1,..,N quad j=1,...,M
$$
All the numbers are coming out wrong the eigenvalues are nonsensical and do not converge as the matrices get larger, they just get larger in turn, and I am trying to troubleshoot. I noticed that $[F]$ always is singular. I added some "salt" ($1e-10$) so that the program did not rebel on me but I am thinkning that this might indicate some deeper issue about my problem formulation/computation, although I am not really sure what.
So my question is: Does the fact that $[F]$ is singular, point to any such problems and if yes how should I go about correcting it? Furthermore any advice on where to focus the troubleshooting?
Cheers
eigenvalues-eigenvectors programming
asked Nov 22 at 11:17
Tristan Greenwood
1
|
show 8 more comments
up vote
0
down vote
favorite
I am constructing an eigenvalue problem of the form
$$[R]{c} = lambda [F]{c}$$
The matrices are populated by the results of some integrals
$$
I_{i,j} = int f(x,y,i,j) dxdy quad for quad i=1,..,N quad j=1,...,M
$$
All the numbers are coming out wrong the eigenvalues are nonsensical and do not converge as the matrices get larger, they just get larger in turn, and I am trying to troubleshoot. I noticed that $[F]$ always is singular. I added some "salt" ($1e-10$) so that the program did not rebel on me but I am thinkning that this might indicate some deeper issue about my problem formulation/computation, although I am not really sure what.
So my question is: Does the fact that $[F]$ is singular, point to any such problems and if yes how should I go about correcting it? Furthermore any advice on where to focus the troubleshooting?
Cheers
eigenvalues-eigenvectors programming
asked Nov 22 at 11:17
Tristan Greenwood
1
Your issue is about the so-called "generalized eigenvalue problem". See paragraph 7.3 in en.wikipedia.org/wiki/Eigendecomposition_of_a_matrix.
– Jean Marie
Nov 22 at 11:25
I would try at first to convert your issue into the eigenvalue problem $F^+Rc=lambda c$ where $F^+$ is the pseudo-inverse of $F$.
– Jean Marie
Nov 22 at 11:29
1
I am trying to find the lowest value that would cause buckling on a composite plate. This would be the lowest eigenvalue, and the program should converge on that solution as the matrices get larger. Here is a sample $R$ matrix: Here an F matrix:
– Tristan Greenwood
Nov 24 at 17:25
1
$$R = begin{array}{cccc} 0.8334971 & 4.17815877 & 13.86869435 & 34.56910269\ 4.63449654 & 13.83872632 & 33.52342807 & 69.92804483\ 16.28196565 & 35.93753681 & 69.83052589 & 126.14788633\ 41.73478583 & 78.23144509 & 132.15137615 & 216.12985001\ end{array} $$
– Tristan Greenwood
Nov 24 at 17:39
1
Typical (singular) F matrix $$ F = begin{array}{cccc} 2.4674011 & 2.4674011 & 2.4674011 & 2.4674011\ 9.8696044 & 9.8696044 & 9.8696044 & 9.8696044\ 22.2066099 & 22.2066099 & 22.2066099 & 22.2066099\ 39.4784176 & 39.4784176 & 39.4784176 & 39.4784176\ end{array} $$
– Tristan Greenwood
Nov 24 at 17:54
|
show 8 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am constructing an eigenvalue problem of the form
$$[R]{c} = lambda [F]{c}$$
The matrices are populated by the results of some integrals
$$
I_{i,j} = int f(x,y,i,j) dxdy quad for quad i=1,..,N quad j=1,...,M
$$
All the numbers are coming out wrong the eigenvalues are nonsensical and do not converge as the matrices get larger, they just get larger in turn, and I am trying to troubleshoot. I noticed that $[F]$ always is singular. I added some "salt" ($1e-10$) so that the program did not rebel on me but I am thinkning that this might indicate some deeper issue about my problem formulation/computation, although I am not really sure what.
So my question is: Does the fact that $[F]$ is singular, point to any such problems and if yes how should I go about correcting it? Furthermore any advice on where to focus the troubleshooting?
Cheers
eigenvalues-eigenvectors programming
asked Nov 22 at 11:17
Tristan Greenwood
1
I am constructing an eigenvalue problem of the form
$$[R]{c} = lambda [F]{c}$$
The matrices are populated by the results of some integrals
$$
I_{i,j} = int f(x,y,i,j) dxdy quad for quad i=1,..,N quad j=1,...,M
$$
All the numbers are coming out wrong the eigenvalues are nonsensical and do not converge as the matrices get larger, they just get larger in turn, and I am trying to troubleshoot. I noticed that $[F]$ always is singular. I added some "salt" ($1e-10$) so that the program did not rebel on me but I am thinkning that this might indicate some deeper issue about my problem formulation/computation, although I am not really sure what.
So my question is: Does the fact that $[F]$ is singular, point to any such problems and if yes how should I go about correcting it? Furthermore any advice on where to focus the troubleshooting?
Cheers
eigenvalues-eigenvectors programming
eigenvalues-eigenvectors programming
asked Nov 22 at 11:17
Tristan Greenwood
1
asked Nov 22 at 11:17
Tristan Greenwood
1
asked Nov 22 at 11:17
Tristan Greenwood
1
asked Nov 22 at 11:17
Tristan Greenwood
1
asked Nov 22 at 11:17
Tristan Greenwood
1
1
Your issue is about the so-called "generalized eigenvalue problem". See paragraph 7.3 in en.wikipedia.org/wiki/Eigendecomposition_of_a_matrix.
– Jean Marie
Nov 22 at 11:25
I would try at first to convert your issue into the eigenvalue problem $F^+Rc=lambda c$ where $F^+$ is the pseudo-inverse of $F$.
– Jean Marie
Nov 22 at 11:29
1
I am trying to find the lowest value that would cause buckling on a composite plate. This would be the lowest eigenvalue, and the program should converge on that solution as the matrices get larger. Here is a sample $R$ matrix: Here an F matrix:
– Tristan Greenwood
Nov 24 at 17:25
1
$$R = begin{array}{cccc} 0.8334971 & 4.17815877 & 13.86869435 & 34.56910269\ 4.63449654 & 13.83872632 & 33.52342807 & 69.92804483\ 16.28196565 & 35.93753681 & 69.83052589 & 126.14788633\ 41.73478583 & 78.23144509 & 132.15137615 & 216.12985001\ end{array} $$
– Tristan Greenwood
Nov 24 at 17:39
1
Typical (singular) F matrix $$ F = begin{array}{cccc} 2.4674011 & 2.4674011 & 2.4674011 & 2.4674011\ 9.8696044 & 9.8696044 & 9.8696044 & 9.8696044\ 22.2066099 & 22.2066099 & 22.2066099 & 22.2066099\ 39.4784176 & 39.4784176 & 39.4784176 & 39.4784176\ end{array} $$
– Tristan Greenwood
Nov 24 at 17:54
|
show 8 more comments
Your issue is about the so-called "generalized eigenvalue problem". See paragraph 7.3 in en.wikipedia.org/wiki/Eigendecomposition_of_a_matrix.
– Jean Marie
Nov 22 at 11:25
I would try at first to convert your issue into the eigenvalue problem $F^+Rc=lambda c$ where $F^+$ is the pseudo-inverse of $F$.
– Jean Marie
Nov 22 at 11:29
1
I am trying to find the lowest value that would cause buckling on a composite plate. This would be the lowest eigenvalue, and the program should converge on that solution as the matrices get larger. Here is a sample $R$ matrix: Here an F matrix:
– Tristan Greenwood
Nov 24 at 17:25
1
$$R = begin{array}{cccc} 0.8334971 & 4.17815877 & 13.86869435 & 34.56910269\ 4.63449654 & 13.83872632 & 33.52342807 & 69.92804483\ 16.28196565 & 35.93753681 & 69.83052589 & 126.14788633\ 41.73478583 & 78.23144509 & 132.15137615 & 216.12985001\ end{array} $$
– Tristan Greenwood
Nov 24 at 17:39
1
Typical (singular) F matrix $$ F = begin{array}{cccc} 2.4674011 & 2.4674011 & 2.4674011 & 2.4674011\ 9.8696044 & 9.8696044 & 9.8696044 & 9.8696044\ 22.2066099 & 22.2066099 & 22.2066099 & 22.2066099\ 39.4784176 & 39.4784176 & 39.4784176 & 39.4784176\ end{array} $$
– Tristan Greenwood
Nov 24 at 17:54
Your issue is about the so-called "generalized eigenvalue problem". See paragraph 7.3 in en.wikipedia.org/wiki/Eigendecomposition_of_a_matrix.
– Jean Marie
Nov 22 at 11:25
Your issue is about the so-called "generalized eigenvalue problem". See paragraph 7.3 in en.wikipedia.org/wiki/Eigendecomposition_of_a_matrix.
– Jean Marie
Nov 22 at 11:25
I would try at first to convert your issue into the eigenvalue problem $F^+Rc=lambda c$ where $F^+$ is the pseudo-inverse of $F$.
– Jean Marie
Nov 22 at 11:29
I would try at first to convert your issue into the eigenvalue problem $F^+Rc=lambda c$ where $F^+$ is the pseudo-inverse of $F$.
– Jean Marie
Nov 22 at 11:29
1
1
I am trying to find the lowest value that would cause buckling on a composite plate. This would be the lowest eigenvalue, and the program should converge on that solution as the matrices get larger. Here is a sample $R$ matrix: Here an F matrix:
– Tristan Greenwood
Nov 24 at 17:25
I am trying to find the lowest value that would cause buckling on a composite plate. This would be the lowest eigenvalue, and the program should converge on that solution as the matrices get larger. Here is a sample $R$ matrix: Here an F matrix:
– Tristan Greenwood
Nov 24 at 17:25
1
1
$$R = begin{array}{cccc} 0.8334971 & 4.17815877 & 13.86869435 & 34.56910269\ 4.63449654 & 13.83872632 & 33.52342807 & 69.92804483\ 16.28196565 & 35.93753681 & 69.83052589 & 126.14788633\ 41.73478583 & 78.23144509 & 132.15137615 & 216.12985001\ end{array} $$
– Tristan Greenwood
Nov 24 at 17:39
$$R = begin{array}{cccc} 0.8334971 & 4.17815877 & 13.86869435 & 34.56910269\ 4.63449654 & 13.83872632 & 33.52342807 & 69.92804483\ 16.28196565 & 35.93753681 & 69.83052589 & 126.14788633\ 41.73478583 & 78.23144509 & 132.15137615 & 216.12985001\ end{array} $$
– Tristan Greenwood
Nov 24 at 17:39
1
1
Typical (singular) F matrix $$ F = begin{array}{cccc} 2.4674011 & 2.4674011 & 2.4674011 & 2.4674011\ 9.8696044 & 9.8696044 & 9.8696044 & 9.8696044\ 22.2066099 & 22.2066099 & 22.2066099 & 22.2066099\ 39.4784176 & 39.4784176 & 39.4784176 & 39.4784176\ end{array} $$
– Tristan Greenwood
Nov 24 at 17:54
Typical (singular) F matrix $$ F = begin{array}{cccc} 2.4674011 & 2.4674011 & 2.4674011 & 2.4674011\ 9.8696044 & 9.8696044 & 9.8696044 & 9.8696044\ 22.2066099 & 22.2066099 & 22.2066099 & 22.2066099\ 39.4784176 & 39.4784176 & 39.4784176 & 39.4784176\ end{array} $$
– Tristan Greenwood
Nov 24 at 17:54
|
show 8 more comments
1 Answer
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If $F$ is rank one (as in the example you gave) the generalized eigenproblem
$$Rc=lambda Fc tag{1}$$
has indeed (according to you terms) a "deeper" issue.
Indeed, if $F$ is rank one, we can write it under the form :
$$F=C mathbb{U}^T tag{2}$$
with $C$ any column of $F$, (e.g. $2.4, 9.8,22.2,39.4$ in the example you have given) and $ mathbb{U}$ the column vector of $mathbb{R^4}$ with null entries.
Thus (1) becomes $Rc=lambda C(mathbb{U}^T c)$ ; as parentheses enclose in fact a number, one gets $Rc=mu C$ for a certain $mu$ ; otherwise said (provided $R$ is invertible):
$$c=mu R^{-1}C tag{3}$$
giving a a unique family of (generalized) eigenvectors ($mu$ has no constraint on it).
Having an eigenvector, it is of course easy to get the corresponding eigenvalue.
Remark : in fact, of course, this reasoning works as well in nD.
answered Nov 25 at 11:18
Jean Marie
28.6k41849
$ mu = lambda (U^{T}c)$ ?
– Tristan Greenwood
Nov 26 at 11:47
Yes, exactly...
– Jean Marie
Nov 26 at 18:18
add a comment |
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If $F$ is rank one (as in the example you gave) the generalized eigenproblem
$$Rc=lambda Fc tag{1}$$
has indeed (according to you terms) a "deeper" issue.
Indeed, if $F$ is rank one, we can write it under the form :
$$F=C mathbb{U}^T tag{2}$$
with $C$ any column of $F$, (e.g. $2.4, 9.8,22.2,39.4$ in the example you have given) and $ mathbb{U}$ the column vector of $mathbb{R^4}$ with null entries.
Thus (1) becomes $Rc=lambda C(mathbb{U}^T c)$ ; as parentheses enclose in fact a number, one gets $Rc=mu C$ for a certain $mu$ ; otherwise said (provided $R$ is invertible):
$$c=mu R^{-1}C tag{3}$$
giving a a unique family of (generalized) eigenvectors ($mu$ has no constraint on it).
Having an eigenvector, it is of course easy to get the corresponding eigenvalue.
Remark : in fact, of course, this reasoning works as well in nD.
answered Nov 25 at 11:18
Jean Marie
28.6k41849
$ mu = lambda (U^{T}c)$ ?
– Tristan Greenwood
Nov 26 at 11:47
Yes, exactly...
– Jean Marie
Nov 26 at 18:18
add a comment |
up vote
0
down vote
If $F$ is rank one (as in the example you gave) the generalized eigenproblem
$$Rc=lambda Fc tag{1}$$
has indeed (according to you terms) a "deeper" issue.
Indeed, if $F$ is rank one, we can write it under the form :
$$F=C mathbb{U}^T tag{2}$$
with $C$ any column of $F$, (e.g. $2.4, 9.8,22.2,39.4$ in the example you have given) and $ mathbb{U}$ the column vector of $mathbb{R^4}$ with null entries.
Thus (1) becomes $Rc=lambda C(mathbb{U}^T c)$ ; as parentheses enclose in fact a number, one gets $Rc=mu C$ for a certain $mu$ ; otherwise said (provided $R$ is invertible):
$$c=mu R^{-1}C tag{3}$$
giving a a unique family of (generalized) eigenvectors ($mu$ has no constraint on it).
Having an eigenvector, it is of course easy to get the corresponding eigenvalue.
Remark : in fact, of course, this reasoning works as well in nD.
answered Nov 25 at 11:18
Jean Marie
28.6k41849
$ mu = lambda (U^{T}c)$ ?
– Tristan Greenwood
Nov 26 at 11:47
Yes, exactly...
– Jean Marie
Nov 26 at 18:18
add a comment |
up vote
0
down vote
up vote
0
down vote
If $F$ is rank one (as in the example you gave) the generalized eigenproblem
$$Rc=lambda Fc tag{1}$$
has indeed (according to you terms) a "deeper" issue.
Indeed, if $F$ is rank one, we can write it under the form :
$$F=C mathbb{U}^T tag{2}$$
with $C$ any column of $F$, (e.g. $2.4, 9.8,22.2,39.4$ in the example you have given) and $ mathbb{U}$ the column vector of $mathbb{R^4}$ with null entries.
Thus (1) becomes $Rc=lambda C(mathbb{U}^T c)$ ; as parentheses enclose in fact a number, one gets $Rc=mu C$ for a certain $mu$ ; otherwise said (provided $R$ is invertible):
$$c=mu R^{-1}C tag{3}$$
giving a a unique family of (generalized) eigenvectors ($mu$ has no constraint on it).
Having an eigenvector, it is of course easy to get the corresponding eigenvalue.
Remark : in fact, of course, this reasoning works as well in nD.
answered Nov 25 at 11:18
Jean Marie
28.6k41849
If $F$ is rank one (as in the example you gave) the generalized eigenproblem
$$Rc=lambda Fc tag{1}$$
has indeed (according to you terms) a "deeper" issue.
Indeed, if $F$ is rank one, we can write it under the form :
$$F=C mathbb{U}^T tag{2}$$
with $C$ any column of $F$, (e.g. $2.4, 9.8,22.2,39.4$ in the example you have given) and $ mathbb{U}$ the column vector of $mathbb{R^4}$ with null entries.
Thus (1) becomes $Rc=lambda C(mathbb{U}^T c)$ ; as parentheses enclose in fact a number, one gets $Rc=mu C$ for a certain $mu$ ; otherwise said (provided $R$ is invertible):
$$c=mu R^{-1}C tag{3}$$
giving a a unique family of (generalized) eigenvectors ($mu$ has no constraint on it).
Having an eigenvector, it is of course easy to get the corresponding eigenvalue.
Remark : in fact, of course, this reasoning works as well in nD.
answered Nov 25 at 11:18
Jean Marie
28.6k41849
edited Nov 25 at 17:41
answered Nov 25 at 11:18
Jean Marie
28.6k41849
answered Nov 25 at 11:18
Jean Marie
28.6k41849
answered Nov 25 at 11:18
Jean Marie
28.6k41849
28.6k41849
$ mu = lambda (U^{T}c)$ ?
– Tristan Greenwood
Nov 26 at 11:47
Yes, exactly...
– Jean Marie
Nov 26 at 18:18
add a comment |
$ mu = lambda (U^{T}c)$ ?
– Tristan Greenwood
Nov 26 at 11:47
Yes, exactly...
– Jean Marie
Nov 26 at 18:18
$ mu = lambda (U^{T}c)$ ?
– Tristan Greenwood
Nov 26 at 11:47
$ mu = lambda (U^{T}c)$ ?
– Tristan Greenwood
Nov 26 at 11:47
Yes, exactly...
– Jean Marie
Nov 26 at 18:18
Yes, exactly...
– Jean Marie
Nov 26 at 18:18
add a comment |
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Your issue is about the so-called "generalized eigenvalue problem". See paragraph 7.3 in en.wikipedia.org/wiki/Eigendecomposition_of_a_matrix.
– Jean Marie
Nov 22 at 11:25
I would try at first to convert your issue into the eigenvalue problem $F^+Rc=lambda c$ where $F^+$ is the pseudo-inverse of $F$.
– Jean Marie
Nov 22 at 11:29
1
I am trying to find the lowest value that would cause buckling on a composite plate. This would be the lowest eigenvalue, and the program should converge on that solution as the matrices get larger. Here is a sample $R$ matrix: Here an F matrix:
– Tristan Greenwood
Nov 24 at 17:25
1
$$R = begin{array}{cccc} 0.8334971 & 4.17815877 & 13.86869435 & 34.56910269\ 4.63449654 & 13.83872632 & 33.52342807 & 69.92804483\ 16.28196565 & 35.93753681 & 69.83052589 & 126.14788633\ 41.73478583 & 78.23144509 & 132.15137615 & 216.12985001\ end{array} $$
– Tristan Greenwood
Nov 24 at 17:39
1
Typical (singular) F matrix $$ F = begin{array}{cccc} 2.4674011 & 2.4674011 & 2.4674011 & 2.4674011\ 9.8696044 & 9.8696044 & 9.8696044 & 9.8696044\ 22.2066099 & 22.2066099 & 22.2066099 & 22.2066099\ 39.4784176 & 39.4784176 & 39.4784176 & 39.4784176\ end{array} $$
– Tristan Greenwood
Nov 24 at 17:54