Eigenvalue problem - Right hand matrix is singular












1












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    Your issue is about the so-called "generalized eigenvalue problem". See paragraph 7.3 in en.wikipedia.org/wiki/Eigendecomposition_of_a_matrix.
    $endgroup$
    – Jean Marie
    Nov 22 '18 at 11:25












  • $begingroup$
    I would try at first to convert your issue into the eigenvalue problem $F^+Rc=lambda c$ where $F^+$ is the pseudo-inverse of $F$.
    $endgroup$
    – Jean Marie
    Nov 22 '18 at 11:29






  • 1




    $begingroup$
    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:
    $endgroup$
    – Tristan Greenwood
    Nov 24 '18 at 17:25








  • 1




    $begingroup$
    $$R = begin{array}{cccc} 0.8334971 & 4.17815877 & 13.86869435 & 34.56910269\ 4.63449654 & 13.83872632 & 33.52342807 & 69.92804483\ 16.28196565 & 35.93753681 & 69.83052589 & 126.14788633\ 41.73478583 & 78.23144509 & 132.15137615 & 216.12985001\ end{array} $$
    $endgroup$
    – Tristan Greenwood
    Nov 24 '18 at 17:39








  • 1




    $begingroup$
    Typical (singular) F matrix $$ F = begin{array}{cccc} 2.4674011 & 2.4674011 & 2.4674011 & 2.4674011\ 9.8696044 & 9.8696044 & 9.8696044 & 9.8696044\ 22.2066099 & 22.2066099 & 22.2066099 & 22.2066099\ 39.4784176 & 39.4784176 & 39.4784176 & 39.4784176\ end{array} $$
    $endgroup$
    – Tristan Greenwood
    Nov 24 '18 at 17:54


















1












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    Your issue is about the so-called "generalized eigenvalue problem". See paragraph 7.3 in en.wikipedia.org/wiki/Eigendecomposition_of_a_matrix.
    $endgroup$
    – Jean Marie
    Nov 22 '18 at 11:25












  • $begingroup$
    I would try at first to convert your issue into the eigenvalue problem $F^+Rc=lambda c$ where $F^+$ is the pseudo-inverse of $F$.
    $endgroup$
    – Jean Marie
    Nov 22 '18 at 11:29






  • 1




    $begingroup$
    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:
    $endgroup$
    – Tristan Greenwood
    Nov 24 '18 at 17:25








  • 1




    $begingroup$
    $$R = begin{array}{cccc} 0.8334971 & 4.17815877 & 13.86869435 & 34.56910269\ 4.63449654 & 13.83872632 & 33.52342807 & 69.92804483\ 16.28196565 & 35.93753681 & 69.83052589 & 126.14788633\ 41.73478583 & 78.23144509 & 132.15137615 & 216.12985001\ end{array} $$
    $endgroup$
    – Tristan Greenwood
    Nov 24 '18 at 17:39








  • 1




    $begingroup$
    Typical (singular) F matrix $$ F = begin{array}{cccc} 2.4674011 & 2.4674011 & 2.4674011 & 2.4674011\ 9.8696044 & 9.8696044 & 9.8696044 & 9.8696044\ 22.2066099 & 22.2066099 & 22.2066099 & 22.2066099\ 39.4784176 & 39.4784176 & 39.4784176 & 39.4784176\ end{array} $$
    $endgroup$
    – Tristan Greenwood
    Nov 24 '18 at 17:54
















1












1








1





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 9 at 11:02







Tristan Greenwood

















asked Nov 22 '18 at 11:17









Tristan GreenwoodTristan Greenwood

266




266












  • $begingroup$
    Your issue is about the so-called "generalized eigenvalue problem". See paragraph 7.3 in en.wikipedia.org/wiki/Eigendecomposition_of_a_matrix.
    $endgroup$
    – Jean Marie
    Nov 22 '18 at 11:25












  • $begingroup$
    I would try at first to convert your issue into the eigenvalue problem $F^+Rc=lambda c$ where $F^+$ is the pseudo-inverse of $F$.
    $endgroup$
    – Jean Marie
    Nov 22 '18 at 11:29






  • 1




    $begingroup$
    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:
    $endgroup$
    – Tristan Greenwood
    Nov 24 '18 at 17:25








  • 1




    $begingroup$
    $$R = begin{array}{cccc} 0.8334971 & 4.17815877 & 13.86869435 & 34.56910269\ 4.63449654 & 13.83872632 & 33.52342807 & 69.92804483\ 16.28196565 & 35.93753681 & 69.83052589 & 126.14788633\ 41.73478583 & 78.23144509 & 132.15137615 & 216.12985001\ end{array} $$
    $endgroup$
    – Tristan Greenwood
    Nov 24 '18 at 17:39








  • 1




    $begingroup$
    Typical (singular) F matrix $$ F = begin{array}{cccc} 2.4674011 & 2.4674011 & 2.4674011 & 2.4674011\ 9.8696044 & 9.8696044 & 9.8696044 & 9.8696044\ 22.2066099 & 22.2066099 & 22.2066099 & 22.2066099\ 39.4784176 & 39.4784176 & 39.4784176 & 39.4784176\ end{array} $$
    $endgroup$
    – Tristan Greenwood
    Nov 24 '18 at 17:54




















  • $begingroup$
    Your issue is about the so-called "generalized eigenvalue problem". See paragraph 7.3 in en.wikipedia.org/wiki/Eigendecomposition_of_a_matrix.
    $endgroup$
    – Jean Marie
    Nov 22 '18 at 11:25












  • $begingroup$
    I would try at first to convert your issue into the eigenvalue problem $F^+Rc=lambda c$ where $F^+$ is the pseudo-inverse of $F$.
    $endgroup$
    – Jean Marie
    Nov 22 '18 at 11:29






  • 1




    $begingroup$
    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:
    $endgroup$
    – Tristan Greenwood
    Nov 24 '18 at 17:25








  • 1




    $begingroup$
    $$R = begin{array}{cccc} 0.8334971 & 4.17815877 & 13.86869435 & 34.56910269\ 4.63449654 & 13.83872632 & 33.52342807 & 69.92804483\ 16.28196565 & 35.93753681 & 69.83052589 & 126.14788633\ 41.73478583 & 78.23144509 & 132.15137615 & 216.12985001\ end{array} $$
    $endgroup$
    – Tristan Greenwood
    Nov 24 '18 at 17:39








  • 1




    $begingroup$
    Typical (singular) F matrix $$ F = begin{array}{cccc} 2.4674011 & 2.4674011 & 2.4674011 & 2.4674011\ 9.8696044 & 9.8696044 & 9.8696044 & 9.8696044\ 22.2066099 & 22.2066099 & 22.2066099 & 22.2066099\ 39.4784176 & 39.4784176 & 39.4784176 & 39.4784176\ end{array} $$
    $endgroup$
    – Tristan Greenwood
    Nov 24 '18 at 17:54


















$begingroup$
Your issue is about the so-called "generalized eigenvalue problem". See paragraph 7.3 in en.wikipedia.org/wiki/Eigendecomposition_of_a_matrix.
$endgroup$
– Jean Marie
Nov 22 '18 at 11:25






$begingroup$
Your issue is about the so-called "generalized eigenvalue problem". See paragraph 7.3 in en.wikipedia.org/wiki/Eigendecomposition_of_a_matrix.
$endgroup$
– Jean Marie
Nov 22 '18 at 11:25














$begingroup$
I would try at first to convert your issue into the eigenvalue problem $F^+Rc=lambda c$ where $F^+$ is the pseudo-inverse of $F$.
$endgroup$
– Jean Marie
Nov 22 '18 at 11:29




$begingroup$
I would try at first to convert your issue into the eigenvalue problem $F^+Rc=lambda c$ where $F^+$ is the pseudo-inverse of $F$.
$endgroup$
– Jean Marie
Nov 22 '18 at 11:29




1




1




$begingroup$
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:
$endgroup$
– Tristan Greenwood
Nov 24 '18 at 17:25






$begingroup$
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:
$endgroup$
– Tristan Greenwood
Nov 24 '18 at 17:25






1




1




$begingroup$
$$R = begin{array}{cccc} 0.8334971 & 4.17815877 & 13.86869435 & 34.56910269\ 4.63449654 & 13.83872632 & 33.52342807 & 69.92804483\ 16.28196565 & 35.93753681 & 69.83052589 & 126.14788633\ 41.73478583 & 78.23144509 & 132.15137615 & 216.12985001\ end{array} $$
$endgroup$
– Tristan Greenwood
Nov 24 '18 at 17:39






$begingroup$
$$R = begin{array}{cccc} 0.8334971 & 4.17815877 & 13.86869435 & 34.56910269\ 4.63449654 & 13.83872632 & 33.52342807 & 69.92804483\ 16.28196565 & 35.93753681 & 69.83052589 & 126.14788633\ 41.73478583 & 78.23144509 & 132.15137615 & 216.12985001\ end{array} $$
$endgroup$
– Tristan Greenwood
Nov 24 '18 at 17:39






1




1




$begingroup$
Typical (singular) F matrix $$ F = begin{array}{cccc} 2.4674011 & 2.4674011 & 2.4674011 & 2.4674011\ 9.8696044 & 9.8696044 & 9.8696044 & 9.8696044\ 22.2066099 & 22.2066099 & 22.2066099 & 22.2066099\ 39.4784176 & 39.4784176 & 39.4784176 & 39.4784176\ end{array} $$
$endgroup$
– Tristan Greenwood
Nov 24 '18 at 17:54






$begingroup$
Typical (singular) F matrix $$ F = begin{array}{cccc} 2.4674011 & 2.4674011 & 2.4674011 & 2.4674011\ 9.8696044 & 9.8696044 & 9.8696044 & 9.8696044\ 22.2066099 & 22.2066099 & 22.2066099 & 22.2066099\ 39.4784176 & 39.4784176 & 39.4784176 & 39.4784176\ end{array} $$
$endgroup$
– Tristan Greenwood
Nov 24 '18 at 17:54












2 Answers
2






active

oldest

votes


















2












$begingroup$

I don't see how I can mark a comment as the correct answer but after much pain and tears, using the pseudo-inverse matrix as suggested by Jean Marie yielded the best results. Not completely there yet, the resulting eigenvalues are always what I am looking for divided by 2 for some reason, but it's getting there.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    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.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      $ mu = lambda (U^{T}c)$ ?
      $endgroup$
      – Tristan Greenwood
      Nov 26 '18 at 11:47










    • $begingroup$
      Yes, exactly...
      $endgroup$
      – Jean Marie
      Nov 26 '18 at 18:18










    • $begingroup$
      Has my answer been satisfying for you ?
      $endgroup$
      – Jean Marie
      Dec 19 '18 at 21:19










    • $begingroup$
      The final answer has not been much of a help to be honest, but the suggestion of the pseudoinverse has helped a lot, mainly since I was not aware of its existence sadly. After a lot more tweaking on different parts of the code it yielded the best results. Thank you very much for taking all this time to help
      $endgroup$
      – Tristan Greenwood
      Jan 9 at 10:55












    • $begingroup$
      I understand. Nevertheless, there was another side in my contribution : the remark that your matrix has identical columns (= rank one).
      $endgroup$
      – Jean Marie
      Jan 9 at 11:12











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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    I don't see how I can mark a comment as the correct answer but after much pain and tears, using the pseudo-inverse matrix as suggested by Jean Marie yielded the best results. Not completely there yet, the resulting eigenvalues are always what I am looking for divided by 2 for some reason, but it's getting there.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      I don't see how I can mark a comment as the correct answer but after much pain and tears, using the pseudo-inverse matrix as suggested by Jean Marie yielded the best results. Not completely there yet, the resulting eigenvalues are always what I am looking for divided by 2 for some reason, but it's getting there.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        I don't see how I can mark a comment as the correct answer but after much pain and tears, using the pseudo-inverse matrix as suggested by Jean Marie yielded the best results. Not completely there yet, the resulting eigenvalues are always what I am looking for divided by 2 for some reason, but it's getting there.






        share|cite|improve this answer









        $endgroup$



        I don't see how I can mark a comment as the correct answer but after much pain and tears, using the pseudo-inverse matrix as suggested by Jean Marie yielded the best results. Not completely there yet, the resulting eigenvalues are always what I am looking for divided by 2 for some reason, but it's getting there.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 9 at 11:07









        Tristan GreenwoodTristan Greenwood

        266




        266























            1












            $begingroup$

            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.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              $ mu = lambda (U^{T}c)$ ?
              $endgroup$
              – Tristan Greenwood
              Nov 26 '18 at 11:47










            • $begingroup$
              Yes, exactly...
              $endgroup$
              – Jean Marie
              Nov 26 '18 at 18:18










            • $begingroup$
              Has my answer been satisfying for you ?
              $endgroup$
              – Jean Marie
              Dec 19 '18 at 21:19










            • $begingroup$
              The final answer has not been much of a help to be honest, but the suggestion of the pseudoinverse has helped a lot, mainly since I was not aware of its existence sadly. After a lot more tweaking on different parts of the code it yielded the best results. Thank you very much for taking all this time to help
              $endgroup$
              – Tristan Greenwood
              Jan 9 at 10:55












            • $begingroup$
              I understand. Nevertheless, there was another side in my contribution : the remark that your matrix has identical columns (= rank one).
              $endgroup$
              – Jean Marie
              Jan 9 at 11:12
















            1












            $begingroup$

            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.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              $ mu = lambda (U^{T}c)$ ?
              $endgroup$
              – Tristan Greenwood
              Nov 26 '18 at 11:47










            • $begingroup$
              Yes, exactly...
              $endgroup$
              – Jean Marie
              Nov 26 '18 at 18:18










            • $begingroup$
              Has my answer been satisfying for you ?
              $endgroup$
              – Jean Marie
              Dec 19 '18 at 21:19










            • $begingroup$
              The final answer has not been much of a help to be honest, but the suggestion of the pseudoinverse has helped a lot, mainly since I was not aware of its existence sadly. After a lot more tweaking on different parts of the code it yielded the best results. Thank you very much for taking all this time to help
              $endgroup$
              – Tristan Greenwood
              Jan 9 at 10:55












            • $begingroup$
              I understand. Nevertheless, there was another side in my contribution : the remark that your matrix has identical columns (= rank one).
              $endgroup$
              – Jean Marie
              Jan 9 at 11:12














            1












            1








            1





            $begingroup$

            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.






            share|cite|improve this answer











            $endgroup$



            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.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Nov 25 '18 at 17:41

























            answered Nov 25 '18 at 11:18









            Jean MarieJean Marie

            30.3k42051




            30.3k42051












            • $begingroup$
              $ mu = lambda (U^{T}c)$ ?
              $endgroup$
              – Tristan Greenwood
              Nov 26 '18 at 11:47










            • $begingroup$
              Yes, exactly...
              $endgroup$
              – Jean Marie
              Nov 26 '18 at 18:18










            • $begingroup$
              Has my answer been satisfying for you ?
              $endgroup$
              – Jean Marie
              Dec 19 '18 at 21:19










            • $begingroup$
              The final answer has not been much of a help to be honest, but the suggestion of the pseudoinverse has helped a lot, mainly since I was not aware of its existence sadly. After a lot more tweaking on different parts of the code it yielded the best results. Thank you very much for taking all this time to help
              $endgroup$
              – Tristan Greenwood
              Jan 9 at 10:55












            • $begingroup$
              I understand. Nevertheless, there was another side in my contribution : the remark that your matrix has identical columns (= rank one).
              $endgroup$
              – Jean Marie
              Jan 9 at 11:12


















            • $begingroup$
              $ mu = lambda (U^{T}c)$ ?
              $endgroup$
              – Tristan Greenwood
              Nov 26 '18 at 11:47










            • $begingroup$
              Yes, exactly...
              $endgroup$
              – Jean Marie
              Nov 26 '18 at 18:18










            • $begingroup$
              Has my answer been satisfying for you ?
              $endgroup$
              – Jean Marie
              Dec 19 '18 at 21:19










            • $begingroup$
              The final answer has not been much of a help to be honest, but the suggestion of the pseudoinverse has helped a lot, mainly since I was not aware of its existence sadly. After a lot more tweaking on different parts of the code it yielded the best results. Thank you very much for taking all this time to help
              $endgroup$
              – Tristan Greenwood
              Jan 9 at 10:55












            • $begingroup$
              I understand. Nevertheless, there was another side in my contribution : the remark that your matrix has identical columns (= rank one).
              $endgroup$
              – Jean Marie
              Jan 9 at 11:12
















            $begingroup$
            $ mu = lambda (U^{T}c)$ ?
            $endgroup$
            – Tristan Greenwood
            Nov 26 '18 at 11:47




            $begingroup$
            $ mu = lambda (U^{T}c)$ ?
            $endgroup$
            – Tristan Greenwood
            Nov 26 '18 at 11:47












            $begingroup$
            Yes, exactly...
            $endgroup$
            – Jean Marie
            Nov 26 '18 at 18:18




            $begingroup$
            Yes, exactly...
            $endgroup$
            – Jean Marie
            Nov 26 '18 at 18:18












            $begingroup$
            Has my answer been satisfying for you ?
            $endgroup$
            – Jean Marie
            Dec 19 '18 at 21:19




            $begingroup$
            Has my answer been satisfying for you ?
            $endgroup$
            – Jean Marie
            Dec 19 '18 at 21:19












            $begingroup$
            The final answer has not been much of a help to be honest, but the suggestion of the pseudoinverse has helped a lot, mainly since I was not aware of its existence sadly. After a lot more tweaking on different parts of the code it yielded the best results. Thank you very much for taking all this time to help
            $endgroup$
            – Tristan Greenwood
            Jan 9 at 10:55






            $begingroup$
            The final answer has not been much of a help to be honest, but the suggestion of the pseudoinverse has helped a lot, mainly since I was not aware of its existence sadly. After a lot more tweaking on different parts of the code it yielded the best results. Thank you very much for taking all this time to help
            $endgroup$
            – Tristan Greenwood
            Jan 9 at 10:55














            $begingroup$
            I understand. Nevertheless, there was another side in my contribution : the remark that your matrix has identical columns (= rank one).
            $endgroup$
            – Jean Marie
            Jan 9 at 11:12




            $begingroup$
            I understand. Nevertheless, there was another side in my contribution : the remark that your matrix has identical columns (= rank one).
            $endgroup$
            – Jean Marie
            Jan 9 at 11:12


















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