Positive definiteness of a Runge-Kutta method












0












$begingroup$


A Runge-Kutta method can be characterized by the $s times s$ matrix $A$ and the $s$-element column vectors $mathbf{b}$ and $mathbf{c}$.



In this paper, a special type of Runge-Kutta method is introduced that satisfy the following particular assumptions on $A$, $mathbf{b}$ and $mathbf{c}$:





  1. $A$ is invertibile,

  2. The elements of $mathbf{b}$ are positive,

  3. The last row of $A$ is given by $mathbf{b}^top$,


  4. $mathbf{b}^top mathbf{1} = 1$, where $mathbf{1}$ is the $s$-element vector of ones,


  5. $A mathbf{1} = mathbf{c}$,

  6. The matrix $C = BA + A^top B - mathbf{b}mathbf{b}^top - mathbf{d}mathbf{d}^top$ is positive semi-definite, where $B = mathrm{diag}(mathbf{b})$ and $mathbf{d} = A^top B A^{-1}mathbf{1}$.


It is claimed without proof that under the assumptions 1-6, the matrix $A$ is positive definite, here meaning that $mathrm{Re} , lambda_j > 0$, where $lambda_j, , j = 1, dots, s$ are the eigenvalues of $A$.



My question is, how can this be seen?



It follows from the sixth assumption that $mathrm{Re} , lambda_j geq 0$, and the first assumption excludes the possibility of a zero eigenvalue, but how do we see that there are no non-zero imaginary eigenvalues?










share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    A Runge-Kutta method can be characterized by the $s times s$ matrix $A$ and the $s$-element column vectors $mathbf{b}$ and $mathbf{c}$.



    In this paper, a special type of Runge-Kutta method is introduced that satisfy the following particular assumptions on $A$, $mathbf{b}$ and $mathbf{c}$:





    1. $A$ is invertibile,

    2. The elements of $mathbf{b}$ are positive,

    3. The last row of $A$ is given by $mathbf{b}^top$,


    4. $mathbf{b}^top mathbf{1} = 1$, where $mathbf{1}$ is the $s$-element vector of ones,


    5. $A mathbf{1} = mathbf{c}$,

    6. The matrix $C = BA + A^top B - mathbf{b}mathbf{b}^top - mathbf{d}mathbf{d}^top$ is positive semi-definite, where $B = mathrm{diag}(mathbf{b})$ and $mathbf{d} = A^top B A^{-1}mathbf{1}$.


    It is claimed without proof that under the assumptions 1-6, the matrix $A$ is positive definite, here meaning that $mathrm{Re} , lambda_j > 0$, where $lambda_j, , j = 1, dots, s$ are the eigenvalues of $A$.



    My question is, how can this be seen?



    It follows from the sixth assumption that $mathrm{Re} , lambda_j geq 0$, and the first assumption excludes the possibility of a zero eigenvalue, but how do we see that there are no non-zero imaginary eigenvalues?










    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      A Runge-Kutta method can be characterized by the $s times s$ matrix $A$ and the $s$-element column vectors $mathbf{b}$ and $mathbf{c}$.



      In this paper, a special type of Runge-Kutta method is introduced that satisfy the following particular assumptions on $A$, $mathbf{b}$ and $mathbf{c}$:





      1. $A$ is invertibile,

      2. The elements of $mathbf{b}$ are positive,

      3. The last row of $A$ is given by $mathbf{b}^top$,


      4. $mathbf{b}^top mathbf{1} = 1$, where $mathbf{1}$ is the $s$-element vector of ones,


      5. $A mathbf{1} = mathbf{c}$,

      6. The matrix $C = BA + A^top B - mathbf{b}mathbf{b}^top - mathbf{d}mathbf{d}^top$ is positive semi-definite, where $B = mathrm{diag}(mathbf{b})$ and $mathbf{d} = A^top B A^{-1}mathbf{1}$.


      It is claimed without proof that under the assumptions 1-6, the matrix $A$ is positive definite, here meaning that $mathrm{Re} , lambda_j > 0$, where $lambda_j, , j = 1, dots, s$ are the eigenvalues of $A$.



      My question is, how can this be seen?



      It follows from the sixth assumption that $mathrm{Re} , lambda_j geq 0$, and the first assumption excludes the possibility of a zero eigenvalue, but how do we see that there are no non-zero imaginary eigenvalues?










      share|cite|improve this question









      $endgroup$




      A Runge-Kutta method can be characterized by the $s times s$ matrix $A$ and the $s$-element column vectors $mathbf{b}$ and $mathbf{c}$.



      In this paper, a special type of Runge-Kutta method is introduced that satisfy the following particular assumptions on $A$, $mathbf{b}$ and $mathbf{c}$:





      1. $A$ is invertibile,

      2. The elements of $mathbf{b}$ are positive,

      3. The last row of $A$ is given by $mathbf{b}^top$,


      4. $mathbf{b}^top mathbf{1} = 1$, where $mathbf{1}$ is the $s$-element vector of ones,


      5. $A mathbf{1} = mathbf{c}$,

      6. The matrix $C = BA + A^top B - mathbf{b}mathbf{b}^top - mathbf{d}mathbf{d}^top$ is positive semi-definite, where $B = mathrm{diag}(mathbf{b})$ and $mathbf{d} = A^top B A^{-1}mathbf{1}$.


      It is claimed without proof that under the assumptions 1-6, the matrix $A$ is positive definite, here meaning that $mathrm{Re} , lambda_j > 0$, where $lambda_j, , j = 1, dots, s$ are the eigenvalues of $A$.



      My question is, how can this be seen?



      It follows from the sixth assumption that $mathrm{Re} , lambda_j geq 0$, and the first assumption excludes the possibility of a zero eigenvalue, but how do we see that there are no non-zero imaginary eigenvalues?







      matrices eigenvalues-eigenvectors positive-definite runge-kutta-methods






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      asked Dec 26 '18 at 13:56









      ekkilopekkilop

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