Positive definiteness of a Runge-Kutta method
$begingroup$
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 $mathbf{c}$:
$A$ is invertibile,- The elements of $mathbf{b}$ are positive,
- The last row of $A$ is given by $mathbf{b}^top$,
$mathbf{b}^top mathbf{1} = 1$, where $mathbf{1}$ is the $s$-element vector of ones,
$A mathbf{1} = mathbf{c}$,- 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
asked Dec 26 '18 at 13:56
ekkilopekkilop
1,714519
$endgroup$
add a comment |
$begingroup$
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 $mathbf{c}$:
$A$ is invertibile,- The elements of $mathbf{b}$ are positive,
- The last row of $A$ is given by $mathbf{b}^top$,
$mathbf{b}^top mathbf{1} = 1$, where $mathbf{1}$ is the $s$-element vector of ones,
$A mathbf{1} = mathbf{c}$,- 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
asked Dec 26 '18 at 13:56
ekkilopekkilop
1,714519
$endgroup$
add a comment |
$begingroup$
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 $mathbf{c}$:
$A$ is invertibile,- The elements of $mathbf{b}$ are positive,
- The last row of $A$ is given by $mathbf{b}^top$,
$mathbf{b}^top mathbf{1} = 1$, where $mathbf{1}$ is the $s$-element vector of ones,
$A mathbf{1} = mathbf{c}$,- 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
asked Dec 26 '18 at 13:56
ekkilopekkilop
1,714519
$endgroup$
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 $mathbf{c}$:
$A$ is invertibile,- The elements of $mathbf{b}$ are positive,
- The last row of $A$ is given by $mathbf{b}^top$,
$mathbf{b}^top mathbf{1} = 1$, where $mathbf{1}$ is the $s$-element vector of ones,
$A mathbf{1} = mathbf{c}$,- 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
matrices eigenvalues-eigenvectors positive-definite runge-kutta-methods
asked Dec 26 '18 at 13:56
ekkilopekkilop
1,714519
asked Dec 26 '18 at 13:56
ekkilopekkilop
1,714519
asked Dec 26 '18 at 13:56
ekkilopekkilop
1,714519
asked Dec 26 '18 at 13:56
ekkilopekkilop
1,714519
asked Dec 26 '18 at 13:56
ekkilopekkilop
1,714519
1,714519
add a comment |
add a comment |
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