Eigenvalue analysis, subject to boundary conditions
For some non-linear finite element program I have a Tangent stiffness matrix $textbf{K} in mathbb{R}^{ntimes n}$, which is symmetric. I want to find the Eigenvector corresponding to the smallest Eigenvalue of this matrix, such that each component of this Eigenvector is non-negative. I know the smallest Eigenpair can be found via different methods (e.g. Inverste iterative method, or Rayleigh quotient iteration, Lanczos method etc.). But, is there a method to extract the smallest Eigenpair of the tangent stiffness matrix, such that all the components of the Eigenvector are non-negative? Is this even possible?
eigenvalues-eigenvectors nonlinear-optimization
asked Nov 27 at 11:06
Randy
163
add a comment |
For some non-linear finite element program I have a Tangent stiffness matrix $textbf{K} in mathbb{R}^{ntimes n}$, which is symmetric. I want to find the Eigenvector corresponding to the smallest Eigenvalue of this matrix, such that each component of this Eigenvector is non-negative. I know the smallest Eigenpair can be found via different methods (e.g. Inverste iterative method, or Rayleigh quotient iteration, Lanczos method etc.). But, is there a method to extract the smallest Eigenpair of the tangent stiffness matrix, such that all the components of the Eigenvector are non-negative? Is this even possible?
eigenvalues-eigenvectors nonlinear-optimization
asked Nov 27 at 11:06
Randy
163
The Perron-Frobenius theorem is one tool to guarantee positive eigenvectors. However they correspond to the largest eigenvalue of the matrix. You could maybe consider $K^{-1}$. The theorem also states that the eigenvalue is simple, so any eigenvector you compute will have either negative or positive components. The matrix need to have positive entries, I don't know if that's the case for you.
– Olivier Moschetta
Nov 27 at 11:18
In my case it is not guaranteed that the matrix has positive entries. Is there any option to guarantee non-negative eigenvectors without the property that the matrix is positive?
– Randy
Nov 27 at 11:54
Moreover, because of non-linearity I have $textbf{K} = textbf{K}(textbf{x})$, i.e. the tangent stiffness matrix depends on the Eigenvector
– Randy
Nov 27 at 11:58
add a comment |
For some non-linear finite element program I have a Tangent stiffness matrix $textbf{K} in mathbb{R}^{ntimes n}$, which is symmetric. I want to find the Eigenvector corresponding to the smallest Eigenvalue of this matrix, such that each component of this Eigenvector is non-negative. I know the smallest Eigenpair can be found via different methods (e.g. Inverste iterative method, or Rayleigh quotient iteration, Lanczos method etc.). But, is there a method to extract the smallest Eigenpair of the tangent stiffness matrix, such that all the components of the Eigenvector are non-negative? Is this even possible?
eigenvalues-eigenvectors nonlinear-optimization
asked Nov 27 at 11:06
Randy
163
For some non-linear finite element program I have a Tangent stiffness matrix $textbf{K} in mathbb{R}^{ntimes n}$, which is symmetric. I want to find the Eigenvector corresponding to the smallest Eigenvalue of this matrix, such that each component of this Eigenvector is non-negative. I know the smallest Eigenpair can be found via different methods (e.g. Inverste iterative method, or Rayleigh quotient iteration, Lanczos method etc.). But, is there a method to extract the smallest Eigenpair of the tangent stiffness matrix, such that all the components of the Eigenvector are non-negative? Is this even possible?
eigenvalues-eigenvectors nonlinear-optimization
eigenvalues-eigenvectors nonlinear-optimization
asked Nov 27 at 11:06
Randy
163
asked Nov 27 at 11:06
Randy
163
asked Nov 27 at 11:06
Randy
163
asked Nov 27 at 11:06
Randy
163
asked Nov 27 at 11:06
Randy
163
163
The Perron-Frobenius theorem is one tool to guarantee positive eigenvectors. However they correspond to the largest eigenvalue of the matrix. You could maybe consider $K^{-1}$. The theorem also states that the eigenvalue is simple, so any eigenvector you compute will have either negative or positive components. The matrix need to have positive entries, I don't know if that's the case for you.
– Olivier Moschetta
Nov 27 at 11:18
In my case it is not guaranteed that the matrix has positive entries. Is there any option to guarantee non-negative eigenvectors without the property that the matrix is positive?
– Randy
Nov 27 at 11:54
Moreover, because of non-linearity I have $textbf{K} = textbf{K}(textbf{x})$, i.e. the tangent stiffness matrix depends on the Eigenvector
– Randy
Nov 27 at 11:58
add a comment |
The Perron-Frobenius theorem is one tool to guarantee positive eigenvectors. However they correspond to the largest eigenvalue of the matrix. You could maybe consider $K^{-1}$. The theorem also states that the eigenvalue is simple, so any eigenvector you compute will have either negative or positive components. The matrix need to have positive entries, I don't know if that's the case for you.
– Olivier Moschetta
Nov 27 at 11:18
In my case it is not guaranteed that the matrix has positive entries. Is there any option to guarantee non-negative eigenvectors without the property that the matrix is positive?
– Randy
Nov 27 at 11:54
Moreover, because of non-linearity I have $textbf{K} = textbf{K}(textbf{x})$, i.e. the tangent stiffness matrix depends on the Eigenvector
– Randy
Nov 27 at 11:58
The Perron-Frobenius theorem is one tool to guarantee positive eigenvectors. However they correspond to the largest eigenvalue of the matrix. You could maybe consider $K^{-1}$. The theorem also states that the eigenvalue is simple, so any eigenvector you compute will have either negative or positive components. The matrix need to have positive entries, I don't know if that's the case for you.
– Olivier Moschetta
Nov 27 at 11:18
The Perron-Frobenius theorem is one tool to guarantee positive eigenvectors. However they correspond to the largest eigenvalue of the matrix. You could maybe consider $K^{-1}$. The theorem also states that the eigenvalue is simple, so any eigenvector you compute will have either negative or positive components. The matrix need to have positive entries, I don't know if that's the case for you.
– Olivier Moschetta
Nov 27 at 11:18
In my case it is not guaranteed that the matrix has positive entries. Is there any option to guarantee non-negative eigenvectors without the property that the matrix is positive?
– Randy
Nov 27 at 11:54
In my case it is not guaranteed that the matrix has positive entries. Is there any option to guarantee non-negative eigenvectors without the property that the matrix is positive?
– Randy
Nov 27 at 11:54
Moreover, because of non-linearity I have $textbf{K} = textbf{K}(textbf{x})$, i.e. the tangent stiffness matrix depends on the Eigenvector
– Randy
Nov 27 at 11:58
Moreover, because of non-linearity I have $textbf{K} = textbf{K}(textbf{x})$, i.e. the tangent stiffness matrix depends on the Eigenvector
– Randy
Nov 27 at 11:58
add a comment |
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The Perron-Frobenius theorem is one tool to guarantee positive eigenvectors. However they correspond to the largest eigenvalue of the matrix. You could maybe consider $K^{-1}$. The theorem also states that the eigenvalue is simple, so any eigenvector you compute will have either negative or positive components. The matrix need to have positive entries, I don't know if that's the case for you.
– Olivier Moschetta
Nov 27 at 11:18
In my case it is not guaranteed that the matrix has positive entries. Is there any option to guarantee non-negative eigenvectors without the property that the matrix is positive?
– Randy
Nov 27 at 11:54
Moreover, because of non-linearity I have $textbf{K} = textbf{K}(textbf{x})$, i.e. the tangent stiffness matrix depends on the Eigenvector
– Randy
Nov 27 at 11:58