Skip to main content
MathOverflow

Return to Question

added 10 characters in body
Source Link
Dirk
Dirk
  • 12.7k
  • 6
  • 54
  • 97

Let say we have a symmetric matrix $A(\omega)$ depending smoothly on some variables $\omega \in \Omega$ with $\Omega \subset \mathbb{R}^d$ a $d$-dimensional parameterspace (this means the eigenvalues are real). For calculating the eigenvalues we can make use of the characteristic polynomial $\rho(A( \omega))$ and by searching for the roots of this polynomial, we find the eigenvalues I am wondering if these eigenvalues are always differentiable with respect to the variables $\omega$. If the eigenvalue has multiplicity > 1, then we can’t be sure. For example, if we take the matrix $$A(\omega) = \begin{bmatrix} \omega_1+1 & \omega_2 \\ \omega_2 & -\omega_1+1 \end{bmatrix}$$ Then the characteristic polynomial $\rho(A(\omega)) = (\lambda-1)^2 - \omega_1 - \omega_2$$\rho(A(\omega)) = (\lambda-1)^2 - \omega_1^2 - \omega_2^2$. If we set this equal to 0, we get that the eigenvalues fulfil $ (\lambda-1)^2 = \omega_1 + \omega_2$$ (\lambda-1)^2 = \omega_1^2 + \omega_2^2$ so $\lambda_1(\omega)= \sqrt{ \omega_1^2 + \omega_2^2} + 1$ enand $\lambda_2(\omega) = -\sqrt{ \omega_1^2 + \omega_2^2} + 1$. This is not differentiable in $\omega_1 = \omega_2 = 0$ for both derivatives $\dfrac{\partial \lambda_i(\omega)}{\partial \omega_j} , i,j = 1,2$ and has there the same eigenvalues $\lambda_1 = \lambda_2 = 1$. In the case of simple eigenvalues (= multiplicity = 1), do we always have that the eigenvalues are differentiable?

Let say we have a symmetric matrix $A(\omega)$ depending smoothly on some variables $\omega \in \Omega$ with $\Omega \subset \mathbb{R}^d$ a $d$-dimensional parameterspace (this means the eigenvalues are real). For calculating the eigenvalues we can make use of the characteristic polynomial $\rho(A( \omega))$ and by searching for the roots of this polynomial, we find the eigenvalues I am wondering if these eigenvalues are always differentiable with respect to the variables $\omega$. If the eigenvalue has multiplicity > 1, then we can’t be sure. For example, if we take the matrix $$A(\omega) = \begin{bmatrix} \omega_1+1 & \omega_2 \\ \omega_2 & -\omega_1+1 \end{bmatrix}$$ Then the characteristic polynomial $\rho(A(\omega)) = (\lambda-1)^2 - \omega_1 - \omega_2$. If we set this equal to 0, we get that the eigenvalues fulfil $ (\lambda-1)^2 = \omega_1 + \omega_2$ so $\lambda_1(\omega)= \sqrt{ \omega_1^2 + \omega_2^2} + 1$ en $\lambda_2(\omega) = -\sqrt{ \omega_1^2 + \omega_2^2} + 1$ This is not differentiable in $\omega_1 = \omega_2 = 0$ for both derivatives $\dfrac{\partial \lambda_i(\omega)}{\partial \omega_j} , i,j = 1,2$ and has there the same eigenvalues $\lambda_1 = \lambda_2 = 1$. In the case of simple eigenvalues (= multiplicity = 1), do we always have that the eigenvalues are differentiable?

Let say we have a symmetric matrix $A(\omega)$ depending smoothly on some variables $\omega \in \Omega$ with $\Omega \subset \mathbb{R}^d$ a $d$-dimensional parameterspace (this means the eigenvalues are real). For calculating the eigenvalues we can make use of the characteristic polynomial $\rho(A( \omega))$ and by searching for the roots of this polynomial, we find the eigenvalues I am wondering if these eigenvalues are always differentiable with respect to the variables $\omega$. If the eigenvalue has multiplicity > 1, then we can’t be sure. For example, if we take the matrix $$A(\omega) = \begin{bmatrix} \omega_1+1 & \omega_2 \\ \omega_2 & -\omega_1+1 \end{bmatrix}$$ Then the characteristic polynomial $\rho(A(\omega)) = (\lambda-1)^2 - \omega_1^2 - \omega_2^2$. If we set this equal to 0, we get that the eigenvalues fulfil $ (\lambda-1)^2 = \omega_1^2 + \omega_2^2$ so $\lambda_1(\omega)= \sqrt{ \omega_1^2 + \omega_2^2} + 1$ and $\lambda_2(\omega) = -\sqrt{ \omega_1^2 + \omega_2^2} + 1$. This is not differentiable in $\omega_1 = \omega_2 = 0$ for both derivatives $\dfrac{\partial \lambda_i(\omega)}{\partial \omega_j} , i,j = 1,2$ and has there the same eigenvalues $\lambda_1 = \lambda_2 = 1$. In the case of simple eigenvalues (= multiplicity = 1), do we always have that the eigenvalues are differentiable?

Post Closed as "Duplicate" by Francois Ziegler, j.c., Christian Remling, Mikhail Katz, Jan-Christoph Schlage-Puchta
Source Link
Koen
Koen
  • 143
  • 4

when is an eigenvalue differentiable with respect to a parameter?

Let say we have a symmetric matrix $A(\omega)$ depending smoothly on some variables $\omega \in \Omega$ with $\Omega \subset \mathbb{R}^d$ a $d$-dimensional parameterspace (this means the eigenvalues are real). For calculating the eigenvalues we can make use of the characteristic polynomial $\rho(A( \omega))$ and by searching for the roots of this polynomial, we find the eigenvalues I am wondering if these eigenvalues are always differentiable with respect to the variables $\omega$. If the eigenvalue has multiplicity > 1, then we can’t be sure. For example, if we take the matrix $$A(\omega) = \begin{bmatrix} \omega_1+1 & \omega_2 \\ \omega_2 & -\omega_1+1 \end{bmatrix}$$ Then the characteristic polynomial $\rho(A(\omega)) = (\lambda-1)^2 - \omega_1 - \omega_2$. If we set this equal to 0, we get that the eigenvalues fulfil $ (\lambda-1)^2 = \omega_1 + \omega_2$ so $\lambda_1(\omega)= \sqrt{ \omega_1^2 + \omega_2^2} + 1$ en $\lambda_2(\omega) = -\sqrt{ \omega_1^2 + \omega_2^2} + 1$ This is not differentiable in $\omega_1 = \omega_2 = 0$ for both derivatives $\dfrac{\partial \lambda_i(\omega)}{\partial \omega_j} , i,j = 1,2$ and has there the same eigenvalues $\lambda_1 = \lambda_2 = 1$. In the case of simple eigenvalues (= multiplicity = 1), do we always have that the eigenvalues are differentiable?