when is an eigenvalue differentiable with respect to a parameter? [duplicate]

Question

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?

Answer 1

When the roots are simple, they can be chosen as smooth functions of $\omega$ if the matrix $A$ is smooth of $\omega$ ; both "smooth" above can be replaced by "analytic". This is a consequence of the implicit function theorem, since the characteristic polynomial is $P_{A(\omega)}(\lambda)$ and the simplicity of a given root gives $$ \frac{\partial P_{A(\omega)}(\lambda)}{\partial \lambda}\not=0, $$ so that the IFT provides that the equation $P_{A(\omega)}(\lambda)=0$ is equivalent to $\lambda=\alpha_j(\omega)$ with smooth $\alpha_j$, $1\le j\le $degree $P$.

The matters get awfully more complicated when the roots can be multiple ; in Kato's book, Perturbation Theory, you will find that the roots can be chosen Hölderian $C^{1/\mu}$ if the maximal multiplicity is $\mu$. However there is a drastic improvement for multiple roots of hyperbolic polynomial, i.e. a polynomial with real roots, which is within the framework of your question: then the roots can be chosen Lipschitz-continuous as proven by Bronshtein's Theorem. I recommend the paper https://www.mat.univie.ac.at/~armin/publ/hyperbolic.pdf on this topic.