Is spectral decomposition a differentiable function?

Question

Given a symmetric $n\times n$ matrix $A$, we can write it as: $A=U\operatorname{diag}(\lambda_1, \dots, \lambda_n)U^T$. So the spectral decomposition of $A$ is a function of that maps $A$ to $U(A)$ and the eigenvalues of $A$. Is U(A) differentiable in $A$?

Answer 1

Idea: near a point (matrix) $A_0$ with $n$ different eigenvalues, apply the implicit function theorem to the equation $$\det(A - \lambda I) = 0.$$