Do eigenvalues of a Hermitian matrix depend smoonthly on its entries?

Question

Let's consider the following maps: $$\Lambda_i:S\to\Bbb R, \quad A\mapsto \Lambda_i(A),\quad i=1,\cdots,n$$ where $S$ is the vector space of all $n\times n$ Hermitian matrices under the Frobenius norm and $\Lambda_i(A)$ is the $i$-th greatest eigenvalue of $A$ (since all eigenvalues of $A$ are real this map is obviously well defined). Given any matrix $A\in S$, does there always exist an $\epsilon>0$ such that $\Lambda_i$ is smooth on $B_\epsilon(A)$? In the event of distinct eigenvalues, the answer is yes as a direct corollary from this post. But what about general $A$ i.e. not necessarily with distinct eigenvalues?

Answer 1

No, you can already check that this isn't true for diagonal matrices. Consider, for example, the smooth path

$$M(t) = \left[ \begin{array}{cc} t & 0 \\ 0 & -t \end{array} \right]$$

through $S$, where $t \in (-\epsilon, \epsilon)$. The eigenvalues of this matrix, in order of size, are $|t|$ and $-|t|$, so neither $\Lambda_1(M(t))$ nor $\Lambda_2(M(t))$ are smooth at $t = 0$, where the two eigenvalues of $M(t)$ collide.