Let $A$ be a matrix of $n\times n$ whose element is continous functiions on $\mathbb{R}^n$. Assume $A$ is an Hermitian matrix at every point of $\mathbb{R}^n$, which means all its $n$ eigenvalues are real mumbers . Let $\lambda_{1}(x) \leqslant \ldots \leqslant \lambda_{n}(x)$ be the eigenvalues of $A$
My qustion is :Is $\lambda_i$ continous for arbitrary $i\in\{1,\ldots,n\}$?
clue: In Chapter 9 of Linear Algebra and Its Applications__Wiley-Interscience__Peter D. Lax. It seems to be a standard result. But here I want to make sure.
Any ideas are appreciated!
The characteristic polynomial has coefficients which are affine functions of the individual matrix elements, hence this polynomial is a continuous function of all variables.
But the real roots in $\lambda$ can appear or disappear in pairs, causing discontinuities in the numbering.