Let $f: \mathbb{R}^n \rightarrow \mathbb{R}$ be a smooth function, which is symmetric under the action of the symmetric group (acting on $\mathbb{R}^n$ by permuting the variables).


Let $M\_{n\times n}$ be the space of $n\times n$ matrices. Using $f$, we can form a function $F: M\_{n\times n} \rightarrow \mathbb{R}$ by setting
$$ F(A) = f\bigl(\lambda\_1(A), \dots, \lambda\_n(A)\bigr),$$
where $\lambda\_1(A), \dots, \lambda\_n(A)$ are the eigenvalues, repeated with multiplicity (where the ordering is irrelevant as $f$ is symmetric).


**Q:** Is $F$ always smooth? How do I show this?


This is obvious if $f$ is a polynomial, since then I can write $f = g\circ (e\_1, \dots, e\_n)$, where $e\_1, \dots, e\_n$ are the elementary symmetric polynomials, and that the functions $e\_j(\lambda\_1(A), \dots, \lambda\_n(A))$ are smooth is well-known (they are polynomials in the entries of $A$). This also implies the same result if $f$ is analytic. But what is $f$ is merely smooth?



Symmetric functions of eigenvalues
See section 7 (pages 85-97) of [here](http://www.mat.univie.ac.at/~michor/dgbook.pdf) for full proofs.


