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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?

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    $\begingroup$ A smooth symmetric function can be expressed as a smooth function of the elementary symmetric polynomials: see this Encyclopedia article and the references therein. $\endgroup$
    – abx
    Jan 7, 2018 at 11:54

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See section 7 (pages 85-97) of here for full proofs.

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