Sym_n/O(n) is a closed Weyl chamber

Idea. The spectral theorem says two symmetric matrices are orthogonally conjugate iff they have the same eigenvalues, so the orbits are the fibres of $\lambda$.

Step 1: fibres. If $S=Q\Lambda Q^T$ and $T=Q'\Lambda Q'^T$ with the same ordered diagonal $\Lambda$, then $T=(Q'Q^T)S(Q'Q^T)^T$ is a conjugate of $S$ by an orthogonal matrix. Conversely conjugation preserves eigenvalues.

Step 2: surjectivity. For $\mu\in\mathcal C$, $\lambda(\operatorname{diag}\mu)=\mu$.

Step 3: continuity. By the Courant-Fischer formula
$$\lambda_k(S)=\max_{\dim V=k}\ \min_{x\in V,\ |x|=1}x^TSx.$$
If $\|S-T\|\le\delta$ (operator norm), then $|x^TSx-x^TTx|\le\delta$ for unit $x$, so each min changes by at most $\delta$, and so does the max: $|\lambda_k(S)-\lambda_k(T)|\le\|S-T\|$ (Weyl's inequality). So $\lambda$ is $1$-Lipschitz in each coordinate.

Step 4: closedness. Let $F\subset\mathrm{Sym}_n$ be closed with $\lambda(S_m)\to\mu$, $S_m\in F$. Then $\|S_m\|=\max_i|\lambda_i(S_m)|$ is bounded, so $S_{m_j}\to S\in F$ along a subsequence and $\lambda(S)=\mu$ by continuity. So $\lambda(F)$ is closed.

Step 5: conclusion. $\lambda$ is a continuous closed surjection with fibres equal to the orbits, so it induces a homeomorphism $\mathrm{Sym}_n/O(n)\to\mathcal C$.

Remark. The same pattern works for $\mathrm{Herm}_n/U(n)$ and, with singular values in place of eigenvalues, for the next problem. The chamber is a cone with non-empty interior; the interior corresponds to matrices with simple spectrum.

Lesson. "Eigenvalues are a complete conjugacy invariant" is a topological statement once properness is checked.