Orbits of real-diagonalisable matrices are closed

Setup. Let $\lambda_1,\dots,\lambda_r$ be the distinct eigenvalues of $A$ (real), and $p(X)=\prod_{i=1}^r(X-\lambda_i)$. Since $A$ is diagonalisable, $p$ is its minimal polynomial and $p(A)=0$.

Claim. $\mathcal{O}(A)=\mathcal{F}$ where $\mathcal{F}=\{B\in M_n(\mathbb{R}):p(B)=0\text{ and }\chi_B=\chi_A\}$.

*($\subseteq$)* If $B=PAP^{-1}$, then $p(B)=Pp(A)P^{-1}=0$ and $\chi_B=\chi_A$.

*($\supseteq$)* Let $B\in\mathcal{F}$. The minimal polynomial of $B$ divides $p$, which has simple real roots, so $B$ is diagonalisable over $\mathbb{R}$. Its eigenvalues with multiplicities are read from $\chi_B=\chi_A$, so $B$ is similar over $\mathbb{R}$ to the same diagonal matrix as $A$, i.e. $B=PAP^{-1}$ with $P$ real.

Closedness. $\{B:p(B)=0\}$ is the zero set of the polynomial map $B\mapsto p(B)$, hence closed; $\{B:\chi_B=\chi_A\}$ is the preimage of a point under the continuous map $B\mapsto\chi_B$ (coefficients are polynomial in the entries). The intersection $\mathcal{F}=\mathcal{O}(A)$ is closed.

Remark. Together with the next problem this shows: *the orbit of a real matrix is closed iff the matrix is diagonalisable over $\mathbb{C}$.* Here the "if" for complex eigenvalues needs a similar argument with the real polynomial $p=\prod(X-\lambda_i)$ over the distinct complex eigenvalues. Lesson. To prove an orbit closed, describe it by closed algebraic conditions (minimal polynomial + characteristic polynomial).