The interior of the real-diagonalisable matrices

Idea. ($\supset$) simple real eigenvalues persist under perturbation. ($\subset$) a diagonalisable matrix with a repeated eigenvalue can be perturbed into a non-diagonalisable one.

Let $\mathcal S$ be the set of matrices with $n$ distinct real eigenvalues. Matrices in $\mathcal S$ are diagonalisable, so $\mathcal S\subset\mathcal D_{\mathbb{R}}$.

Step 1: $\mathcal S$ is open. Let $A\in\mathcal S$ with eigenvalues $r_1<\dots<r_n$. Choose reals $s_0<r_1<s_1<r_2<\dots<r_n<s_n$. The values $\chi_A(s_0),\dots,\chi_A(s_n)$ are non-zero and alternate in sign (a monic polynomial with simple roots changes sign at each root). Each $B\mapsto\chi_B(s_i)$ is continuous, so for $B$ close to $A$ the signs are the same. By the intermediate value theorem $\chi_B$ has a root in each interval $(s_{i-1},s_i)$, giving $n$ distinct real roots; as $\deg\chi_B=n$, these are all of them. So $B\in\mathcal S$. Hence $\mathcal S\subset\operatorname{int}\mathcal D_{\mathbb{R}}$.

Step 2: a diagonalisable matrix with a repeated eigenvalue is not interior. Let $A\in\mathcal D_{\mathbb{R}}$ with $A=PDP^{-1}$, $D=\operatorname{diag}(d_1,\dots,d_n)$, and suppose some value $d$ occurs $m\ge2$ times; reorder so that $d_1=d_2=d$. Put
$$A_\varepsilon=P(D+\varepsilon E_{12})P^{-1}.$$
Then $A_\varepsilon\to A$ as $\varepsilon\to0$ (conjugation is continuous). We show $A_\varepsilon$ is not diagonalisable. Let $M=D+\varepsilon E_{12}$, so $A_\varepsilon=PMP^{-1}$. The matrix $M$ is upper triangular (the extra entry lies above the diagonal) with diagonal $d_1,\dots,d_n$, so $\chi_M=\prod(t-d_i)$ and the eigenvalue $d$ has algebraic multiplicity $m$. The geometric multiplicity is $n-\operatorname{rank}(M-dI)$. We have $\operatorname{rank}(D-dI)=n-m$ (it has $n-m$ non-zero diagonal entries), and $M-dI=(D-dI)+\varepsilon E_{12}$ where column $2$ of $D-dI$ is zero (as $d_2=d$) and the new entry $\varepsilon$ lies in column $2$ and in row $1$, a row where $D-dI$ is zero. So the rank increases by $1$: $\operatorname{rank}(M-dI)=n-m+1$ and the geometric multiplicity is $m-1<m$. Hence $M$, and therefore $A_\varepsilon$, is not diagonalisable.

So $A\notin\operatorname{int}\mathcal D_{\mathbb{R}}$.

Step 3: conclusion. If $A\in\operatorname{int}\mathcal D_{\mathbb{R}}$ then $A\in\mathcal D_{\mathbb{R}}$ has all eigenvalues real, and by Step 2 no repeated eigenvalue, so $A\in\mathcal S$. Together with Step 1, $\operatorname{int}\mathcal D_{\mathbb{R}}=\mathcal S$.

Numeric check. $D=\operatorname{diag}(2,2,2,5)$, $M=D+0.1E_{12}$: $\operatorname{rank}(M-2I)=2=n-m+1$ with $n=4$, $m=3$.

Remark.
- Combined with the problem on closures (next), we see $\mathcal D_{\mathbb{R}}$ is "squeezed" between an open set $\mathcal S$ and a closed set $\overline{\mathcal D_{\mathbb{R}}}\supsetneq\mathcal S$.
- Over $\mathbb{C}$ the analogous interior is the set of matrices with $n$ distinct complex eigenvalues.