Matrices with a repeated eigenvalue are rare

Idea. $\mathcal D$ is an *algebraic hypersurface*: the zero set of a single non-zero polynomial, hence closed and nowhere dense.

Step 1: $\mathcal D$ is the zero set of a polynomial. As in the problem on distinct eigenvalues, $\Delta(A)=\operatorname{disc}(\chi_A)$ is a polynomial in the entries of $A$, and $\chi_A$ has a repeated root iff $\Delta(A)=0$. So $\mathcal D=\Delta^{-1}(\{0\})$ is closed.

Step 2: $\Delta\not\equiv0$. For $A_0=\operatorname{diag}(1,2,\dots,n)$ the roots of $\chi_{A_0}$ are $1,\dots,n$, distinct, so $\Delta(A_0)\neq0$.

Step 3: a non-zero polynomial does not vanish on a ball. Suppose a polynomial $p$ in $N=n^2$ variables vanished on a ball $B(a,r)$. Then all partial derivatives of $p$ at $a$ vanish (they are limits of difference quotients of the zero function inside the ball). A polynomial is equal to its Taylor expansion at $a$, which has all coefficients $\frac{\partial^\alpha p(a)}{\alpha!}=0$. So $p\equiv0$.

Step 4: conclusion. By Steps 2 and 3, $\Delta$ does not vanish identically on any ball, so $\mathcal D$ contains no ball: $\operatorname{int}\mathcal D=\emptyset$.

Remark.
- For $n=1$ there are no repeated roots, so $\mathcal D=\emptyset$.
- The argument is a template: *the zero set of a non-zero polynomial on $\mathbb{R}^N$ is closed and nowhere dense.* It covers the set of singular matrices ($\det=0$), the set $\{\operatorname{rank}\le r\}$, the set of matrices with a repeated eigenvalue, and so on.
- For $n=2$, $\mathcal D=\{\Delta=(\operatorname{tr})^2-4\det=0\}$ is the boundary of the set of matrices with a real eigenvalue (see the problem on real eigenvalues).