Closure of the matrices with a non-real eigenvalue

1. $n=2$. For $A=\begin{pmatrix}a&b\\c&d\end{pmatrix}$, $\chi_A=X^2-(a+d)X+(ad-bc)$ has a non-real root iff its discriminant $\Delta(A)=(a+d)^2-4(ad-bc)=(a-d)^2+4bc$ is negative. So $\mathcal{N}_2=\{\Delta<0\}$.

*Claim:* $\overline{\mathcal{N}_2}=\{\Delta\le0\}$.
- "$\subseteq$": $\{\Delta\le0\}$ is closed and contains $\mathcal{N}_2$.
- "$\supseteq$": let $\Delta(A)=0$. Consider $\Delta$ as a function of $(x,b,c)$ where $x=a-d$: $\Delta=x^2+4bc$, with gradient $(2x,4c,4b)$. If $(x,b,c)\ne0$, the gradient is nonzero and $\Delta$ decreases strictly along $-\nabla\Delta$ near this point, so there are points with $\Delta<0$ arbitrarily close. If $x=b=c=0$ (so $A=aI$), the matrices $A_\varepsilon=aI+\begin{pmatrix}0&-\varepsilon\\\varepsilon&0\end{pmatrix}$ have $\Delta=-4\varepsilon^2<0$ and tend to $A$.

2. $n=3$. Let $A_0=\operatorname{diag}(1,2,3)$ and $\rho=\frac13$. Take $A$ with $\|A-A_0\|$ small enough that each of the intervals around $1,2,3$ of radius $\rho$ contains an eigenvalue of $A$ (by the inequality $\min_i|\mu-\lambda_i(A)|\le|\chi_A(\mu)|^{1/3}\to0$ for $\mu=1,2,3$). The three eigenvalues are in disjoint discs, hence distinct, and they are all the eigenvalues of $A$ (there are only $3$). A non-real eigenvalue $\lambda$ of the real matrix $A$ would have the conjugate $\bar\lambda\ne\lambda$ as another eigenvalue in the *same* disc (the discs are centred on the real axis and stable under conjugation), contradicting the fact that each disc contains exactly one eigenvalue. So all eigenvalues of $A$ are real.

Thus a neighbourhood of $\operatorname{diag}(1,2,3)$ misses $\mathcal{N}_3$, so $\operatorname{diag}(1,2,3)\notin\overline{\mathcal{N}_3}$.

Remark. For $n=2$, scalar matrices $aI$ are on the boundary. In odd dimension the characteristic polynomial always has a real root, but that is not what we use. Lesson. Open conditions (distinct real eigenvalues) show a set is not dense.