Nonderogatory matrices form an open dense set

Open. Let $V(A)=\bigl[\operatorname{vec}I\,|\,\operatorname{vec}A\,|\dots|\operatorname{vec}A^{n-1}\bigr]\in M_{n^2,n}(\mathbb{R})$. The columns are independent iff $\operatorname{rank}V(A)=n$ iff at least one of the $n\times n$ minors $m_S(A)$ of $V(A)$ (one for each choice $S$ of $n$ rows) is nonzero. Each $m_S$ is a polynomial in the entries of $A$, hence continuous. So
$$\mathcal{C}=\bigcup_S\{m_S\ne0\}$$
is a union of open sets.

Dense. If $A$ has $n$ distinct eigenvalues $\lambda_i$ then $I,A,\dots,A^{n-1}$ are independent: a relation $\sum_jc_jA^j=0$ means the polynomial $q=\sum c_jX^j$, of degree $<n$, satisfies $q(A)=0$; in an eigenbasis this gives $q(\lambda_i)=0$ for all $i$, i.e. $n$ roots for a polynomial of degree $<n$, so $q=0$. The set of matrices with distinct eigenvalues is dense, hence so is $\mathcal{C}$.

Equivalence with the minimal polynomial. Independence of $I,\dots,A^{n-1}$ means no nonzero polynomial of degree $<n$ annihilates $A$, i.e. $\deg\mu_A\ge n$; since $\mu_A\mid\chi_A$ and $\deg\chi_A=n$, this means $\mu_A=\chi_A$.

Remark. Matrices in $\mathcal{C}$ are called nonderogatory (or cyclic): they have a cyclic vector. The complement, e.g. the scalar matrices, is a closed subset with empty interior. Lesson. "Some minor $\ne0$" is a typical way to see openness of a rank condition.