Setup. Let $k$ be a field of characteristic $0$ with an algebraic closure $\overline{k}$, and $V$ a finite dimensional $k$-vector space. We fix a linear map $a: V \to V$ and write $A_a \subset \text{End}_kV$ for the $k$-subalgebra generated by $a$ (that is, $A_a$ is the $k$-span of $1, a, a^2, \ldots$). The imbedding $A_a \hookrightarrow \text{End}_kV$ makes $V$ an $A_a$-module.
The following $4$ properties of the operator $a$ are equivalent.
- $V$ is a cyclic $A_a$-module.
- We have $\dim_k(\text{End}_{A_a}V) = \dim_k(V)$.
- We have $\text{End}_{A_a}V = A_a$.
- In Jordan normal form of $a$ over $\overline{k}$, any two different Jordan blocks have different diagonal entries.
Any element $a \in \text{End}_kV$ that satisfies the equivalent properties above is called regular.
Now, let $k = \mathbb{R}$ and view $\text{M}_m(\mathbb{R})$ as a topological space using the natural identification$$\text{M}_m(\mathbb{R}) \cong \mathbb{R}^{m^2}.$$ Let$$\mathbb{O}_a = \{g \cdot a \cdot g^{-1} : g \in \text{GL}_m(\mathbb{R})\}$$denote the conjugacy class of an element $a \in \text{M}_m(\mathbb{R})$.
Question. For any $a \in \text{M}_m(\mathbb{R})$ does there exist a regular element $b \in \text{M}_m(\mathbb{R})$ such that the set $\mathbb{O}_a$ is contained in the closure of the set $\mathbb{O}_b$?
The answer is yes.
We may assume that $a$ is in Jordan normal form. Moreover, it suffices to show the result when $a$ is in the form $a=\operatorname{diag}(J_{m_1},\cdots,J_{m_k})$, where $J_p$ is the nilpotent Jordan block of dimension $p$. Let $b=J_r$ where $r=m_1+\cdots+m_k$. We consider the continuous curve $t\in [0,1]\rightarrow b_t=(1-t)b+ta$ ; note that, when $t<1$, $b_t\in O_b$ ($\operatorname{dim}(\ker(b_t))=1$) and $b_1=a$. Thus $a\in \overline{O_b}$ where $b$ is cyclic and we are done when the underlying field is $\mathbb{C}$.
EDIT. The real case. It suffices to show the result when $spectrum(a)=\{q\times i,q\times (-i)\}$ or, more precisely, when $a$ is similar to the Jordan form $a=\operatorname{diag}(iI_{m_1}+J_{m_1},\cdots,iI_{m_k}+J_{m_k},-iI_{m_1}+J_{m_1},\cdots,-iI_{m_k}+J_{m_k})$; in particular, we may assume that $a$ is the associated real Frobenius form: $a=\operatorname{diag}(F_{2m_1},\cdots,F_{2m_k})$. Let $b$ be the Frobenius block (the entries of the first subdiagonal are 1) of dimension $m=2m_1+\cdots+2m_k$ with eigenvalues $\{q\times i,q\times (-i)\}$. Note that $b$ is cyclic.
We construct a continuous curve $t\in[0,1]\rightarrow a_t\in M_m$ as follows:
Case $k=2$. Let $a_t=\begin{pmatrix}F_{2m_1}&0\\(1-t)U&F_{2m_2}\end{pmatrix}$. Assume that $t<1$; $\ker(a_t-iI_{m})$ is given by $F_{2m_1}=ix,(1-t)Ux+F_{2m_2}y=iy$; assume that $x\not= 0$; since the kernel of $F_{2m_1}-iI_{m_1}$ is in the form $span(x_0)$, we may assume that $x=x_0$; we choose $U$ s.t. $Ux_0\notin im(F_{2m_2}-iI_{2m_2})$, that gives a contradiction. Then $x=0$ and $\operatorname{dim}(\ker(a_t-iI_m))=1$, that implies $a_t\in O_b$. Finally $a_1=a\in\overline{O_b}$ and we are done.
Case $k>2$. Use the same method. Let $a_t=\begin{pmatrix}F_{2m_1}&0&0\\(1-t)U&F_{2m_2}&0\\0&(1-t)V&F_{2m_3}\end{pmatrix}$. Constuct, as above, the matrices $U,V$ so that, when $t<1$, $x=0,y=0$ and, consequently, $\operatorname{dim}(\ker(a_t-iI_m))=1$, and so on...