Let $m,n$ be positive integers with $m \leqslant n$, and denote by $\mu\_M$ the minimal polynomial of a matrix.


Do we know for which $m$ the set $E\_m$ of $M \in \mathfrak{M}\_n(\mathbb{R})$ such that $\deg(\mu\_M) = m$ is connected?



Connected subset of matrices ?
The answer is always yes. Indeed the set is path-connected.


Let $C(f)$ denote the companion matrix associated to the monic
polynomial $f$. Every matrix $A$ is similar to a matrix in rational
canonical form:
$$B=C(f\_1)\oplus C(f\_1 f\_2)\oplus\cdots\oplus C(f\_1 f\_2,\cdots f\_k)$$
where here $\oplus$ denotes diagonal sum. Then $m$ is the degree of
$f\_1 f\_2\cdots f\_k$. Starting with $B$ deform each $f\_i$ into a power
of $x$. We get a path from $B$ to
$$B'=C(x^{a\_1})\oplus C(x^{a\_2})\oplus\cdots\oplus C(x^{a\_k})$$
inside $E\_m$. There's a path from $B'$ in $E\_m$ given by
$$(1-t)C(x^{a\_1})\oplus (1-t)C(x^{a\_2})\oplus\cdots\oplus C(x^{a\_k})$$
ending at
$$B\_m=O\oplus C(x^m).$$
Thus there is a path in $E\_m$ from $A$ to $UB\_mU^{-1}$
where $U$ is a nonsingular matrix. If $\det(U)\ne0$ then there is
a path in $GL\_n(\mathbf{R})$ from $U$ to $I$ and so a path in $E\_m$
from $A$ to $B\_m$. If $m< n$ then there is a matrix $V$ of negative determinant
with $VB\_m V^{-1}=B\_m$ so that we may take $U$ to have positive determinant.


The only case that remains is when $m=n$. In this case $E\_m$ contains
diagonal matrices with distinct entries, and each of these commutes
with a matrix of negative determinant.


