Let $k$ be an arbitrary field (in my case $k = \mathbb Q_p$), and $G\subset \mathrm{GL}(n)_{/k}$ a reductive group. Let $G^0$ be its identity connected component.
Suppose that $G^0(k)$ contains an element with pairwise distinct eigenvalues (in the natural representation $G(k)\subset \mathrm{GL}(n, k)$), and hence a Zariski-dense subset of such elements.
Is it the case that every connected component of $G(k)$ contains an element with pairwise distinct eigenvalues?
No. Consider the group $H$ of $2\times 2$ matrices of the form $d_t=\begin{pmatrix}t & 0 \\ 0 & t^{-1}\end{pmatrix}$ or $c_t=\begin{pmatrix} 0 & u^{-1} \\ u & 0\end{pmatrix}$ with $t\in k^*$. Let $G$ be the group of $4\times 4$ matrices of the form either $D_{t,u}=\begin{pmatrix}d_t & 0 \\ 0 & d_u\end{pmatrix}$ or $C_{t,u}=\begin{pmatrix}c_t & 0 \\ 0 & c_u\end{pmatrix}$, $t,u\in k^*$.
Its identity component contains a matrix with no double eigenvalue (any $D_{t,u}$ with $t,u,t^{-1},u^{-1}$ pairwise distinct. On the other hand, the other component consists of the $C_{t,u}$, which have the spectrum $(1,-1,1,-1)$.
