Simple Lie algebras: making subspaces 'very transversal' Let $G$ be a Lie group or group of Lie type whose Lie algebra $\mathfrak{g}$ is simple. Because the Lie algebra is simple, for any proper subspace $V\subset \mathfrak{g}$,
there is a $g\in G$ such that $g V g^{-1} \ne V$. Is it the case that there is a $g\in G$ such that $V$ and $g V g^{-1}$ are "as transversal as possible", meaning that $\dim(V + g V g^{-1}) = \min(2 \dim(V), \dim(G))$? Is this the case, at least, for $G$ a classical group?
 A: This is not true. Namely, in the 8-dimensional $\mathfrak{sl}_3$, consider the 4-dimensional Lie subalgebra $\mathfrak{v}$ consisting of matrices of the form 
$$\begin{pmatrix} a & x & z\\ 0 & -2a & y\\ 0 & 0 & a\end{pmatrix}.$$
(This is the centralizer of $E_{13}$ in $\mathfrak{sl}_3$.) I claim that $g\mathfrak{v}g^{-1}\cap \mathfrak{v}\neq 0$ for every $g\in \mathrm{GL}_3$.
Indeed, let $\mathfrak{b}$ be the normalizer of $\mathfrak{v}$, namely the Lie subalgebra of upper triangular matrices of trace zero. Let $\mathfrak{d}$ be the subalgebra of diagonal matrices: it is a Cartan subalgebra in both $\mathfrak{b}$ and $\mathfrak{sl}_3$. It is known that the intersection of any two Borel subalgebras contains a Cartan subalgebra $\mathfrak{d}'$. This applies to the intersection $\mathfrak{b}\cap g\mathfrak{b}g^{-1}$. Conjugating by some upper triangular matrix conjugating $\mathfrak{d}'$ into $\mathfrak{d}$, we can suppose that $\mathfrak{d}'=\mathfrak{d}$. So $g\mathfrak{b}g^{-1}$ is one of the 6 Borel subalgebras containing $\mathfrak{d}$ (images of $\mathfrak{b}$ by the Weyl group $\mathfrak{S}_3$). The condition $[\mathfrak{b},\mathfrak{b}]\cap g[\mathfrak{b},\mathfrak{b}]g^{-1}=0$ forces $g\mathfrak{b}g^{-1}$ to be the opposite Borel subalgebra $\mathfrak{b}_-$, that is, the Lie subalgebra of lower triangular matrices. So $g$ maps the unique flag preserved by $\mathfrak{b}$ to the unique flag preserved by $\mathfrak{b}_-$. So $g$ is "south-east"-triangular. Right-multiplying $g$ by a suitable element of $[\mathfrak{b},\mathfrak{b}]$, we can suppose that $g$ is an anti-diagonal matrix. Then we see that $g$ centralizes $\mathfrak{d}\cap \mathfrak{v}$, hence conjugates $\mathfrak{d}\cap \mathfrak{v}$ into itself and this contradicts $\mathfrak{v}\cap g\mathfrak{v}g^{-1}=0$.
