Let $\mathcal{S}_{n}(\mathbb{R})$ be the real vector space of symmetric $n\times n$ traceless matrices with real entries and let $L\subset \mathcal{S}_{n}(\mathbb{R})$ be a linear subspace. Noticing that every matrix $A\in L$ is diagonalizable, suppose that for all non-zero $A\in L,$ the matrix $A$ has distinct eigenvalues.
Question: What is the optimal upper bound on the dimension of the real vector space $L?$
There is an obvious upper bound which follows from the following observation (I'll omit the numbers to be brief). Let $\Delta\subset \mathcal{S}_{n}(\mathbb{R})$ denote the $(n-1)$-dimensional linear subspace of diagonal matrices. Then, we must have that $L\cap \Delta$ has dimension at most one. Moreover, there is nothing special about $\Delta,$ given any $g\in SO(n),$ it also needs to be true that $L\cap (g\cdot \Delta\cdot g^{-1})$ has dimension at most one. This latter condition leads to an equivalent reformulation of the question.
Question: Suppose for every $g\in SO(n),$ the intersection $L\cap (g\cdot \Delta\cdot g^{-1})$ has dimension at most one. What is the optimal upper bound on the dimension of $L?$
For Lie-theoretic minded people, there is an obvious generalization of this question. Let $\mathfrak{g}$ be a real (semi)-simple Lie algebra and $\mathfrak{g}=\mathfrak{k}\oplus \mathfrak{m}$ a Cartan decomposition. Let $L\subset \mathfrak{m}$ be a linear subspace, and suppose that for any non-zero $X\in L,$ the vector $X$ is regular. By regular, I mean that $X$ belongs to a unique Cartan subspace $\mathfrak{a}\subset \mathfrak{m}.$ Then the question is again, how big can the dimension of $L$ be?
