# Linearly independent skew symmetric complex matrices having the least eigenvalues

Question: Let $A$, $B$ be two $5 \times 5$ (or $7 \times 7$) skew-symmetric complex matrices (i.e. $A^t = -A$), and suppose that $$\forall t,s \in \mathbb{C}, \quad M(t,s):=(tA+sB)^*(tA+sB) \text{ has at most one non-zero eigenvalue } \lambda > 0.$$ Then, how to prove that $A$ and $B$ must in fact be linearly dependent?

Why $5$ or $7$? The reason for the strange numbers is the following: a complex skew-symmetric $n\times n$ matrix $A$ enjoys the Youla decomposition (see the answer to my previous question here: Inequality of Frobenius norm for skew matrices ), which implies that $A^*A$ has (non-negative) eigenvalues coming in pairs. I am interested in when they are as homogeneous as possible, and for odd $n$ this means $n-1$ copies of $\lambda$ and one of $0$. I have a proof of the above fact for $n \geq 9$: First remark that hypothesis and thesis are invariant under common conjugation by a unitary matrix. Let $V(t,s)$ be the $\lambda(t,s)$-eigenspace of $M(t,s)$, and consider $$W = V(1,0) \cap V(0,1) \cap V(1,1) \cap V(1,i).$$ Take an orthonormal basis of $\mathbb{C}^n$ such that the first $n-4$ vectors form a basis for $W$. In this basis, the first $n-4$ column vectors of $A$ are orthogonal, and the same holds for $B$. Some easy computations with $A+B$ and $A+iB$ allows to show that in fact they are all mutually orthogonal, thus giving $2n-8$ orthogonal vectors. Hence $2n - 8 < n$, that is, $n < 8$, unless some of them vanish.

The reason why $3$ is excluded is that the hypothesis is empty in that case, and in fact taking any two linearly independent matrices gives a counterexample. This line of ideas would give the conclusion if one could prove that $W$ has lower codimension than expected (because diagonalizing $A^*A$ and $B^*B$ somehow diagonalizes also $(A+B)^*(A+B)$ and $(A+iB)^*(A+iB)$), but I wasn't able to do that.

Motivation: These strange matrices arise from studying the subspace of the symmetric space $\mathcal{X}_n$ associated to the Lie group $SO^*(2n)$ maximizing the holomorphic sectional curvature. The above "critical" cases are $SO^*(10)$ and $SO^*(14)$. For higher (or even) $n$, this proves that there is no complex plane in any holomorphic tangent space $T_x^{1,0}\mathcal{X}_n$ maximizing the holomorphic sectional curvature. For $n = 3$, $\mathcal{X}_3 \cong \mathbb{B}^3$, the complex $3$-ball, which has constant holomorphic sectional curvature, hence there is indeed a complex $3$-space doing that (the whole tangent space).