# Elementary proof of an inequality for the Radon-Hurwitz numbers

Edit:
In all likelihood, the original question does not have a positive answer (see comment by abx).

Modified question: Let $$\rho_H(n)$$ be the maximal dimension of a space of symmetric real matrices contained in $$\mathrm{GL}_n(\mathbb{R})\cup\{0\}$$. Are there any upper bounds on $$\rho_H(n)$$, in terms of $$\rho_H(m)$$ and $$\rho(m)$$, that can be derived by elementary methods?

For example, in On Matrices Whose Real Linear Combinations are Nonsingular (Proc. Amer. Math. Soc., 16(2), 1965), Adams, Lax and Phillips prove that $$\rho_H(n) \le \rho(8n)-7$$.
(This, together with $$\rho(16n)\le \rho(n)+8$$ implies $$\rho_H(n)\le \rho(n/2)+1$$, which was the original motivation for my question.)

Original post:
For the purpose of this question, let us define the Radon-Hurwitz number $$\rho(n)$$ to be the maximal dimension of a subspace $$W$$ of the the real vector space $$\mathbb{R}^{n\times n}$$ of $$n\times n$$ matrices, such that $$W\subset\mathrm{GL}_n(\mathbb{R})\cup \{0\}$$.

Question: Is there an elementary proof of the inequality $$\rho(16n) \le \rho(n) + 8$$?

1. By "elementary" proof I mean one that does not rely, directly or indirectly, on $$K$$-theory.
2. From the general formula for $$\rho(n)$$ it is plain that $$\rho(16n) = \rho(n) + 8$$. My question is if the inequality can be derived without using the formula, and ideally only using methods from linear algebra and/or elementary topology.
• This inequality implies that there is no division algebra over $\mathbb{R}$ of dimension $2^n$ with $n\geq 4$. I don't think that there exists an ''elementary" (in your sense) proof of that fact.