# Given a symmetric bilinear map $B:\mathbb{R}^8 \times \mathbb{R}^8 \to \mathbb{R}$ and a subspace $U,\ \dim{U}=7$ then $rk(B|_U)\geq6$

Given a symmetric bilinear map $B:\mathbb{R}^8 \times \mathbb{R}^8 \to \mathbb{R}$ and a subspace $U,\ \dim{U}=7$ with the signature $(5,3)$ then $rk(B|_U)\geq6$?

Meaning if we look at $B|_{U \times U}$ is it possible to find such subspace on which the rank will be less then 6 (i.e a counter example)? Or is this statement true?

I've tried constructing all kind of "counter examples" (for instance choosing 3 vectors that span a 3 dimensional sub-space $W$ then $\forall u,v B(u,v) =0$ but then when I've tried completing it to a base of a $7$ dimensional space I've always ended up with rank 6 or seven). Is there some property I'm missing? If there is, I'd really like a direction...

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You wrote “$\geq 6$”, followed by “less than $6$.” Which is it ? – Ewan Delanoy Jun 12 '13 at 11:11
@EwanDelanoy sorry maybe I was not clear, the statements says that if the conditions that I described are true then $rk(B|_{U \times U})\geq6$ , the sentance you refer to just asks whether there might be a counter example e.g we can find such $B$ with the signature $(5,3)$ but $rk(B|_{U \times U})<6$ thus proving it wrong. But I'm not at all sure it's wrong because as I stated I've tried looking at different $7$ dimensional sub-spaces and $B$ was always of $rk\geq6$ but that of course does not prove anything... – SadStudent Jun 12 '13 at 11:20