Show that a nonsingular symmetric or skew-symmetric bilinear pairing over a field \(F\), of the form \(F^n \times F^n \to F\), cannot be identically zero when restricted to all pairs of vectors \(\nu, w\) in a \(k\)-dimensional subspace \(V \subset F^n\) if \(k > n/2\).
Solution
1. Let \(B:F^n\times F^n\to F\) be nonsingular, with matrix \(A\) (\(B(x,y)=x^tAy\)).
choose basis.
2. Suppose \(V\subset F^n\) is \(k\)-dimensional and \(B|_{V\times V}=0\).
assume for contradiction \(k>n/2\).
3. The map \(F^n\to V^*\) given by \(x\mapsto B(x,\cdot)|_V\) has kernel \(V^\perp\).
definition of orthogonal.
4. \(\dim V^\perp = n-\dim V = n-k\) (nonsingularity).
3.
5. \(V\cap V^\perp=0\) would imply \(B|_{V\times V}\) nondegenerate, but \(B|_{V\times V}=0\).
2.
6. \(\dim(V+V^\perp)=\dim V+\dim V^\perp = k+n-k=n\), so \(V\oplus V^\perp =F^n\) if \(V\cap V^\perp=0\).
dimension formula.
7. \(B|_{V\times V}=0\) forces \(V\subset V^\perp\), so \(k\le n-k\), i.e. \(k\le n/2\).
\(B(v,w)=0\) for all \(v,w\in V\) means \(V\subset V^\perp\).
8. Contradicting \(k>n/2\).
7.
9. Q.E.D.
8.