A null vector is a nonzero vector that is orthogonal to
itself.
If W is a subspace of V,let $W^{\perp}$ = [$v{\in}$ W : $v{\perp}$W].
$W^{\perp}$ is a subspace of V called W perp.
A subspace W of V is called nondegenerate if g | W is nondegenerate.
The nondegeneracy of g on the whole space V is equivalent to
$v^{\perp}$ = 0.
A bilinear form on V is an
R-bilinear function b:$V{\times}$ $V {\to}$ R,
and we consider only the symmetric ciase:
b{v, w) = fo(w, v) for all v, w.
A symmetric bilinear form b on V is :
nondegenerate provided b{v, w) = 0 for all $w {\in}$V implies v=0.
A scalar product g on a vector space V is a nondegenerate symmetria bilinear form on V.
How do I show the $v^{\perp}$ is a degenerate subspace if v is a null vector and the operator F:$v^{\perp}$${\to}$$v^{\perp}$?