let N be the intersection of the sets $\displaystyle \{x:<v_i,x>=0\}$,i=1,2...m.
Prove or disprove that a vector v can be expressed as a linear combination of vectors $\displaystyle v_i$,i=1,...m iff for any x in N, $\displaystyle <v,x>=0$.
let N be the intersection of the sets $\displaystyle \{x:<v_i,x>=0\}$,i=1,2...m.
Prove or disprove that a vector v can be expressed as a linear combination of vectors $\displaystyle v_i$,i=1,...m iff for any x in N, $\displaystyle <v,x>=0$.
The first direction is easy. If v is a linear combination of $\displaystyle v_i$'s then it's clear that <v,x>=0 for any x in N.
The other direction is the tricky one. Assume that <v,x>=0 for any x in N. First, define a subspace U as follows:
$\displaystyle U:=span\{v_1, ..., v_m\}$
It's clear that $\displaystyle N=U^{\perp }$ (I'll leave you to check that if you're not convinced). So in particular, N has a basis, say $\displaystyle \{x_1, ..., x_n\}$. We can also assume, without loss of generality, that this basis is orthonormal.
Now, we use the fact that U+N gives the whole vector space (I'm assuming this is a fact you know). Then, v must be a linear combination of $\displaystyle v_i$'s and $\displaystyle x_j$'s. So write:
$\displaystyle v=a_1v_1+...+a_mv_m+b_1x_1+...+b_nx_n$
Take any one of the $\displaystyle x_j$.
$\displaystyle <v,x_j>=b_j$
Since this must be zero, we get that $\displaystyle b_j=0$ and since our choice of $\displaystyle x_j$ was arbitrary, all the b's must be zero. So we're left with:
$\displaystyle v=a_1v_1+...+a_mv_m$
And so v is indeed a linear combination of the $\displaystyle v_i$'s