# Thread: Linear combination and dot product

1. ## Linear combination and dot product

let N be the intersection of the sets $\{x:=0\}$,i=1,2...m.
Prove or disprove that a vector v can be expressed as a linear combination of vectors $v_i$,i=1,...m iff for any x in N, $=0$.

2. Originally Posted by Shanks
let N be the intersection of the sets $\{x:=0\}$,i=1,2...m.
Prove or disprove that a vector v can be expressed as a linear combination of vectors $v_i$,i=1,...m iff for any x in N, $=0$.
The first direction is easy. If v is a linear combination of $v_i$'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:

$U:=span\{v_1, ..., v_m\}$

It's clear that $N=U^{\perp }$ (I'll leave you to check that if you're not convinced). So in particular, N has a basis, say $\{x_1, ..., x_n\}$. 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 $v_i$'s and $x_j$'s. So write:

$v=a_1v_1+...+a_mv_m+b_1x_1+...+b_nx_n$

Take any one of the $x_j$.

$=b_j$

Since this must be zero, we get that $b_j=0$ and since our choice of $x_j$ was arbitrary, all the b's must be zero. So we're left with:

$v=a_1v_1+...+a_mv_m$

And so v is indeed a linear combination of the $v_i$'s