How Can Gram-Schmidt Prove dim(w)=dim(V)-1?

[theFuture]

So I'm to show that the non-zero vector w={v e V|<v,x> = 0} for all x in V that dim(w)=dim(V)-1.

It recommends using the Gram-Schmidt process to prove this but I tried to work it out and I couldn't make any sense of it. Any suggestions on how to start this out?

[edit]: nevermind, I got it. If you were curious, start by saying x is an element of the set S that is linearly independent and spans V. Then do G-S on V and you find that you lose an element of the set, so there you have it.

 

[HallsofIvy]

I think what you mean to say is "prove that the subspace consisting of all vectors v such that <v, x>= 0 (where x is non-zero) has dimension dim(V)- 1."
Select a basis for V containing x and perform a "Gram-Schmidt" starting the process with x as the first vector (so Gram-Schmidt will give a basis containing one basis vector in the same direction as x).

 

[theFuture]

That's definitely what I meant, but I copied it verbatim from my crappy book (Messer's Linear Algebra) starting with "show the vector . . ." It's good with proofs, not so good with the math.