Dimension of subspace of V^n with orthogonal vectors

[raphael3d]

in a space V^n, prove that the set of all vectors {v1,v2,..}, orthogonal to any v≠0, form a subspace V^(n-1).

i know that a subspace of V^n must be at least one dimension less and the set of vector v1,v2,... build a orthogonal basis, but how can one show with this preconditions that the subspace has to be of dimension (n-1)?

 

[rasmhop]

For a fixed non-zero vector v extend (v) to a basis (v,b_2,b_3,\ldots,b_n) of V^n. Now what is the dimension of \text{Span}(b_2,\ldots,b_n) and how this space relate to all vectors orthogonal to v?

 

[raphael3d]

so would v1+ span(b2,...,bn) be a subspace of v^n ?
but why should (v,b2,b3,...,bn) be a basis, when v isn't normalized to a unit vector?

thanks rasmhop for your reply

 

[rasmhop]

so would v1+ span(b2,...,bn) be a subspace of v^n ?
but why should (v,b2,b3,...,bn) be a basis, when v isn't normalized to a unit vector?

thanks rasmhop for your reply

I'm not talking about an orthonormal basis or anything like that, just a basis. In general in a vector space V of dimension n, for m linearly independent vectors b_1,\ldots,b_m we can find vectors b_{m+1},\ldots,b_n such that b_1,\ldots,b_n is a basis for V.

The idea of my first reply was that we let W = \text{Span}(b_2,\ldots,b_n) and let W' be the vector space of vectors orthogonal to v. Now it's possible to show W=W' by showing that if a vector is in W', then it must be in W; and if a vector is in W, then it is in W'.