Proof about inner product spaces

[evilpostingmong]

Homework Statement

Suppose V is a real inner-product space and (v1, . . . , vm) is a
linearly independent list of vectors in V. Prove that there exist
exactly 2^m orthonormal lists (e1, . . . , em) of vectors in V such
that
span(v1, . . . , vj) = span(e1, . . . , ej)
for all j ∈ {1, . . . , m}.

Homework Equations

The Attempt at a Solution

Alright, just to be sure I have the right idea (this is not the proof attempt)
I'll consider an orthogonal basis {u1} with j=1 so the non orthogonal basis for V
is {v1}.
I guess that if j=1 then let u1=v1 so e1=u1/llu1ll so two possible orthonormal lists are {e1} and{-e1} then after adding an additional element to the basis {v1} (call it
v2) j becomes 2, the non-orthogonal becomes {v1,v2}, the orthogonal basis is {u1,u2}
and u2=v2-proj_u2(v2) and e2=u2/llu2ll (with the other possible orthonormal basis
being -e2) but since e1 and -e1 cannot share
the same list, they must be put in separate (but similar) lists which double
the amount of lists from when j was 1 so now we have {e1, e2} and {e1, -e2}
and {-e1, e2} finally {-e1, -e2} four possible orthonormal lists. If m=j, that's it,
but if not, eventually when it does, we will keep multiplying 2^j by 2 as each vj is added to the basis {v1..vj-1}.
This isn't the proof, just want to see if I'm seeing this right.

 

[Hurkyl]

Are you sure you stated the problem right? Consider, for example, that any two-dimensional real inner-product space has infinitely many orthonormal bases.

(edit: aha, nevermind)

 

[Office_Shredder]

Yeah, that's the basic idea. Given $e_1,..., e_{j-1},$ there are two choices for $e_j$ and they are additive opposites of each other