Hey I have a question:
Say I have a euclidian Vectorspace $(V,\langle\cdot,\cdot\rangle)$ (finite dimension, lets say $n$). And now say that I have an ordered subset of $V$, say $S = (v_1 , ..., v_k)$.
Then I can expand this subset using Steinitz into a Basis of $V$ and then using the Gram-Schmidt process I can turn this basis into a orthogonal and then a orthonormal basis $S = (v_1,...,v_k, v_{k+1},...,v_n)$ of $V$.
Now if have to prove that if the subspace $W=\operatorname{span}(v_1,...,v_k)$ then a $F = (v_{k+1},...,v_n)$
is a Basis of $W^⊥$ \begin{aligned}[]W^{\perp }=\{ u\in V\mid \forall v\in W:\langle u,v\rangle =0\} .\end{aligned}
Question 1: Since I have a Orthonormal Basis then all Elements in this Basis are Orthonormal to each other and only by defining the subspace $W$ can I then define the subspace $W^⊥$? I am pretty sure this is true but then something does not add up in question 2.
Question 2: idea of proof for $W^⊥ \subseteq\operatorname{span}(F)$: Since I have a Orthonormal Basis $S$ then
for all $v\in V: v = \sum_{i=1}^n \langle v_i,v\rangle v_i $
now if $v\in W^⊥: v = \sum_{i=k+1}^n \langle v_i,v\rangle v_i \in \operatorname{span}(F)$
For $1\leq i \leq k$ the inner product is 0 since they are Orthogonal and in $W$. But aren't my other vectors (in $W^⊥$ also orthogonal to each other) meaning that say $v$ the sum would only consist of one vector where $v$ is not orthogonal to it times a scalar? Or is this only true if I can't write $v$ as a linear combination of my elements in $F$ but only as a multiple of one element? And if choose a $v$ as a linear combination of elements in $F$ then all $k+1 \leq i \leq n$ dissapear in the inner product except those used in the linear combination?
Thank You!