Problem with the proof for the Gram-Schmidt orthonormalisation-process I have an exam coming up next month and I'm trying to understand all the proofs in my book.
I got stuck on the Gram-Schmidt proof and would really appreciate some help.
I understand everything except for small 2 conclusions (actual questions are bolded), but first things first, let my write down the proof:

The field $K$ = $\mathbb{R}$ or $\mathbb{C}$. $(V,<.,.>)$ is a finite dimensional inproduct-space over $K$. $W$ is a subspace of $V$ with dim $W = d \geq$ 1. 
  This means $W$ has an orthonormal basis $\{b1,...,bd\}$.  
The proof is based on induction towards d.
Assume that $d=1$ and $0 \neq w \in W$. Then $\{b_1\}$ is the orthonormal basis where $b_1 = \frac{w}{||w||}$.

Now in my book, we need to assume that $d \geq 1$; the induction hypotheses implies that the claim from the first quoted paragraph is true for subspaces of dimension $d-1$.
Is this just always true for proofs where induction is used or am I missing something?
Moving on, here is the rest of the proof:

We will now prove the claim for an arbitrary subspace $W$ of dimension $d$.
  Define the 1-dimensional subspace $W' = span(\{w\})$ where $0 \neq w \in W$.
  Consider the linear form:
$<.,w> : W \rightarrow K : v \mapsto <v,w>$
This linear form is surjective since $<w,w> \neq 0$.

Why does the fact that $<w,w> \neq 0$ imply surjectivity?
Thanks in advance!
(The rest of the proof has been left out.)
 A: Answer 1: the proof may have assumed that the theorem is vacuously true if $\;\dim = 0\;$ and from there to continue, or more probably imo, it assumed the theorem is trivial if $\;\dim =1\;$ and from here it takes on.
Answer 2: this is a basic theorem: any non-zero linear functional (or linear form), i.e. a linear map $\;V_{\Bbb F}\to \Bbb F\;,\;\;V_{\Bbb F}\;$ a vector space over a field $\;\Bbb F\;$, is automatically surjective. Try to prove it, it is very easy. Any problem write back.
A: More thoughts on induction
Proof by induction works like this:
$P$ is a proposition on a natural number $n$ (like a n-dimension subspace...)
Show that is true that $P(n) \Rightarrow P(n+1)$ ( *)
Then, if you show it is indeed true for $P(1)$, then by ( * ) we can say that is true for $P(2)$. Then by (*) we can say it's true for $P(3)$, and so forth. So you have $P(n)$ is  true $\forall n$
2)
As wikipedia put it:
Any linear functional L is either trivial (equal to 0 everywhere) or surjective onto the scalar field.  Indeed, this follows since just as the image of a vector subspace under a linear transformation is a subspace, so is the image of V under L.  But the only subspaces (i.e., k-subspaces) of k are {0} and k itself.
http://en.wikipedia.org/wiki/Linear_form
