Proofs of vector space, inner product

Hi MHF,

I don't have a concrete idea about how do perform these 2 proofs.

Let $\displaystyle V$ be a vector space having an inner product and whose dimension is n. Let $\displaystyle v\in V$ be a non zero vector and $\displaystyle W=\{ w\in V \text{ such that w is orthogonal to v} \}$.

a)Prove that $\displaystyle W$ is a subspace of $\displaystyle V$.

b)Prove that $\displaystyle \dim W=n-1$.

My attempt : for a)

First I notice that $\displaystyle W \subset V$. Now I must show that the addition of 2 vectors in $\displaystyle W$ remains in $\displaystyle W$ and the multiplication of a vector by a scalar remains in $\displaystyle W$. I understand it's obvious because multiplying a vector $\displaystyle w$ which is orthogonal to $\displaystyle v$ will give another vector orthogonal to $\displaystyle v$.

Let $\displaystyle p$ be the inner product function. We have the property : $\displaystyle p(w,v)=0$. And I see that $\displaystyle p(kw,v)=kp(w,v)=k\cdot 0=0$.

Now I must show that $\displaystyle (w_1+w_2)\in W$ $\displaystyle \forall w_i \in W$.

$\displaystyle p(w_i+w_j, v)=p(w_i, v)+p(w_j, v)=0+0=0$.

Wow, it seems I've show part a)! Eventually I must show other properties of $\displaystyle W$, like that the zero vector is in... and so on.

I'll try part b).

I'm unable. I wish $\displaystyle v\in W^{\perp}$. It would be easy, because as $\displaystyle W\subset V$, $\displaystyle W^{\perp} \subset V$ and $\displaystyle \dim V=\dim W+ \dim W^{\perp}$ and I think that $\displaystyle \dim v =1$, so it would have been done.

Did I do well part a)? How can I show part b)? (a tip would be welcome).