# Proofs of vector space, inner product

• Feb 21st 2009, 04:59 PM
arbolis
Proofs of vector space, inner product
Hi MHF,
I don't have a concrete idea about how do perform these 2 proofs.
Let $V$ be a vector space having an inner product and whose dimension is n. Let $v\in V$ be a non zero vector and $W=\{ w\in V \text{ such that w is orthogonal to v} \}$.
a)Prove that $W$ is a subspace of $V$.
b)Prove that $\dim W=n-1$.
My attempt : for a)
First I notice that $W \subset V$. Now I must show that the addition of 2 vectors in $W$ remains in $W$ and the multiplication of a vector by a scalar remains in $W$. I understand it's obvious because multiplying a vector $w$ which is orthogonal to $v$ will give another vector orthogonal to $v$.
Let $p$ be the inner product function. We have the property : $p(w,v)=0$. And I see that $p(kw,v)=kp(w,v)=k\cdot 0=0$.
Now I must show that $(w_1+w_2)\in W$ $\forall w_i \in W$.
$p(w_i+w_j, v)=p(w_i, v)+p(w_j, v)=0+0=0$.
Wow, it seems I've show part a)! Eventually I must show other properties of $W$, like that the zero vector is in... and so on.
I'll try part b).
I'm unable. I wish $v\in W^{\perp}$. It would be easy, because as $W\subset V$, $W^{\perp} \subset V$ and $\dim V=\dim W+ \dim W^{\perp}$ and I think that $\dim v =1$, so it would have been done.

Did I do well part a)? How can I show part b)? (a tip would be welcome).
• Feb 21st 2009, 06:32 PM
HallsofIvy
Quote:

Originally Posted by arbolis
Hi MHF,
I don't have a concrete idea about how do perform these 2 proofs.
Let $V$ be a vector space having an inner product and whose dimension is n. Let $v\in V$ be a non zero vector and $W=\{ w\in V \text{ such that w is orthogonal to v} \}$.
a)Prove that $W$ is a subspace of $V$.
b)Prove that $\dim W=n-1$.

Be careful! If n= 1, these are NOT true!

Quote:

My attempt : for a)
First I notice that $W \subset V$. Now I must show that the addition of 2 vectors in $W$ remains in $W$ and the multiplication of a vector by a scalar remains in $W$. I understand it's obvious because multiplying a vector $w$ which is orthogonal to $v$ will give another vector orthogonal to $v$.
Let $p$ be the inner product function. We have the property : $p(w,v)=0$. And I see that $p(kw,v)=kp(w,v)=k\cdot 0=0$.
Now I must show that $(w_1+w_2)\in W$ $\forall w_i \in W$.
$p(w_i+w_j, v)=p(w_i, v)+p(w_j, v)=0+0=0$.
Wow, it seems I've show part a)! Eventually I must show other properties of $W$, like that the zero vector is in... and so on.
No, you don't. You don't need to show that the zero vector is in the subset because you have already shown if v is any vector in the set and a any number, av is in the set. Take a= 0. (You DO have to show that the set is non-empty. That's why you need n>1.) You don't need to show, for example, that addition is associative, commutative, etc. because you are given that V is a vector space so those must be true.

Quote:

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

Did I do well part a)? How can I show part b)? (a tip would be welcome).
Y are given the single vector v. You can always choose a basis containing that vector. Can you use the Gram-Schmidt process to construct an orthonormal basis from that? Once again, for this to be true you must have n> 1.
• Feb 21st 2009, 06:49 PM
arbolis
Ok, thanks a lot!
I'll try part b) following your tip. And ok, they should have said n>1 but they didn't.
• Feb 22nd 2009, 02:03 PM
arbolis
Quote:

You are given the single vector v. You can always choose a basis containing that vector. Can you use the Gram-Schmidt process to construct an orthonormal basis from that? Once again, for this to be true you must have n> 1.
I must say that I'm still unable to do the job. Let $\bold B$ be a base of $V$, such that $\bold B=\{ v, v_1,...v_{n-1} \}$. Note that $\dim \bold B =n$. Now I must orthogonalize $\bold B$ in order to find a base of $W$, right? I call this base $\bold C$.
(Gram-Schmidt)I set $w_0=v$.
$w_1=v_1- \frac{\langle w_0 | v_1 \rangle}{\langle w_0 | w_0 \rangle}w_0$. And so on, but I don't see why $\dim \bold C=n-1$...