# Inner Product

1. Jul 17, 2009

1. The problem statement, all variables and given/known data
Prove that if u and v are given non-zero vectors in the arbitrary inner-product space V, and are such that <u,v>=0, then {u,v} is a linearly independent subset of V.

2. Relevant equations

3. The attempt at a solution
I have no idea where to start. It's hard to prove because the inner-product could be defined in various ways. I can't just pick one and go with it, since another equally valid definition may not work.

I am still working on this and I will post my attempt as soon as I have one, but until then I was hoping to get some ideas for a starting point.

Last edited: Jul 17, 2009
2. Jul 17, 2009

### lanedance

Hi Daniel - I would try starting with the definition of linear independence

u, v are linearly independent if the only solution to the equation
c1.u + c2.v = 0, for scalars c1, c2, is c1=c2=0

then you could try writing an arbitrary vector as a linear combination of u&v, and look at the inner product with u or v

3. Jul 17, 2009

Thanks for the input! It was very helpful. Here's what I have so far. I'm not sure where the inner-product comes into play, though:

Let $$\pmb{u}=\left(u_1,u_2,u_3\right)$$ and let $$\pmb{v}=\left(v_1,v_2,v_3\right)$$.
If {u, v} is a linearly independent subset of V, then the only solution to $$c_1\pmb{u}+c_2\pmb{v}=0$$ is when $$c_1=c_2=0$$.

Thus, $$c_1\pmb{u}+c_2\pmb{v}=c_1\left(u_1,u_2,u_3\right)+c_2\left(v_1,v_2,v_3\right)=0$$.

Since $$c_1$$ and $$c_2$$ are coefficients for separate terms, and since it is given that u and v are non-zero vectors, the only way the above equation could be true is if $$c_1=c_2=0$$
QED

Okay, this doesn't seem like much of a proof. Any ideas?

Last edited: Jul 17, 2009
4. Jul 17, 2009

### Dick

It's not much of a proof. You don't need to split u and v into components. If c1*u+c2*v=0, what conclusions can you draw from <u,c1*u+c2*v>=0 and <v,c1*u+c2*v>=0, remembering <u,v>=0? That's what lanedance was suggesting. Use some properties of the inner product.

Last edited: Jul 17, 2009
5. Jul 17, 2009

Okay, this proof is actually easier than I thought. Tell me what you think. (I am using a and b for the constants to make it easier to type).

If {u, v} is a linearly independent subset of V, then the only solution to au+bv=0 is when a=b=0. We seek to prove that this is in fact the case.

Let u, v, and w be non-zero vectors.
The properties of inner products require that <au+bv,w>=a<u,w>+b<v,w>=0.
Since a and b are coefficients for separate terms, we know that a=b=0.

6. Jul 17, 2009

### Dick

Gack! No, that doesn't tell me anything since w could be anything. Put w=u. Try 0=<au+bv,u>=a<u,u>+b<v,u>. <v,u>=0. So 0=a<u,u>. What do the properties of the inner product tell you about <u,u> if u is nonzero?

7. Jul 17, 2009

Let's try it again, then:

If {u, v} is a linearly independent subset of V, then the only solution to au+bv=0 is when a=b=0. We seek to prove that this is in fact the case.

The properties of inner products require that <au+bv,u>=a<u,u>+b<v,u>=0.
The properties of inner products also require that <u,v>=<v,u>, so <v,u>=0.
We are left with a<u,u>=0.
However, the properties of inner products also require that <u,u> be greater than zero for a nonzero vector u.
Thus, we know that <u,u> is greater than zero, which means a=0.

Now we repeat this process, replacing u with v in the first argument:

The properties of inner products require that <au+bv,v>=a<u,v>+b<v,v>=0.
Since <u,v>=0, we are left with b<v,v>=0.
However, the properties of inner products also require that <v,v> be greater than zero for a nonzero vector v.
Thus, we know that <v,v> is greater than zero, which means b=0.

Thus, both a and b must be equal to zero.

QED

Last edited: Jul 17, 2009
8. Jul 17, 2009

### Dick

Yes, yes, yes, yes. That's it. Exactly.

9. Jul 17, 2009