# Inner Product once again :S

1. Nov 2, 2005

### playboy

Let V be a vector in an inner product space V

show that ||v|| >= ||proj(u) v|| holds for all finitie dimensional subspaces of U.

Hint: Pythagorean Theorm.

Okay... where on earth do i begin?

I thought perhaps I should expand the RIGHT side of the equation, but that dosn't seem to be getting me anywhere really.

The LEFT side seems pretty useless too, so im stuck trying to show that the RIGHT side is >= ...

anybody have any ideas?

Thanks

2. Nov 2, 2005

### NateTG

Do you mean
$\vec{V}$ is a vector in an inner product space $U$?
otherwise that doesn't make sense?
Maybe you can find some easy cases, and work from there?

3. Nov 2, 2005

### amcavoy

If you draw these two vectors out, it should be very obvious. The magnitude of the projection is just the component:

$$\left|\text{proj}_{u}v\right|=\frac{\left(u,v\right)}{u}$$

...which is really just saying v*cosθ. What do you know about the cosine function? Well, for one, it is less than or equal to 1 for all values of θ. Does this help?

4. Nov 2, 2005

### Gokul43201

Staff Emeritus
apmcavoy : There are no "two vectors". A vector v in U is being projected onto some subspace (W, say) of U. However, with some additional construction, your method would work.

playboy : Call the subspace W, and write the projection down in terms of the basis vectors $w_i$ of W. similarly write down the expansion of the norm in terms of the basis vectors $u_i$ of U.

What do you know about these two sets of basis vectors ?

Last edited: Nov 2, 2005
5. Nov 2, 2005

### playboy

okay thanks for the help everyone!

Gokul43201: ill try your metod and get give back my results!

6. Nov 2, 2005

### playboy

Now i am even more lost than ever :S

For the RHS....i get...

<v, w1> + <v, w2> + ....
||w1||.....||w2||

For the left hand side...

<u,u>^0.5

Am i on the right track?

7. Nov 2, 2005

### Gokul43201

Staff Emeritus
First, we need to confirm that we're all solving the same problem. Where did U come from. Did you mean to write (in the OP) : "...for all finite dimensional subspaces U of the space V" ?
The length (or norm) is not equal to the sum of its components.

$$proj_U(v) = \sum _{i=1}^k \langle v,w_i \rangle w_i$$

$$\implies ||proj_U(v)|| = \sqrt{\sum _{i=1}^k \langle v,w_i \rangle ^2}$$

Expand ||v|| similarly.

8. Nov 2, 2005

### playboy

Noooo.... this is why its not making sense.

Isn't the projection....

<v wi>
______ wi
<wi wi>

and not

<v wi> wi

And on the LHS... (v is just a vecotr)
so ||v|| = (v1^2 + v2^2 + ..... + vn^2)^0n5

9. Nov 2, 2005

### amcavoy

Yes, right. I'm sorry if that confused anyone who read this. I guess I was just assuming some vector u in U.

10. Nov 3, 2005

### Gokul43201

Staff Emeritus
If $w_i$ is a basis vector, what's $\langle w_i,w_i \rangle$ ?
What are v1, v2, etc ?

PS : Why are you not addressing the problem that no one really knows what the actual question is? The question as stated in the OP is incorrect and needs to be fixed.

Last edited: Nov 3, 2005
11. Nov 3, 2005

### playboy

Let v be a vector in an inner product space V

show that ||v|| >= ||proj u (v)|| holds for all finitie dimensional subspaces of U.

Hint: Pythagorean Theorm.
_____________________________________________
how i approached it:

Let (e1,.....,en) be an orthonognal basis:

||proj u (V)||^2 = <proj u (V), proj u (V)>
||proj u (V)|| = ( <v e1>^2/<e1 e1> + .... + <v en>^2/<en en>)^0.5

Note that i did not show my full computation from the beginning, that would
just be too long to type out.

and ||v|| = (v1^2 + .... + Vn^2)^0.5 since v is a Vector