# Homework Help: Orthogonal projection

1. Feb 17, 2012

### Shackleford

I found a final answer online, but my vector is slightly different. I haven't been able to catch my mistake.

I'm supposed to find the orthogonal projection of the given vector on the given subspace W of the the inner product space V.

P1 has dimension 2 and basis = {1,x}.

http://i111.photobucket.com/albums/n149/camarolt4z28/File2.png [Broken]

Last edited by a moderator: May 5, 2017
2. Feb 17, 2012

### Fredrik

Staff Emeritus
I haven't thought it through to the end, but doesn't the formula you're trying to use for the orthogonal projection onto W require that you use an orthonormal basis for W? (Check what your book says. I was too lazy to look it up myself or think about it).

3. Feb 17, 2012

### Shackleford

I didn't normalize {1,x} for W. That's the problem. Good catch. Thanks.

4. Feb 19, 2012

### Shackleford

I normalized the basis {1,x} for W: {1, sqrt(3)x}

I'm getting for the projection (29/6) + (15/2)x.

5. Feb 19, 2012

### vela

Staff Emeritus
You normalized the basis vectors, but they're still not orthogonal.

6. Feb 20, 2012

### Shackleford

The orthogonality slipped my mind. I suppose I should use Gram-Schdmit for that.

{1, x-(1/2)}

7. Feb 20, 2012

### Shackleford

I don't know why this problem is giving me trouble.

I used Gram-Schmidt on the standard basis for P1 and got {1, x-(1/2)}. I normalized this basis for W and got

u1 = 1
u2 = sqrt(12)(x-(1/2)).

8. Feb 20, 2012

### vela

Staff Emeritus
Looks good.

9. Feb 20, 2012

### Shackleford

When I use the projection formula I get x - (62/66).

The answer I found has x - (13/3).

10. Feb 20, 2012

### vela

Staff Emeritus
I get neither. Show us your calculations.

11. Feb 20, 2012

### Shackleford

http://i111.photobucket.com/albums/n149/camarolt4z28/File21.png [Broken]

Last edited by a moderator: May 5, 2017
12. Feb 20, 2012

### vela

Staff Emeritus
You just added incorrectly when evaluating the last integral. You should get $1/\sqrt{12}$.

13. Feb 20, 2012

### Shackleford

Shoot. You're right. I should have had a +1/2, not -1/2. The last integral should actually be sqrt(12)/12. I do end up with x + 26/6.

Thanks for the help! I'm all finished with this assignment. I'm graduating in May, so I'm counting down the weeks and assignments! Haha.

14. Feb 20, 2012

### vela

Staff Emeritus
Ah, yes, it is x+13/3. I didn't actually multiply it out here.