Orthogonal Projection in Inner Product Space with Dimension 2 and Basis {1,x}

[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

 

[Fredrik]

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).

 

[Shackleford]

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).

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

 

[Shackleford]

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

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

 

[vela]

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

 

[Shackleford]

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

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

{1, x-(1/2)}

 

[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)).

 

[vela]

Looks good.

 

[Shackleford]

Looks good.

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

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

 

[vela]

I get neither. Show us your calculations.

 

[Shackleford]

I get neither. Show us your calculations.

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

 

[vela]

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

 

[Shackleford]

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

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.

 

[vela]

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