Finding the Projection onto Subspaces

[shaon0]

Homework Statement

See attachment

The Attempt at a Solution

How should I approach these questions? By using the projection formula?

 

[HallsofIvy]

Yes, since the problem asks for projections, use the projection formula!

 

[shaon0]

Yes, since the problem asks for projections, use the projection formula!

Will I have to do two different calculations ie one with v and the first vector in W and then v with the second vector in W? Also, how would I find P?

 

[vela]

W is a plane. The question is asking you to find the projection of the vector v onto that plane.

As far as finding the matrix P goes, the projection onto W is a linear mapping. How do you find the matrix representing a linear mapping?

 

[shaon0]

W is a plane. The question is asking you to find the projection of the vector v onto that plane.

As far as finding the matrix P goes, the projection onto W is a linear mapping. How do you find the matrix representing a linear mapping?

For P, the projection of what vector onto W? Would I just span the vectors I've found so, P=span{v1,v2} and find the co-effs s.t Basis1=a1.v1+a2.v2 where v1,v2 are the vectors I've found using the projection formula and a1,a2 are constants which will give me the 1st column of P?

 

[vela]

You might find the page helpful:

http://www.cliffsnotes.com/study_guide/Projection-onto-a-Subspace.topicArticleId-20807,articleId-20792.html

 

[shaon0]

You might find the page helpful:

http://www.cliffsnotes.com/study_guide/Projection-onto-a-Subspace.topicArticleId-20807,articleId-20792.html

I've tried to use the proj_Wv=proj_v1v+proj_v2v where v1=(-2,,1,-2)^T and v2=(1,4,-8)^T but i don't get the correct answer. Maybe because v1, v2 are not mutually orthogonal?

 

[unlearned]

Since any vector v can be written as a linear combination of vectors of a basis of \mathbb{R}^3, if you can find the projection of each of the vectors of the canonical basis of \mathbb{R}^3, you can then write a projection matrix using the results.

 

[vela]

I've tried to use the proj_Wv=proj_v1v+proj_v2v where v1=(-2,,1,-2)^T and v2=(1,4,-8)^T but i don't get the correct answer. Maybe because v1, v2 are not mutually orthogonal?

Yes, that's why it didn't work.

 

[shaon0]

Since any vector v can be written as a linear combination of vectors of a basis of \mathbb{R}^3, if you can find the projection of each of the vectors of the canonical basis of \mathbb{R}^3, you can then write a projection matrix using the results.

So; proj_e1v1? for all combinations of e1,e2,e3 and v1, v2?

 

[vela]

If you look at the first picture on that web page, it illustrates that you can write
\vec{v} = (\mathrm{proj}_W\ \vec{v}) + \vec{v}_\perpThat is, you can resolve any vector \vec{v} into a piece that lies in the plane W and a piece that's perpendicular to W. Solving for the projection, you get
\mathrm{proj}_W\ \vec{v} = \vec{v} - \vec{v}_\perpSo if you can figure out how to find \vec{v}_\perp, which is very likely a problem you solved before, you can then find the projection of \vec{v} onto W. Hint: think about the normal to the plane.

If you don't want to use that approach, you can go with the method you tried. But as you noted, you need an orthogonal basis for W. You've been given a basis. You just need to make it orthogonal.

 

[shaon0]

If you look at the first picture on that web page, it illustrates that you can write
\vec{v} = (\mathrm{proj}_W\ \vec{v}) + \vec{v}_\perpThat is, you can resolve any vector \vec{v} into a piece that lies in the plane W and a piece that's perpendicular to W. Solving for the projection, you get
\mathrm{proj}_W\ \vec{v} = \vec{v} - \vec{v}_\perpSo if you can figure out how to find \vec{v}_\perp, which is very likely a problem you solved before, you can then find the projection of \vec{v} onto W. Hint: think about the normal to the plane.

If you don't want to use that approach, you can go with the method you tried. But as you noted, you need an orthogonal basis for W. You've been given a basis. You just need to make it orthogonal.

Ok, thanks. I think I've got it.