Linear Algebra - Orthogonal Projections

[steelphantom]

Homework Statement

Prove that if P in L(V) is such that P² = P and every vector in Ker(P) is orthogonal to every vector in Im(P), then P is an orthogonal Projection.

Homework Equations

Orthogonal projections have the following properties:

1) Im(P) = U
2) Ker(P) = Uperp
3) v - P(v) is in Uperp for every v in V.
4) P² = P
5) ||P(v)|| <= ||v|| for every v in V.

The Attempt at a Solution

Property 4) is given, so that's done. I proved property 3) by using the equation v = P(u) + w, where w is in Uperp. Rearranging, I get v - P(u) = w, which is in Uperp.

Property 1) was proven in class: Clearly Im(P) is a subset of U. But for every u in U, P(u) = u since u = u + 0. Hence every element of U is in the image, and so Im(P) = U.

To prove property 2), I'm not really sure how to do this. My prof said it's basically the same thing as property 1, but I'm not really seeing it.

For property 5), I know that P(v) = <v, e_1>e_1 + ... + <v, e_m>e_m, where (e_1, ... e_m) is an orthonormal basis of U. I know that ||v|| = sqrt(<v, v>). Not sure how to compare these to come up with the inequality.

Thanks for your help!

 

[steelphantom]

Can anyone help on proving 2) and 5)? Does 2) directly follow from 1) since every vector in Ker(P) is orthogonal to every vector in Im(P)? Thanks.

 

[Dick]

I don't see how you can 'prove' 1) unless you define what U is. I thought U was just another name for Im(P). If so then the premises only tell you that Ker(P) is a subset of Uperp. To prove they are equal use dim(Im(P))+dim(Ker(P))=dim(V), as it is for any linear transformation. Now all you need to do is show that the intersection of Im(P) and Ker(P) is {0}. Then you have that V=Im(P)+Ker(P) as a direct sum, so you can decompose any element of v into an image part and a kernel part. Then for 5) think about the triangle inequality.

 

[steelphantom]

U is just any subspace of V. V = U \oplus Uperp. Thanks for your help.