Thread: Prove the Orthogonal Projection Matrix

Show that if {v1,..., vk} is an orthonormal basis of a subspace E of C^n, then the orthogonal projection matrix P_E:C^n --> C^n is P =the sum as i goes from 1 to k of (vi)(vi bar)^t.

An orthonormal basis is if <vi, vj> = 0 when i does not equal j and if the length of vi is 1. vi bar is the complex conjugate of vi and (vi bar)^t is the transpose of that.

Would I show this by showing that the projection of v onto itself is SUM(<vi, vi> vi)?

Originally Posted by letitbemww

Show that if {v1,..., vk} is an orthonormal basis of a subspace E of C^n, then the orthogonal projection matrix P_E:C^n --> C^n is P =the sum as i goes from 1 to k of (vi)(vi bar)^t.

An orthonormal basis is if <vi, vj> = 0 when i does not equal j and if the length of vi is 1. vi bar is the complex conjugate of vi and (vi bar)^t is the transpose of that.

Would I show this by showing that the projection of v onto itself is SUM(<vi, vi> vi)?

You need to show that Pv = v for all v in E, and also that Pv = 0 if v is in the orthogonal complement of E.

To show that Pv = v for v in E, it is sufficient to check that Pv_j = v_j for j = 1,2,...,k. So you need to evaluate

For all :



Now, using that is an orthonormal basis, determine .


Edited : Sorry, I didn't see Opalg's post.