# Prove the Orthogonal Projection Matrix

• Apr 12th 2011, 07:20 PM
letitbemww
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)?
• Apr 13th 2011, 12:11 AM
Opalg
Quote:

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 $\displaystyle \sum_{i=1}^kv_i\overline{v_i}^\textsc t}v_j.$
• Apr 13th 2011, 12:12 AM
FernandoRevilla
For all $\displaystyle x\in \mathbb{C}^n$:

$\displaystyle Px\in E\Rightarrow Px=\sum_{j=1}^k\alpha_jv_j\Rightarrow <x-\sum_{j=1}^k\alpha_jv_j,v_s>=0\;(\forall s=1,\ldots,k)$

Now, using that $\displaystyle \{v_j\}$ is an orthonormal basis, determine $\displaystyle \alpha_j$ .

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