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Math Help - Prove the Orthogonal Projection Matrix

  1. #1
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    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)?
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  2. #2
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    Quote Originally Posted by letitbemww View Post
    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 \sum_{i=1}^kv_i\overline{v_i}^\textsc t}v_j.
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  3. #3
    MHF Contributor FernandoRevilla's Avatar
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    For all x\in \mathbb{C}^n:

    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 \{v_j\} is an orthonormal basis, determine \alpha_j .


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