Orthogonal projectors (minimization and variational problem)

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Homework Statement


S1 is in subspace of C^n. P unique orthogonal projector P : C^n -> S1, and x is in range of C^n. Show that the

minimization problem: y in range of S1 so that:
2norm(x-y) = min 2norm(x-z)
where z in range of S1

and

variational problem: y in range of S1 so that:
(x-y)*z = 0 (* is the hermitian so this is an inner product)
for each z in range of S1

have the same unique solution y = Px


Homework Equations


P^2-P = 0
Range(P) = S1 ?


The Attempt at a Solution


I have no idea what's going on here. I am using Numerical Linear Algebra by Trefethen and Bau but there's nothing about minimization or variational problems in the book =/ any hint to get me started would be much appreciated.

For the variational problem this might be one way but it doesn't show that it's unique and I'm not even sure it's correct:
(x-y)*z = 0
Set y = Px and multiply by P
(Px-P^2x)*z = 0
but P^2 = P so (Px-P^2x) = 0 and therefore LHS = RHS

Clueless on the first one.
 

Answers and Replies

  • #2
Dick
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I think the fact you might be missing is that x-Px is orthogonal to any element y of S1. So (x-Px)*y=0 for all y in S1. That's what it means to be an ORTHOGONAL projector. For the first part, write norm(x-z)=norm((x-Px)+(Px-z)). Can you see the Pythagorean theorem in there?
 
  • #3
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I think the fact you might be missing is that x-Px is orthogonal to any element y of S1. So (x-Px)*y=0 for all y in S1. That's what it means to be an ORTHOGONAL projector. For the first part, write norm(x-z)=norm((x-Px)+(Px-z)). Can you see the Pythagorean theorem in there?
Wow, I didn't know maths could be fun :P thanks a lot.
 

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