275A-Solutions-4

275A-Solutions-4 - ECE 275A Homework #4 Solutions – Fall...

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Unformatted text preview: ECE 275A Homework #4 Solutions – Fall 2009 Homework Solutions 1. (a) We have ℓ ( x ) = x H Π x − 2Re x H By + y H Wy = x H Π x − x H By − y H B H x + y H Wy = x H Π x − x H Π Π − 1 By − y H B H Π − 1 Π x + y H Wy = ( x − Π − 1 By ) H Π ( x − Π − 1 By ) + y H Wy − y H B H Π − 1 By = ( x − Π − 1 By ) H Π ( x − Π − 1 By ) + y H ( W − B H Π − 1 B ) y Thus for all x ℓ ( x ) ≥ y H ( W − B H Π − 1 B ) y with equality if and only if x = Π − 1 By . Thus we have proved that ˆ x = Π − 1 By = arg min x ℓ ( x ) ℓ (ˆ x ) = y H ( W − B H Π − 1 B ) y = min x ℓ ( x ) (b) It is straightforward to apply this result to the full column-rank, weighted least- squares problem. ℓ ( x ) = ∥ y − Ax ∥ 2 W = ( y − Ax ) H W ( y − Ax ) = x H A H WA | {z } Π x − x H A H W | {z } B y − y H WA |{z} B H x + y H Wy = x H Π x − x H By − y H B H x + y H Wy = x H Π x − 2Re x H By + y H Wy With A full column rank and W = W H > 0, the matrix Π is Hermitian and full rank. Thus the weighted least-squares estimate of x is ˆ x = Π − 1 By = ( A H WA ) − 1 A H Wy with optimal (minimal) least-squares cost ℓ (ˆ x ) = y H ( W − B H Π − 1 B ) y = y H ( W − WA ( A H WA ) − 1 A H W ) y Comment. Suppose that ⟨ y 1 , y 2 ⟩ = y H 1 Wy 2 and ⟨ x 1 , x 2 ⟩ = x H 1 x 2 (i.e., Ω = I ) 1 Then A ∗ = A H W A + = ( AA ∗ ) − 1 A ∗ = ( A H WA ) − 1 A H W P R ( A ) = AA + = A ( A H WA ) − 1 A H W P N ( A * ) = I − P R ( A ) This shows that the optimal cost can be rewritten as ℓ (ˆ x ) = y H W ( I − P R ( A ) ) y = y H WP N ( A * ) y = ⟨ y, P 2 N ( A * ) ⟩ = ⟨ P ∗ N ( A * ) y, P N ( A * ) ⟩ = ⟨ P N ( A * ) y, P N ( A * ) or ℓ (ˆ x ) = ∥ P N ( A * ) y ∥ 2 = ∥ P N ( A * ) y ∥ 2 W What is the optimal cost if y ∈ R ( A )? Does this make sense?...
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This note was uploaded on 01/14/2011 for the course ECE 210a taught by Professor Chandrasekara during the Fall '08 term at UCSB.

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275A-Solutions-4 - ECE 275A Homework #4 Solutions – Fall...

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