Lecture8

# Lecture8 - Lecture 8 Orthogonal Projections 8.1...

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Unformatted text preview: Lecture 8 Orthogonal Projections 8.1 Introduction We will now come back to our original aim: Given a vector space V , a subspace W , and a vector v V , find the vector w W which is closest to v . First let us clarify what the closest means. The tool to measure distance is the norm , so we want k v- w k to be as small as possible. Thus our problem is: Find a vector w W such that k v- w k k v- u k for all u W . Now let us recall that if W = R w 1 is a line, then the vector w on the line W is the one with the property that v- w W . We will start by showing that this is always the case. 8.2 w W is closest to v iff v- w W Theorem. Let V be a vector space with inner product ( , ). Let W V be a subspace and v V . If v- w W , then k v- w k k v- u k for all u W and k v- w k = k v- u k if and only if w = u . Thus w is the member of W closest to v . 47 48 LECTURE 8. ORTHOGONAL PROJECTIONS Proof. First we remark that k v- w k k v- u k if and only if k v- w k 2 k v- u k 2 . Now we simply calculate k v- u k 2 = k ( v- w ) + ( w- u ) k 2 = k v- w k 2 + k w- u k 2 because v- w W and w- u W ( * ) k v- w k 2 because k w- u k 2 So k v- u k k v- w k . If k v- u k 2 = k v- w k 2 , then we see - using ( * )- that k w- u k 2 = 0, or w = u . As k v- w k = k v- u k if u = w , we have shown that the statement is correct. Theorem. Let V be a vector space with inner product ( , ). Let W V be a subspace and v V . If w W is the closest to v , then v- w W . Proof. We know that k v- w k 2 k v- u k 2 for all u W . Therefore the function f : R- R F ( t ) := k v- w + tx k 2 ( x W ) has a minimum at t = 0. We have F ( t ) = ( v- w + tx,v- w + tx ) = ( v- w,v...
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## This note was uploaded on 11/23/2011 for the course MATH 2025 taught by Professor Staff during the Spring '08 term at LSU.

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Lecture8 - Lecture 8 Orthogonal Projections 8.1...

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