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MATB24H3 (13)
Lecture

# Lecture14.pdf

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Department
Mathematics
Course Code
MATB24H3
Professor
Sophie Chrysostomou

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University of Toronto at Scarborough Department of Computer & Mathematical Sciences MAT B24S Fall 2011 MAT B24 Lectures 14. 1 Orthogonality 1.1 Projections Review: In R with the dot product as an inner product: consider the ques- tion of ”decomposing” a vector b into a sum of two terms: one parallel to a speciﬁed nonzero vector a and the other perpendicular to a. We will call the component parallel to the vector a the orthogonal projection of b on a and denote it by proa b. Let p = proa b, then we want to ﬁnd p and v = such that v is orthogonal to a and b = p + v b · a In A23 we have shown that p = projab = 2a Let us remind ourselves ||a|| of the proof: Suppose that b = p + v where p = ka and v is orthogonal to a. Then : b · a = (p + v) · a = p · a + v · a = (ka) · a + 0 = k(a · a) Thus k = b · a= b · a a · a ||a|| and p = b · aa ||a|| and v = b − p 1 EXAMPLE: Find p, the orthogonal projection of b = [3,1,−7] on a = [1,0,5] and v, the vector component of b orthogonal to a. solution. If p is the orthogonal projection of b on a, then: ▯ ▯ b · a 3 · 1 + 1 · 0 + (−7) · −32 −16 −80 p = projab = 2a = a = [1,0,5] = ,0, ||a|| 1 + 25 26 13 13 and if v is the vector component of b orthogonal to a, then: ▯−16 −80 ▯ ▯23 −171 ▯ v = b − p = [3,1,−7] − ,0, = ,1, 13 13 13 13 We will expand the above to R In the previous example, we may think of p as a vector in the subspace W = sp(a) and of v as a vector from a set of all vectors perpendicular to every vector in W. DEFINITION: Suppose that W is a subspace of the vector space V with inner product , then the set W ⊥ = {v ∈ V | < v,w >= 0 ∀ w ∈ W} is called the orthogonal complement of W. With the above deﬁnition then, our previous example can be viewed as ⊥ b = p + v where p ∈ W = sp(a) and v ∈ W Suppose that b ∈ R and W is a subspace of R and with dot product ⊥ again as the inner product we get W . We will show that we can ”decom- pose” b into the sum of two unique vectoWs and bW⊥ such that W ∈ W and bW ⊥. We will call the vectoW the projection of b on W. 2 ⊥ 1.2 To ﬁnd W First we establish a method to get W : Given a basi1 v ,·k· ,v of W we ⊥ know that every vector v ∈ W satisﬁes thativ · v = 0. If we construct the matrix   v  1   v 2   ·  A =    ·   ·  v k ⊥ Then Av = 0 for all v ∈ W. Thus, the nullspace of A is W . EXAMPLE: Suppose that W = sp(
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