# MATB24H3 Lecture Notes - Orthogonal Complement, Dot Product, Proa

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Published on 21 Apr 2013
School
Department
Course
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 R3with the dot product as an inner product: consider the ques-
tion of ”decomposing” a vector binto a sum of two terms: one parallel to a
speciﬁed nonzero vector aand the other perpendicular to a. We will call the
component parallel to the vector athe orthogonal projection of bon aand
denote it by projab. Let p=projab, then we want to ﬁnd pand v= such
that vis orthogonal to aand b=p+v
In A23 we have shown that p=projab=b·a
||a||2aLet us remind ourselves
of the proof:
Suppose that b=p+vwhere p=kaand vis orthogonal to a. Then :
b·a= (p+v)·a=p·a+v·a= (ka)·a+ 0 = k(a·a)
Thus
k=b·a
a·a=b·a
||a||2
and
p=b·a
||a||2a
and v=bp
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EXAMPLE: Find p, the orthogonal projection of b= [3,1,7] on a=
[1,0,5] and v, the vector component of borthogonal to a.
solution. If pis the orthogonal projection of bon a, then:
p=projab=b·a
||a||2a=3·1 + 1 ·0 + (7) ·5
1 + 25 a=32
26 [1,0,5] = 16
13 ,0,80
13
and if vis the vector component of borthogonal to a, then:
v=bp= [3,1,7] 16
13 ,0,80
13 =23
13,1,171
13
We will expand the above to Rn
In the previous example, we may think of pas a vector in the subspace
W=sp(a) and of vas a vector from a set of all vectors perpendicular to
every vector in W.
DEFINITION: Suppose that Wis a subspace of the vector space Vwith
inner product <, >, then the set
W={vV|<v,w>= 0 wW}
is called the orthogonal complement of W.
With the above deﬁnition then, our previous example can be viewed as
b=p+vwhere pW=sp(a) and vW
Suppose that bRnand Wis a subspace of Rnand with dot product
again as the inner product we get W. We will show that we can ”decom-
pose” binto the sum of two unique vectors bWand bWsuch that bWW
and bW. We will call the vector bWthe projection of bon W.
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## Document Summary

We will call the component parallel to the vector a the orthogonal projection of b on a and denote it by projab. Let p = projab, then we want to nd p and v = such that v is orthogonal to a and b = p + v. In a23 we have shown that p = projab = ||a||2 a let us remind ourselves of the proof: Suppose that b = p + v where p = ka and v is orthogonal to a. A = (p + v) a = p a + v a = (ka) a + 0 = k(a a) Thus and and v = b p k = 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: p = projab =