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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