Page 389 - Matrix Analysis & Applied Linear Algebra

5.9 Complementary Subspaces 385

Projection
Suppose that V = X⊕ Y so that for each v ∈V there are unique
vectors x ∈X and y ∈Y such that v = x + y.
• The vector x is called the projection of v onto X along Y.
• The vector y is called the projection of v onto Y along X.

It’s clear that if X⊥ Y in Figure 5.9.1, then this notion of projection
agrees with the concept of orthogonal projection that was discussed on p. 322.
The phrase “oblique projection” is sometimes used to emphasize the fact that
X and Y are not orthogonal subspaces. In this text the word “projection” is
synonymous with the term “oblique projection.” If it is known that X⊥ Y, then
we explicitly say “orthogonal projection.” Orthogonal projections are discussed
in detail on p. 429.
n
Given a pair of complementary subspaces X and Y of and an arbitrary
n
vector v ∈ = X⊕ Y, how can the projection of v onto X be computed?
One way is to build a projector (a projection operator) that is a matrix P n×n
n
with the property that for each v ∈ , the product Pv is the projection of
v onto X along Y. Let B X = {x 1 , x 2 ,..., x r } and B Y = {y 1 , y 2 ,..., y n−r }
n
be respective bases for X and Y so that B X ∪B Y is a basis for —recall
(5.9.4). This guarantees that if the x i ’s and y i ’s are placed as columns in

B n×n = x 1 x 2 ··· x r | y 1 y 2 ··· y n−r = X n×r | Y n×(n−r) ,
then B is nonsingular. If P n×n is to have the property that Pv is the pro-
n
jection of v onto X along Y for every v ∈ , then (5.9.3) implies that
Px i = x i ,i =1, 2,...,r and Py j = 0,j =1, 2,...,n − r, so

PB = P X | Y = PX | PY = X | 0
and, consequently,

−1 I r 0 −1
P = X | 0 B = B B . (5.9.5)
0 0
To argue that Pv is indeed the projection of v onto X along Y, set x = Pv
and y =(I − P)v and observe that v = x + y, where
−1
x = Pv = X | 0 B v ∈ R (X)= X (5.9.6)
and

0 0 −1 −1
y =(I − P)v = B B v = 0 | Y B v ∈ R (Y)= Y. (5.9.7)
0I n−r