Page 392 - Matrix Analysis & Applied Linear Algebra
P. 392
388 Chapter 5 Norms, Inner Products, and Orthogonality
Furthermore, R (P) ∩ N (P)= 0 because
2
x ∈ R (P) ∩ N (P)=⇒ x = Py and Px = 0 =⇒ x = Py = P y = 0,
and thus (5.9.14) is established. Now that we know R (P) and N (P) are com-
plementary, we can conclude that P is a projector because each v ∈V can be
uniquely written as v = x + y, where x ∈ R (P) and y ∈ N (P), and (5.9.15)
guarantees Pv = x.
Notice that there is a one-to-one correspondence between the set of idem-
potents (or projectors) defined on a vector space V and the set of all pairs of
complementary subspaces of V in the following sense.
• Each idempotent P defines a pair of complementary spaces—namely, R (P)
and N (P).
• Every pair of complementary subspaces X and Y defines an idempotent—
namely, the projector onto X along Y.
Example 5.9.1
3
Problem: Let X and Y be the subspaces of that are spanned by
1 0 1
,
B X = −1 1 and B Y = −1 ,
−1 −2 0
respectively. Explain why X and Y are complementary, and then determine
T
the projector onto X along Y. What is the projection of v =( −213 )
onto X along Y? What is the projection of v onto Y along X?
Solution: B X and B Y are linearly independent, so they are bases for X and
Y, respectively. The spaces X and Y are complementary because
1 0 1
rank [X | Y]= rank −1 1 −1 =3
−1 −2 0
3
insures that B X ∪B Y is a basis for —recall (5.9.4). The projector onto X
along Y is obtained from (5.9.12) as
1 0 0 −2 −2 −1 −2 −2 −1
−1
P = X | 0 X | Y = −1 1 0 1 1 0 = 3 3 1 .
−1 −20 3 2 1 0 0 1
Youmay wish to verify that P is indeed idempotent. The projection of v onto
X along Y is Pv, and, according to (5.9.9), the projection of v onto Y along
X is (I − P)v.
