Page 391 - Matrix Analysis & Applied Linear Algebra

5.9 Complementary Subspaces 387

The validity of (5.9.9) is established by observing that v = x + y = Pv + y
implies y = v − Pv =(I − P)v. The properties in (5.9.11) and (5.9.10) are
immediate consequences of the definition. Formula (5.9.12) is the result of the
arguments that culminated in (5.9.5), but it can be more elegantly derived by
making use of the material in §4.7 and §4.8. If B X and B Y are bases for X
and Y, respectively, then B = B X ∪B Y = {x 1 , x 2 ,..., x r , y 1 , y 2 ,..., y n−r } is
a basis for V, and (4.7.4) says that the matrix of P with respect to B is

1 2

[P] B = [Px 1 ] B ··· [Px r ] B [Py 1 ] B ··· [Py n−r ] B
1 2

= [x 1 ] B ··· [x r ] B [0] B ··· [0] B

1 2 0

= e 1 ··· e r 0 ··· 0 = I r .
0 0
If S is the standard basis, then (4.8.5) says that [P] B = B −1 [P] S B in which
1 2

B =[I] BS = [x 1 ] S ··· [x r ] S [y 1 ] S ··· [y n−r ] S = X | Y ,
0 −1

and therefore [P] S = B[P] B B −1 = X | Y I r X | Y .
0 0
In the language of §4.8, statement (5.9.12) says that P is similar to the
I 0
diagonal matrix . In the language of §4.9, this means that P must be
0 0
the matrix representation of the linear operator that when restricted to X is
the identity operator and when restricted to Y is the zero operator.
Statement (5.9.8) says that if P is a projector, then P is idempotent
2
( P = P ). But what about the converse—is every idempotent linear operator
necessarily a projector? The following theorem says, “Yes.”
Projectors and Idempotents
2
A linear operator P on V is a projector if and only if P = P. (5.9.13)
Proof. The fact that every projector is idempotent was proven in (5.9.8). The
proof of the converse rests on the fact that

2
P = P =⇒ R (P) and N (P) are complementary subspaces. (5.9.14)
To prove this, observe that V = R (P)+ N (P)because for each v ∈V,

v = Pv +(I − P)v, where Pv ∈ R (P) and (I − P)v ∈ N (P). (5.9.15)