Page 390 - Matrix Analysis & Applied Linear Algebra

386 Chapter 5 Norms, Inner Products, and Orthogonality

Is it possible that there can be more than one projector onto X along
Y ? No, P is unique because if P 1 and P 2 are two such projectors, then for

i =1, 2, we have P i B = P i X | Y = P i X | P i Y = X | 0 , and this implies
P 1 B = P 2 B, which means P 1 = P 2 . Therefore, (5.9.5) is the projector onto
X along Y, and this formula for P is independent of which pair of bases for X
and Y is selected. Notice that the argument involving (5.9.6) and (5.9.7) also
establishes that the complementary projector—the projector onto Y along
X —must be given by

−1 0 0 −1
Q = I − P = 0 | Y B = B B .
0I n−r
Below is a summary of the basic properties of projectors.

Projectors

Let X and Y be complementary subspaces of a vector space V so that
each v ∈V can be uniquely resolved as v = x + y, where x ∈X and
y ∈Y. The unique linear operator P defined by Pv = x is called the
projector onto X along Y, and P has the following properties.
2
• P = P ( P is idempotent). (5.9.8)
• I − P is the complementary projector onto Y along X. (5.9.9)

• R (P)= {x | Px = x} (the set of “fixed points” for P ). (5.9.10)
• R (P)= N (I − P)= X and R (I − P)= N (P)= Y. (5.9.11)
n n
• If V = or C , then P is given by

−1 I 0 −1
P = X | 0 X | Y = X | Y X | Y , (5.9.12)
00
where the columns of X and Y are respective bases for X and Y.
Other formulas for P are given on p. 634.

Proof. Some of these properties have already been derived in the context of
n
. But since the concepts of projections and projectors are valid for all vector
n
spaces, more general arguments that do not rely on properties of will be
provided. Uniqueness is evident because if P 1 and P 2 both satisfy the defining
condition, then P 1 v = P 2 v for every v ∈V, and thus P 1 = P 2 . The linearity
of P follows because if v 1 = x 1 +y 1 and v 2 = x 2 +y 2 , where x 1 , x 2 ∈X and
y 1 , y 2 ∈Y, then P(αv 1 + v 2 )= αx 1 + x 2 = αPv 1 + Pv 2 . To prove that P is
idempotent, write
2 2
P v = P(Pv)= Px = x = Pv for every v ∈V =⇒ P = P.