Page 438 - Matrix Analysis & Applied Linear Algebra

434 Chapter 5 Norms, Inner Products, and Orthogonality

P is symmetric, then it must be an orthogonal projector because (5.11.5) on
p. 405 allows us to write

P = P T =⇒ R (P)= R P T =⇒ R (P) ⊥ N (P).
To see why (5.13.10) characterizes projectors that are orthogonal, refer back
to Example 5.9.2 on p. 389 (or look ahead to (5.15.3)) and note that P =
2
1/ sin θ, where θ is the angle between R (P) and N (P). This makes it clear
that P ≥ 1 for all projectors, and P =1 if and only if θ = π/2, (i.e., if
2 2
and only if R (P) ⊥ N (P)).
Example 5.13.3
m×n
Problem: For A ∈ such that rank (A)= r, describe the orthogonal
projectors onto each of the four fundamental subspaces of A.
Solution 1: Let B m×r and N n×n−r be matrices whose columns are bases for
R (A) and N (A), respectively—e.g., B might contain the basic columns of

⊥
A. The orthogonal decomposition theorem on p. 405 says R (A) = N A T

⊥
and N (A) = R A T , so, by making use of (5.13.3) and (5.13.6), we can write
T −1 T
P R(A) = B B B B ,
T −1 T
P N(A T ) = P ⊥ = I − P R(A) = I − B B B B ,
R(A)
T −1 T
P N(A) = N N N N ,
T −1 T
P R(A T ) = P ⊥ = I − P N(A) = I − N N N N .
N(A)
Note: If rank (A)= n, then all columns of A are basic and

T −1 T
P R(A) = A A A A . (5.13.11)
Solution 2: Another way to describe these projectors is to make use of the
Moore–Penrose pseudoinverse A (p. 423). Recall that if A has a URV factor-
†
ization
−1
C0 T C 0 T
†
A = U V , then A = V U ,
0 0 0 0

are orthogonal matrices in which the
where U = U 1 | U 2 and V = V 1 | V 2

columns of U 1 and V 1 constitute orthonormal bases for R (A) and R A T ,

respectively, and the columns of U 2 and V 2 are orthonormal bases for N A T
†
and N (A), respectively. Computing the products AA and A A reveals
†

I 0 T T I 0 T T
†
AA = U U = U 1 U 1 and A A = V V = V 1 V ,
†
1
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