Page 458 - Matrix Analysis & Applied Linear Algebra

454 Chapter 5 Norms, Inner Products, and Orthogonality

Similarly, δ(N, M)= (I − P M )P N = P N (I − P M ) . If U = U 1 | U 2
2
2

are the orthogonal matrices introduced on p. 451, then
and V = V 1 | V 2

T

T
δ(M, N)= P M (I − P N ) = U 1 U V 1 V T

1
2 1 1 2 = U V 1 2
and (5.15.11)

T

T
δ(N, M)= (I − P M )P N = U 2 U V 2 V T

.

2
2 2 2 2 = U V 2 2
Combining these observations with (5.15.7) leads us to conclude that
5 6

T

T
P M − P N = max
2
2 U V 1 2 , U V 2 2
1

= max δ(M, N),δ(N, M) (5.15.12)
= gap (M, N).
Below is a summary of these and other properties of the gap measure.
Gap Properties
n
The following statements are true for subspaces M, N⊆ .
• gap (M, N)= P M − P N .
2

• gap (M, N)= max (I − P N )P M , (I − P M )P N .
2 2
• gap (M, N)= 1 whenever dim M = dim N. (5.15.13)
• If dim M = dim N, then δ(M, N)= δ(N, M), and
⊥
⊥
gap (M, N)= 1 when M ∩N (or M∩N ) = 0, (5.15.14)
gap (M, N) < 1 when M ∩N (or M∩N )= 0. (5.15.15)
⊥
⊥
Proof of (5.15.13). Suppose that dim M = r and dim N = k, where r< k.
⊥
Notice that this implies that M ∩N = 0, for otherwise the formula for the
dimension of a sum (4.4.19) yields
⊥
⊥
n ≥ dim(M + N)= dim M + dim N = n − r + k> n,
which is impossible. Thus there exists a nonzero vector x ∈M ∩N, and by
⊥
normalization we can take x =1. Consequently, (I−P M )x = x = P N x, so
2
(I − P M )P N x =1. This insures that (I − P M )P N =1, which implies
2 2
δ(N, M)=1.
Proof of (5.15.14). Assume dim M = dim N = r, and use the formula for the
dimension of a sum along with (M∩N ) = M + N (Exercise 5.11.5) to
⊥ ⊥
⊥
conclude that

⊥ ⊥ ⊥
dim M ∩N = dim M + dim N− dim M + N
⊥
⊥ ⊥
=(n − r)+ r − dim M∩N = dim M∩N .