Page 455 - Matrix Analysis & Applied Linear Algebra

5.15 Angles between Subspaces 451

This together with (5.2.9) and the fact that P M x ∈M and P N y ∈N means

T
T
cos θ min = max v u = max v u
u∈M, v∈N u∈M, v∈N
u = v =1 u ≤1, v ≤1
2 2 2 2
T
= max y P N P M x = P N P M .
2
x ≤1, y ≤1
2 2

Proof of (5.15.3). Let U = U 1 | U 2 and V = V 1 | V 2 be orthogonal
matrices in which the columns of U 1 and U 2 constitute orthonormal bases for
⊥ ⊥
M and M , respectively, and V 1 and V 2 are orthonormal bases for N
T
T
and N, respectively, so that U U i = I and V V i = I for i =1, 2, and
i i
T
T
T
T
P M = U 1 U , I − P M = U 2 U , P N = V 2 V , I − P N = V 1 V .
1 2 2 1
As discussed on p. 407, there is a nonsingular matrix C such that

C 0 T T
P MN = U V = U 1 CV . (5.15.6)
1
0 0
T
Notice that P 2 = P MN implies C = CV U 1 C, which in turn insures
MN 1
T
C −1 = V U 1 . Recall that XAY = A whenever X has orthonormal
1 2 2
columns and Y has orthonormal rows (Exercise 5.6.9). Consequently,
1 1
P MN = C =

=

(recall (5.2.6)).
2 2 T
x
min
C −1
min
V U 1 x
1
x =1 2 x =1 2
2
2
Combining this with (5.15.2) produces (5.15.3) by writing

2
T
2
2
sin θ min =1 − cos θ min =1 − P N P M =1 − V 2 V U 1 U T
2 2

2
1
2

2

2
T T

=1 − (I − V 1 V )U 1
=1 − max
(I − V 1 V )U 1 x
1 2 1 2
x =1
2

2
T
T
T
T
=1 − max x U (I − V 1 V )U 1 x =1 − max 1 − V U 1 x

1
1
1
x =1 x =1 2
2 2

T
2 1
=1 − 1 − min
V U 1 x
= .
1
x =1 2 P MN 2
2
2
Proof of (5.15.4). Observe that
T
U 1
T
T
T
U (P M − P N )V = (U 1 U − V 2 V ) V 1 | V 2
1
2
U T (5.15.7)
2

T
U V 1 0
1
= T ,
0 −U V 2
2