Page 388 - Matrix Analysis & Applied Linear Algebra

384 Chapter 5 Norms, Inner Products, and Orthogonality

Prove these by arguing (5.9.2) =⇒ (5.9.3) =⇒ (5.9.4) =⇒ (5.9.2).
Proof of (5.9.2) =⇒ (5.9.3). First recall from (4.4.19) that
dim V = dim (X + Y)= dim X + dim Y− dim (X∩ Y) .
If V = X⊕ Y, then X∩ Y = 0, and thus dim V = dim X + dim Y. To prove
(5.9.3), suppose there are two ways to represent a vector v ∈V as “something
from X plus something from Y. ”If v = x 1 + y 1 = x 2 + y 2 , where x 1 , x 2 ∈X
and y 1 , y 2 ∈Y, then
 
 x 1 − x 2 ∈X 
x 1 − x 2 = y 2 − y 1 =⇒ and =⇒ x 1 − x 2 ∈X ∩ Y.
 
x 1 − x 2 ∈Y
But X∩ Y = 0, so x 1 = x 2 and y 1 = y 2 .
Proof of (5.9.3) =⇒ (5.9.4). The hypothesis insures that V = X + Y, and we
know from (4.1.2) that B X ∪B Y spans X +Y, so B X ∪B Y must be a spanning
set for V. To prove B X ∪B Y is linearly independent, let B X = {x 1 , x 2 ,..., x r }
and B Y = {y 1 , y 2 ,..., y s } , and suppose that
r s

0 = α i x i + β j y j .
i=1 j=1
This is one way to express 0 as “something from X plus something from Y, ”
while 0 = 0 + 0 is another way. Consequently, (5.9.3) guarantees that
r s

α i x i = 0 and β j y j = 0,
i=1 j=1
and hence α 1 = α 2 = ··· = α r =0 and β 1 = β 2 = ··· = β s =0 because
B X and B Y are both linearly independent. Therefore, B X ∪B Y is linearly
independent, and hence it is a basis for V.
Proof of (5.9.4) =⇒ (5.9.2). If B X ∪B Y is a basis for V, then B X ∪B Y is a
linearly independent set. This together with the fact that B X ∪B Y always spans
X + Y means B X ∪B Y is a basis for X + Y as well as for V. Consequently,
V = X + Y, and hence
dim X + dim Y = dim V = dim(X + Y)= dim X + dim Y− dim (X∩ Y) ,
so dim (X∩ Y)= 0 or, equivalently, X∩ Y = 0.
If V = X⊕ Y, then (5.9.3) says there is one and only one way to resolve
each v ∈V into an “X -component” and a “Y-component” so that v = x + y.
These two components of v have a definite geometrical interpretation. Look
3
back at Figure 5.9.1 in which = X⊕ Y, where X is a plane and Y is a line
outside the plane, and notice that x (the X -component of v )is the result of
projecting v onto X along a line parallel to Y, and y (the Y-component of
v )is obtained by projecting v onto Y along a line parallel to X. This leads
to the following formal definition of a projection.