Page 268 - 35Linear Algebra

268 Orthonormal Bases and Complements

 
0
1
 
Notice that the addends have elements in common;   is in both addends. Even
1
 
0
though both of the addends are 2-dimensional their sum is not 4-dimensional.
In the special case that U and V do not have any non-zero vectors in
common, their sum is a vector space with dimension dim U + dim V .

Definition If U and V are subspaces of a vector space W such that U ∩ V = {0 W }
then the vector space

U ⊕ V := span(U ∪ V ) = {u + v | u ∈ U, v ∈ V }

is the direct sum of U and V .

Remark
• When U ∩ V = {0 W }, U + V = U ⊕ V.

• When U ∩ V 6= {0 W }, U + V 6= U ⊕ V .

This distinction is important because the direct sum has a very nice property:

Theorem 14.6.1. If w ∈ U ⊕ V then there is only one way to write w as
the sum of a vector in U and a vector in V .

0
0
0
0
Proof. Suppose that u + v = u + v , with u, u ∈ U, and v, v ∈ V . Then we
0
0
0
0
could express 0 = (u − u ) + (v − v ). Then (u − u ) = −(v − v ). Since U
0
0
and V are subspaces, we have (u − u ) ∈ U and −(v − v ) ∈ V . But since
0
these elements are equal, we also have (u−u ) ∈ V . Since U ∩V = {0}, then
0
0
0
0
(u − u ) = 0. Similarly, (v − v ) = 0. Therefore u = u and v = v , proving
the theorem.
Reading homework: problem 3
Here is a sophisticated algebra question:
Given a subspace U in W, what are the solutions to
U ⊕ V = W.
That is, how can we write W as the direct sum of U and some-
thing?

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