Page 66 - Advanced Linear Algebra

50 Advanced Linear Algebra

for if the vectors in can be expressed as finite linear combinations of the
9
Z
Z
vectors in a proper subset of , then spans , which is not the case.
8
8
=
8
Since ( (< L for all  9 , Theorem 0.17 implies that
9
9
8
( (
( ( ~0  L (( ~ ((
8
9
8 (( 
But we may also reverse the roles of and , to conclude that 9 ( ( and so
¨
the Schroder–Bernstein theorem implies that (( ~ ( 9 . (
8
Theorem 1.12 allows us to make the following definition.
Definition A vector space is finite-dimensional if it is the zero space ¸ ¹ , or
=
if it has a finite basis. All other vector spaces are infinite-dimensional . The
dimension of the zero space is and the dimension of any nonzero vector

space is the cardinality of any basis for . If a vector space has a basis of
=
=
=
cardinality , we say that is -dimensional and write dim ² = ³ ~ .
=

:
It is easy to see that if is a subspace of = , then dim ²:³  dim ²= ³ . If in
addition, dim²:³ ~ dim²= ³  B , then : ~ = .

Theorem 1.13 Let be a vector space.
=
)
=
1 If is a basis for and if ~ 8 8 8 r and 8 8 q ~ J , then
8
=~ º 8 » l º 8 »
)
2 Let =~ : l ; . If 8 is a basis for and 8 : is a basis for , then
;
=
8 8 q ~ J 8 and 8 ~ 8 r is a basis for .
Theorem 1.14 Let and be subspaces of a vector space . Then

:
=
;
dim²:³ b dim²;³ ~ dim²: b ;³ b dim²: q ;³
In particular, if is any complement of in , then
=
;
:
dim²:³ b dim²;³ ~ dim²= ³
that is,
dim²: l ;³ ~ dim²:³ b dim²;³
Proof. Suppose that 8 ~¸ “ 0¹ is a basis for : q ; . Extend this to a basis

8
8
7 8 r : for where 7 ~ ¸ “  1 ¹ is disjoint from . Also, extend to a
;
basis 8 9 r for where 9 ~ ¸ “  2 ¹ is disjoint from . We claim that
8
7 8 r 9 r is a basis for : b ; . It is clear that 7 º 8 r 9 r » ~ : b ; .
To see that 7 8 r 9 r is linearly independent, suppose to the contrary that