Page 64 - Advanced Linear Algebra
P. 64
48 Advanced Linear Algebra
Theorem 1.9 Let be a nonzero vector space. Let be a linearly independent
0
=
:
=
0
set in and let be a spanning set in containing . Then there is a basis 8
=
for for which 0 8 : . In particular,
=
)
1 Any vector space, except the zero space ¸ ¹ , has a basis.
2 Any linearly independent set in is contained in a basis.
)
=
)
3 Any spanning set in contains a basis.
=
Proof. Consider the collection 7 of all linearly independent subsets of =
containing and contained in . This collection is not empty, since 0 7 .
0
:
Now, if
9 ~¸0 2¹
is a chain in , then the union
7
<~ 0
2
is linearly independent and satisfies 0 < : , that is, < 7 . Hence, every
chain in has an upper bound in and according to Zorn's lemma, must
7
7
7
contain a maximal element , which is linearly independent.
8
Now, is a basis for the vector space º:» ~ = , for if any : is not a linear
8
8
combination of the elements of , then 8 r¸ ¹ : is linearly independent,
8
8
contradicting the maximality of . Hence : º» and so = ~ º:» º» .
8
The reader can now show, using Theorem 1.9, that any subspace of a vector
space has a complement.
The Dimension of a Vector Space
The next result, with its classical elegant proof, says that if a vector space has
=
a finite spanning set , then the size of any linearly independent set cannot
:
exceed the size of .
:
=
Theorem 1.10 Let be a vector space and assume that the vectors # ÁÃÁ#
=
are linearly independent and the vectors Á Ã Á span . Then .
Proof. First, we list the two sets of vectors: the spanning set followed by the
linearly independent set:
Á ÃÁ Â# ÁÃÁ#
Then we move the first vector to the front of the first list:
#
#Á Á Ã Á Â #Á Ã Á #
span , is a linear combination of the 's. This implies that
=
Since Á Ã Á #
we may remove one of the 's, which by reindexing if necessary can be ,
from the first list and still have a spanning set
#Á Á Ã Á Â #Á Ã Á #
