A linearly independent set of vectors which
spans a subspace S of 
is called a basis
for S.
Examples
is spanned by 
,
V is not a basis for S,
since we have shown earlier that
V is dependent.
spans
and is a basis for T
since W is independent.
Independence of W follows from
= (0,0,0)
.
Proof
Let 
be a basis for given subspace 
.
By Prop. 
, 
,
(according to Prop. )
,
(according to Propositions and )
.
Proof Left as an exercise.
Proof Left as an exercise.
Proof
This follows immediately from Prop.
and Prop. .
The set of coordinate vectors
is called the
standard basis for .
Proof
Follows immediately from Propositions , , and .
Proof
Let
S be a nontrivial subspace of .
Then there is some vector 
such that 
.
Now
.
If 
, 
is a basis for S.
Otherwise,
there is some 
such that 
.
By Prop. ,
is independent.
Let
.
If 
, 
is a basis for S.
Otherwise,
there is some 
such that 
and we continue the above process,
next forming 
and 
.
This process must eventually terminate
with an independent set
which is a basis for
,
since by Prop. we cannot produce
an independent set 
with m > n.
Proof
Let S be a subspace of
and let
and 
be arbitrary bases for S.
Since V spans S and W is independent it follows
from Prop. that 
.
Since W spans S and V is independent it follows
from Prop. that 
.
Thus j = k
all bases for S
have the same number of vectors.
The number of basis vectors in a basis of a
subspace as specified in Prop.
is called the dimension of that subspace.
The dimension of subspace S will be denoted by
.
Since the standard basis for has n vectors,
has dimension n.
Proof
By Prop. , has dimension n.
Let S be a subspace of
with
dimension n.
By Propositions and ,
there is a basis 
for S
V is independent and ,
(according to Prop. ).
Certainly, .
[Show 
.]
Assume, to the contrary, that there is a vector 
such that 
.
By Prop. , 
is independent
there is an independent set in with n+1 elements,
contradicting Prop. !
Thus .
Proof
Follows immediately from Prop. .
Proof Left as an exercise.
Proof
Let S, V, and
b be given
with
and
W defined as above
.
Assume W is a basis for S.
Assume, to the contrary, that 
.
Then
.
Rearranging, we obtain
.
Thus
is a nontrivial representation of 
in terms of W, a contradiction
to W being a basis!
Thus 
.
Assume .
Then
or
.
Then any vector of S
representable in terms of the vectors of V
is also representable in terms of the vectors of W.
Thus W spans S.
[We now show that W is independent.]
Let
.
Assume, to the contrary, that
are not all zero.
Case 1 Assume 
and not all of
are zero
we have a nontrivial representation of
in terms of V,
a contradiction to V being a basis!
Case 2 Assume 
.
Thus
.
Now 
so that
is a nontrivial representation of
in terms of V, a contradiction to V being a basis!
Thus
all of
must be zero
there is no nontrivial representation of
in terms of W
W is linearly independent.
Hence W is a basis for S.
Given a
,
the set 
is said to be a maximal independent subset of W
if V is independent
and
is dependent.
Any basis for a subspace of
is a maximal independent subset of that subspace.
Proof Left as an exercise.
