A set of vectors 
is said to span
a subspace S of 
If V spans S it is also said to be a spanning set for V.
Note that a spanning set V for subspace S
must be a subset of S itself and all
vectors of S other that 
must have a representation in terms of V.
Representation of is generally a
trivial matter and
we do not require its representation
in condition (2),
which was carefully crafted
so that the trivial subspace
of will have a spanning set.
In cases
other than that of the trivial subspace,
it is operationally simpler
to
make use of the following Proposition and
show that
.
Proof
Let
S be a given nontrivial subspace of .
Assume
.
Certainly, 
and .
Now and 
V spans S.
Assume V spans S
and .
Since S is a nontrivial subspace of , 
.
Also,
there is some 
.
Certainly, 
.
And,
and 
.
Examples
is a subspace of 
.
is a subset of S.
Since
,
any element of S has
a representation in terms of V
V spans S .
is a subspace of .
is a subset of T.
Since
,
any element of T has
a representation in terms of W
W spans T .
Proof
Certainly, 
.
By Prop. 
,
spans , by Prop. .
Proof
Assume
and
.
Certainly 
, since all n-vectors are in .
By Prop. , 
.
Now and
.
Proof
Follows immediately since 
and 
.
Proof Left as an exercise.
Proof Left as an exercise.
Proof
Assume
spans subspace S of
and
is independent.
By Prop. ,
is dependent and spans S.
By Prop. ,
there is some vector in
which is a linear combination of
the preceeding vectors in 
.
Let 
be that vector.
By Prop. ,
spans S.
Repeating the above argument,
we have that
is dependent and spans S
and
some vector in
is a linear combination of the preceeding vectors.
This vector cannot be one of
since
W is independent and Prop.
is independent.
Let that vector be 
,
which must be one of the vectors in
.
Then
spans S.
This process can be continued as long as 
,
at each step
forming a new spanning set for S,
replacing a vector of V by a new vector from W
in the previouly constructed spanning set
so that
spans S.
[We now show that .]
Assume, to the contrary, that j>k.
Then after only k of the above replacement steps,
is obtained
as a spanning set for S.
But this would imply that 
are linear combinations of
,
contradicting that W is independent!
Thus .
Proof
Assume 
is independent.
Now is a subspace of .
By Prop. , the set
spans .
Certainly, .
By Prop. , 
.
