Linear Algebra, Theory And Applications, 2012a

60 MATRICES AND LINEAR TRANSFORMATIONS
Definition 2.4.8 A finite set of vectors, {x 1 , ··· , x r } is a basis for a subspace V of F n if
span (x 1 , ··· , x r )=V and {x 1 , ··· , x r } is linearly independent.
Corollary 2.4.9 Let {x 1 , ··· , x r } and {y 1 , ··· , y s } be two bases for V .Thenr = s.
Proof: From the exchange theorem, r ≤ s and s ≤ r.
Definition 2.4.10 Let V be a subspace of F n . Then dim (V ) read as the dimension of V
is the number of vectors in a basis.
Of course you should wonder right now whether an arbitrary subspace even has a basis.
In fact it does and this is in the next theorem. First, here is an interesting lemma.
Lemma 2.4.11 Suppose v /∈ span (u 1 , ··· , u k ) and {u 1 , ··· , u k } is linearly independent.
Then {u 1 , ··· , u k , v} is also linearly independent.
Proof: Suppose ∑ k
i=1 c iu i + dv = 0. It is required to verify that each c i = 0 and
that d =0. But if d ≠0, then you can solve for v as a linear combination of the vectors,
{u 1 , ··· , u k },
k∑ ( ci
)
v = − u i
d
i=1
contrary to assumption. Therefore, d =0. But then ∑ k
i=1 c iu i = 0 and the linear independence
of {u 1 , ··· , u k } implies each c i = 0 also.
Theorem 2.4.12 Let V be a nonzero subspace of F n .ThenV has a basis.
Proof: Let v 1 ∈ V where v 1 ≠ 0. If span {v 1 } = V, stop. {v 1 } is a basis for V .
Otherwise, there exists v 2 ∈ V which is not in span {v 1 } . By Lemma 2.4.11 {v 1 , v 2 } is a
linearly independent set of vectors. If span {v 1 , v 2 } = V stop, {v 1 , v 2 } is a basis for V. If
span {v 1 , v 2 } ̸= V, then there exists v 3 /∈ span {v 1 , v 2 } and {v 1 , v 2 , v 3 } is a larger linearly
independent set of vectors. Continuing this way, the process must stop before n + 1 steps
because if not, it would be possible to obtain n + 1 linearly independent vectors contrary to
the exchange theorem.
In words the following corollary states that any linearly independent set of vectors can
be enlarged to form a basis.
Corollary 2.4.13 Let V be a subspace of F n and let {v 1 , ··· , v r } be a linearly independent
set of vectors in V . Then either it is a basis for V or there exist vectors, v r+1 , ··· , v s such
that {v 1 , ··· , v r , v r+1 , ··· , v s } is a basis for V.
Proof: This follows immediately from the proof of Theorem 2.4.12. You do exactly the
same argument except you start with {v 1 , ··· , v r } rather than {v 1 }.
It is also true that any spanning set of vectors can be restricted to obtain a basis.
Theorem 2.4.14 Let V be a subspace of F n and suppose span (u 1 ··· , u p ) = V where
the u i are nonzero vectors. Then there exist vectors {v 1 ··· , v r } such that {v 1 ··· , v r }⊆
{u 1 ··· , u p } and {v 1 ··· , v r } is a basis for V .
Proof: Let r be the smallest positive integer with the property that for some set
{v 1 ··· , v r }⊆{u 1 ··· , u p } ,
span (v 1 ··· , v r )=V.
Then r ≤ p anditmustbethecasethat{v 1 ··· , v r } is linearly independent because if it
were not so, one of the vectors, say v k would be a linear combination of the others. But
then you could delete this vector from {v 1 ··· , v r } and the resulting list of r − 1 vectors
would still span V contrary to the definition of r.