Page 43 - Linear Algebra Done Right

insures that no vector in B is in the span of the previous ones. Thus B
is linearly independent, by the linear dependence lemma (2.4). Hence
B is a basis of V. Bases 29
Consider the list

(1, 2), (3, 6), (4, 7), (5, 9) ,
2
which spans F . To make sure that you understand the last proof, you

should verify that the process in the proof produces (1, 2), (4, 7) ,a
2
basis of F , when applied to the list above.
Our next result, an easy corollary of the last theorem, tells us that
every finite-dimensional vector space has a basis.

2.11 Corollary: Every finite-dimensional vector space has a basis.

Proof: By definition, a finite-dimensional vector space has a span-
ning list. The previous theorem tells us that any spanning list can be
reduced to a basis.

We have crafted our definitions so that the finite-dimensional vector
space {0} is not a counterexample to the corollary above. In particular,
the empty list () is a basis of the vector space {0} because this list has
been defined to be linearly independent and to have span {0}.
Our next theorem is in some sense a dual of 2.10, which said that
every spanning list can be reduced to a basis. Now we show that given
any linearly independent list, we can adjoin some additional vectors so
that the extended list is still linearly independent but also spans the
space.

2.12 Theorem: Every linearly independent list of vectors in a finite- This theorem can be
dimensional vector space can be extended to a basis of the vector space. used to give another
proof of the previous
Proof: Suppose V is finite dimensional and (v 1 ,...,v m ) is linearly corollary. Specifically,
independent in V. We want to extend (v 1 ,...,v m ) to a basis of V.We suppose V is finite
do this through the multistep process described below. First we let dimensional. This
(w 1 ,...,w n ) be any list of vectors in V that spans V. theorem implies that
the empty list () can be
Step 1
extended to a basis
If w 1 is in the span of (v 1 ,...,v m ), let B = (v 1 ,...,v m ).If w 1 is
of V. In particular, V
not in the span of (v 1 ,...,v m ), let B = (v 1 ,...,v m ,w 1 ). has a basis.