Basic Theorems about Independence, Spanning, Basis,
and Dimension
Theorem 1 If v1 ; : : : ; vm span V; then every set of vectors in V with more than m vectors
must be linearly dependent.
Proof. Let w1 ; : : : ; wn be vectors in V with n > m: We will show that w1 ; : : : ; wn are
linearly dependent. Consider the canonical equation
c1 w1 + : : : + cn wn = 0
(1)
We will show that there exist c1 ; : : : ; cn not all zero satisfying the canonical equation
above. Since v1 ; : : : ; vm span V we can write each wi in as a linear combination of
v1 ; : : : ; vm : Then equation (1) becomes
c1 (a11 v1 + : : : + am1 vm ) + : : : + cn (a1n v1 + : : : + amn vm ) = 0
(2)
Rearranging the terms gives
(a11 c1 + : : : + a1n cn )v1 + : : : + (am1 c1 + : : : + amn cn )vm = 0
(3)
Now consider the homogeneous system of equations shown below.
a11 c1 + : : : + a1n cn = 0
..
.. ..
(4)
.
. .
am1 c1 + : : : + amn cn = 0
A previous theorem implies there exist solutions c1 ; : : : ; cn not all zero to system because
n > m; that is, there are more unknowns than equations. These same c1 ; : : : ; cn (not all
zero) satisfy equations (3) and (2) and (1). Because there exist c1 ; : : : ; cn not all zero
satisfying the canonical equation (1) the vectors w1 ; : : : ; wn are linearly dependent.
Theorem 2 If v1 ; : : : ; vn are linearly independent in V; then every set of vectors in V
with fewer than n vectors cannot span V:
Proof. (by contradiction) Assume w1 ; : : : ; wm (with m < n) span V: Then by Theorem
1 every set of vectors in V with more than m vectors must be linearly dependent. This
contradicts our hypothesis that v1 ; : : : ; vn are linearly independent.
Theorem 3 If v1 ; : : : ; vn is a basis of V and w1 ; : : : ; wm is a basis of V; then n = m:
Proof. By de nition of basis v1 ; : : : ; vn span V and w1 ; : : : ; wm are linearly independent,
so theorem 1 says m
n: Similarly, w1 ; : : : ; wm span V and v1 ; : : : ; vn are linearly
independent, so theorem 1 implies n m: The. only possibility is that n = m:
Theorem 3 says that if V has one basis with n elements, then all bases of V have n
elements. The number n is intrinsic to the space itself and is independent of which basis
you choose. We make the following de nition.
De nition Let V be a vector space with a basis of n vectors. The number n is called the
dimension of V:
Theorem 4 Suppose v1 ; : : : ; vk are linearly independent. If vk+1 2
= spanfv1 ; : : : ; vk g;
then v1 ; : : : ; vk ; vk+1 are linearly independent.
Proof. Consider the canonical equation
c1 v1 + : : : + ck vk + ck+1 vk+1 = 0
We rst claim that ck+1 = 0: For if ck+1 6= 0; the we can solve the canonical equation for
vk+1 to get
1
vk+1 =
( c1 v1
ck vk )
ck+1
This contradicts the fact that vk+1 2
= spanfv1 ; : : : ; vk g: Hence, ck+1 = 0 as claimed. The
canonical equation now simpli es to
c1 v1 + : : : + ck vk = 0
Now, since v1 ; : : : ; vk are linearly independent we have c1 = 0; : : : ; ck = 0 (in addition
to ck+1 = 0): Hence, the vectors v1 ; : : : ; vk ; vk+1 are linearly independent, which is what
we wanted to show.
Theorem 5 Suppose v1 ; : : : ; vk span V: If vk 2 spanfv1 ; : : : ; vk 1 g; then v1 ; : : : ; vk
span V:
Proof. Choose an arbitrary vector v in V: and show that v can be written as a linear
combination of v1 ; : : : ; vk 1 : By hypothesis we can write v as a linear combination of
v1 ; : : : ; vk ; That is,
v = c1 v1 + : : : + ck 1 vk 1 + ck vk
But since vk 2 spanfv1 ; : : : ; vk 1 g we can write
vk = b1 v1 + : : : + bk 1 vk 1
Substituting this expression for vk into the previous expression for v gives
v = c1 v1 + : : : + ck 1 vk 1 + ck (b1 v1 + : : : + bk 1 vk 1 )
v = (c1 + ck b1 )v1 + : : : + (ck 1 + ck bk 1 )vk 1
Thus we see that v is a linear combination of v1 ; : : : ; vk 1 : Since v was an
arbitrary vector in V; v1 ; : : : ; vk 1 span V; as claimed
1
Theorem 6 (The Cut-Your-Work-In-Half Theorem)
(Part I) If V is of dimension n and v1 ; : : : ; vn are linearly independent, then v1 ; : : : ; vn
must span V (and hence form a basis of V ):
(Part II) If V is of dimension n and v1 ; : : : ; vn span V; then v1 ; : : : ; vn must be linearly
independent (and hence form a basis of V ):
Proof. (of Part I by contradiction) We want to show v1 ; : : : ; vn span V: To arrive at a
contradiction, suppose they do not span V: That means there is some vector in V which
is not in the span of v1 ; : : : ; vn ; call it vn+1 : By Theorem 4
the vectors v1 ; : : : ; vn ; vn+1 are linearly independent:
Now our hypothesis that V is of dimension n says V has a basis w1 ; : : : ; wn and these n
basis vectors span V: This is now a contradiction to Theorem 1 which says every set with
more than n vectors must be linearly dependent. This completes the proof of part I.
(of Part II by contradiction) We want to show that v1 ; : : : ; vn are linearly independent.
To arrive at a contradiction suppose they are linearly dependent. This means that one of
v1 ; : : : ; vn is a linear combination of the others. Without loss of generality, suppose
vn = c1 v1 + : : : + cn 1 vn 1
Then by Theorem 6 it must the case that v1 ; : : : ; vn 1 span V: Now V has a basis
w1 ; : : : ; wn and these n vectors are linearly independent. This is now a contradiction to
theorem 2 which says that fewer than n vectors cannot span V:
Theorem 7 Suppose V is of dimension n: Then every linearly independent set with fewer
than n vectors can be extended to a basis of V:
Proof. Suppose Sk is a set of k linearly independent vectors with k < n: By Theorem
2 Sk does not span V: So there exists a vector vk+1 2
= span(Sk ): By Theorem 4
Sk+1 = Sk [ fvk+1 g must be linearly independent. In this way, we can continue to
append vectors to the original set Sk until we arrive at a set Sn of n linearly independent
vectors. By Theorem 6 (Part I), Sn must be a basis of V
Theorem 8 Suppose V is of dimension n: Then every spanning set with more than n
vectors can be reduced to a basis of V:
Proof. Suppose Sk is a set of k vectors which spans V with k > n: By Theorem 1
Sk must be linearly dependent, meaning one is a linear combination of the others. By
Theorem 5 that one vector may be discarded from the set and the remaining set Sk 1
of k 1 vectors still spans V: In this way, we can continue to discard vectors from the
original set Sk until we arrive at a set Sn of n vectors which spans V: By Theorem 6 (Part
II) Sn must be a basis of V: