Page 45 - Linear Algebra Done Right

Dimension
a 1 u 1 +· · ·+ a m u m − b 1 w 1 −· · ·− b n w n = 0.
Because (u 1 ,...,u m ,w 1 ,...,w n ) is linearly independent, this implies
that a 1 =· · ·= a m = b 1 = ··· = b n = 0. Thus v = 0, completing the 31
proof that U ∩ W ={0}.
Dimension

Though we have been discussing finite-dimensional vector spaces,
we have not yet defined the dimension of such an object. How should
dimension be defined? A reasonable definition should force the dimen-
n
sion of F to equal n. Notice that the basis

(1, 0,..., 0), (0, 1, 0,..., 0),...,(0,..., 0, 1)
has length n. Thus we are tempted to define the dimension as the
length of a basis. However, a finite-dimensional vector space in general
has many different bases, and our attempted definition makes sense
only if all bases in a given vector space have the same length. Fortu-
nately that turns out to be the case, as we now show.

2.14 Theorem: Any two bases of a finite-dimensional vector space
have the same length.

Proof: Suppose V is finite dimensional. Let B 1 and B 2 be any two
bases of V. Then B 1 is linearly independent in V and B 2 spans V, so the
length of B 1 is at most the length of B 2 (by 2.6). Interchanging the roles
of B 1 and B 2 , we also see that the length of B 2 is at most the length
of B 1 . Thus the length of B 1 must equal the length of B 2 , as desired.

Now that we know that any two bases of a finite-dimensional vector
space have the same length, we can formally define the dimension of
such spaces. The dimension of a finite-dimensional vector space is
defined to be the length of any basis of the vector space. The dimension
of V (if V is finite dimensional) is denoted by dim V. As examples, note
n
that dim F = n and dim P m (F) = m + 1.
Every subspace of a finite-dimensional vector space is finite dimen-
sional (by 2.7) and so has a dimension. The next result gives the ex-
pected inequality about the dimension of a subspace.