Page 200 - Matrix Analysis & Applied Linear Algebra

4.4 Basis and Dimension 195

Spaces that possess a basis containing an infinite number of vectors are
referred to as infinite-dimensional spaces, and those that have a finite basis
are called finite-dimensional spaces. This is often a line of demarcation in
the study of vector spaces. A complete theoretical treatment would include the
analysis of infinite-dimensional spaces, but this text is primarily concerned with
finite-dimensional spaces over the real or complex numbers. It can be shown that,
n n
in effect, this amounts to analyzing or C and their subspaces.
The original concern of this section was to try to eliminate redundancies
from spanning sets so as to provide spanning sets containing a minimal number
of vectors. The following theorem shows that a basis is indeed such a set.

Characterizations of a Basis
m
Let V be a subspace of , and let B = {b 1 , b 2 ,..., b n }⊆V. The
following statements are equivalent.
• B is a basis for V. (4.4.1)
• B is a minimal spanning set for V. (4.4.2)
• B is a maximal linearly independent subset of V. (4.4.3)

Proof. First argue that (4.4.1) =⇒ (4.4.2) =⇒ (4.4.1), and then show (4.4.1)
is equivalent to (4.4.3).
Proof of (4.4.1) =⇒ (4.4.2). First suppose that B is a basis for V, and
prove that B is a minimal spanning set by using an indirect argument—i.e.,
assume that B is not minimal, and show that this leads to a contradiction. If
X = {x 1 , x 2 ,..., x k } is a basis for V in which k< n, then each b j can be
written as a combination of the x i ’s. That is, there are scalars α ij such that

k

b j = α ij x i for j =1, 2,...,n. (4.4.4)
i=1
If the b ’s and x ’s are placed as columns in matrices

B m×n = b 1 | b 2 |···| b n and X m×k = x 1 | x 2 |···| x k ,
then (4.4.4) can be expressed as the matrix equation

B = XA, where, A k×n =[α ij ] .

Since the rank of a matrix cannot exceed either of its size dimensions, and since
k< n, we have that rank (A) ≤ k< n, so that N (A) = {0} —recall (4.2.10).
If z = 0 is such that Az = 0, then Bz = 0. But this is impossible because