CHAPTER II DIMENSION In the present chapter we investigate ...

As a result of this corollary, we know:
Corollary 1.4.3. Every finite dimensional vector space has a basis.
Next, assume that a set of linearly independent vectors w1, w2, . . . , wp in a finite
dimensional space V is given. Let v1, v2, . . . , vr be any spanning set of V . Now
put these two sets of vectors together to form
w1, w2, . . . , wp, v1, v2, . . . , vr
The above list of vectors certainly spans V . By our previous argument, we see that the
leading vectors in this list form a basis of V , say B. Since the linearly independent vectors
w1, w2, . . . , wp appear at the beginning of this list, all of them are leading vectors. So
they belong to the basis B. we have shown
Corollary 1.4.4. A finite set of linearly independent vectors in a finite dimensional
linear space V can be enlarged to a basis of V .
1.5. Given a finite set of vectors v1, v2, . . . , vn in a finite dimensonal linear space
V , how do we find the leading vectors and how do we express other vectors as linear
combinations of the leading vectors, as suggested by Theorem 1.4.1 above?
First, by using a coordinate system, we can convert these vectors into column vectors
in space F m (where F is the field we work with and m is the number of coordinates),
say C1, C2, . . . , Cn. In this way we turn the original problem about vectors into the one
about column vectors. We form an m × n matrix A by using these vectors as columns:
A = [C1 C2 · · · Cn]
Now we apply elementary row operations to this matrix to its reduced row echelon form.
Recall that elementary row operations either exchange two rows, or adding a scalar multiple
of one row to another, or multiply one row by a nonzero scalar. An elementary row
operation has the same effect as multiplying an invertible matrix on the left. For example,
given a 2 × 2 matrix with A, exchanging two rows (adding 2 × the second row to the first,
resp.) has the same effect as multiplying A on the left by
as we can check that
0
1 a
b
1 0 c d
=
0 1
1 0
c d
,
a b
(by
1 2
0 1
resp.)
1 2 a b
=
0 1 c d
8
a + 2c b + 2d
.
c d