Linear Algebra, 2020a

Section III. Basis and Dimension 137
3.4 Lemma The nonzero rows of an echelon form matrix make up a linearly
independent set.
Proof Lemma One.III.2.5 says that no nonzero row of an echelon form matrix
is a linear combination of the other rows. This result restates that using this
chapter’s terminology.
QED
Thus, in the language of this chapter, Gaussian reduction works by eliminating
linear dependences among rows, leaving the span unchanged, until no nontrivial
linear relationships remain among the nonzero rows. In short, Gauss’s Method
produces a basis for the row space.
3.5 Example From any matrix, we can produce a basis for the row space by
performing Gauss’s Method and taking the nonzero rows of the resulting echelon
form matrix. For instance,
⎛
⎜
1 3 1
⎞
⎛
⎟
⎝1 4 1⎠ −ρ 1+ρ 2 6ρ 2 +ρ 3 ⎜
1 3 1
⎞
⎟
−→ −→ ⎝0 1 0⎠
−2ρ 1 +ρ 3
2 0 5
0 0 3
produces the basis 〈(1 3 1), (0 1 0), (0 0 3)〉 for the row space. This is a basis
for the row space of both the starting and ending matrices, since the two row
spaces are equal.
Using this technique, we can also find bases for spans not directly involving
row vectors.
3.6 Definition The column space of a matrix is the span of the set of its columns.
The column rank is the dimension of the column space, the number of linearly
independent columns.
Our interest in column spaces stems from our study of linear systems. An
example is that this system
c 1 + 3c 2 + 7c 3 = d 1
2c 1 + 3c 2 + 8c 3 = d 2
c 2 + 2c 3 = d 3
4c 1 + 4c 3 = d 4
has a solution if and only if the vector of d’s is a linear combination of the other
column vectors, ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞
1 3 7 d 1
2
c 1 ⎜ ⎟
⎝0⎠ + c 3
2 ⎜ ⎟
⎝1⎠ + c 8
3 ⎜ ⎟
⎝2⎠ = d 2
⎜ ⎟
⎝d 3 ⎠
4 0 4 d 4