Linear Algebra, 2020a

136 Chapter Two. Vector Spaces
III.3
Vector Spaces and Linear Systems
We will now reconsider linear systems and Gauss’s Method, aided by the tools
and terms of this chapter. We will make three points.
For the first, recall the insight from the Chapter One that Gauss’s Method
works by taking linear combinations of rows — if two matrices are related by
row operations A −→ ··· −→ B then each row of B is a linear combination of
the rows of A. Therefore, the right setting in which to study row operations in
general, and Gauss’s Method in particular, is the following vector space.
3.1 Definition The row space of a matrix is the span of the set of its rows. The
row rank is the dimension of this space, the number of linearly independent
rows.
3.2 Example If
( )
2 3
A =
4 6
then Rowspace(A) is this subspace of the space of two-component row vectors.
{c 1 · (2 3)+c 2 · (4 6) | c 1 ,c 2 ∈ R}
The second row vector is linearly dependent on the first and so we can simplify
the above description to {c · (2 3) | c ∈ R}.
3.3 Lemma If two matrices A and B are related by a row operation
A
ρ i↔ρ j
kρ i
kρ i +ρ j
−→ B or A −→ B or A −→
B
(for i ≠ j and k ≠ 0) then their row spaces are equal. Hence, row-equivalent
matrices have the same row space and therefore the same row rank.
Proof Corollary One.III.2.4 shows that when A −→ B then each row of B is a
linear combination of the rows of A. That is, in the above terminology, each row
of B is an element of the row space of A. Then Rowspace(B) ⊆ Rowspace(A)
follows because a member of the set Rowspace(B) is a linear combination of the
rows of B, so it is a combination of combinations of the rows of A, and by the
Linear Combination Lemma is also a member of Rowspace(A).
For the other set containment, recall Lemma One.III.1.5, that row operations
are reversible so A −→ B if and only if B −→ A. Then Rowspace(A) ⊆
Rowspace(B) follows as in the previous paragraph.
QED
Of course, Gauss’s Method performs the row operations systematically, with
the goal of echelon form.