Linear Algebra

54 Chapter 1. Linear Systems
x1 has been removed from x5’s equation. That is, Gauss’ method has made x5’s
row independent of x1’s row.
Independence of a collection of row vectors, or of any kind of vectors, will
be precisely defined and explored in the next chapter. But a first take on it is
that we can show that, say, the third row above is not comprised of the other
rows, that ρ3 �= c1ρ1 + c2ρ2 + c4ρ4. For, suppose that there are scalars c1, c2,
and c4 such that this relationship holds.
�
0 0 0 3 3
� �
0 = c1 2 3 7 8 0
�
0
� �
+ c2 0 0 1 5 1 1
� �
+ c4 0 0 0 0 2 1
The first row’s leading entry is in the first column and narrowing our consideration
of the above relationship to consideration only of the entries from the first
column 0 = 2c1+0c2+0c4 gives that c1 = 0. The second row’s leading entry is in
the third column and the equation of entries in that column 0 = 7c1 +1c2 +0c4,
along with the knowledge that c1 = 0, gives that c2 = 0. Now, to finish, the
third row’s leading entry is in the fourth column and the equation of entries
in that column 3 = 8c1 +5c2 +0c4, along with c1 =0andc2 = 0, gives an
impossibility.
The following result shows that this effect always holds. It shows that what
Gauss’ linear elimination method eliminates is linear relationships among the
rows.
2.5 Lemma In an echelon form matrix, no nonzero row is a linear combination
of the other rows.
Proof. Let R be in echelon form. Suppose, to obtain a contradiction, that
some nonzero row is a linear combination of the others.
ρi = c1ρ1 + ...+ ci−1ρi−1 + ci+1ρi+1 + ...+ cmρm
We will first use induction to show that the coefficients c1, ... , ci−1 associated
with rows above ρi are all zero. The contradiction will come from consideration
of ρi and the rows below it.
The base step of the induction argument is to show that the first coefficient
c1 is zero. Let the first row’s leading entry be in column number ℓ1 be the
column number of the leading entry of the first row and consider the equation
of entries in that column.
ρi,ℓ1
= c1ρ1,ℓ1 + ...+ ci−1ρi−1,ℓ1 + ci+1ρi+1,ℓ1 + ...+ cmρm,ℓ1
The matrix is in echelon form so the entries ρ2,ℓ1 , ... , ρm,ℓ1 , including ρi,ℓ1 ,are
all zero.
0=c1ρ1,ℓ1 + ···+ ci−1 · 0+ci+1 · 0+···+ cm · 0
Because the entry ρ1,ℓ1 is nonzero as it leads its row, the coefficient c1 must be
zero.