3.5. LINEAR ALGEBRA OVER
Z
191
Proof.
Suppose that
A
has an entry
ˇ
is in the first column, in the
.i; 1/
position and that
ˇ
is not divisible by
˛
. Write
ˇ
D
˛q
C
r
where
0 < r <
j
˛
j
. A row operation of type 1,
a
i
!
a
i
qa
1
produces a matrix with
r
in the
.i; 1/
position. Then transposing rows
1
and
i
yields a matrix with
r
in the
.1; 1/
position. The case that
A
has an entry in the first row that is
not divisible by
˛
is handled similarly, with column operations rather than
row operations.
If
˛
divides all the entries in the first row and column, then row and
column operations of type 1 can be used to replace the nonzero entries by
zeros.
n
Proof of Proposition
3.5.9
.
If
A
is the zero matrix, there is nothing to
do. Otherwise, we proceed as follows:
Step 1.
There is a nonzero entry of minimum size. By row and column
permutations, we can put this entry of minimum size in the
.1; 1/
position.
Denote the
.1; 1/
entry of the matrix by
˛
. According to Lemma
3.5.10
,
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '08
 EVERAGE
 Linear Algebra, Algebra, Lemma, ﬁrst row

Click to edit the document details