MATH 102
SOLUTIONS TO HW #3
Section 2.4, problem 8.
In this case the rank of the matrix is
r
=
n
.
Recall that in general the rank
r
is
{
# of pivot variables
}
, while the dimension of the null space is
n

r
=
{
# of free variables
}
.
In this case the dimension of the nullspace is 0.
So the
number of free variables must also be zero. Thus we have that
n

r
= 0.
Let us write the matrix
A
in terms of column vectors as follows
A
= [
v
1
. . . v
n
].
Then the nullspace of
A
consists of all
n
×
1 vectors of the from
c
T
= (
c
1
, . . . , c
n
),
which give the sum:
c
1
v
1
+
. . .
+
c
n
v
n
= 0
.
Since the nullspace is only the zero vector, then the only possible sum of this form
has
all
c
i
= 0. That is the same as saying that the collection of vectors
{
v
1
, . . . , v
n
}
are linearly independent.
Section 2.4, problem 18.
TO BE POSTED SOON.
Section 2.4, problem 6.
In this case the incidence matrix is:
A
=

1
1
0
0

1
0
1
0
0

1
1
0
0

1
0
1

1
0
0
1
0
0

1
1
.
In this case, the three elementary loop vectors are associated with the loops:
(1) Loop #1:
y
5
→
y
4
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 Fall '08
 SZYPOWSK
 Math, Linear Algebra, Algebra, Vectors, Space, free variables

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