M341 Homework Solution 12
Section 4.6:
1(c):
Using the Simpli ed Span Method,
0
1
1
0
6
2
−
1
0
−
2
1
−
1
2
1
1
2
3
−
2
0
−
2
−
3
1
1
1
−
1
4
2
−
1
−
1
1
3
is reduced to
1
0
0
0
0
0
1
0
0
0
0
0
1
0
0
0
0
0
1
0
0
0
0
0
1
0
0
0
0
0
Therefore
{
[1
,
0
,
0
,
0
,
0]
,
[0
,
1
,
0
,
0
,
0]
,
[0
,
0
,
1
,
0
,
0]
,
[0
,
0
,
0
,
1
,
0]
,
[0
,
0
,
0
,
0
,
1]
}
is a basis for
V
.
4(f):
We can use Independence Test Method to nd a basis.
3
2
0
0
6
1
1
−
1
0
5
2
5
0
7
0
−
21
0
7
is reduced to
1
0
0
2
2
0
0
1
0
−
3
0
0
0
0
0
0
0
1
The pivot columns are 1st, 2nd and 6th columns. Therefore
B
=
{
[3
,
1
,
0]
,
[2
,
−
1
,
7]
,
[1
,
5
,
7]
}
is a basis
for
V
by independence test method.
4(g):
Consider
B
=
{
[1
,
0
,
0]
,
[0
,
0
,
1]
}
. I will show that
B
is a basis for span
(
S
) =
V
. It is obvious
that
B
is a subset of
S
and
B
is linearly independent. It remains to show that span
(
B
)
=
V
. First
we have span
(
B
)
⊂
span
(
S
) =
V
, since
B
⊂
S
. For any vector
v
in
S
,
v
is of the form
v
= [
a,
0
, b
]
for some
a, b
∈
R
. Hence we have
v
=
a
[1
,
0
,
0] +
b
[0
,
0
,
1]
and therefore
v
∈
span
(
B
)
. It follows
S
⊂
span
(
B
)
and
V
=
span
(
S
)
⊂
span
(
B
)
. This shows span
(
B
)
=
V
and thus
B
is a basis for
V
.
5(f):
Consider
B
=
{
8
x
3
+ 1
,
8
x
3
+
x,
8
x
3
+
x
2
,
8
x
3
}
. It is obviously that
B
is a subset of
S
. To see
B
is a basis for
V
, we check if
B
is linearly independent rst. Suppose there exists
a
1
, a
2
, a
3
, a
4
∈
R
such that
a
1
(8
x
3
+ 1) +
a
2
(8
x
3
+
x
) +
a
3
(8
x
3
+
x
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 Spring '08
 HIETMANN
 Linear Algebra, basis, Row echelon form, independence test method

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