T
HE
U
NIVERSITY OF
S
YDNEY
P
URE
M
ATHEMATICS
Linear Mathematics
2012
Tutorial 4 (week 5) — Solutions
1.
In each of the following find a basis for
Span(
X
)
, and say whether or not
Span(
X
)
is equal to
the vector space
R
n
of which
X
is a subset.
a
)
X
=
(
1
5
)
,
(
2
6
)
,
(
3
7
)
,
(
4
8
)
⊆
R
2
.
b
)
X
=
1
2
3
,
4
5
6
,
0
0
0
⊆
R
3
.
c
)
X
=
1
2
3
4
,
2
4
6
8
,
0
0
0
0
⊆
R
4
.
d
)
X
=
1
1
1
1
1
,

1

1

1

1

1
,
4
4
4
4
4
⊆
R
5
.
Solution
a
)
Since
{
(
1
5
)
,
(
2
6
)
}
is linearly independent, and
dim(
R
2
) = 2
,
{
(
1
5
)
,
(
2
6
)
}
is a basis for
Span(
X
)
, and
Span(
X
) =
R
2
.
(Indeed, any subset of
X
containing two vectors is linearly
independent, and is a basis for
Span(
X
)
, as is the standard basis
{
(
1
0
)
,
(
0
1
)
}
.)
b
) Span(
X
) =
a
1
2
3
+
b
4
5
6
+
c
0
0
0

a, b, c
∈
R
=
a
1
2
3
+
b
4
5
6

a, b
∈
R
,
and
1
2
3
,
4
5
6
is linearly independent. Hence this set is a basis for
Span(
X
)
, which
therefore has dimension 2, and is not equal to
R
3
.
c
)
The vectors
2
4
6
8
and
0
0
0
0
are scalar multiples of
1
2
3
4
, and so a basis for
Span(
X
)
is
1
2
3
4
.
Span(
X
)
therefore has dimension 1, and is not equal to
R
4
.
d
)
The vectors

1

1

1

1

1
and
4
4
4
4
4
are scalar multiples of
1
1
1
1
, and so a basis for
Span(
X
)
is
1
1
1
1
1
.
Span(
X
)
therefore has dimension 1, and is not equal to
R
5
.
2.
Find a basis for the subspace
V
=
x
y
z
∈
R
3
2
x
+
y
= 3
z
of
R
3
.
Solution
An arbitrary element
x
y
z
of
V
can be written in the form
x
y
z
=
x
3
z

2
x
z
, since
x
y
z
∈
V
if and only if
y
= 3
z

2
x
.
Now,
x
3
z

2
x
z
=
x
1

2
0
+
z
0
3
1
, showing that
1

2
0
,
0
3
1
spans
V
. Since
1

2
0
is not a scalar multiple of
0
3
1
, this set is clearly linearly
independent and is a basis for
V
, which is a subspace of
R
3
of dimension
2
.
Math 2061: Tutorial 4 (week 5) — Solutions
S.B and A.M 20/3/2012
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Linear Mathematics
Tutorial 4 (week 5) — Solutions
Page 2
3.
Let
X
=
6
3
2
,
2
3
1
,
4

6
0
. Find the dimension of
Span(
X
)
.
Solution
Note that the subset containing the first two vectors of
X
is linearly independent
(since the vectors are not scalar multiples of each other), and so the dimension of
Span(
X
)
is
at least 2. Hence, the dimension is either 2 or 3, depending on whether or not
X
is linearly
independent.
Suppose that
a
6
3
2
+
b
2
3
1
+
c
4

6
0
=
0
0
0
,
for some
a, b, c
∈
R
. To find all possible
a, b, c
we have to solve the system of linear equations
corresponding to the following augmented matrix:
6
2
4
0
3
3

6
0
2
1
0
0
Row reduce
→
1
0
2
0
0
1

4
0
0
0
0
0
.
(Note that the row echelon form of a matrix is not unique, as it depends upon the sequence of row
operations that you choose; the reduced row echelon form is, however, unique. Consequently,
you might obtain a different row echelon form for the augmented matrix above. However, your
row echelon form should still give the same set of solutions as the one above. This comment
applies to all of the questions below which compute row echelon forms.)
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 One '08
 Cartwright
 Linear Algebra, Algebra, Vector Space, basis, Row echelon form, — Solutions, Linear Mathematics

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