3
3. Find a basis for the subspace
U
of
R
5
defined by
U
=
{
(
x
1
,x
2
,...,x
5
)

x
1
= 3
x
2
,x
3
= 7
x
4
}
.
Solution
:
Proposition.
Let
U
be the subspace of
R
5
defined by
U
=
{
(
x
1
,x
2
,...,x
5
)

x
1
= 3
x
2
,x
3
= 7
x
4
}
.
Then a basis for
U
is given by the list
S
of vectors in
R
5
, where
S
= ((3
,
1
,
0
,
0
,
0)
,
(0
,
0
,
7
,
1
,
0)
,
(0
,
0
,
0
,
0
,
1))
.
Proof.
Let
U
and
S
be as given.
Then we need to check two things in order to
prove that
S
is a basis for
U
.
(a) First, we must show that
U
= span(
S
).
In other words, given any vector
u
∈
U
, we must show that there exist scalars
a
1
,a
2
,a
3
∈
R
such that
u
=
a
1
(3
,
1
,
0
,
0
,
0) +
a
2
(0
,
0
,
7
,
1
,
0) +
a
3
(0
,
0
,
0
,
0
,
1)
Let
u
∈
U
. Then, to see that such scalars exist, note that
u
∈
U
implies that
there are scalars
x,y,z
∈
R
such that
u
= (3
x,x,
7
y,y,z
). It follows that
u
=
x
(3
,
1
,
0
,
0
,
0) +
y
(0
,
0
,
7
,
1
,
0) +
z
(0
,
0
,
0
,
0
,
1)
,
and so
u
∈
span(
S
).