Math 136
Assignment 5 Solutions
Section 34 Exercises
B3. a)
S
is not a subspace of
R
4
since it does not contain the zero vector.
b)
We ﬁrst observe
T
contains the zero vector and hence is a nonempty subset of
R
4
.
Let
x, y
∈
T
then (
x
1
, x
2
, x
3
, x
4
) + (
y
1
, y
2
, y
3
, y
4
) = (
x
1
+
y
1
, x
2
+
y
2
, x
3
+
y
3
, x
4
+
y
4
) and
(
x
1
+
y
1
) + 2(
x
3
+
y
3
) = (
x
1
+ 2
x
3
) + (
y
1
+ 2
y
3
) = 0
and
(
x
1
+
y
1
)

3(
x
4
+
y
4
) = (
x
1

3
x
4
) + (
y
1

3
y
4
) = 0
So
T
is closed under addition.
For
k
∈
R
and
x
∈
T
We have
k
(
x
1
, x
2
, x
3
, x
4
) = (
kx
1
, kx
2
, kx
3
, kx
4
) and
(
kx
1
) + 2(
kx
3
) =
k
(
x
1
+ 2
x
3
) = 0
and
(
kx
1
)

3(
kx
4
) =
k
(
x
1

3
x
4
) = 0
,
so
T
is closed under scalar multiplication. Therefore
T
is a subspace of
R
4
.
c)
Two elements of
U
are
x
= (1
,
1
,
0
,
1) and
y
= (1
,
1
,
1
,
0) then
x
+
y
= (2
,
2
,
1
,
1) and
(
x
1
+
y
1
)
2
= (
x
2
+
y
2
)
2
= 4, but (
x
3
+
y
3
)
2
+ (
x
4
+
y
4
)
2
= 1 + 1 = 2. Thus
x
+
y
is not in
U
. Hence
U
is not closed under addition and therefore is not a subspace of
R
4
.
Note:
Any example which shows that it is not closed under addition or is not closed under
scalar multiplication is ﬁne.
B4. a)
If (

4
,

2
,
2
,

6) can be expressed as a linear combination of vectors of
V
then
there exists scalars
x
1
,
x
2
,
and
x
3
such that
x
1
(1
,
1
,
0
,
1) +
x
2
(2
,
0
,
0
,
2) +
x
3
(0
,
2
,

1
,
1) =
(

4
,

2
,
2
,

6). Thus we are trying to solve the system with augmented matrix
1 2
0

4
1 0
2

2
0 0

1
2
1 2
1

6
.
Rowreducing gives
1 0 0
2
0 1 0

3
0 0 1

2
0 0 0
0
.
Hence the system is consistent and we have
x
1
= 2,
x
2
=

3 and
x
3
=

2. i.e.
2(1
,
1
,
0
,
1)

3(2
,
0
,
0
,
2)

2(0
,
2
,

1
,
1) = (

4
,

2
,
2
,

6)
.