Math 136
Assignment 2 Solutions
1.
Determine, with proof, which of the following are subspaces of
R
3
and which are not.
a)
S
1
=
x
1
0
x
2
∈
R
3

x
1

x
2
= 0
Solution: By definition
S
1
is a subset of
R
3
. Also,
0
0
0
∈
S
1
since 0

0 = 0. Thus,
S
1
is a
nonempty subset of
R
3
, so we can apply the Subspace Test.
Let
~x
=
x
1
0
x
2
,
~
y
=
y
1
0
y
2
∈
S
1
. Then
x
1

x
2
= 0 and
y
1

y
2
= 0. Then we have
~x
+
~
y
=
x
1
+
y
1
0
x
2
+
y
2
and
(
x
1
+
y
1
)

(
x
2
+
y
2
) =
x
1

x
2
+
y
1

y
2
= 0

0 = 0
So,
~x
+
~
y
satisfies the condition of
S
1
, so
~x
+
~
y
∈
S
1
.
Similarly,
c~x
=
cx
1
0
cx
2
and
cx
1

cx
2
=
c
(
x
1

x
2
) =
c
(0) = 0
so
c~x
∈
S
1
.
Thus, by the Subspace Test,
S
1
is a subspace of
R
3
.
b)
S
2
=
x
1
x
2
x
3
∈
R
3

x
1
, x
2
, x
3
∈
Z
Solution: The definition of
R
n
allows multiplication by any real scalar. Thus, the fact that
we are restricting the entries of the vectors in
S
2
to be integers makes us believe that the set
is not a subspace. For example if
c
=
√
2 and
~x
=
1
1
1
, then
c~x
=
√
2
√
2
√
2
6∈
S
2
. Therefore,
S
2
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 Winter '10
 cormack
 Linear Algebra, Vector Space, Subspace Test

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