Math 320  Fall 2010
HW #6 Selected Solutions
Problem 4.1.32
Show that the given set
V
is closed under addition and under multiplication by scalars and is therefore
a subspace of
R
3
.
V
is the set of all (
x, y, z
) such that
z
= 2
x
+ 3
y
As instructed, to show
V
is a subspace of
R
3
, it suffices to check that
V
is closed under addition and
scalar multiplication. Let’s check addition first. Suppose
v
1
and
v
2
are two vectors in
V
(fancy notation:
v
1
, v
2
∈
V
). We need to check that
v
1
+
v
2
is in
V
. Let’s write
v
1
= (
x
1
, y
1
, z
1
) and
v
2
= (
x
2
, y
2
, z
2
), where
the
x
1
, x
2
, y
1
, y
2
, z
1
, z
2
are all arbitrary real numbers.
Note that at this point, we haven’t yet used the definition of the subset
V
. Since
v
1
and
v
2
are vectors
in
V
, we know that
z
1
= 2
x
1
+ 3
y
1
and
z
2
= 2
x
2
+ 3
y
2
. Now, let’s perform the addition:
v
1
+
v
2
= (
x
1
, y
1
,
2
x
1
+ 3
y
1
) + (
x
2
, y
2
,
2
x
2
+ 3
y
2
) = (
x
1
+
x
2
, y
1
+
y
2
,
(2
x
1
+ 3
y
1
) + (2
x
2
+ 3
y
2
))
How do we check that the sum
v
1
+
v
2
is in
V
? Well, the ticket for admission into
V
is that
z
= 2
x
+ 3
y
. In
other words, to check that a vector is in
V
, we need to check that the
z
component is equal to twice the
x
component plus three times the
y
component. Let’s check:
2(
x
1
+
x
2
)

{z
}
twice
x
component of
v
1
+
v
2
+
3(
y
1
+
y
2
)

{z
}
three times
y
component of
v
1
+
v
2
= 2
x
1
+ 2
x
2
+ 3
y
1
+ 3
y
2
= (2
x
1
+ 3
y
1
) + (2
x
2
+ 3
y
2
)

{z
}
z
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 Spring '08
 Neely
 Linear Algebra, Vectors, Vector Space, scalar multiplication

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