Problem Set 1  Solutions
J. Scholtz
October 1, 2006
Question 1 (1.1/2a)
Let
~a
= (3
,

2
,
4) and
~
b
= (

5
,
7
,
1). Then
~
b

~a
= (

8
,
9
,

3) and so the
equation of the line is:
l
:
~v
= (
~
b

~a
)
t
+
~a
= (

8
,
9
,

3)
t
+ (3
,

2
,
4).
Question 2 (1.1/3a)
This is almost the same as previous question only in one more dimension: The
can be parametrized as
P
:
~v
= (
~
b

~a
)
t
+ (
~
c

~a
)
s
+
~a
. Plugging in the numbers
gives us:
~v
= (

2
,
9
,
7)
t
+ (

5
,
12
,
2)
s
+ (2
,

5
,

1).
Question 3 (1.2/8)
Given that
x,y
∈
V
and
a,b
∈
F
then we know we can use various Vectorspace
and Field axioms to manipulate this expression. Since
a
+
b
∈
F
then we can
use VS7 to transform (
a
+
b
)(
x
+
y
) = (
a
+
b
)
x
+ (
a
+
b
)
y
. Then by applying
VS8 twice we get
ax
+
ay
+
bx
+
by
. Finally by VS1 we can rearange to get
ax
+
ay
+
bx
+
by
.
Question 4 (1.2/16)
If
V
is a set of all
m
×
n
matrices with real coeﬃcients is
V
a vectorspace over
Q
? We need to check whether all the axioms are satisﬁed. However, from the
example 2 of the book we know that real matrices form a vectorspace over real
numbers so we need to check a little less:
VS1, VS2, VS3, VS4 form an abelian group
VS5: 1
∈
Q
VS6, VS7, VS8: Since
Q
⊂
R
as a subﬁeld then these axioms are satisﬁed.
Question 5 (1.2/19)
This construction does not form a vectorspace because it violates VS8. Consider:
(
a
+
b
)(
x,y
) = ((
a
+
b
)
x,
y
a
+
b
). However, by VS7 and VS8, (
a
+
b
)(
x,y
) =
a
(
x,y
) +
b
(
x,y
) = (
ax,y/a
) + (
bx,y/b
) = ((
a
+
b
)
x,
y
a
+
y
b
)
6
= ((
a
+
b
)
x,
y
a
+
b
).
1
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View Full DocumentQuestion 6 (1.3/8 c,e)
In order to check that a set is a subspace we need to check whether it is closed
under vector addition, scalar multiplication, existence of zero and additive in
verses.
c)
W
=
{
(
a
1
,a
2
,a
3
)
,
2
a
1

7
a
2
+
a
3
= 0
}
W
=
{
(
a
1
,a
2
,a
3
)
,
2
a
1

7
a
2
+
a
3
= 0
}
is a subspace. To make all the notation
shorter we can note that if
~
z
= (2
,

7
,
1), then
~v
∈
V
iﬀ
~
z.~v
= 0 (meaning
standard dot product). Then it follows that if both
~a.~
z
=
~
b.~
z
= 0 then also
(
~a
+
~
b
)
.~
z
= 0. Similarly if
~a.~
z
= 0 then (
c~a
)
.~
z
=
c
(
~a.~
z
) =
c.
0 = 0. Hence
W
is closed under vector addition and scalar multiplication. Note that
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