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
This preview has intentionally blurred sections. Sign up to view the full version.
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
This is the end of the preview. Sign up
to
access the rest of the document.
 Spring '08
 GUREVITCH
 Addition, Vector Space, scalar multiplication, additive inverse

Click to edit the document details