Linear Algebra I
Solutions of Exercise 1.2
#1,8,13,17,18,19,21
1. (a) True. (b) False. (c) False. If
x
is the zero vector then this is false. (d) False. If
a
= 0 then this is false. (e) True. (f) False. (g) False. (h) False. If
f
(
x
) =
x
2
+ 1 and
g
(
x
) =

x
2
+ 1 then deg
f
= deg
g
=2. But deg(
f
+
g
)= 0. (j) True. (k) True.
8.
(
a
+
b
)(
x
+
y
) = (
a
+
b
)
x
+ (
a
+
b
)
y
(by VS 7)
=
ax
+
bx
+
ay
+
by
(by VS 8)
=
ax
+
ay
+
bx
+
by
(by VS 1)
13. Let (
x, y
) be the zero vector.
Then by VS 3, we must have: (
a, b
) + (
x, y
) = (
a, b
)
for all (
a, b
) in
V
. Since (
a, b
) + (
x, y
) = (
a
+
b, by
) must be equal to (
a, b
), we have
(
x, y
) = (0
,
1). On the other hand, by VS 4, for any vector (
a, b
) there must be a vector
(
a , b
) such that (
a, b
) + (
a , b
) = (0
,
1). Since (
a, b
) + (
a , b
) = (
a
+
a , bb
) we have
a
=

a
and
b
=
1
b
. Hence if we take
b
= 0 then for any
a
, (
a,
0) has no such (
a , b
).
Therefore,
V
is not a vector space.
17. By Theorem 1.2 (a), 0(
a, b
) = (
a,
0) must be the zero vector. But then (
x, y
)+(
a,
0) =
(
x
+
a, y
) = (
x, y
). Hence (
a,
0) cannot be the zero vector. This shows
V
is not a vector
space under the operations.
18. (
a
1
, a
2
) + (
b
1
, b
2
) = (
a
1
+ 2
b
1
, a
2
+ 3
b
2
) and (
b
1
, b
2
) + (
a
1
, a
2
) = (
b
1
+ 2
a
1
, b
2
+ 3
a
2
).
Hence (
a
1
, b
1
)+(
a
2
, b
2
) = (
a
2
, b
2
)+(
a
1
, b
1
) which says that our operation violates VS1.
Accordingly,
V
is not a vector space.
19. Let
c, d
be nonzero scalars such that
c
+
d
= 0. Write
x
= (
a
1
, a
2
). Then (
c
+
d
)
x
=
((
c
+
d
)
a
1
,
a
2
c
+
d
) and
cx
+
dx
= ((
c
+
d
)
a
1
, a
2
(
1
c
+
1
d
)). Therefore (
c
+
d
)
x
=
cx
+
dx
.
Therefore
V
is not a vector space.