Chapter 2: FiniteDimensional Vector Spaces
2.1
Let
v
∈
V
; since (
v
1
,...,v
n
) spans
v
, there exist scalars
a
1
,...,a
n
∈
F
such that
v
=
∑
n
i
=1
a
n
v
n
.
We wish to ﬁnd
b
1
,...,b
n
∈
F
such that
v
=
b
1
(
v
1

v
2
) +
···
+
b
n

1
(
v
n

1

v
n
) +
b
n
v
n
(since
v
is an
arbitrary element of
V
, doing so ensures that the new list also spans
V
). To do so, set
b
1
=
a
1
,
b
n
=
a
n
,
and
b
i

b
i

1
=
a
i
otherwise. Then we have
b
1
(
v
1

v
2
) +
···
+
b
n

1
(
v
n

1

v
n
) +
b
n
v
n
=
b
1
v
1
+ (
b
2

b
1
)
v
2
+
···
+ (
b
n

1

b
n

2
)
v
n

1
+
b
n
v
n
=
a
1
v
1
+
a
2
v
2
+
···
+
a
n
v
n
=
v
Since
a
i
,...,a
n
are determined for each
v
∈
V
, we have shown that these deﬁnitions for
b
1
,
···
,b
n
determine
a linear combination expressing
v
in terms of the new list. Hence, that list spans
V
.
2.2
Suppose that 0 =
a
1
(
v
1

v
2
) +
···
+
a
n
v
n
; we wish to show that
a
1
=
···
=
a
n
= 0. To do so, we will
use the fact that (
v
1
,...,v
n
) is a linearly independent in list in
V
. To wit, we expand out this expression:
0 =
a
1
(
v
1

v
2
) +
···
+
a
n
v
n
=
a
1
v
1
+ (
a
2

a
1
)
v
2
+
···
+
a
n
v
n
The independence of (
v
1
,...,v
n
) implies that
a
1
= 0
,a
2

a
1
= 0
,...,a
n
= 0. Starting from the second
expression we have that
a
2
=
a
1
= 0; continuing as such down the list shows that
a
1
=
a
2
=
···
=
a
n
= 0,
just as desired. Hence, the list (
v
1

v
2
,...,v
n

1

v
n
,v
n
) is linearly independent in
V
.
2.3
Once again, let 0 =
a
1
(
v
1
+
w
) +
···
+
a
n
(
v
n
+
w
). If we expand this expression, we obtain
0 =
a
1
(
v
1
+
w
) +
···
+
a
n
(
v
n
+
w
) =
a
1
v
1
+
···
+
a
n
v
n
+ (
a
1
+
···
+
a
n
)
w
=
⇒ 
(
a
1
+
···
+
a
n
)
w
=
a
1
v
1
+
···
+
a
n
v
n
Now, since (
v
1
,
···
,v
n
) is a linearly independent list in
V
, we have that
α
=
∑
n
i
=1
a
i
6
= 0. Hence, we can
divide both sides of our equality by this sum to obtain
w
=

a
1
α
v
1
 ··· 
a
n
α
v
n
Since we have expressed
w
as a linear combination of (
v
1
,...,v
n
), we must have
w
∈
span (
v
1
,...,v
n
).
2.4
The answer is no. Consider the following polynomials, both of which are in the described subset
P
m
(
F
):
p
+
(
z
) =
a
0
+
a
1
z
+
···
+
a
m
z
m
and
p

(
z
) =
a
0
+
a
1
z
+
··· 
a
m
z
m
If we take the sum, we have
p
+
(
z
) +
p

(
z
) =
a
0
+
a
1
z
+
···
+
a
m

1
z
m

1
6
=
P
m
(
F
), for the degree is not