MATH 223, Linear Algebra
Winter, 2008
Solutions to Assignment 3
1. Let
V
=
P
2
(
x
) be the vector space of polynomials with real coeﬃcients of
degree at most two. For each of the following subsets of
V
, decide if it is
independent, if it is a spanning set for
V
, and if it is a basis for
V
. Justify
your answers.
(a)
S
1
=
{
1 +
x,
1 +
x
2
,x
+
x
2
}
Solution: This is independent, and a spanning set, and hence a basis
for
V
. For independence, suppose that
a
(1 +
x
) +
b
(1 +
x
2
) +
c
(
x
+
x
2
) = 0. Then
a
+
b
=
a
+
c
=
b
+
c
= 0; this easily implies that
a
=
b
=
c
= 0.
Since
P
2
(
x
) is 3dimensional, this implies it is a basis, but let’s check
from the deﬁnition that it’s also a spanning set for
V
. Given any
polynomial
a
0
+
a
1
x
+
a
2
x
2
∈
V
, let
α
=
a
0
+
a
1

a
2
2
,
β
=
a
0
+
a
2

a
1
2
and
γ
=
a
1
+
a
2

a
0
2
; it is easily seen that
α
(1+
x
)+
β
(1+
x
2
)+
γ
(
x
+
x
2
) =
a
0
+
a
1
x
+
a
2
x
2
.
(b)
S
2
=
{
1 +
x

2
x
2
,
3

2
x
+
x
2
,
5

3
x
2
,
13

2
x

5
x
2
}
.
Solution: This set cannot possibly be independent, as it has four
polynomials (vectors) in it. Also, 2(1 +
x

2
x
2
) + 1(3

2
x
+
x
2
) =
5

3
x
2
and 4(1 +
x

2
x
2
) + 3(3

2
x
+
x
2
) = 13

2
x

5
x
2
. So
S
2
spans the same subspace as
{
1 +
x

2
x
2
,
3

2
x
+
x
2
}
; it cannot be
a spanning set for all of
V
either.
But let’s just see directly that the constant polynomial 1 is not in
Span
(
S
2
). If it were, we could solve
x
1
(1 +
x

2
x
2
) +
x
2
(3

2
x
+
x
2
)+
x
3
(5

3
x
2
)+
x
4
(13

2
x

5
x
2
) = 1, or
x
1
+3
x
2
+5
x
3
+13
x
4
=
1,
x
1

2
x
2
+ 0
x
3

2
x
4
= 0

2
x
1
+
x
2

3
x
3

5
x
4
= 0. The
corresponding augamented matrix is
1
3
5
13

1
1

2
0

2

0

2
1

3

5

0
.
It rowreduces to
1 0 2 4

2
5
0 1 1 3

1
5
0 0 0 0

2
5
, so there is no solution.
(c)
S
3
=
{
1

x
2
,
2 +
x
+ 3
x
2
,
5

x
+ 4
x
2
,
2
x

7
x
2
}
.
Solution: Again, this set cannot be independent. Speciﬁcally, if we
try to solve
x
1
(1

x
2
)+
x
2
(2+
x
+3
x
2
)+
x
3
(5

x
+4
x
2
) = 2
x

7
x
2
,
we ﬁnd we do have a solution
x
1
=
63
14
,
x
2
=
11
14
,
x
3
=

17
14
.
In fact for any