MATH 223, Linear Algebra
Fall, 2007
Assignment 4 Solutions
1. Consider the vector space
V
=
P
5
(
R
) of polynomials with real coefficients (in one variable
t
) of
degree at most 5 (including the zero polynomial). Show that if
c
∈ R
is any real number, then the
set
{
1
, t

c,
(
t

c
)
2
,
(
t

c
)
3
,
(
t

c
)
4
,
(
t

c
)
5
}
is a basis of
V
.
Solution:
Since the set
{
1
, t, t
2
, t
3
, t
4
, t
5
}
is visibly a basis of
V
, we know that this vector space has
dimension 6. To show that the 6 vectors
{
1
, t

c,
(
t

c
)
2
,
(
t

c
)
3
,
(
t

c
)
4
,
(
t

c
)
5
}
are a basis of
V
, it therefore suffices to show that they are linearly independent (as spanning is then automatic).
So suppose that for some scalars
a
0
, a
1
, a
2
, a
3
, a
4
, a
5
we have
a
0
+
a
1
(
t

c
) +
a
2
(
t

c
)
2
+
a
3
(
t

c
)
3
+
a
4
(
t

c
)
4
+
a
5
(
t

c
)
5
= 0;
we wish to show that each
a
i
must be zero. Comparing coefficients of
t
5
on both sides, we see that
we must have
a
5
= 0.
Comparing coefficients of
t
4
on both sides of the resulting equation then
forces
a
4
= 0 as well. Continuing in this manner, we find that
a
3
=
a
2
=
a
1
=
a
0
= 0, as desired.
2. Let
V
be as in problem (1). Let
W
be the subset of
V
consisting of those polynomials
p
(
t
) such
that
p
(0) = 0, and let
U
be the subset of polynomials
p
(
t
) that are
even
, i.e. such that
p
(
t
) =
p
(

t
).
(a) Show that
U
and
W
are subspaces of
V
.
(b) Compute dim
V
, dim
W
, dim
U
, dim
U
∩
W
.
Solution:
(a) It suffices to show that each set
U, W
contains the zero vector and is closed under addition
and scalar multiplication.
Since the value of the zero polynomial at zero is indeed zero,
W
contains zero; similarly, as 0 =

0, we see that
U
contains zero as well. Now suppose that
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 Fall '07
 Loveys
 Linear Algebra, Algebra, Polynomials, Vector Space, basis

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