Section 1.5
18.
This is a
bonus problem
, with very little partial credit given.
Suppose
S
consists of polynomials of
p
1
, p
2
, . . .
, of degrees 0
≤
d
1
< d
2
< . . .
(we
can relabel the polynomials to enumerate them in the order of decreasing degree).
That is, we can write
p
k
(
x
) =
c
0
k
+
c
1
k
x
+
. . .c
d
k
k
x
d
k
,
with
c
d
k
k
n
= 0. Suppose, for the sake of contradiction, that the set
S
is linearly
dependent. Then there exists
N
∈
N
and scalars
a
1
, . . ., a
N
, not all of them equal
0, such that
a
1
p
1
+
. . .
+
a
N
p
N
= 0. Let
n
be the highest number for which
a
n
n
= 0.
Then the coeFcient of
x
d
n
in
a
1
p
1
+
. . .
+
a
N
p
N
equals
a
n
c
d
n
n
n
= 0, contradicting the
assumption that
a
1
p
1
+
. . .
+
a
N
p
N
= 0.
Section 1.6
2.
By Corollary 2 (pp. 4748), a set of
n
elements in an
n
dimensional space is a
basis i± it is linearly independent. In the examples below, we deal with sets in the
3dimensional space
R
3
, containing 3 elements each. To determine whether such a
set is a basis, we only need to test the sets for linear independence. The sets in the
examples below contain 3 elements each, which equals to the dimension of
R
3
.
(b) No
, the set
{
(2
,
−
4
,
1)
,
(0
,
3
,
−
1)
,
(6
,
0
,
−
1)
}
is not a basis for
R
3
. Indeed,
{
(2
,
−
4
,
1)
,
(0
,
3
,
−
1)
,
(6
,
0
,
−
1)
}
is linearly independent i± (0
,
0
,
0) is the only so
lution for the equation
(1)
t
1
(2
,
−
4
,
1) +
t
2
(0
,
3
,
−
1) +
t
3
(6
,
0
,
−
1) = (0
,
0
,
0)
,
or equivalently, the system
2
t
1
+ 6
t
3
= 0
−
4
t
1
+ 3
t
2
= 0
t
1
−
t
2
−
t
3
= 0
has only the trivial solution. By placing the last equation ²rst, and eliminating
t
1
from the other two equations, we obtain the system
t
1
−
t
2
−
t
3
= 0
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 Spring '10
 FUCKHEAD
 Linear Algebra, Linear Independence, Vector Space, basis

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