y
−
x
=
2
β
β
=
y
−
x
2
(
x, y
)
=
x
+
y
2
(1
,
1) +
y
−
x
2
(
−
1
,
1)
Since (
x, y
) is an arbitrary element of
R
2
,
{
(1
,
1)
,
(
−
1
,
1)
}
spans
R
2
. If (
x, y
) = (0
,
0),
α
=
0 + 0
2
= 0
,
β
=
0
−
0
2
= 0
so the coeﬃcients are all zero, so
{
(1
,
1)
,
(
−
1
,
1)
}
is linearly independent. Since it is linearly independent
and spans
R
2
, it is a basis.
Example:
{
(1
,
0
,
0)
,
(0
,
1
,
0)
}
is not a basis of
R
3
, because it does not span.
Example:
{
(1
,
0)
,
(0
,
1)
,
(1
,
1)
}
is not a basis for
R
2
.
1(1
,
0) + 1(0
,
1) + (
−
1)(1
,
1) = (0
,
0)
so the set is not linearly independent.
Theorem 2 (1.2’, see Corrections handout)
Let
V
be a Hamel basis for
X
. Then every vector
x
∈
X
has a
unique
representation as a linear combination (with
all
coeﬃcients nonzero) of a finite number of
elements of
V
.
(
Aside:
the unique representation of 0 is 0 =
∑
i
∈∅
α
i
b
i
.)
Proof:
Let
x
∈
X
. Since
V
spans
X
, we can write
x
=
s
∈
S
1
α
s
v
s
where
S
1
is finite,
α
s
∈
F
,
α
s
= 0,
v
s
∈
V
for
s
∈
S
1
. Now, suppose
x
=
s
∈
S
1
α
s
v
s
=
s
∈
S
2
β
s
v
s
2