36
LECTURE 6. GENERATING SETS AND BASES
We have similar situation for
R
3
and all of the spaces
R
n
.
In the case of
R
3
, for example, every vector can be written as combinations
of (1
,
0
,
0)
,
(0
,
1
,
0) and (0
,
0
,
1), i.e.,
(
x,y,z
) =
x
(1
,
0
,
0) +
y
(0
,
1
,
0) +
z
(0
,
0
,
1)
.
Or, as a combination of (1
,

1
,
0)
,
(1
,
1
,
1) and (0
,
1
,

1), that is:
(
x,y,z
) =
a
(1
,

1
,
0) +
b
(1
,
1
,
1) +
c
(0
,
1
,

1)
.
The latter gives three equation:
a
+
b
=
x
(1)

a
+
b
+
c
=
y
(2)
b

c
=
z
(3)
.
(2) + (3) gives:

a
+ 2
b
=
y
+
z
(4)
(4) + (1) gives:
3
b
=
x
+
y
+
z
or
b
=
x
+
y
+
z
3
.
Then (1) gives:
a
=
x

b
=
x

x
+
y
+
z
3
=
2
x

y

z
3
.
Finally, (3) gives:
c
=
b

z
=
x
+
y

2
z
3
Hence, we get:
(
x,y,z
) =
2
x

y

z
3
(1
,

1
,
0) +
x
+
y
+
z
3
(1
,
1
,
1) +
x
+
y

2
z
3
(0
,
1
,

1)
.