Unit 20
Independence and Basis in
<
n
The idea of dimension is fairly intuitive. Consider a vector in
<
n
,(
a
1
,a
2
3
, ..., a
n
).
Each of the
n
components are independent
i.e.
choosing a single compo-
nent in no way eFects the choice for the other components. ±or this reason
we say that
<
n
has
n
degrees of freedom
or equivalently has dimension
n
,
i.e.
<
3
has dimension 3,
<
2
has dimension 2 etc.
Consider now the “subspace” (soon to be de²ned) of
<
3
consisting of all
vectors (
x, y, z
)w
ith
x
+
y
+
z
= 0. We may now only choose 2 components
freely and these determine the third. So this subspace is said to have dimen-
sion 2.
Defnition 22.1.
By a
linear combination
of the vectors
~v
1
,~v
2
, .., ~v
k
in
<
n
we mean an expression of the form
c
1
1
+
c
2
2
+
...
+
c
k
k
when each
c
i
is a
scalar.
Example 1.
Let
v
1
=(
−
2
,
0
,
1),
2
=(1
,
−
1
,
2) and
3
=(4
,
−
2
,
3).
Express
3
as a linear combination of
1
and
2
.
Solution:
We must ²nd scalars
c
1
and
c
2
so that
3
=
c
1
1
+
c
2
2
we set
(4
,
−
2
,
3
)=
c
1
(
−
2
,
0
,
1) +
c
2
(1
,
−
1
,
2)
⇒
(4
,
−
2
,
3
)=(
−
2
c
1
,
0
,c
1
)+(
c
2
,
−
c
2
,
2
c
2
)
⇒
(4
,
−
2
,
3
−
2
c
1
+
c
2
,
−
c
2
1
−
2
c
2
)
1