2Vectors
IA Vectors and Matrices
2.6 Spanning sets and bases
2.6.1 2D space
Definition
(Spanning set)
.
A set of vectors
{a, b}
spans
R
2
if for all vectors
r ∈ R
2
, there exist some λ, µ ∈ R such that r = λa + µb.
In R
2
, two vectors span the space if a × b 6= 0.
Theorem. The coefficients λ, µ are unique.
Proof.
Suppose that
r
=
λa
+
µb
=
λ
0
a
+
µ
0
b
. Take the vector product with
a
on both sides to get (
µ − µ
0
)
a × b
=
0
. Since
a × b 6
=
0
, then
µ
=
µ
0
. Similarly,
λ = λ
0
.
Definition
(Linearly independent vectors in
R
2
)
.
Two vectors
a
and
b
are
linearly independent if for
α, β ∈ R
,
αa
+
βb
=
0
iff
α
=
β
= 0. In
R
2
,
a
and
b
are linearly independent if a ×b 6= 0.
Definition
(Basis of
R
2
)
.
A set of vectors is a basis of
R
2
if it spans
R
2
and
are linearly independent.
Example. {
ˆ
i,
ˆ
j}
=
{
(1
,
0)
,
(0
,
1)
}
is a basis of
R
2
. They are the standard basis
of R
2
.
2.6.2 3D space
We can extend the above definitions of spanning set and linear independent set
to R
3
. Here we have
Theorem.
If
a, b, c ∈ R
3
are non-coplanar, i.e.
a ·
(
b × c
)
6
= 0, then they form
a basis of R
3
.
Proof.
For any
r
, write
r
=
λa
+
µb
+
νc
. Performing the scalar product
with
b × c
on both sides, one obtains
r · (b × c)
=
λa · (b × c)
+
µb · (b × c)
+
νc · (b × c)
=
λ[a, b, c]
. Thus
λ
=
[r, b, c]/[a, b, c]
. The values of
µ
and
ν
can
be found similarly. Thus each
r
can be written as a linear combination of
a, b
and c.
By the formula derived above, it follows that if
αa
+
βb
+
γc
=
0
, then
α = β = γ = 0. Thus they are linearly independent.
Note that while we came up with formulas for
λ, µ
and
ν
, we did not actually
prove that these coefficients indeed work. This is rather unsatisfactory. We
could, of course, expand everything out and show that this indeed works, but
in IB Linear Algebra, we will prove a much more general result, saying that if
we have an
n
-dimensional space and a set of
n
linear independent vectors, then
they form a basis.
In R
3
, the standard basis is
ˆ
i,
ˆ
j,
ˆ
k, or (1, 0, 0), (0, 1, 0) and (0, 0, 1).
2.6.3 R
n
space
In general, we can define
Definition
(Linearly independent vectors)
.
A set of vectors
{v
1
, v
2
, v
3
···v
m
}
is linearly independent if
m
X
i=1
λ
i
v
i
= 0 ⇒ (∀i) λ
i
= 0.
Definition
(Spanning set)
.
A set of vectors
{u
1
, u
2
, u
3
···u
m
} ⊆ R
n
is a
spanning set of R
n
if
(∀x ∈ R
n
)(∃λ
i
)
m
X
i=1
λ
i
u
i
= x
Definition
(Basis vectors)
.
A basis of
R
n
is a linearly independent spanning
set. The standard basis of
R
n
is
e
1
= (1
,
0
,
0
, ···
0)
, e
2
= (0
,
1
,
0
, ···
0)
, ···e
n
=
(0, 0, 0, ··· , 1).
Definition
(Orthonormal basis)
.
A basis
{e
i
}
is orthonormal if
e
i
· e
j
= 0 if
i 6= j and e
i
· e
i
= 1 for all i, j.
Using the Kronecker Delta symbol, which we will define later, we can write
this condition as e
i
· e
j
= δ
ij
.
Definition
(Dimension of vector space)
.
The dimension of a vector space is
the number of vectors in its basis. (Exercise: show that this is well-defined)
We usually denote the components of a vector
x
by
x
i
. So we have
x
=
(x
1
, x
2
, ··· , x
n
).
Definition
(Scalar product)
.
The scalar product of
x, y ∈ R
n
is defined as
x · y =
P
x
i
y
i
.
The reader should check that this definition coincides with the
|x||y|cos θ
definition in the case of R
2
and R
3
.
2.6.4 C
n
space
C
n
is very similar to
R
n
, except that we have complex numbers. As a result, we
need a different definition of the scalar product. If we still defined
u ·v
=
P
u
i
v
i
,
then if we let
u
= (0
, i
), then
u · u
=
−
1
<
0. This would be bad if we want to
use the scalar product to define a norm.
Definition
(
C
n
)
. C
n
=
{
(
z
1
, z
2
, ··· , z
n
) :
z
i
∈ C}
. It has the same standard
basis as
R
n
but the scalar product is defined differently. For
u, v ∈ C
n
,
u · v
=
P
u
∗
i
v
i
. The scalar product has the following properties:
(i) u · v = (v · u)
∗
(ii) u · (λv + µw) = λ(u · v) + µ(u · w)
(iii) u · u ≥ 0 and u · u = 0 iff u = 0
Instead of linearity in the first argument, here we have (
λu
+
µv
)
· w
=
λ
∗
u · w + µ
∗
v · w.
Example.
4
X
k=1
(−i)
k
|x + i
k
y|
2
=
X
(−i)
k
hx + i
k
y | x + i
k
yi
=
X
(−i)
k
(hx + i
k
y | xi + i
k
hx + i
k
y | yi)
=
X
(−i)
k
(hx | xi + (−i)
k
hy | xi + i
k
hx | yi + i
k
(−i)
k
hy | yi)
=
X
(−i)
k
[(|x|
2
+ |y |
2
) + (−1)
k
hy | xi + hx | yi]
= (|x|
2
+ |y |
2
)
X
(−i)
k
+ hy | xi
X
(−1)
k
+ hx | y i
X
1
= 4hx | yi.
We can prove the Cauchy-Schwarz inequality for complex vector spaces using
the same proof as the real case, except that this time we have to first multiply
y
by some
e
iθ
so that
x ·
(
e
iθ
y
) is a real number. The factor of
e
iθ
will drop off at
the end when we take the modulus signs.