Vector spaces are the principal object of study
in linear algebra. A vector space is always defined with respect
to a field of scalars.
A field is a set FF equipped with two operations, addition and
mulitplication, and containing two special members 0 and 1
(
0≠1
0
1
), such that for all
abc∈F
a
b
c
F
-
-
a+b∈F
a
b
F
-
a+b=b+a
a
b
b
a
-
(
a
+
b
)
+
c
=
a
+
(
b
+
c
)
(
a
+
b
)
+
c
a
+
(
b
+
c
)
-
a+0=a
a
0
a
- there exists
-a
a
such that
a+-a=0
a
a
0
-
-
ab∈F
a
b
F
-
ab=ba
a
b
b
a
-
abc=abc
a
b
c
a
b
c
-
a
·
1
=a
a
·
1
a
- there exists
a-1
a
such that
aa-1=1
a
a
1
-
ab+c=ab+ac
a
b
c
a
b
a
c
More concisely
- FF is an
abelian group under addition
- FF is an
abelian group under multiplication
- multiplication distributes over addition
Let FF be
a field, and VV a
set. We say VV
is a vector space over
F
F if there exist two operations, defined for all
a∈F
a
F
,
u∈V
u
V
and
v∈V
v
V
:
-
vector addition: (uu, vv) →
u+v∈V
u
v
V
-
scalar multiplication:
(aa,vv) →
av∈V
a
v
V
and if there exists an element denoted
0∈V
0
V
, such that the following hold for all
a∈F
a
F
,
b∈F
b
F
, and
u∈V
u
V
,
v∈V
v
V
, and
w∈V
w
V
-
-
u
+
(
v
+
w
)
=
(
u
+
v
)
+
w
u
+
(
v
+
w
)
(
u
+
v
)
+
w
-
u+v=v+u
u
v
v
u
-
u+0=u
u
0
u
- there exists
-u
u
such that
u+-u=0
u
u
0
-
-
au+v=au+av
a
u
v
a
u
a
v
-
a+bu=au+bu
a
b
u
a
u
b
u
-
abu=abu
a
b
u
a
b
u
-
1
·
u
=u
1
·
u
u
More concisely,
-
VV is an abelian
group under plus
- Natural properties of scalar multiplication
-
ℝN
N
is a vector space over ℝ
-
ℂN
N
is a vector space over ℂ
-
ℂN
N
is a vector space over ℝ
-
ℝN
N
is not a vector space
over ℂ
The elements of
VV
are called
vectors.
Throughout this course we will think of a signal
as a vector
x=
x
1
x
2
⋮
x
N
=
x
1
x
2
…
x
N
T
x
x
1
x
2
⋮
x
N
x
1
x
2
…
x
N
The samples
x
i
x
i
could be samples from a finite duration, continuous
time signal, for example.
A signal will belong to one of two vector spaces:
Let VV be a vector
space over FF.
A subset
S⊆V
S
V
is called a subspace of VV if SS is a vector space over
FF in its own right.
V=ℝ2
V
2
,
F=ℝ
F
,
S=any line though the origin
S
any line though the origin
.
Are there other subspaces?
S⊆V
S
V
is a subspace if and only if for all
a∈F
a
F
and
b∈F
b
F
and for all
s∈S
s
S
and
t∈S
t
S
,
as+bt∈S
a
s
b
t
S
Let
u
1
,
…
,
u
k
∈V
u
1
,
…
,
u
k
V
.
We say that these vectors are linearly
dependent if there exist scalars
a
1
,
…
,
a
k
∈F
a
1
,
…
,
a
k
F
such that
∑i=1k
a
i
ui=0
i
1
k
a
i
u
i
0
(1)
and at least one
a
i
≠0
a
i
0
.
If Equation 1 only holds for the case
a
1
=…=
a
k
=0
a
1
…
a
k
0
, we say that the vectors are linearly
independent.
11-12−2-230+1-57-2=0
1
1
-1
2
2
-2
3
0
1
-5
7
-2
0
so these vectors are linearly dependent in
ℝ3
3
.
Consider the subset
S=v1v2…vk
S
v
1
v
2
…
v
k
. Define the span of SS
<
S
>
≡spanS≡{∑i=1k
a
i
vi|
a
i
∈F}
<
S
>
span
S
i
1
k
a
i
v
i
a
i
F
Fact:
<
S
>
<
S
>
is a subspace of VV.
V=ℝ3
V
3
,
F=ℝ
F
,
S=v1v2
S
v
1
v
2
,
v1=100
v
1
1
0
0
,
v2=010
v
2
0
1
0
⇒
<
S
>
=xy-plane
<
S
>
xy-plane
.
If SS is infinite, the notions of
linear independence and span are easily generalized:
We say SS is linearly independent if, for
every finite collection
u1
,
…
,
uk
∈S
u
1
,
…
,
u
k
S
, (kk arbitrary) we
have
∑i=1k
a
i
ui=0⇒∀i:
a
i
=0
i
1
k
a
i
u
i
0
i
a
i
0
The span of SS is
<
S
>
={∑i=1k
a
i
ui|
a
i
∈F∧ui∈S∧k<∞}
<
S
>
i
1
k
a
i
u
i
a
i
F
u
i
S
k
In both definitions, we only consider
finite sums.
A set
B⊆V
B
V
is called a basis for VV over
FF if and only if
-
B
B is linearly independent
-
<
B
>
=V
<
B
>
V
Bases are of fundamental importance in signal processing. They
allow us to decompose a signal into building blocks (basis
vectors) that are often more easily understood.
VV = (real or complex) Euclidean
space,
ℝN
N
or
ℂN
N
.
B=e1…eN≡standard basis
B
e
1
…
e
N
standard basis
ei=0⋮1⋮0
e
i
0
⋮
1
⋮
0
where the 1 is in the
ith
i
th
position.
V=ℂN
V
N
over ℂ.
B=u1…uN
B
u
1
…
u
N
which is the DFT basis.
uk=1ⅇ-ⅈ2πkN⋮ⅇ-ⅈ2πkNN−1
u
k
1
2
k
N
⋮
2
k
N
N
1
where
ⅈ=-1
-1
.
If BB
is a basis for VV,
then every
v∈V
v
V
can be written uniquely (up to order of terms) in
the form
v=∑i=1N
a
i
vi
v
i
1
N
a
i
v
i
where
a
i
∈F
a
i
F
and
vi∈B
v
i
B
.
- If SS is a
linearly independent set, then SS can be extended to a basis.
- If
<
S
>
=V
<
S
>
V
, then SS contains a basis.
Let VV
be a vector space with basis BB. The dimension of VV, denoted
dimV
dim
V
, is the cardinality of BB.
Every vector space has a basis.
Every basis for a vector space has
the same cardinality.
⇒dimV
dim
V
is well-defined.
If
dimV<∞
dim
V
, we say VV
is finite dimensional.
Table 1
| vector space |
field of scalars |
dimension |
|
ℝN
N
|
ℝ
|
|
ℂN
N
|
ℂ
|
|
ℂN
N
|
ℝ
|
Every subspace is a vector space, and
therefore has its own dimension.
Suppose
S=u1…uk⊆V
S
u
1
…
u
k
V
is a linearly independent set. Then
dim
<
S
>
=
dim
<
S
>
- If SS
is a subspace of VV,
then
dimS≤dimV
dim
S
dim
V
.
- If
dimS=dimV<∞
dim
S
dim
V
, then
S=V
S
V
.
Let VV
be a vector space, and let
S⊆V
S
V
and
T⊆V
T
V
be subspaces.
We say VV is the direct sum of
SS and TT, written
V=S⊕T
V
⊕
S
T
, if and only if for every
v∈V
v
V
, there exist unique
s∈S
s
S
and
t∈T
t
T
such that
v=s+t
v
s
t
.
If
V=S⊕T
V
⊕
S
T
, then
TT is called a
complement of SS.
V=
C
′
=
{
f
:
ℝ
→
ℝ
|
f
is continuous
}
V
C
′
{
f
:
→
|
f
is continuous
}
S=
even funcitons in
C
′
S
even funcitons in
C
′
T=
odd funcitons in
C
′
T
odd funcitons in
C
′
ft=12ft+f-t+12ft−f-t
f
t
1
2
f
t
f
t
1
2
f
t
f
t
If
f=g+h=
g
′
+
h
′
f
g
h
g
′
h
′
,
g∈S
g
S
and
g
′
∈S
g
′
S
,
h∈T
h
T
and
h
′
∈T
h
′
T
, then
g−
g
′
=
h
′
−h
g
g
′
h
′
h
is odd and even, which implies
g=
g
′
g
g
′
and
h=
h
′
h
h
′
.
- Every subspace has a complement
-
V=S⊕T
V
⊕
S
T
if and only if
-
S⋂T=0
S
T
0
-
<
S
,
T
>
=V
<
S
,
T
>
V
- If
V=S⊕T
V
⊕
S
T
, and
dimV<∞
dim
V
, then
dimV=dimS+dimT
dim
V
dim
S
dim
T
Let VV
be a vector space over FF. A norm
is a mapping
V→F
→
V
F
, denoted by
∥·∥
·
, such that forall
u∈V
u
V
,
v∈V
v
V
, and
λ∈F
λ
F
-
∥u∥>0
u
0
if
u≠0
u
0
-
∥λu∥=|λ|∥u∥
λ
u
λ
u
-
∥u+v∥≤∥u∥+∥v∥
u
v
u
v
Euclidean norms:
x∈ℝN
x
N
:
∥x∥=∑i=1N
x
i
212
x
i
1
N
x
i
2
1
2
x∈ℂN
x
N
:
∥x∥=∑i=1N|
x
i
|212
x
i
1
N
x
i
2
1
2
Every norm induces a metric on VV
duv≡∥u−v∥
d
u
v
u
v
which leads to a notion of "distance" between vectors.
Let
V
V be a vector space over FF,
F=ℝ
F
or
ℂ. An inner product is a mapping
V×V
→
F
V
V
→
F
, denoted
<·,·>
·
·
, such that
-
<v,v>≥0
v
v
0
, and
<v,v>=0⇔v=0
⇔
v
v
0
v
0
-
<u,v>=<v,u>¯
u
v
v
u
-
<au+bv,w>=a<u,w>+b<v,w>
a
u
b
v
w
a
u
w
b
v
w
ℝN
N
over ℝ:
<x,y>=xTy=∑i=1N
x
i
y
i
x
y
x
y
i
1
N
x
i
y
i
ℂN
N
over ℂ:
<x,y>=xHy=∑i=1N
x
i
¯
y
i
x
y
x
y
i
1
N
x
i
y
i
If
x=
x
1
…
x
N
T∈ℂ
x
x
1
…
x
N
, then
xH≡
x
1
¯⋮
x
N
¯T
x
x
1
⋮
x
N
is called the "Hermitian," or "conjugate
transpose" of xx.
If we define
∥u∥=<u,u>
u
u
u
, then
∥u+v∥≤∥u∥+∥v∥
u
v
u
v
Hence, every inner product induces a norm.
For all
u∈V
u
V
,
v∈V
v
V
,
|<u,v>|≤∥u∥∥v∥
u
v
u
v
In inner product spaces, we have a notion of the
angle between two vectors:
∠uv=arccos<u,v>∥u∥∥v∥∈02π
∠
u
v
u
v
u
v
0
2
uu
and vv are
orthogonal if
<u,v>=0
u
v
0
Notation:
u⊥v
⊥
u
v
.
If in addition
∥u∥=∥v∥=1
u
v
1
, we say uu and vv are orthonormal.
In an orthogonal (orthonormal)
set, each pair of vectors is orthogonal
(orthonormal).
An Orthonormal basis is a basis
vi
v
i
such that
<vi,vi>=
δ
i
j
=1ifi=j0ifi≠j
v
i
v
i
δ
i
j
1
i
j
0
i
j
The standard basis for
ℝN
N
or
ℂN
N
The normalized DFT basis
uk=1N1ⅇ-ⅈ2πkN⋮ⅇ-ⅈ2πkNN−1
u
k
1
N
1
2
k
N
⋮
2
k
N
N
1
If the representation of vv with respect to
vi
v
i
is
v=∑
a
i
vi
v
i
a
i
v
i
then
a
i
=<vi,v>
a
i
v
i
v
Every inner product space has an orthonormal
basis. Any (countable) basis can be made orthogonal by the
Gram-Schmidt orthogonalization process.
Let
S⊆V
S
V
be a subspace. The orthogonal
compliment SS is
S
⊥
={u|u∈V∧<u,v>=0∧∀v:v∈S}
S
⊥
u
u
V
u
v
0
v
v
S
S
⊥
S
⊥
is easily seen to be a subspace.
If
dimv<∞
dim
v
, then
V=S⊕
S
⊥
V
⊕
S
S
⊥
.
If
dimv=∞
dim
v
, then in order to have
V=S⊕
S
⊥
V
⊕
S
S
⊥
we require VV to be a Hilbert
Space.
Loosely speaking, a linear transformation is a
mapping from one vector space to another that
preserves vector space operations.
More precisely, let VV, WW be vector spaces over the same
field FF. A linear
transformation is a mapping
T
:
V
→
W
T
:
V
→
W
such that
Tau+bv=aTu+bTv
T
a
u
b
v
a
T
u
b
T
v
for all
a∈F
a
F
,
b∈F
b
F
and
u∈V
u
V
,
v∈V
v
V
.
In this class we will be concerned with linear
transformations between (real or complex) Euclidean
spaces, or subspaces thereof.
imageT={w|w∈W∧
Tv=w
for some
v
}
T
w
w
W
T
v
w
for some
v
Also known as the kernel:
kerT={v|v∈V∧Tv=0}
ker
T
v
v
V
T
v
0
Both the image and the nullspace are easily seen
to be subspaces.
rankT=dimimageT
rank
T
dim
T
nullT=dimkerT
null
T
dim
ker
T
rankT+nullT=dimV
rank
T
null
T
dim
V
Every linear transformation
TT has a matrix
representation. If
T
:
𝔼
N
→
𝔼
M
T
:
𝔼
N
→
𝔼
M
,
𝔼=ℝ
𝔼
or ℂ, then
TT is represented by an
M×N
M
N
matrix
A=
a
1
1
…
a
1
N
⋮⋱⋮
a
M
1
…
a
M
N
A
a
1
1
…
a
1
N
⋮
⋱
⋮
a
M
1
…
a
M
N
where
a
1
i
…
a
M
i
T=Tei
a
1
i
…
a
M
i
T
e
i
and
ei=0…1…0T
e
i
0
…
1
…
0
is the
ith
i
th
standard basis vector.
A linear transformation can be represented
with respect to any bases of
𝔼N
𝔼
N
and
𝔼M
𝔼
M
, leading to a different AA. We will always represent a
linear transformation using the standard bases.
colspanA=
<
A
>
=imageA
colspan
A
<
A
>
A
If
A
:
ℝ
N
→
ℝ
M
A
:
N
→
M
, then
ker⊥A=imageAT
ker
A
⊥
A
If
A
:
ℂ
N
→
ℂ
M
A
:
N
→
M
, then
ker⊥A=imageAH
ker
A
⊥
A
The linear transformation/matrix
AA is
invertible if and only if there exists a matrix
BB such that
AB=BA=I
A
B
B
A
I
(identity).
Only square matrices
can be invertible.
Let
A
:
𝔽
N
→
𝔽
N
A
:
𝔽
N
→
𝔽
N
be linear,
𝔽=ℝ
𝔽
or ℂ. The
following are equivalent:
- AA is
invertible (nonsingular)
-
rankA=N
rank
A
N
-
nullA=0
null
A
0
-
detA≠0
A
0
- The columns of AA form a basis.
If
A-1=AT
A
A
(or
AH
A
in the complex case), we say AA is orthogonal (or
unitary).