Download Linear codes. Groups, fields and vector spaces

Information and Coding
Theory
Linear Codes.
Groups, fields and vector spaces - a brief
survey.
Codes defined as vector subspaces. Dual
codes.
Juris Viksna, 2017
Codes – how to define them?
In most cases it would be natural to use binary block codes that maps
input vectors of length k to output vectors of length n.
For example: 101101
110011
etc.


11100111
01000101
Thus we can define code as an injective mapping from vector space V
with dimension k to vector space W with dimension n.
V
W
Such definition essentially is used in the original Shanon’s theorem.
Codes – how to define them?
Simpler to define and use are linear codes that can be defined by
multiplication with matrix of size kn (called generator matrix).
What should be the elements of
vector spaces V and W?
In principle in most cases it will be sufficient to have just 0-s and 1-s,
however, to define vector space in principle we need a field – an
algebraic system with operations “+” and “” defined and having similar
properties as we have in ordinary arithmetic (think of real numbers).
Field with just “0” and “1” may look very simple, but it turns out that to
get some real progress we will need more complicated fields, just that
elements of these fields themselves will be regarded as (most often)
binary vectors.
Groups - definition
Consider set G and binary operator +.
Definition
Pair (G,+) is a group, if there is eG such that for all a,b,cG:
1) a+bG
2) (a+b)+c = a+(b+c)
3) a+e = a and e+a = a
4) there exists inv(a) such that a+ inv(a)= e and inv(a)+a = e
5) if additionally a+b = b+a, group is commutative (Abelian)
If group operation is denoted by “+” then e is usually denoted by 0 and
inv(a) by a.
If group operation is denoted by “” hen e is usually denoted by 1 and
inv(a) by a1 (and ab are usually written as ab).
It is easy to show that e and inv(a) are unique.
Groups - definition
Definition
Pair (G,+) is a group, if there is eG such that for all a,b,cG:
1)
2)
3)
4)
a+bG
(a+b)+c = a+(b+c)
a+e = a and e+a = a
there exists inv(a) such that a+ inv(a)= e and inv(a)+a = e
Examples:
(Z,+), (Q,+), (R,+)
(Z2,+), (Z3,+), (Z4,+)
(Q{0},), (R{0},) (but not (Z{0},))
(Z2{0},), (Z3{0},) (but not (Z4{0},))
A simple non-commutative group:
x,y, where x: abccab ( rotation) and y: abcacb ( inversion).
Groups - definition
(H,+) is a subgroup of (H,+) if HG and (H,+) is a group.
H<G - notation that H is a subgroup of G.
x1, x1,,xk - a subgroup of G generated by x1, x1,,xk G.
o(G) - number of elements of group G.
For first few lectures we just need to remember group definition.
Other facts will become important when discussing finite fields.
We will consider only commutative and finite groups!
Lagrange theorem
Theorem
If H < G then o(H) | o(G).
Proof
• For all aG consider sets aH = {ah | hH} (these are called cosets).
• All elements of a coset aH are distinct and |aH| = o(H).
• Each element gG belongs to some coset aH (for example ggH).
• Two cosets aH and bH are either identical or disjoint.
• Thus G is a union of disjoint cosets, each having o(H) elements, hence
o(H) | o(G).
Fields - definition
Consider set F and binary operators “+” and “”.
Definition
Triple (F,+, ) is a field, if 0,1F such that for all a,b,c,dF and d0:
1) a+bF and ab F
2) a+b=b+a and ab= ba
3)(a+b)+c=a+(b+c) and (ab)c=a(bc)
4) a+0=a and a1 = a
5) there exist a,d1F such that a+( a)= 0 and dd1=1
6) a(b+c)=ab+bc
We can say that F is a field if both (F,+) and (F{0}, ) are commutative
groups and operators “+” and “” are linked with property (6).
Examples:
(Q,+,), (R,+,), (Z2,+,), (Z3,+,)
(but not (Z4, ,+,))
Finite fields - some examples
[Adapted from V.Pless]
Finite fields - some examples
Note that we could use alternative notation and consider the field
elements as 2-dimensional binary vectors (and usual vector addition
corresponds to operation “+” in field!) :
0=(0 0), 1=(0 1), =(1 0),  =(1 1).
[Adapted from V.Pless]
Vector spaces - definition
What we usually understand by vectors?
In principle we can say that vectors are n-tuples of the form:
(x1,x2,,xn)
and operations of vector addition and multiplication by scalar are defined
and have the following properties:
(x1,x2,,xn)+(y1,y2,,yn)=(x+y1,x+y2,,x+yn)
a(x1,x2,,xn)=(ax1,ax2,,axn)
The requirements actually are a bit stronger – elements a and xi should
come from some field F.
We might be able to live with such a definition, but then we will link a
vector space to a unique and fixed basis and often this will be technically
very inconvenient.
Vector spaces - definition
Let (V,+) be a commutative group with identity element 0, let F be a
field with multiplicative identity 1 and let “” be an operator mapping
FV to V.
Definition
4-tuple (V,F,+,) is a vector space if (V,+) is a commutative group with
identity element 0 and for all u,vV and all a,bF:
1)
2)
3)
4)
a(u+v)=au+av
(a+b)v=av+bv
a(bv)=(ab)v
1v=v
The requirement that (V,+) is a group is used here for brevity of
definition – the structure of (V,+) will largely depend from F.
Note that operators “+” and “” are “overloaded” – the same symbols are
used for field and vector space operators.
Vector spaces - definition
Definition
4-tuple (V,F,+,) is a vector space if (V,+) is a commutative group with
identity element 0 and for all u,vV and all a,bF:
1)
2)
3)
4)
a(u+v)=au+av
(a+b)v=av+bv
a(bv)=(ab)v
1v=v
Usually we will represent vectors as n-tuples of the form (x1,x2,,xn),
however such representations will not be unique and will depend from a
particular basis of vector space, which we will chose to use (but 0 will
always be represented as n-tuple of zeroes (0,0,,0)).
Vector spaces - some terminology
A subset S of vectors from vector space V is a subspace of V if S is
itself a vector space. We denote this by S⊑V.
A linear combination of v1,,vkV (V being defined over a field F) is a
vector of the form a1v1++akvk where aiF.
For given v1,,vkV a set of all linear combinations of these vectors are
denoted by v1,,vk.
It is easy to show that v1,,vk⊑V.
We also say that vectors v1,,vk span the subspace v1,,vk.
A set of vectors v1,,vk is (linearly) independent if a1v1++akvk  0
for all a1,,akF such that ai  0 for at least one index i.
Independent vectors - some properties
A set of vectors v1,,vk is (linearly) independent if a1v1++akvk  0
for all a1,,akF such that ai  0 for at least one index i.
Assume that vectors v1,v2,v3,,vk are independent. Then independent
will be also the following:
• any permutation of vectors v1,v2,v3,,vk (this doesn’t change the set :)
• av1,v2,v3,,vk , if a  0
• v1+v2,v2,v3,,vk
Moreover, these operations do not change subspace spanned by initial
vectors v1,,vk.
Vector spaces - some results
Theorem 1
If v1,,vk V span the entire vector space V and w1,,wr V is an
independent set of vectors then r  k.
Proof
Suppose this is not the case and r  k+1. Since v1,,vk=V we have:
w1 = a11v1+  +a1kvk
w2 = a21v1+  +a2kvk
..........................................
wk = ak1v1+  +akkvk
wk+1 = a(k+1)1v1++a(k+1)kvk
We can assume that a11 ≠ 0 (at least some scalar should be non-zero due
to independence of wi's).
Vector spaces - some results
Thus, we have b1w1 = v1++a1kvk for some non-zero b1.
Subtracting the value b1w1ai1 from all wi's with i >1 gives us a new
system of equations with independent set of k+1 vectors on left side.
By repeating this process k+1 time by starting with the i-th equation in
each iteration we end up with set of k+1 independent vectors, where the
i-th vector is expressed as a sum of no more than k+1–i vi's. However,
this means that k+1-st vector will be 0 and this contradicts that set of
k+1 vectors is independent.
Vector spaces - basis
Theorem 2
If two finite sets of independent vectors span a space V, then there are
the same number of vectors in each set.
Proof
Let k be a number of vectors in the first set and r the number of vectors
in the second set. By Theorem 1 we have k  r and r  k.
We will be interested in vector spaces that are spanned by a finite
number of vectors, so we assume this from now.
We say that a set of vectors v1,,vkV is a basis of V if they are
independent and span V.
Vector spaces - basis
Theorem 3
Let V be a vector space over a field F. Then the following hold:
1) V has a basis.
2) Any two bases of V contain the same number of vectors.
3) If B is a basis of V then every vector in V has a unique
representation as a linear combination of vectors in B.
Proof
1. Let v1V and v1 ≠ 0. If V = v1 we are finished. If not, there is a
vector v2Vv1 and v2 ≠ 0. If V = v1,v2 we are finished. If not,
continue until we obtain a basis for V.
The process must terminate by our assumption that V is spanned by a
finite number k of vectors and the result given by Theorem 1 that the
basis cannot have more than k vectors.
Vector spaces - basis
Theorem 3
Let V be a vector space over a field F. Then the following hold:
1) V has a basis.
2) Any two bases of V contain the same number of vectors.
3) If B is a basis of V then every vector in V has a unique
representation as a linear combination of vectors in B.
Proof
2. This is a direct consequence from Theorem 2.
3. Since the vectors in B span V, any vV can be written as
v = a1b1+...+akbk .
If we also have v = c1b1+...+ckbk then a1b1+...+akbk = c1b1+...+ckbk
and, since bi's are independent, ai = ci for all i.
Vector spaces - dimension
The dimension of a vector space V, denoted by dim V, is the number of
vectors in any basis of V. By Theorem 3 the dimension of V is well
defined.
Independent vectors – few more properties
Assume that set of vectors v1,,vk V is independent and in some fixed
basis b1,,bkV we have representations vi = ai1b1+...+aikbk. Then
independent will be also the following sets of vectors obtained from
v1,,vk by:
• for some j,k swapping all aij-s with aik-s
• for some j and non-zero cF replacing all aij-s with caij-s (*)
• for some k replacing all aij-s with aij+aik-s (**)
Note however that these operations can change subspace spanned by
initial vectors v1,,vk.
Vector spaces and rangs of matrices
If M is a matrix whose elements are contained in a field F, then the row
rank of M, denoted by rr(M), is defined to be dimension of the subspace
spanned by rows of M.
The column rank of M , denoted by rc(M), is defined to be dimension
of the subspace spanned by columns of M.
An nn matrix A is called nonsingular if rr(A) = n. This means that the
rows of A are linearly independent.
Vector spaces and rangs of matrices
If M is a matrix, the following operations on its rows are called
elementary row operations:
1) Permuting rows.
2) Multiplying a row by a nonzero scalar.
3) Adding a scalar multiple of one row to another row.
Similarly, we can define the following to be elementary column
operations:
4) Permuting columns.
5) Multiplying a column by a nonzero scalar.
6) Adding a scalar multiple of one column to another column.
As we saw, none of these operations affects rr(M) or rc(M), although
1,2,3 and 4,5,6 correspondingly could change M column/row spaces.
Vector spaces - matrix row-eschelon form
A matrix M is said to be in row-echelon form if, after possibly
permuting its columns,
I
M = 
A
 0 0 


where I is kk identity matrix.
Lemma
If M is any matrix, then M can by applying elementary operations 1,2,3,4
it can be transformed into a matrix M in row echelon form.
Vector spaces - matrix row-eschelon form
Lemma
If M is any matrix, then M can by applying elementary operations 1,2,3,4
it can be transformed into a matrix M in row echelon form.
Proof (sketch)
Proceed iteratively in steps 1...k as follows:
Step i
• Perform row permutation (if such exists) that places in i-th position
row with a non-zero element in i-th column. Let it be row v. If there
are no appropriate permutation, perform first column permutation that
places non-zero column in i-th position.
• Add to each of other rows vector av, where a is chosen such that after
the addition the row has value 0 in i-th position.
Vector spaces - matrix row-eschelon form
Theorem
The row rank rr(M) of matrix M, equals its column rank rc(M).
Proof
Transform M into M in row echelon form applying elementary
operations 1,2,3,4. This doesn’t change either rr(M) or rc(M), it is also
obvious that for matrix M we have rr(M) = rc(M) = k, where k is the
size of identity matrix I.
I
M = 
0
A

0
Linear codes
Message
source
Encoder
Channel
x = x1,...,xk
message
c = c1,...,cn
codeword
y=c+e
received
vector
Decoder
Receiver
x'
estimate of
message
e = e1,...,en
error from
noise
Generally we will define linear codes as vector spaces – by taking C to
be a k-dimensional subspace of some n-dimensional space V.
Codes and linear codes
Let V be an n-dimensional vector space over a finite field F.
Definition
A code is any subset CV.
Definition
A linear (n,k) code is any k-dimensional subspace C⊑V.
Whilst this definition is largely “standard”, it doesn’t distinguish any
particular basis of V. However the properties and parameters of code
will vary greatly with selection of particular basis. So in fact we
assume that V is given already with some fixed basis b1,,bnV and
will assume that all elements of V and C will be represented in this
particular basis. At the same time we might be quite flexible in
choosing a specific basis for C.
Codes and linear codes
Let V be an n-dimensional vector space over a finite field F.
Definition
A linear (n,k) code is any k-dimensional subspace C⊑V.
Example (choices of bases for V and code C):
Basis of V (fixed):
Set of V elements:
Set of C elements:
2 alternative bases
for code C:
001,010,100
{000,001,010,011,100,101,110,111}
{000,001,010,011}
001,010
001,011
Essentially, we will be ready to consider alternative bases, but will stick
to “main one” for representation of V elements.
Hamming code [7,4]
What to do, if there are errors?
- we assume that the number of errors is as small as possible - i.e.
we can find the code word c (and the corresponding x) that is
closest to received vector y (using Hamming distance)
- consider vectors a = 0001111, b = 0110011 and c = 1010101,
-- if y is received, compute ya, yb and yc (inner products), e.g.,
for y = 1010010 we obtain ya = 1, yb = 0 and yc = 0.
-- this represents a binary number (100 or 4 in example above) and
we conclude that error is in 4th digit, i.e. x = 1011010.
Easy, bet why this method work?
Vector spaces – dot (scalar) product
Let V be a k-dimensional vector space over field F. Let b1,,bkV be
some basis of V. For a pair of vectors u,vV, such that u=a1b1+...+akbk
and v=c1b1+...+ckbk their dot (scalar) product is defined by:
The question whether for an inner
product defined in abstract way there
Thus operator “” maps VV to will
F. be a basis that will allow to
compute it as a scalar product by
formula above is somewhat tricky and
Lemma
depend
from particular
For u,v,wV and all a,bF the may
following
properties
hold: vector
space V.
u·v = a1·c1 +...+ ak·ck
1) u·v = v·u.
2) (au+bv)·w = a(u·v)+b(v·w).
3) If u·v = 0 for all v in V, then u = 0.
Note. The Lemma above can also be used as a more abstract definition of
inner product.
Vector spaces – dot (scalar) product
Let V be a k-dimensional vector space over field F. Let b1,,bkV be
some basis of V. For a pair of vectors u,vV, such that u=a1b1+...+akbk
and v=c1b1+...+ckbk their dot (scalar) product is defined by:
u·v = a1·c1 +...+ ak·ck
Two vectors u and v are said to be orthogonal if u·v = 0. If C is a
subspace of V then it is easy to see that the set of all vectors in V that are
orthogonal to each vector in C is a subspace, which is called the space
orthogonal to C and denoted by C.
For us important and not too obvious result!
Theorem
If C is a subspace of V, then dim C + dim C = dim V.
Vector spaces and linear transformations
Definition
Let V be a vector space over field F. Function f : VV is called a linear
transformation, if for all u,vV and all aF the following hold:
1)af(u) = f(au).
2)f(u)+f(v) = f(u+v).
The kernel of f is defined as ker f ={vV | f(v) = 0}.
The range of f is defined as range f ={f(v) | vV}.
It is quite obvious property of f linearity that vector sums and scalar
products doesn't leave ker f or range f .
Thus ker f ⊑V and range f ⊑V.
Vector spaces and linear transformations
Theorem (rank-nullity theorem)
dim (ker f) + dim (range f) = dim V.
ker f
range f
v0
0
Proof?
Chose some basis u1,,uk of ker f and some basis w1,,wn of range f.
Then try to show that u1,,uk, v1,,vn is a basis for V, where wi = f(vi)
(vectors vi may not be uniquely defined, but is sufficient to chose
arbitrary pre-images of wi -s).
Rank-nullity theorem
Theorem
dim (ker f) + dim (range f) = dim V.
[Adapted from R.Milson]
Rank-nullity theorem
[Adapted from R.Milson]
Dimensions of orthogonal vector spaces
Theorem
If C is a subspace of V, then dim C + dim C = dim V.
C
C
0
Proof?
We could try to reduce this to “similarly looking” equality dim V = dim
(ker f) + dim (range f).
However how we can define a linear transformation from dot product?
Dimensions of orthogonal vector spaces
Theorem
If C is a subspace of V, then dim C + dim C = dim V.
Proof
However how we can define a linear transformation from dot product?
Let u1,,uk be some basis of C. We define transformation f as follows:
for all vV: f(v) = (vu1) u1 +  + (vuk) uk
Note that vui F, thus f(v)C. Therefore we have:
• ker f = C (this directly follows form definition of C)
• range f = C (this follows form definition of f)
Thus from rank-nullity theorem: dim C + dim C = dim V.
Linear codes
Message
source
Encoder
Channel
x = x1,...,xk
message
c = c1,...,cn
codeword
y=c+e
received
vector
Decoder
Receiver
x'
estimate of
message
e = e1,...,en
error from
noise
Generally we will define linear codes as vector spaces – by taking C to
be a k-dimensional subspace of some n-dimensional space V.
Codes and linear codes
Let V be an n-dimensional vector space over a finite field F.
Definition
A code is any subset CV.
Definition
A linear (n,k) code is any k-dimensional subspace C⊑V.
Whilst this definition is largely “standard”, it doesn’t distinguish any
particular basis of V. However the properties and parameters of code
will vary greatly with selection of particular basis. So in fact we
assume that V is given already with some fixed basis b1,,bnV and
will assume that all elements of V and C will be represented in this
particular basis. At the same time we might be quite flexible in
choosing a specific basis for C.
Codes and linear codes
Let V be an n-dimensional vector space over a finite field F together
with some fixed basis b1,,bnV.
Definition
A linear (n,k) code is any k-dimensional subspace C⊑V.
Definition
The weight wt(v) of a vector vV is a number of nonzero components of
v in its representation as a linear combination v = a1b1+...+anbn.
Definition
The distance d(v,w) between vectors v,wV is a number of distinct
components in their representation in given basis.
Definition
The minimum weight of code C⊑V is defined as minvC,v0 wt(v).
Codes and linear codes
Theorem
Linear (n,k,d) code can correct any number of errors not exceeding
t = (d1)/2.
Proof
The distance between any two codewords is at least d.
So, if the number of errors is smaller than d/2 then the closest
codeword to the received vector will be the transmitted one
However a far less obvious problem: how to find which codeword is the
closest to received vector?
Coding theory - the main problem
A good (n,k,d) code has small n, large k and large d.
The main coding theory problem is to optimize one of the parameters n,
k, d for given values of the other two.
Generator matrices
Definition
Consider (n,k) code C⊑V. G is a generator matrix of code C, if
C = {vG | vV} and all rows of G are independent.
It is easy to see that generator matrix exists for any code – take any
matrix G rows of which are vectors v1,,vk (represented as n-tuples in
the initially agreed basis of V) that form a basis of C. By definition G
will be a matrix of size kn.
Obviously there can be many different generator matrices for a given
code. For example, these are two alternative generator matrices for the
same (4,3) code:
 1 0 0 1
0 0 1 1




G1   0 1 0 1 G2   1 1 1 1 
 0 0 1 1
1 0 1 0




Equivalence of codes
Definition
Codes C1,C2⊑V. are equivalent, if a generator matrix G2 of C2 can be
obtained from a generator matrix G1 of C1 by a sequence of the following
operations:
1)permutation of rows
2)multiplication of a row by a non-zero scalar
3)addition of one row to another
4)permutation of columns
5)multiplication of a column by a non-zero scalar
Note that operations 1-3 actually doesn’t change the code C1. Applying
operations 4 and 5 C1 could be changed to a different subspace of V,
however the weight distribution of code vectors remains the same.
In particular, if C1 is (n,k,d) code so is C2. In binary case vectors of C1
and C2 would differ only by permutation of positions.
Generator matrices
Definition
A generator matrix G of (n,k) code C⊑V is said to be in standard form
if G = (I,A), where I is kk identity matrix.
Theorem
For code C⊑V there is an equivalent code C that has a generator matrix
in standard form.
Proof
We have already shown that each matrix can be transformed in rowechelon form by applying the same operations that define equivalent
codes. Since for (n,k) code generator matrix must have rank k, we obtain
that in this case I should be kk identity matrix (i.e. we can’t have rows
with all zeroes).
Dual codes
Definition
Consider code C⊑V. A dual or orthogonal code of C is defined as
C = {vV |  wC: vw = 0}.
It is easy check that C  ⊑V, i.e. C is a code. Note that actually this is
just a re-statement of definition of orthogonal vector spaces we have
already seen.
Remember theorem we have proved shortly ago:
If C is a subspace of V, then dim C + dim C = dim V.
Thus, if C is (n,k) code then C is (nk,k) code and vice versa.
There are codes that are self-dual, i.e. C = C.
Dual codes - some examples
For the (n,1) -repetition code C, with the generator matrix
G = (1 1 … 1)
the dual code C is (n, n1) code with the generator matrix
G, described by:
 1 1 0 0 ... 0 


 1 0 1 0 ... 0 

G 

..


 1 0 0 0 ... 1 


Dual codes - some examples
[Adapted from V.Pless]
Dual codes – parity checking matrices
Definition
Let code C⊑V and let C be its dual code. A generator matrix H of C is
called a parity checking matrix of C.
Theorem
If kn generator matrix of code C⊑V is in standard form if G = (I,A)
then (kn)n matrix H = (AT,I) is a parity checking matrix of C.
Proof
It is easy to check that any row of G is orthogonal to any row of H (each
dot product is a sum of only two non-zero scalars with opposite signs).
Since dim C + dim C = dim V, i.e. k + dim C = n we have to
conclude that H is a generator matrix of C.
Note that in binary vector spaces H = (AT,I) = (AT,I).
Dual codes – parity checking matrices
Theorem
If kn generator matrix of code C⊑V is in standard form if G = (I,A)
then (kn)n matrix H = (AT,I) is a parity checking matrix of C.
So, up to the equivalence of codes we have an easy way to obtain a
parity check matrix H from a generator matrix G in standard form and
vice versa.
Example of generator and parity check matrices in standard form:
1 0 0 1 1 0 0


G   0 1 0 0 1 1 0
0 0 1 0 0 1 1


1

1
H 
0

0

0
1
1
0
0
0
1
1
1
0
0
0
0
1
0
0
0
0
1
0
0

0
0

1 