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Explicit and implicit description
There are two standard ways to describe a subspace:
Explicitly by giving a basis orImplicitly by the solution spaceof a set of homogeneous linear equations
hi1X1 + · · · + hinXn = 0 for i = 1, . . . ,m
Therefore there are two ways to describe a linear codeExplicitly by a generator matrixImplicitly by the null space of a matrixH with entries hij
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Parity check matrix
Let C be an Fq -linear [n, k ] codeSuppose that H is an m × n matrix with entries in Fq
Let C be the null space of HSo C is the set of all c ∈ Fn
q such that HcT = 0These are called parity check equations or parity checks
The dimension k of C is at least n −mIf there are dependent rows in the matrix H , that is if k > n −mthen we can delete a few rows until we obtain an (n − k )× n matrix H ′
with independent rows and with the same null space as HSo H ′ has rank n − kSuch a matrix is called a parity check matrix of C
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Error detection
The parity check matrix of a code can be used for error detectionUseful in a communication channel where one asks for retransmission
C is a linear code of minimum distance d with parity check matrix HLet c be the codeword transmitted and r = c+ e is receivedThen e is the error vector and wt(e) the number of errors
H rT = 0 if there is no error andH rT 6= 0 for all e such that 0 < wt(e) < d
One can detect any pattern of t errors with t < d , but not moreIf the error vector is equal to a nonzero codeword of minimal weight dthen the receiver would assume that no errors have been madeH rT is called the syndromeof the received word r
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Proposition
Suppose C is an [n, k ] code
Let Ik be the k × k identity matrixLet P be a k × (n − k )matrix
Then(Ik |P ) is a generator matrix of C
if and only if
(−P T|In−k ) is a parity check matrix of C
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Proof
Every codeword c is of the form form c = (m, r) withm ∈ Fk
q and r = mP ∈ Fn−kq
The following statements are equivalent:
c is a codeword
−mP + r = 0
−P TmT+ rT = 0(
−P T|In−k
)(m, r)T = 0(
−P T|In−k
)cT = 0
Hence(−P T|In−k
)is a parity check matrix of C
The converse is proved similarly.
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Example 1 - trivial codes
The linear codes with parameters [n,0, n + 1] and [n, n,1]are the trivial codes {0} and Fn
q
Have the n × n identity matrix In andthe empty matrixas parity check matrix, respectively
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Example 2 - binary even weight code
The generator matrix of the binary even weight code
G =
1 0 . . . 0 0 10 1 . . . 0 0 1...
.... . .
......
...
0 0 . . . 1 0 10 0 . . . 0 1 1
Gives the parity check matrix
H = ( 1 1 · · · 1| 1 )
which is the generator matrix of the repetition code
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Example 3 - ISBN code
The ISBN code over the finite field Z/11Zthat satisfies the parity check equation:
b1 + 2b2 + 3b3 + · · · + 10b10 ≡ 0 mod 11
Hence has parity check matrix
H = ( 1 2 3 4 5 6 7 8 9 10 )
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Example 4 - Hamming code
The generator matrix G of the Hamming code
G =
1 0 0 0 0 1 10 1 0 0 1 0 10 0 1 0 1 1 00 0 0 1 1 1 1
has parity check matrix
H =
0 1 1 1 1 0 01 0 1 1 0 1 01 1 0 1 0 0 1
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Example 5
Consider the code C with generator matrix
G =
1 0 1 0 1 0 1 00 1 1 0 0 1 1 00 0 0 1 1 1 1 00 0 0 0 0 0 0 1
Has parity check matrix
H =
1 1 1 0 0 0 0 01 0 0 1 1 0 0 00 1 0 1 0 1 0 01 1 0 1 0 0 1 0
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Proposition
Let H be a parity check matrix of a code C
Then the minimum distance d of Cis the smallest integer d such thatd columns of H are linearly dependent
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Proof
Let h1, . . . ,hn be the columns of HLet c be a nonzero codeword of weight wLet supp(c) = {j1, . . . , jw} with 1 ≤ j1 < · · · < jw ≤ nThen HcT = 0, so
cj1 hj1 + · · · + cjw hjw = 0
with cji 6= 0 for all i = 1, . . . ,wTherefore the columns hj1, . . . ,hjw are dependent
Converse is proved similarly
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Cases d = 1, d = 2 and d ≥ 3
The minimum distance of code is 1 if and only if H has a zero column
Now suppose that H has no zero columnthen the minimum distance of C is at least 2
The minimum distance is equal to 2 if and only if H has two columnssay hj1,hj2 that are dependent
In the binary case that means hj1 = hj2
In other words the minimum distance of a binary code is at least 3if and only ifH has no zero columns and all columns are mutually distinctThis is the case for the [7,4,3] Hamming code
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Maximal length for d = 3
For a given redundancy r the length of a binary linear code Cof minimum distance 3 is at most 2r
− 1the number of all nonzero binary columns of length r
For arbitrary Fq
the number of nonzero columns with entries in Fq is q r− 1
Two such columns are dependentif and only ifone is a nonzero multiple of the other
Hence the length of an Fq -linear code code C withd (C ) ≥ 3 and redundancy r is at most
(q r− 1)/(q − 1)
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Projective space
A point of the projective space Pr(F)over a field F of dimension ris a line through the origin in Fr+1
A line in Pr(F) is a plane through the origin in Fr+1
More generally a projective subspace of dimension s in Pr(F)is a linear subspace of dimension s + 1 of the vector space Fr+1
and r − s is called the codimension of the subspace
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Homogeneous coordinates
A point in Pr(F) is denoted by its homogeneous coordinates
(x0 : x1 : · · · : xr)
with x0, x1, . . . , xr ∈ F and not all zero, where λ(x0, x1, . . . , xr), λ ∈ Fis a parametrization of the corresponding line in Fr+1
(x0 : x1 : · · · : xr) and (y0 : y1 : · · · : yr) represent the same pointif and only if(x0, x1, . . . , xr) = λ(y0, y1, . . . , yr) for some λ ∈ F∗
The standard homogeneous coordinates of (x0 : x1 : · · · : xr)is such that there exists a j with xj = 1 and xi = 0 for all i < j
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Projective space over a finite field
Pr(Fq) =(Fr+1q \ {0}
)/F∗q
The number of elements of Pr(Fq) is
q r+1− 1
q − 1= q r
+ q r−1+ · · · + q + 1
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Hamming and simplex code
Let n = (q r− 1)/(q − 1)
Let Hr(q) be a r × n matrix over Fq with nonzero columnssuch that no two columns are dependent
The code H r(q) with Hr(q) as parity check matrixis called a q-ary Hamming code
The code with Hr(q) as generator matrixis called a q-ary simplex code and is denoted by Sr(q)
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Parameters Hamming code
The q-ary Hamming code Hr(q) has parameters[q r− 1
q − 1,q r− 1
q − 1− r,3
]
PROOF:The rank of the matrix Hr(q) is rsince the r standard basis vectors of weight 1are among the columns of the matrixAny 2 columns are independent by constructionAnd a column of weight 2is a linear combination of two columns of weight 1
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Ternary Hamming code with r = 3
The ternary Hamming H3(3) code of redundancy 3has length 32
+ 3+ 1 = 13 and parity check matrix
H3(3) =
1 1 1 1 1 1 1 1 1 0 0 0 02 2 2 1 1 1 0 0 0 1 1 1 02 1 0 2 1 0 2 1 0 2 1 0 1
All rows have weight 9In fact all nonzero codewords of the simplex code S3(3) have weight 9
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Parameters Simplex code
The q-ary Simplex code Sr(q) has parameters[q r− 1
q − 1, r, q r−1
]
Every nonzero codeword of Sr(q) has weight q r−1
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Proof
Hr(q) has rank r
Let c be a nonzero codeword of the simplex codeThen c = mHr(q) for some nonzero m ∈ Fr
q
Let hTj be the j -th column of Hr(q)
Then cj = 0 if and only if m · hj = 0Now m · x = 0 is a nontrivial homogeneous linear equationThis equation has q r−1 solutions x ∈ Fr
qit has q r−1
− 1 nonzero solutionsIt has (q r−1
− 1)/(q − 1) solutions x such that xT is a column of Hr(q)since for every nonzero x ∈ Fr
q there is exactly one column in Hr(q)that is a nonzero multiple of xT
So the number of zeros of c is (q r−1− 1)/(q − 1)
Hence the weight of c is the number of nonzeros which is q r−1
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Inner product
The standard inner product is defined by
a · b = a1b1 + · · · + anbn
Is bilinear and non-degeneratebut "positive definite"makes no sense
Two subsets A and B of Fnq are perpendicular:
A ⊥ B if and only if a · b = 0 for all a ∈ A and b ∈ B
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Dual code
Let C be a linear code in Fnq
The dual code is defined by
C⊥ = { x | x · c = 0 for all c ∈ C }
PROPOSITIONLet C be an [n, k ] code with generator matrix GThen C⊥ is an [n, n − k ] code with parity check matrix G
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Dimension dual code
PROOF:The following statements are equivalent:
x ∈ C⊥
c · x = 0 for all c ∈ C
mGxT = 0 for all m ∈ Fkq
GxT = 0
This means that C⊥ is the null space of GG is a k × n matrix of rank kHence C⊥ has dimension n − kand G is a parity check matrix of C⊥
