Isometries of the Hamming Space and Equivalence Relations of Linear Codes Over a Finite Field

Abstract

Detection and error capabilities are preserved when applying to a linear code an isomorphism which preserves Hamming distance. We study here two such isomorphisms: permutation isometries and monomial isometries.

Appendix: Minimal Annihilating Polynomial of Permutation and Monomial Isometries

The minimal annihilating polynomial of a permutation isometry can be determined by the decomposition of the permutation in disjoint cycles. More concretely, if P ₁, P ₂ are two permutation matrices associated to two permutations $\sigma _{2}$, $\sigma _{2}$ with the same cycle type (conjugate in the symmetric group), the minimal annihilating polynomials $M_{P_{1}}(t)$ and $M_{P_{2}}(t)$ coincide.

Proposition 1

Let P be a permutation matrix associated to a permutation with cycle type $1+\mathop{\ldots }\limits^{\mathop{ \smile }\limits^{ m_{1}}} + 1 + 2+\mathop{\ldots }\limits^{\mathop{ \smile }\limits^{ m_{2}}} + 2 + 3+\mathop{\ldots }\limits^{\mathop{ \smile }\limits^{ m_{3}}} + 3 +\ldots +r+\mathop{\ldots }\limits^{\mathop{ \smile }\limits^{ m_{r}}} + r$ . Then

$$\displaystyle{M_{P}(t) = MCM\{(t - 1)^{n_{1} },(t^{2} - 1)^{n_{2} },\ldots,(t^{3} - 1)^{n_{3} },\ldots,(t^{r} - 1)^{n_{r} }\}}$$

where n _i = 0 if m _i = 0 and n _i = 1 if m _i > 0, 1 ≤ i ≤ r (see [2]).

Example 11

Let us consider P = P(2, 3, 4, 1, 6, 5). Then 234165 = (2, 3, 4, 1)(6, 5) has cycle type 2 + 4 and

$$\displaystyle{M_{P}(t) = MCM\{(t^{2} - 1)^{6},(t^{4} - 1)\} = t^{2} - 1.}$$

For any monomial matrix, $M = M(a_{1},\ldots,a_{n};i_{1}\ldots i_{n})$, the characteristic polynomial of the isomorphism g _M can be obtained from the coefficients $a_{1},\ldots,a_{n}$ and the cycle type of the permutation $i_{1}\ldots i_{n}$.

Lemma 4

• (a) The characteristic polynomial of M is a product of factors, each of them corresponding to one of the disjoint cycles in the decomposition of $i_{1}\ldots i_{k}$ .

• (b) Given a matrix $M = M(a_{1},\ldots,a_{k};j_{1},\ldots,j_{k})$ with $j_{1},\ldots,j_{k} = (\,j_{1},\ldots,j_{k})$ a cycle of length k, the characteristic polynomial of M is $t^{k} - a_{1}\cdots a_{k}$ .

Example 12

$$\displaystyle{M = M(1,1,2,2,2,2,2;2,3,1,5,6,7,4) = \left (\begin{array}{ccccccc} 0 & 1 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 2 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 2 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 2 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 2 \\ 0 & 0 & 0 & 2 & 0 & 0 & 0 \end{array} \right )}$$

2 3 1 5 6 7 4 = (2, 3, 1)(5, 6, 7, 4) and then $Q_{M}(t) = (t^{3} - 2)(t^{4} - 16)$.

• In $\mathbb{F}_{5}[t]$: $Q_{M}(t) = (t^{3} - 2)(t^{4} - 1)$.

• In $\mathbb{F}_{7}[t]$: $Q_{M}(t) = (t^{3} - 2)(t^{4} - 2)$.

• In $\mathbb{F}_{11}[t]$: $Q_{M}(t) = (t^{3} - 2)(t^{4} - 5)$.

Let us denote by G _k (t) the GCD of all factors in the characteristic polynomial of $M = M(a_{1},\ldots,a_{n};i_{1},\ldots,i_{n})$ of degree k corresponding to cycles of length k, 1 ≤ k ≤ n.

Proposition 2

The minimal annihilating polynomial of monomial matrix $M = M(a_{1},\ldots,a_{n};i_{1},\ldots,i_{n})$ is:

$$\displaystyle{P_{M}(t) = LCM(G_{1}(t),\ldots,G_{k}(t)).}$$

Example 13

Let us consider

• $\begin{array}{l} M = M(2,3,1,1,3,4,1,1,4;2,3,1,5,4,7,6,8,9) \\ \quad = \left (\begin{array}{ccccccccc} 0&2&0&0&0&0&0&0&0\\ 0 &0 &3 &0 &0 &0 &0 &0 &0 \\ 1&0&0&0&0&0&0&0&0\\ 0 &0 &0 &0 &1 &0 &0 &0 &0 \\ 0&0&0&3&0&0&0&0&0\\ 0 &0 &0 &0 &0 &0 &1 &0 &0 \\ 0&0&0&0&0&1&0&0&0\\ 0 &0 &0 &0 &0 &0 &0 &1 &0 \\ 0&0&0&0&0&0&0&0&4\end{array} \right ) \in M_{9}(\mathbb{F}_{7}) \end{array}$

• 2 3 1 5 4 7 6 8 9 = (2, 3, 1)(5, 4)(7, 6)(8)(9).

• $Q_{M}(t) = (t^{3} - 6)(t^{2} - 3)(t^{2} - 1)(t - 1)(t - 4)$.

• $G_{1}(t) = (t - 1)(t - 4)$, $G_{2}(t) = (t^{2} - 3)(t^{2} - 1)$, $G_{3}(t) = t^{3} - 6$.

• $P_{M}(t) = (t - 1)(t - 3)(t - 4)(t - 5)(t - 6)(t^{2} - 3)$.

An example of invariant subspaces are those spanned by eigenvectors. We can determine the set of eigenvectors of a permutation isomorphism from the decomposition of the permutation associated to it, into disjoint cycles and that of a monomial isomorphism from the decomposition into disjoint cycles and the coefficients in the matrix. The proofs are based on straightforward computations.

Proposition 3

Let P be a permutation matrix associated to a permutation which is a disjoint product of cycles, and $\lambda \in \mathbb{F}_{p}$ an eigenvalue of P, $\lambda$ an mth-root of unity, the vector $(\lambda ^{j_{k}},\ldots,\lambda ^{j_{1}})$ , where $j_{1},\ldots,j_{k}$ is the result of re-ordering the indices of the cycle $(i_{1},\ldots,i_{k})$ in such a way that $j_{1} \leq \ldots \leq j_{k}$ . Then $(\lambda _{j_{k}},\ldots,\lambda _{j_{1}})$ is the eigenvector associated to the eigenvalue $\lambda$ .

Example 14

We will consider p = 5.

• 1. Let us consider the 2-cycle (2, 1) and the 2 × 2-matrix associated to it. Then the eigenvector for the eigenvalue $\lambda \in \mathbb{F}_{5}$, $\lambda ^{2} = 1$, is $(\lambda,1)$. Since the roots of $\lambda ^{2} = 1$ are 1 and 4, there are two linearly independent eigenvectors: (1, 1) and (4, 1).

• 2. If the permutation 3 × 3-matrix is associated to the 2-cycle (2, 3, 1), the eigenvector corresponding to the eigenvalue $\lambda \in \mathbb{F}_{5}$, $\lambda ^{3} = 1$, is $(\lambda,\lambda ^{2},1)$. The equation $\lambda ^{3} = 1$ has only one root, 1, and therefore there is an unique eigenvector is: (1, 1, 1).

  If we consider the 3 × 3-permutation matrix is associated to the 2-cycle (3, 2, 1), there is also an unique eigenvector: (1, 1, 1).

• 3. Let us consider now the case of 4 × 4-permutation matrices associated to 4-cycles. Let $\lambda$ be a 4th-root of unity (there are four 4th-roots of unity: $\lambda _{1} = 1$, $\lambda _{2} = 2$,$\lambda _{3} = 3$ and $\lambda _{4} = 4$).

  • (i) If the 4-cycle is (1, 2, 3, 4), the eigenvector corresponding to the eigenvalue $\lambda \in \mathbb{F}_{5}$, is $(\lambda,\lambda ^{2},\lambda ^{3},1)$. That is to say, there are four linearly independent eigenvectors:

    $$\displaystyle{(1,1,1,1),(2,4,3,1),(3,4,2,1),(4,1,4,1).}$$

  • (ii) If the 4-cycle is (1, 3, 4, 2), the eigenvector corresponding to the eigenvalue $\lambda \in \mathbb{F}_{5}$, $\lambda ^{4} = 1$, is $(\lambda,\lambda ^{3},\lambda ^{2},1)$. That is to say, there are four linearly independent eigenvectors:

    $$\displaystyle{(1,1,1,1),(2,3,4,1),(3,2,4,1),(4,4,1,1).}$$

  • (iii) If the 4-cycle is (1, 2, 3, 4), the eigenvector corresponding to the eigenvalue $\lambda \in \mathbb{F}_{5}$, $\lambda ^{4} = 1$, is $(\lambda,1,\lambda ^{3},\lambda ^{2})$. That is to say, there are four linearly independent eigenvectors:

    $$\displaystyle{(1,1,1,1),(2,1,3,4),(3,1,2,4),(4,1,4,1).}$$

  • (iv) If the 4-cycle is (1, 3, 2, 4), the eigenvector corresponding to the eigenvalue $\lambda \in \mathbb{F}_{5}$, $\lambda ^{4} = 1$, is $(\lambda,\lambda ^{2},1,\lambda ^{3}$. That is to say, there are four linearly independent eigenvectors:

    $$\displaystyle{(1,1,1,1),(2,4,1,3),(3,4,1,2),(4,1,1,4).}$$

  • (v) If the 4-cycle is (1, 4, 2, 3), the eigenvector corresponding to the eigenvalue $\lambda \in \mathbb{F}_{5}$, $\lambda ^{4} = 1$, is $(\lambda,1,\lambda ^{2},\lambda ^{3})$. That is to say, there are four linearly independent eigenvectors:

    $$\displaystyle{(1,1,1,1),(2,1,4,3),(3,1,4,2),(4,1,1,4).}$$

  • (vi) If the 4-cycle is (1, 2, 4, 3), the eigenvector corresponding to the eigenvalue $\lambda \in \mathbb{F}_{5}$, $\lambda ^{4} = 1$, is $(\lambda,\lambda ^{3},1,\lambda ^{2})$. That is to say, there are four linearly independent eigenvectors:

    $$\displaystyle{(1,1,1,1),(2,3,1,4),(3,2,1,4),(4,4,1,1).}$$

• 4. Let us consider the permutation matrix associated to a cycle of type $2 + 2 + 4 + 8$. Then the minimal annihilating polynomial is: $(t^{4} + 1)(t^{2} + 1)(t + 1)(t - 1)$. The eigenvalues in $\mathbb{F}_{5}$ are: $\lambda _{1} = 1$, $\lambda _{2} = 4$, $\lambda _{3} = 2$ and $\lambda _{4} = 3$, being the algebraic multiplicities 4,4,1,1, respectively.

  Let us assume, for example, that the 2-cycles are: (9, 16) and (13, 15), the 4-cycle is (1, 3, 5, 14) and the 8-cycle is (2, 4, 6, 7, 8, 10, 11, 12). Then the following linearly independent eigenvectors are obtained.

  $$\displaystyle{\begin{array}{cc} \lambda _{1} = 1&(0,0,0,0,0,0,0,0,0,0,0,0,1,0,1,0) \\ &(0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,1) \\ &(1,0,1,0,1,0,0,0,0,0,0,0,0,1,0,0) \\ &(0,1,0,1,0,1,1,1,0,1,1,1,0,0,0,0) \\ \lambda _{2} = 4&(0,0,0,0,0,0,0,0,0,0,0,0,4,0,1,0) \\ &(0,0,0,0,0,0,0,0,4,0,0,0,0,0,0,1) \\ &(4,0,1,0,4,0,0,0,0,0,0,0,0,1,0,0) \\ &(0,4,0,1,0,4,1,4,0,1,4,1,0,0,0,0) \\ \lambda _{3} = 2&(2,0,4,0,3,0,0,0,0,0,0,0,0,1,0,0) \\ \lambda _{4} = 3&(3,0,4,0,2,0,0,0,0,0,0,0,0,1,0,0)\end{array} }$$

Let us consider now a monomial matrix Let $M = M(a_{1},\ldots,a_{n};i_{1},\ldots,i_{n})$.

Proposition 4

The eigenvalues of M are the roots of the polynomials $t^{k} - a_{j_{1}}\cdots a_{j_{k}}$ for each cycle $j_{1}\ldots j_{k}$ of length k in the decomposition of the permutation $i_{1}\ldots i_{n}$ into disjoint cycles, being $a_{j_{1}},\ldots,a_{j_{k}}$ the coefficients of M in columns $j_{1},\ldots,j_{k}$ .

Example 15

Let us consider

$$\displaystyle{M = M(1,1,2,2,2,2,2;2,3,1,5,6,7,4) = \left (\begin{array}{ccccccc} 0 & 1 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 2 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 2 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 2 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 2 \\ 0 & 0 & 0 & 2 & 0 & 0 & 0 \end{array} \right )}$$

We have: 2 3 1 5 6 7 4 = (2, 3, 1)(5, 6, 7, 4) and therefore $Q_{M}(t) = (t^{3} - 2)(t^{4} - 16)$.

• In $\mathbb{F}_{5}[t]$: $Q_{M}(t) = (t^{3} - 2)(t^{4} - 1)$ and the eigenvalues of M are: 3(2), 1, 2, 4.

• In $\mathbb{F}_{7}[t]$: $Q_{M}(t) = (t^{3} - 2)(t^{4} - 2)$ and the eigenvalues of M are: 2, 5.

• In $\mathbb{F}_{11}[t]$: $Q_{M}(t) = (t^{3} - 2)(t^{4} - 5)$ and the eigenvalues of M are: 7, 2, 9.

Assume that the permutation $i_{1},\ldots,i_{n}$ splits into m ₁ cycles of length k ₁,…, m _l cycles of length k _l . For any irreducible cycle ( j) of length 1, e _j is an eigenvector. We can generalize this as follows.

Proposition 5

Let $a_{j_{1}}\ldots a_{j_{k}}$ be a cycle of length k ≥ 2 in the decomposition of the characteristic polynomial of M into irreducible factors and $t^{k} - a_{j_{1}}\ldots a_{j_{k}}$ the corresponding factor in Q _M (t). For each root $\lambda$ (in the case where there exists any) of this polynomial we obtain an eigenvector:

$$\displaystyle{(\lambda ^{k-1},a_{ j_{2}}a_{j_{3}}\ldots a_{j_{k}},\lambda a_{j_{3}}\ldots a_{j_{k}},\ldots \lambda ^{k-3}a_{ j_{k}-1}a_{j_{k}},\lambda ^{k-2}a_{ j_{k}})}$$

Example 16

$\begin{array}{rl} M & = M(a_{1},a_{2},a_{3},a_{4},a_{5},a_{6},a_{7},a_{8},a_{9};2,3,4,1,6,7,8,9,5) \\ & = \left (\begin{array}{ccccccccc} 0 &a_{1}& 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 &a_{2}& 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 &a_{3}& 0 & 0 & 0 & 0 & 0 \\ a_{4}& 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 &a_{5}& 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 &a_{6}& 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 &a_{7}& 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 &a_{8} \\ 0 & 0 & 0 & 0 &a_{9}& 0 & 0 & 0 & 0 \end{array} \right ) \end{array}$

the eigenvectors are:

• $(\lambda ^{3},a_{2}a_{3}a_{4},\lambda a_{3}a_{4},\lambda ^{2}a_{4},0,0,0,0,0)$.

• $(\mu ^{4},a_{6}a_{7}a_{8}a_{9},\mu a_{7}a_{8}a_{9},\mu ^{2}a_{8}a_{9},\mu ^{3}a_{9})$.

For each root $\lambda$ of $t^{4} - a_{1}a_{2}a_{3}a_{4}$ and μ of $t^{5} - a_{5}a_{6}a_{7}a_{8}a_{9}$ in $\mathbb{F}_{p}$.