Algebraic structures for pairwise comparison matrices: Consistency, social choices and Arrow’s theorem

Definition 1

A partially ordered vector space G = (G, +, ·, ≤) is a real vector space with an order relation ≤ that is compatible with the algebraic structure of G, that is

1. x ≤ y implies x + z ≤ y + z for each x, y, z ∈ G;

2. x ≤ y implies α x ≤ α y for every x, y ∈ G and α ≥ 0.

Definition 2

A partially ordered vector space G = (G, +, ·, ≤) is a Riesz space (or vector lattice) if the partial order is a lattice order, i.e., every two elements have a unique supremum and a unique infimum.

Examples 1

The Euclidean space ℝⁿ is a Riesz space under the usual ordering, where

whenever x_i ≤ y_i for each i = 1, 2, ⋅, n.

The supremum and infimum of two vectors x and y are given by

and

respectively.

Alo-groups, presented in [11], are examples of Riesz spaces. An Alo-group is a totally ordered lattice group, and hence also an ℓ-group, and by Freudenthal’s theorem (see [20: Theorem 40.2]) every ℓ-group can be embedded into a Riesz space. Let us recall that any archimedean abelian linearly ordered group is isomorphic to a subgroup of ℝ, as Hölder proved. By this, the results contained in this paper generalize the ones contained in [11], i.e., we generalize the additive, the multiplicative and the fuzzy approach (see [5, 26, 23], respectively).

Let G be a Riesz space. For every n ∈ ℕ, Gⁿ is a Riesz space where the ordering is defined coordinate-wise. In particular, the set of square matrices of order n with entries in a Riesz space is a Riesz space, being isomorphic to Gn2.

Both the vector space C(X) of all continuous real functions and the vector space C_b(X) of all bounded continuous real functions on the topological space X are Riesz spaces, when the ordering is defined pointwise.

The space of piecewise linear functions on an interval of the real line, with the usual pointwise ordering, is a Riesz space.

The vector space L_p(μ) (0 ≤ p ≤ ∞) is a Riesz space under the almost everywhere pointwise ordering, i.e., f ≤ g in L_p(μ) if f(x) ≤ g(x) μ-almost everywhere.

The vector spaces ℓ_p (0 < p ≤ ∞) are Riesz spaces under the usual pointwise ordering.

Definition 3

A Riesz space is said to be order complete (or Dedekind complete) if every nonempty subset that is order bounded from above has a supremum, or equivalently if every nonempty subset that is order bounded from below has an infimum.

Definition 4

A vector space X is the direct sum of two subspaces Y and Z if every x ∈ X has a unique decomposition of the form x = y + z, where y ∈ Y and z ∈ Z.

Remark 1

Observe that the sum of any two consistent matrices A = (a_i,j)_i,j and B = (b_i,j)_i,j is still consistent. Indeed, if A and B satisfy condition (3.2), then for every i,j,k ∈ {1, 2, …, n}, we have

getting the consistency of A ⊕ B. Analogously it is possible to see that, if G is a vector space over a field K,A=(ai,j)i,j is consistent and α∈K, then α ⊙ A = (αa_i,j)_i,j is also consistent.

Moreover, if A = (a_i,j)_i,j and B = (b_i,j)_i,j are totally inconsistent, then

for each i ∈ {1, 2, …, n}. Hence, the sum of any two totally inconsistent matrices is still totally inconsistent. Similarly, if G is a vector space over K and α∈K, then, from the equality

we deduce that α ⊙ A is totally inconsistent whenever A is totally inconsistent and α∈K.

Therefore, the sets of all consistent matrices and of all totally inconsistent matrices are two subgroups of Mn, and two subspaces of Mn when Mn is a vector space of K.

Proposition 3.1

Let A = (a_i,j)_i,j. The following results hold.

1. Any two vectors v = ( v 1 , v 2 , … , v n ) , w = ( w 1 , w 2 , … , w n ) , coherent for A, differ by a constant c ∈ G, that is w_i − v_i = c for everyi ∈ {1, 2, …, n}.

2. If v = (v₁, v₂, …, v_n) is a coherent vector forA, thenAis consistent.

3. If A is consistent, then each column vector a ( h ) = a 1 , h a 2 , h … a n , h , h ∈ { 1 , 2 , … , n } , is coherent forA.

4. A matrix A is consistent if and only if there is at least a coherent vector for it.

5. A matrix A is consistent if and only if at least one of their column vectors is coherent for it.

6. If G is a real vector space and α 1 , α 2 , … , α n ∈ R , ∑ r = 1 n α r = 1 , then the vectorv = (v₁,v₂, …, v_n) of the affine combinationsvi=∑r=1nαrai,r,i∈{1,,…,n}, i ∈ {1, 2, …, n}, is coherent forA. Moreover, if G is a vector space over the field ℚ of the rational numbers, then the vectorw = (w₁,w₂, …, w_n) of the meanswi=1n∑r=1nai,r,i∈{1,2,…,n}is coherent for A.

Proof. 3.1.1)

Let v = (v₁,v₂, …, v_n), w = (w₁,w₂, …, w_n) be such that v_i − v_j = w_i − w_j = a_i,j for each i,j ∈ {1, 2, …, n}. Then

If, we denote by c the common value in (3.3), then, we get w_i − v_i = c for any i ∈ {1, 2, …, n}. This proves 3.1.1).

3.1.2) Let v = (v₁,v₂, …, v_n) be as in the hypothesis. For every i,j,k ∈ {1, 2, …, n}, it is

Thus, if a_i,j = v_i − v_j, i,j ∈ {1, 2, …, n}, then from (3.4), we deduce that the matrix A = (a_i,j) is consistent. So, 3.1.2) is proved.

3.1.3) Fix arbitrarily h ∈ {1, 2, …, n}. Since A is consistent, for every i,j ∈ {1, 2, …, n}, we get a_i,h = a_i,j + a_j,h, and hence a_i,h − a_j,h = a_i,j. Thus, 3.1.3) is proved.

3.1.4) and 3.1.5) follow from 3.1.2) and 3.1.3).

3.1.6) Let α₁, α₂, …, α_n be as in the hypothesis. As A is consistent, we get

getting the consistency of v.

The proof of the last assertion is analogous to that of the previous one, by replacing α_r with 1n for each r ∈ {1, 2, …, n}.

Example 1

Given A = (a_i,j)_i,j, for every i,j,k ∈ {1, 2, ⋅, n}, set

and for each i,j ∈ {1, 2, …, n} put

Let E(A)=(ei,j(A))i,j. We prove that E^(A) is skew-symmetric.

Since A is skew-symmetric, for any i,j ∈ {1, 2, …, n} it is

Thus, E^(A) is skew-symmetric.

Now, we prove that E^(A) is totally inconsistent, extending [9: Proposition 11] to the context of arbitrary abelian groups. Choose arbitrarily i ∈ {1, 2, …, n}. Thanks to the skew-symmetry of A and taking into account (3.1), for each i ∈ {1, 2, …, n}, we have

getting the total inconsistency of E^(A).

Lemma 3.1

Let A ~ = ( n ⊙ A ) ⊖ E ( A ) = ( a i , j ~ ) i , j = ( n a i , j − e i , j ( A ) ) i , j , where E^(A) is as in (3.7). ThenA~is consistent.

Proof

First of all, we claim that A~ is skew-symmetric. Indeed, since A and E^(A) are skew-symmetric, for every i,j ∈ {1, 2, …, n} it is

Now, we prove that A~ is consistent. Choose arbitrarily i,j,k ∈ {1, 2, …, n}. Taking into account the skew-symmetry of A, we get

that is the consistency of A~.

Lemma 3.2

Let B = (b_i,j)_i,j, C = (c_i,j)_i,j, D = (d_i,j)_i,j, D = B ⊕ C, where B is totally inconsistent and C is consistent, and let E^(D) be as in (3.7). Then, E^(D) = n ⊙ B.

Proof

For every i,j,k ∈ {1, 2, …, n}, we have d_i,j = b_i,j + c_i,j, and since C is consistent, we obtain

From (3.10), taking into account the skew-symmetry and the total inconsistency of B, we deduce

that is the assertion.

Theorem 3.2

Let G be a vector space over the field ℚ andAbe a skew-symmetric matrix. Then there is a totally inconsistent matrixB₀and a consistent matrixC₀such thatA = B₀ ⊕ C₀.

Moreover, if B₁ is any totally inconsistent matrix and C₁ is any consistent matrix such that A = B₁ ⊕ C₁, then B₁ = B₀ and C₁ = C₀.

Proof

Let E^(A) be as in (3.7), B0=1n⊙E(A),

It is not difficult to check that B₀ is totally inconsistent, since E^(A) is, and that C₀ is consistent, since (n ⊙ A)⊖ E^(A) is.

Moreover, if B₁ and C₁ are as in the hypothesis, then, thanks to Lemma 3.2, we get E^(A) = n ⊙ B₁, and so B₀ = B₁. From this and (3.11), we deduce that C₁ = A ⊖ B₁ = A ⊖ B₀ = C₀. This ends the proof.

Definition 6

Let S be the set of all skew-symmetric n × n-matrices, A=(ai,j)i,j∈S, and v=(v1,v2,…,vn)∈Gn.

We say that v is an ordinal evaluation vector for A if the following implications hold for every i, j ∈ {1, 2, …, n}:

1. [a_{i, j} >0]⇒[v_i > v_j];

2. [a_{i, j}=0]⇒[v_i=v_j].

Remark 3

Observe that condition 6.1) is equivalent to

1. [a_{i, j} <0]⇒[v_i < v_j].

Indeed, suppose that a_{i, j} < 0. Then, by the skew-symmetry of A, we get a_{j, i}=−a_{i, j} >0. By 6.1), we have v_j > v_i, that is v_i < v_j. Thus, 6.1) implies 6.3). The proof of the converse implication is analogous.

Definition 7

A matrix A=(ai,j)i,j∈S is said to be weakly consistent if for every i, j ∈ {1, 2, …, n },

[a_{i, j} > 0] ⇒ [a_{i, k} > a_{j, k} for all k ∈ {1, 2, …, n }], and

[a_{i, j}=0] ⇒ [a_{i, k} = a_{j, k} for any k ∈ {1, 2, …, n }].

Proposition 4.1

1. If A is consistent, then A is weakly consistent.

2. If A is weakly consistent, then every column vector a ( h ) = a 1 , h a 2 , h ⋯ a n , h , h ∈ {1, 2, …, n}, is an ordinal evaluation vector for A.

3. If ϕ: Gⁿ → G is a strictly increasing function, then the vectorw=(w1,w2,…,wn)defined by

   w i = ϕ ( a i , 1 , a i , 2 , … , a i , n ) , i ∈ { 1 , 2 , … , n }

   is an ordinal evaluation vector for A.

Proof

4.1.1) If A is consistent, then for every i, j, k ∈ {1, 2, …, n} it is a_{i, j}+a_{j, k}=a_{i, k}, and hence a_{i, j}=a_{i, k} −a_{j, k}. Thus, if a_{i, j} >0 (resp. a_{i, j}=0), then a_{i, k} > a_{j, k} (resp. a_{i, k}=a_{j, k}). By the arbitrariness of k, we get that A is weakly consistent.

4.1.2) It is a direct consequence of the definitions of weak consistency and ordinal evaluation vector.

4.1.3) Choose arbitrarily i, j ∈ {1, 2, …, n}. By the definition of weak consistency, if a_{i, j} > 0, then a_{i, k} > a_{j, k} for each k ∈ {1, 2, …, n }. Since ϕ is strictly increasing, then

Analogously it is possible to check that, if a_{i, j}=0, then

getting the assertion.

Remark 4

Note that, in general, weak consistency does not imply consistency, and the sum of two weakly consistent matrices is not weakly consistent (see, e.g., [14: Example 4.1], [17: Remark 3]).

The next step is to formulate Arrow’s conditions in the partially ordered space setting, and extend earlier results of [15] and [17].

Let S be as in Definition 6, and ∅≠T⊂Sm. A profile is an element of T. A procedure on T for aggregating and/or synthesizing the preferences of a profile in one matrix is any function Φ:T0→S, where ∅≠T0⊂T.

For every (A1,A2,…,Am)∈Sm and (i, j) ∈ {1, 2, …, n }², set ai,j=(ai,j1,ai,j2,…,ai,jm).

Definition 8

We say that a procedure Φ on T satisfies the condition of unrestricted domain (in short, conditionU*) if T0=T.

A procedure Φ fulfils pairwise unanimity (condition P*) if for every profile (A1,A2,…,Am)∈T0, with As=(ai,js)i,j, s ∈ {1, 2, …, m}, we get that, if ai,js>0 for each s ∈ {1, 2, …, m}, then a~i,j>0, where a~i,j=(Φ(A1,A2,…,Am))i,j, (i, j) ∈ {1, 2, …, n }².

A procedure Φ satisfies the condition of independence from irrelevant alternatives (condition I*) if for each nonempty set Y ⊂ {1, 2, …, n } and for any two profiles (A1,A2,…,Am)=((ai,j1)i,j,(ai,j2)i,j,…,(ai,jm)i,j), (B1,B2,…,Bm)=((bi,j1)i,j,(bi,j2)i,j,…,(bi,jm)i,j), such that

it is (Φ(A₁, A₂, …, A_m))^(Y) = (Φ(B₁, B₂, …, B_m))^(Y).

A procedure Φ satisfies the condition of nondictatorship (condition D*) if there is no element d ∈ {1, 2, …, m} such that Φ(A₁, A₂, …, A_m)=A_d whenever A_i≠ A_j for at least two different i, j ∈ {1, 2, …, n}.

Proposition 4.2

Let T = T 0 = S m , φ: G^m → Gbe an averaging functional andΦ:Sm→Sbe a procedure defined, for each(A1,A2,…,Am)∈T, by

Then Φ satisfies U*, P* and I* on T. Moreover, if G is a partially ordered real vector space and φ is a convex combination as in (4.2), then Φ satisfies alsoD* onT.

Proof

U*) It is readily seen that condition U* is fulfilled, because Φ is defined on the whole on T.

P*) Let (A1,A2,…,Am)∈T, As=(ai,js), s ∈ {1, 2, …, m }, be such that

Since φ is strictly increasing, from (4.5), we obtain

for any i, j ∈ {1, 2, …, n }. Hence, condition P* is fulfilled.

I*) Let (A₁, A₂, …, A_m), (B₁, B₂, …, B_m) ∈ T be as in (4.3), namely such that

for each s ∈ {1, 2, …, m}. This means that

From (4.7) it follows that

Thus, I* is satisfied.

Definition 9

A procedure Ψ on T satisfies the condition of unrestricted domain (in short, condition U**) if T0=T.

A procedure Ψ on T fulfils pairwise unanimity (condition P**) if for every profile (A₁, A₂, …, A_m) ∈T0, with As=(ai,js)i,j, s ∈ {1, 2, …, m}, we get that, if i, j ∈ {1, 2, …, n} are such that ai,js>0 for every s ∈ {1, 2, …, m}, then (Ψ(A₁, A₂, …, A_m))_i > (Ψ(A₁, A₂, …, A_m))_j.