positive sign negative spin
Democrat > Republican > Third Third > Republican > Democrat
Republican > Third > Democrat Democrat > Third > Republican
Linear Algebra/Topic: Voting Paradoxes 163
Third > Democrat > Republican Republican > Democrat > Third
If we conduct the election as just described then after the cancellation of as many opposite pairs of voters as possible, there will be left three sets of preference lists, one set from the first row, one set from the second row, and one set from the third row. We will finish by proving that a voting paradox can happen only if the spins of these three sets are in the same direction. That is, for a voting paradox to occur, the three remaining sets must all come from the left of the table or all come from the right (see Problem 3). This shows that there is some connection between the majority cycle and the decomposition that we are using---a voting paradox can happen only when the tendencies toward cyclic preference reinforce each other.
For the proof, assume that opposite preference orders have been cancelled, and we are left with one set of preference lists from each of the three rows. Consider the sum of these three (here, the numbers , , and could be positive, negative, or zero).
A voting paradox occurs when the three numbers on the right, and and , are all nonnegative or all nonpositive. On the left, at least two of the three numbers, and and , are both nonnegative or both nonpositive. We can assume that they are and . That makes four cases: the cycle is nonnegative and and are nonnegative, the cycle is nonpositive and and are nonpositive, etc. We will do only the first case, since the second is similar and the other two are also easy.
So assume that the cycle is nonnegative and that and are nonnegative. The conditions and add to give that , which implies that is also nonnegative, as desired. That ends the proof.
This result says only that having all three spin in the same direction is a necessary condition for a majority cycle. It is not sufficient; see Problem 4.
Voting theory and associated topics are the subject of current research. There are many intriguing results, most notably the one produced by K. Arrow (Arrow 1963), who won the Nobel Prize in part for this work, showing that no voting system is entirely fair (for a reasonable definition of "fair"). For more information, some good introductory articles are (Gardner 1970), (Gardner 1974), (Gardner 1980), and (Neimi Riker 1976). A quite readable recent book is (Taylor 1995). The long list of cases from recent American political history given in (Poundstone 2008) show that manipulation of these paradoxes is routine in practice (and the author proposes a solution).
This Topic is largely drawn from (Zwicker 1991). (Author's Note: I would like to thank Professor Zwicker for his kind and illuminating discussions.)
the Republican preferred to the Third?
3.
3. Does the table that was just constructed have a cyclic preference order? If not, make one that does.
So it is possible for a voter to have a cyclic preference among candidates. The paradox described above, however, is that even if each voter has a straight-line preference list, a cyclic preference can still arise for the entire group.
Problem 2
Compute the values in the table of decompositions.
Problem 3
Do the cancellations of opposite preference orders for the Political Science class's mock election. Are all the remaining preferences from the left three rows of the table or from the right?
Problem 4
The necessary condition that is proved above—a voting paradox can happen only if all three preference lists remaining after cancellation have the same spin—is not also sufficient.
1. Continuing the positive cycle case considered in the proof, use the two inequalities and
to show that .
2. Also show that , and hence that .
3.
3. Give an example of a vote where there is a majority cycle, and addition of one more voter with the same spin causes the cycle to go away.
4.
4. Can the opposite happen; can addition of one voter with a "wrong" spin cause a cycle to appear?
5.
5. Give a condition that is both necessary and sufficient to get a majority cycle.
Problem 5
A one-voter election cannot have a majority cycle because of the requirement that we've imposed that the voter's list must be rational.
1.
1. Show that a two-voter election may have a majority cycle. (We consider the group preference a majority cycle if all three group totals are nonnegative or if all three are nonpositive---that is, we allow some zero's in the group preference.)
2.
2. Show that for any number of voters greater than one, there is an election involving that many voters that results in a majority cycle.
Problem 6
Let be a subspace of . Prove that the set of vectors that are
perpendicular to each vector in is also a subspace of . Solutions
Linear Algebra/Topic: Voting Paradoxes 165
References
• Arrow, J. (1963), Social Choice and Individual Values, Wiley.
• Gardner, Martin (April 1970), "Mathematical Games, Some mathematical curiosities embedded in the solar system", Scientific American: 108-112.
• Gardner, Martin (October 1974), "Mathematical Games, On the paradoxical situations that arise from nontransitive relations", Scientific American.
• Gardner, Martin (October 1980), "Mathematical Games, From counting votes to making votes count: the mathematics of elections", Scientific American.
• Neimi, G.; Riker, W. (June 1976), "The Choice of Voting Systems", Scientific American: 21-27.
• Poundstone, W. (2008), Gaming the vote, Hill and Wang, ISBN 978-0-8090-4893-9.
• Taylor, Alan D. (1995), Mathematics and Politics: Strategy, Voting, Power, and Proof, Springer-Verlag.
• Zwicker, S. (1991), "The Voters' Paradox, Spin, and the Borda Count", Mathematical Social Sciences 22:
187-227