Linear Algebra

144 Chapter Two. Vector SpacesThe instructor can make any of the other two candidates come out as the winnerby similar manipulations. (Here we will stick to three-candidate elections butthe same thing happens in larger elections.)Mathematicians also study voting paradoxes simply because they are interesting.One interesting aspect is that the group’s overall majority cycle occursdespite that each single voter’s preference list is rational, in a straight-line order.That is, the majority cycle seems to arise in the aggregate without being presentin the components of that aggregate, the preference lists. However we can uselinear algebra to argue that a tendency toward cyclic preference is actuallypresent in each voter’s list and that it surfaces when there is more adding of thetendency than canceling.For this, abbreviating the choices as D, R, and T, we can describe how avoter with preference order D > R > T contributes to the above cycle.−1 voterD1 voterT1 voterR(The negative sign is here because the arrow describes T as preferred to D, butthis voter likes them the other way.) The descriptions for the other preferencelists are in the table on page 146.Now, to conduct the election we linearly combine these descriptions; forinstance, the Political Science mock election5 ·−1TD1R1+ 4 ·−1TD−1R1+ · · · + 2 ·yields the circular group preference shown earlier.Of course, taking linear combinations is linear algebra. The graphical cyclenotation is suggestive but inconvenient so we use column vectors by starting atthe D and taking the numbers from the cycle in counterclockwise order. Thus,we represent the mock election and a single D > R > T vote in this way.⎛⎜7⎞⎟⎝1⎠5and⎛ ⎞−1⎜ ⎟⎝ 1 ⎠1We will decompose vote vectors into two parts, one cyclic and the otheracyclic. For the first part, we say that a vector is purely cyclic if it is in thissubspace of R 3 .⎛⎜k⎞⎛⎟C = { ⎝k⎠ ∣ ⎜1⎞⎟k ∈ R} = {k · ⎝1⎠ ∣ k ∈ R}k11TD−1R−1