Proof that if Voting is Perfect in One Dimension, then the First Eigenvector Extracted from the Double-Centered Transformed Agreement Score Matrix has the Same Rank Ordering as the True Data

Keith T. Poole

University of Houston

24 February 2001

Notation and Definitions

Let the true ideal points of the p legislators be denoted as , , …, . Without loss of generality let the ordering of the true ideal points of the legislators on the dimension from left to right be:

£ £ £ …£

Let q be the number of non-unanimous roll call votes with q > 0 and let the cutpoint for the jth roll call be Z_j. Voting is perfect. That is, all legislators are sincere voters and all legislators to the left of a cutpoint vote for the same alternative and all legislators to right of a cutpoint vote for the oppositive alternative. For example, if all legislators to the left of Z_j vote “Nay”, then all legislators to the right of Z_j vote “Yea”. Without loss of generality we can assume that every legislator to the left of Z_j votes “Yea” and every legislator to the right of Z_j votes “Nay”. That is, the “polarity” of the roll call does not affect the analysis below.

Let k₁ be the number of cutpoints between legislators 1 and 2, k₂ be the number of cutpoints between legislators 2 and 3, and so on, with k_p-1 being the number of cutpoints between legislators p-1 and p. Hence

(1)

The agreement score between two legislators is the simple proportion of roll calls that they vote for the same outcome. Hence, the agreement score between legislators 1 and 2 is simply because 1 and 2 agree on all roll calls except for those with cutpoints between them. Similarly, the agreement score between legislators 1 and 3 is and the agreement score between legislators 2 and 3 is . In general, for two legislators X_a and X_b where a¹b, the agreement score is:

(2)

These definitions allow me to state the following theorem:

Theorem: If Voting is Perfect in One Dimension, then the First Eigenvector Extracted From the Double Centered p by p Matrix of Squared Distances from Equation (3) has at Least the Same Weak Monotone Rank Ordering as the Legislators.

Proof: The agreement scores can be treated as Euclidean distances by simply subtracting them from 1. That is:

(3)

The d’s computed from equation (3) satisfy the three axioms of distance: they are non-negative because by (2) 0 £ A_ab £ 1 so that 0 £ d_ab £ 1; they are symmetric, d_ab = d_ba ; and they satisfy the triangle inequality. To see this, consider any triple of points

X_a < X_b < X_c. The distances are:

and and

Hence

d_ac = d_ab + d_bc (4)

Because all the triangle inequalities are equalities, in Euclidean geometry this implies that X_a, X_b, and X_c all lie on a straight line (Borg and Groenen, 1997, ch. 18).

Because all the triangle inequalities are equalities and all triples of points lie on a straight line, the distances computed from (2) can be directly written as distances between points:

(5)

where d_aa = 0. The p by p matrix of squared distances is:

(6)

To recover the X’s, simply double-center D and perform an eigenvalue-eigenvector decomposition. The first eigenvector is the solution. To see this:

Let the mean of the jth column of D be .

Let the mean of the ith row of D be .

Let the mean of the matrix D be .

Where is the mean of the X_i.

The matrix D is double-centered as follows: from each element subtract the row mean, subtract the column mean, add the matrix mean, and divide by –2; that is,

This produces the p by p symmetric positive semidefinite matrix Y:

(7)

Because Y is symmetric with a rank of one, its eigenvalue-eigenvector decomposition is simply:

(8)

Hence, the solution is

Because, without loss of generality, the origin can be placed at zero, that is, , the solution can also be written as:

(9)

The points from (9) exactly reproduce the distances in (4), the agreement scores in (2), and the original roll call votes. In addition, note that the mirror image of the points in (9) (a multiplication by minus one) also exactly reproduces the original roll call votes. Furthermore, for any pair of true legislator ideal points and with one or more midpoints between them, < Z_j < , the recovered legislator ideal points must have the same ordering, X_a < X_b. If there are no midpoints between and -- that is, their roll call voting pattern is identical -- then the recovered legislator ideal points are identical; X_a = X_b. Hence, if there are cutting points between every pair of adjacent legislators, that is, k_i ³ 1 for i=1, …, p-1, then the rank ordering of the recovered ideal points is the same as the true rank ordering. If some of the k_i = 0, then the recovered ideal points have a weak monotone transformation of the true rank ordering (in other words there are ties, some legislators have the same recovered ideal points).

This completes the proof. QED.

Discussion

Note that an interval level set of points is recovered. However, this is an artifact of the distribution of cutting points. For example, if k₁ > k₂ , this has the effect of making d₁₂ > d₂₃ even if the true coordinates , , were evenly spaced. With perfect one dimensional voting, the legislator configuration is only identified up to a weak monotone transformation of the true rank ordering.

The rank ordering can also be recovered from the matrix Y given in (7) without performing an eigenvalue-eigenvector decomposition. Note that, with the origin at zero, the diagonal elements of Y are simply the legislator coordinates squared. The rank ordering can be recovered by taking the square root of the first diagonal element and then dividing through the first row of the matrix. Note that this sets X₁ > 0 and the remaining points are identified vis a vis X₁.