in the plane, where the three perfect sets are required to be collinear, but the three
A-points are forbidden from being collinear. So we have a triangle4P1P2P3, to- gether with three noncollinear A-points, one on each side of the P triangle. Note that relaxing the collinearity constraint on the A-points gives us back one degree of freedom: While # B22,,11=8, we have # B22,,21=9.
Exercise 5.2-1 Given some complete quadrilateral Q, whose six points represent the budget matroid B2,1, describe what three points must be added to Q in order to represent each of the budgetary matroids B22,,11,,10and B22,,11,,20.
[Answer: In the first case, the three new points must be collinear, but no two may coincide; in the second case, they must not be collinear. In both cases, the three new points must lie in the same plane as the quadrilateral Q, but none of them may lie on the line determined by any two points of Q and the line determined by any two of them must not pass through any point of Q.]
5.3
They really are matroids
It is finally time to verify that the budgetary matroids — and among them, the bud- get matroids — really are matroids.
Proposition 5.3-1 Let b =b1+ · · · +bk be a partition of the total budget b into
nonnegative, integral column budgets (bj), at least two of which are positive. For
each j in [1. .k], let the column dimension dj lie in the half-open interval [bj. .b).
For each i in [1. .b], let Ri denote the ith row of the b-by-k ground matrix e := (ei j)i∈[1..b],j∈[1..k], and, for each j in [1. .k], let Cj denote its jth column. We say
that a subset X of the ground matrix e is perfect when • |X ∩Ri| =1for all i in [1. .b]and
• X ∩Cj=bj for all j in [1. .k].
Finally, we say that a subset X of e is independent when: Ambient Rule |X| ≤b;
Column Rule X ∩Cj≤dj +1, for all j in [1. .k]; and
Perfect Rule X has no perfect subsets.
The ground set e, when equipped with this family of independent subsets, forms a matroid, which we denote Bd1,...,dk
b1,...,bk.
Proof The Empty-Set Axiom is easy: The empty set has zero elements overall, zero elements in each column, and is not perfect, so it satisfies all three rules.
62 CHAPTER 5. THE BUDGETARY MATROIDS
The Subset Axiom is also easy. Deleting an element from an independent set doesn’t increase the number of elements overall, doesn’t increase the number of elements in any column, and doesn’t add any new subsets that might be perfect.
The hard part is verifying the Augmentation Axiom. It is helpful to introduce some nomenclature. Given some set X ⊆ e, we call the ith row of X empty when
X ∩Ri = ∅, and we call the jthcolumn of X
• tight whenX ∩Cj=dj +1;
• snug whenX∩Cj∈[bj . .dj];
• loose whenX∩Cj<bj; and
• baggy whenX ∩Cj<bj −1.
With this nomenclature, the Column Rule says that, for every j , the jthcolumn of
X is either tight, snug, or loose. Some of the loose columns may actually be baggy.
Note that, if a set X has either some row that is empty or some column that is loose, then X can’t have any perfect subsets, so X satisfies the Perfect Rule.
Let X and Y be independent sets with|Y|> |X|. We want to find an element
ei j in Y \ X that can be added to X without destroying independence. We have
|X| < |Y| ≤b, so adding any single element to X won’t cause the augmented X
to have too many elements overall, and hence to violate the Ambient Rule. But we must choose the new element with some care, lest the augmented X violate the Column Rule or the Perfect Rule.
We are going to consider five cases, in sequence. At the outset, note that there must be at least one column of X that is loose, since|X|<b.
Case 1: Suppose that, for some j , the jth column of X is snug and the jth column of Y contains some element that isn’t in X; that is, suppose that bj ≤ X ∩Cj≤dj and that(Y \X)∩Cj 6= ∅. In this case, we can choose any element
in(Y \X)∩Cj and add that element to X without destroying independence. The
jth column of X either remains snug or becomes tight, so the Column Rule still holds. And any column of X that was loose — and there was at least one such — remains loose in the augmented X, so the Perfect Rule also holds.
In what follows, we can assume that the first case does not pertain; that is, for every j for which the jthcolumn of X is snug, we have(Y \X)∩C
j = ∅, which
implies thatY ∩Cj ≤ X ∩Cj. So Y is either snug or loose in such columns.
On the other hand, if the jth column of X is tight, we haveX∩C
j=dj +1 and Y ∩Cj≤dj +1, the latter because Y is independent. But Y has more elements
than X overall. Therefore, for one or more columns j where X is loose, we have
(Y \X)∩Cj 6= ∅.
Case 2: Suppose that X has two loose columns. We choose a column m of X that is loose and in which(Y\X)∩Cm 6= ∅, and we add any element of(Y\X)∩Cm