MATH 5707 HOMEWORK 5 SOLUTIONS 1. Let a Be an N × N

MATH 5707 HOMEWORK 5 SOLUTIONS

CIHAN˙ BAHRAN

1. Let A be an n × n matrix ﬁlled with non-negative real numbers so that every row and every column has sum 1. Such matrices are called doubly stochastic. Show that A is a non-negative linear combination of permutation matrices. Permutation matrices are doubly stochastic matrices with all entries equal to either 0 or 1.

Write N(A) for the number of positive entries of a matrix A. We will prove the statement by induction on N(A). If N(A) = 0, then A = 0 is not doubly stochastic, so the statement “every doubly stochastic A with N(A) = 0 is a non-negative linear combination of permutation matrices” is vacuously true. Now assume that the statement holds for every doubly stochastic matrix with at most k − 1 nonzero entries, where k ≥ 1. For the inductive step, let A be a doubly stochastic matrix with N(A) = k. Writing [n] = {1, . . . , n}, consider the maps γ : 2[n] → 2[n]

R 7→ {j ∈ [n]: Aij 6= 0 for every i ∈ R} and ρ : 2[n] → 2[n]

C 7→ {i ∈ [n]: Aij 6= 0 for every j ∈ C} . ρ should be thought of as taking a set of rows R of A and returning the set of columns which intersect each row in R in a nonzero entry; similarly for γ. We write γ(i) for ρ evaluated at the singleton {i} when i ∈ C. Similarly, we write ρ(j) for ρ({j}). P Since the row sums of A are 1, for every i ∈ [n] we have 1 = j∈C Aij whenever γ(i) ⊆ P C ⊆ [n]. Similarly, for every j ∈ [n] we have 1 = i∈R Aij whenever ρ(j) ⊆ R ⊆ [n]. Fix R ⊆ [n], write Γ(R) = [ γ(i). We get i∈R X X X |R| = 1 = Aij i∈R i∈R j∈Γ(R) X X = Aij j∈Γ(R) i∈R X X X ≤ Aij = 1 = |Γ(R)| . j∈Γ(R) i∈[n] j∈Γ(R) The inequality in the third line above holds because the entries of A are non-negative. Thus, the bipartite graph G, whose vertex classes are two copies of [n] and which has an edge between i ∈ [n] and j ∈ [n] if and only if Aij 6= 0, satisﬁes the conditions of Hall’s marriage theorem. Thus G has a perfect matching, the datum of which is precisely an n × n permutation matrix P such that Pij = 1 implies Aij 6= 0. 1 MATH 5707 HOMEWORK 5 SOLUTIONS 2

Let c = min{Aij : Pij = 1} and B = A − cP . So B has nonnegative entries whose row and column sums are equal to 1 − c. Moreover N(B) < N(A). There are two cases: • c = 1. Then being a matrix with non-negative entries whose row and column sums are zero, B must be identically zero. So A = P and we are done. • 0 < c < 1. Then the matrix 1 Af= B 1 − c is doubly stochastic with N(Af) = N(B) < N(A). Therefore by the inductive hypothesis, Af is a non-negative linear combination of permutation matrices. Since A = cP + (1 − c)Af, so is A.

6. Show that in a stable matching optimal for the boys, at most one boy ends up with his worst choice. (Assume the graph is complete.)

For the sake of contradiction, suppose M is a stable matching optimal for the boys, and a, b are two boys such that aA, bB ∈ M yet A is the worst choice for a and B is the worst choice for b. Let M 0 = (M − {aA, bB}) ∪ {aB, bA}. Clearly M 0 is a matching. Moreover both a and b are better off in M 0 than M and other boys have the same partners in M and M 0. But M is the most optimal stable matching for boys, so M 0 cannot be stable. This means that there exists a boy c and a girl C such that (1) cC∈ / M 0. (2) c prefers C over his partner in M 0. (3) C prefers c over her partner in M 0. Suppose cC∈ / M. Because M is stable, either (2) or (3) should not fail for M. Hence both c and C must have different partners in M (everyone is matched up in any stable matching because the graph is complete) from what they have in M 0; hence c ∈ {a, b} and C ∈ {A, B}. This contradicts cC being neither in M nor M 0. So we must have cC ∈ M. But then cC∈ / M 0 means that c and C have different partners in M and M 0, again yielding a contradiction. 8. Show that if a stable matching contains an edge aA with a and A being worst choices for each other, then every stable matching contains edge aA.

Let M,M 0 be stable matchings with aA ∈ M. Suppose aA∈ / M 0. Then by Corollary 19 in p.88 of the textbook, in M 0 both a and A have mates where one of a and A is better off in M 0 than in M, and the other is worse off. But by assumption, neither a nor A can be worse off in M 0. This is a contradiction, from which we deduce aA ∈ M 0.