By Neil White

A continuation of the speculation of Matroids, (edited via N. White), this quantity comprises a chain of similar surveys via best gurus on coordinatizations, matching conception, transversal and simplicial matroids, and stories of vital matroid versions. a whole bankruptcy is dedicated to matroids in combinatorial optimization, an issue of present curiosity. Care has been taken to make sure a uniform sort all through, and to make a piece that may be used as a reference or as a graduate textbook. Excercises are incorporated.

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White 1971) Let C 1 and C 2 be a modular pair ofcircuits such that C 1 "1= C 2 , C 1 n C 2 "1= 0, and let a be an element of C 1 n ;:: l. There exists a unique circuit C 3 such that and furthermore, C3 :::> C 1 6 C 2• Proof. )=0 because H~nHi=H~nHi=HinHi. So C 3 = E\H~ satisfies the stated conditions. e. 2. Lemma. Given a basis B and a circuit C such that IC\BI ~ 2, there is a modular pair C l , C 2 of circuits such that C l ;6C 2 , C l nC 2 ;60 and C l 6 C2 s;;: C. Proof Let H* = E\C, a hyperplane in M*(E).

2) Given any k circuits CI,Cz, ... ,Ck , their symmetric difference C 16 C z 6 ···6 C k is a disjoint union of circuits (perhaps empty). ,C k , their symmetric difference C I 6 C z 6···6 C k is either empty, or it contains a circuit. (4) Given any two distinct circuits C I , C z, then C I 6 C z contains a circuit. (5) Given any two circuits C 1 , C z, whichform a modular pair, where C I and C z are distinct but not disjoint, then C I 6 C z is a circuit. (6) For every circuit C and every cocircuit (or bond) C*,ICnC*1 is even.

There is a coline L* in M*(E) such that E\(BuC)cL*cH* (otherwise one would have H*=E\(BuC), and Be C, which is impossible). Let el be an element in C\B, let Ht be the hyperplane spanned in M*(e) by L*u{e l }, and choose an element e2 in C\(BuHt). This set C\(BuHt) is non-empty because otherwise the hyperplane Ht in M(E) would contain the basis E\B. Finally, let H! be the hyperplane spanned in M*(E) by L*u{e 2}. Then the circuits C l = E\Ht and C2 = E\H! ), they form a modular pair (because Ht nH!