Fundamentals of Matroid Theory


Matroid, a combinatorial object that was first introduced by Whitney in 1935 and has been studied extensively by mathematicians ever since then, is one of the most important topics and glorious triumphs of combinatorics.

Matroid is thought to be the product when linear algebra meets graph theory. Initially, it was an abstraction and generalization of linear independence of linear algebra. This is also how it got its name, "something similar to matrix". However, it was then found to be strongly connected to graph theory: there are many graph theoretic objects found to be matroids; while graph theory provides metaphors and algorithms, for interpreting matroids and performing matroid operations.

However, despite the fact that linear algebra and graph theory establishes the language of matroid theory, there's no isolated island in mathematics, and matroid is also found to be related to other fields of combinatorics, as well as algebra, geometry, and topology.

On the other hand, matroid has a bunch of published results in combinatorial optimized algorithms, so studying matroid can also improve problem-solving in computer science.

Independent Sets and Bases

As is said, that matroids are abstract linear independence, let's recall what linear independence can do.

First, we know a linearly independent set is a set of vectors such that no vector can be represented by the linear combination of other vectors.

It won't be hard to find any subset of a linearly independent set is also a linearly independent set, since otherwise the linear combination evidence in the subset can also be carried out to the superset, rendering the superset not to be linearly independent. A property present in the superset is said to be hereditary, if it is preserved when pruning to any of its subsets. The linear independence is an example of hereditary property.

The next one is not so obvious but still simple for those familiar with linear algebra, that for two linearly independent sets (of the same vector space), if , then there exists a vector , such that is linearly independent. To prove, it suffices to show some vectors in can't be represented by the linear combination of . This can be simply shown by Gaussian elmination. If not so, let and , represent all as the linear combination of :

This can be viewed as a system of linear equations with variables. We may use Gaussian elimination to wipe the -th equation out, to . But this is accomplished by subtracting coefficients scalar-multiplying the first rows from the -th row, that for some coefficients . So now is a linear combination of the first vectors in , which is a contradiction to is a linearly independent set. The property that a larger linearly independent set can provide some vector to a smaller independent set is called the independent set exchange property (aka. augmentation property).

A matroid is a combinatorial object that captures and generalizes the hereditary property and the independent set exchange property:

Definition (Hereditary Systems and Matroids)

A hereditary system on is a combinatorial object, where is a finite set called the ground set, is a set of subsets of called the independent sets of , such that:

  1. .
  2. .

A matroid on is a combinatorial object such that:

  1. (Hereditary Property) is a hereditary system on .
  2. (Independent Set Exchange Property) .

In a hereditary system or a matroid on , for any subset , it's said to be independent when , otherwise it's said to be dependent when .

For convenience, we may define and use it for retrieving independent set from matroid .

Convention (Set Operation Convention in Matroid Theory)

In matroid theory, we will be taking an element from the ground set , forming a singleton set , and performing set operations with some subset of from time to time. For convenience, we may write:

So the independent set exchange property above may also be written as .

The axioms of matroids above are called the independent set axioms of matroids, and we will see there're many ways to define a matroid axiomatically and equivalently. Here, we define hereditary systems prior to matroids, since we want to reuse it in other axioms of matroids later.

In the case of linearly independent sets, it's okay to let empty set to be linearly independent. However, we will see letting is indispensible if we want the hereditary system to exhibit some natural behaviors, which makes our life easier.

In combinatorics, we are discussing about finite objects in most time, and that's why we require the ground set to be a finite set.

The linearly independent sets itself that motivates us obviously defines a matroid:

Example (Vectorial Matroids)

Let be any finite set of vectors from vector space , and is all linearly independent subsets of . Clearly fulfills the independent set axioms of matroids, and is called a vectorial matroid of .

For example, by letting:

We define a vectorial matroid in real vector space .

However, a matroid over a ground set of vectors with independent sets being linearly independent subsets need not be a vectorial matroid:

Example (Matroid over Vectors but not Vectorial)

In the previous example, the linearly independent subset is dispensible:

And is still a matroid other than . To prove, it suffices to show the independent set exchange property in :

  • For , we can see .
  • For , we can see .

We also pull the independent set exchange property of closer so that we can see what is happening here:

  • For , we can see .
  • For , we can see .
  • For , we can see .

As linearly independent sets are the initial motivation of matroids, playing with them is not very motivating. So let's see an object from graph theory that also turns out to have abstract linear independence:

Example (Cycle Matroids)
Let be a graph, let the ground set be the edge set , and be the set of edge sets such that for any , the spanning subgraph is a spanning forest of . It can be proved that forms a matroid, which is called the cycle matroid of and denoted as .

Proof. We need to show the hereditary property and independent set exchange property of .

For any , it won't be hard to find is obtained by deleting some edges from , and merely deleting edge will not introduce cycle. So the spanning subgraph is also a spanning forest of , and the hereditary property is held.

To prove the independent set exchange property, we need to consider the connected components of their spanning forests. That is, if we add the edges of to one by one, every edge connects two distinct components when added. And for such that , it won't be hard to find the spanning forest contains more connected components than . By pigeonhole principle, there must be one component of containing vertices from two or more components of . Categorize vertices in , and we must find an edge , such that for two distinct components of . As adding to connects and will not introduce cycle, there's , Therefore the independent set exchange property is also held.

In this way, has a finite ground set with hereditary property and independent set exchange property in , and is thus a matroid.

So the abstract linear independence occurs in a seemingly completely unrelated field, the graph theory. Interesting as it is, the purpose of defining matroids is still fuzzy, why would we want to unify the linearly independent sets and the spanning forests? To answer this question, we will have to reveal more properties of matroids.

Bases of Matroids

Again, let's do a quick review about the bases in linear algebra.

For a finite set of vectors , we may use it to span a vector space by the linear combination of vectors in . However, by the theory of linear algebra, some vectors in might be dispensable: let and vector , if is expressible by , then is dispensable and represents the same vector. In this way, the set might be reduced to a basis , a subset of linearly independent vectors, such that .

In the example above, given that , the way of eliminating linearly dependent vectors in is not unique, and we should expect more than one bases of obtainable from eliminations in in general.

For a finite set of vectors and the vector space it spans, a linearly independent set is a basis iff it's a maximal, that is, no proper superset of it is linearly independent. To show the necessity, for a basis and a proper linearly independent superset , there's an element that is not expressible by linear combination of vectors in , but is expected, which is a contradiction. To show the sufficiency, if is maximal linearly independent subset of but not span , there must be a vector that is not spanned by , and thus is linearly independent and larger than , which is contradiction to the maximality of .

And in the case of matroid:

Definition (Bases of Matroids)

A basis of matroid is a maximal independent set of under set inclusion. That is, and .

We shall denote the set of bases of matroid as .

So a basis of matroid captures the maximal property of a basis in linear algebra. However, we will show, it suffices to capture this property in order for a basis matroid to exhibit many behaviors similar to a basis in linear algebra.

We are familiar with the conclusion, that for the -dimensional vector space spanned by , every basis has vectors. And the bases of matroid also possess this property:

Lemma
Proof. Assume the contrary, and without loss of generality, let . By the independent set exchange property, we have , while , which is contradiction to the maximality of .
Lemma
Proof. If , then . Since , by the independent set exchange property, we have while , which is contradiction to the maximality of .

Simple as they are, if we put the proofs above alongside with the proof of every basis of -dimensional vector space has vectors, we will soon find the proofs above are just abstracted versions of the one in linear algebra.

The bases in linear algebra also have the basis exchange property, that is, for two distinct bases of the same vector space, , such that is also a basis. This property also holds for bases of matroids:

Lemma (Weak Basis Exchange Property)
Proof. Since and , must be a proper subset of them while both and are not empty. By for we obtain a set of size , thus . It won't be hard to find and , and we are done with the proof.

The matroid version of basis exchange property is called the weak basis exchange property, in comparision with the strong basis exchange property that . However, the latter one can't be proved by using rudimentary properties of matroids we've known (in this text) so far. We will revisit the proof of the strong basis exchange property once we have discussed sufficiently about properties of matroids to prove.

Matroids and Greedy Algorithms

The matroids are even famous among computer scientists due to their connection with the greedy algorithms, especially the fact that it's possible to generalize the classical Kruskal's MST algorithm over matroids.

The Kruskal's MST algorithm finds the minimum spanning tree of a connected graph, that is, with weights assigned to every edges in graph, the Kruskal's MST algorithm finds a spanning tree minimizing the total weights.

On the other hand, it won't be hard to see the bases of a cycle matroid are the spanning forests such that each component is a spanning tree of a connected component in the original graph . When there's only one component, the bases are the spanning trees of the graph. And when we assign weights to the edges, let , the MST problem can also be viewed as finding among the bases.

And in this section, we would like to show for a general matroid with weights assigned to each elements, a greedy algorithm suffices to solve the problem of finding .

For convenience, we would like to show the following process ends up with finding a basis of a matroid:

Algorithm
The input is a matroid , the output is any basis :

  1. Initialze empty set .
  2. Iterate :
    1. If , update .
  3. Halt with being a basis.

Proof. It won't be hard to find must be independent all along the way, and if , there must be such that .

By independent set exchange lemma, there must be such that . Let's rewind back to the moment when is visited, and let the set be constructed so far in the algorithm. Since , there must be so that is discarded. But , which is a contradiction.

Please notice the order of iterating is arbitrary, and different order of iteration might end up with different basis constructed.

And the algorithm of finding with minimum weight is shown as below:

Algorithm (Greedy Algorithm for Minimum Weighted Basis of Matroid)
The input is a matroid and a weight function , the output is the minimum weighted basis :

  1. Initialze empty set , and sort the elements in into list such that .
  2. Iterate :
    1. If , update .
  3. Halt with being the minimum weighted basis.

Proof. By the previous algorithm we know when the algorithm halts, the result , the only thing remains is to show minimizes .

For convenience, let's sort the elements in by the order of being added to , so and is added after if . Assume the contrary that does not minimize , and let be the one minimizing instead. So we have strictly.

Let be the first element such that while . Such element must exists since otherwise . And we would like to show there's a way to "hybridize" into another basis :

That is, first we initialize with elements from , then we extend with elements from until . For convenience, for the -th element to the -th element in , we sort them by their order in , so that there's and .

Since to in are just some elementes taken from and sorted by their weights afterward, they can maximally be to , and thus . The key to the proof is , in which we have . Therefore is iterated before in . Since is a basis, is independent, however is discarded due to the subset 's being dependent, during the process of constructing , which is a contradiction.

For those who are familiar with the non-matroid version of the proof of Kruskal's MST algorithm, they will quickly find how similar these two proofs are, and no wonder why the algorithm above is considered a generalization of Kruskal's MST algorithm to matroids.

For every matroid, the greedy algorithm above solves the minimum weighted basis problem for it sufficiently. However, it's a common misconception to think every greedy algorithm has necessarily an underlying matroid. The Prim's MST algorithm, which is the twin cousin compared to Kruskal's MST algorithm from time to time, has no underlying matroid inside in this case.

Base System Axioms of Matroids

For every matroid , we can find the bases, a set of maximally and equally-sized independent set via . However, is it possible to specify a set of equally-sized subsets of a ground set , and use it as the set of bases, defining a matroid?

From the previous sections, we know the bases of a matroid have the weak basis exchange property, so if a set of equally-sized subsets would like to define a matroid, they must fulfil this property in the first place. Thinking of extremality, it won't be hard to find the tuple defnies a matroid minimally, and its bases are , so the set of bases can never be empty. But is possessing these properties sufficient, let's give it a try:

Definition (Base Systems)
A base system on is a combinatorial object, where is a finite set called the ground set, and is the set of bases such that:

  1. is non-empty.
  2. .
  3. .

Intuively speaking, in a matroid, every independent set must be a subset of a basis. This inspire us to define:

Definition (Independent Sets Defined by a Set of Subsets)
Lemma
For any non-empty set , defines the independent sets of a hereditary system on .
Proof. Since is not empty, there must be an element , and . And for every , by construction , so . In this way, defines the independent sets of a hereditary system on when is non-empty.

So that we may retrieve the independent sets from a base system through . The only thing left to check is whether the hereditary system also possesses the independent set exchange property, and how are they connected to the way we define the base system above:

Theorem (Base System Axioms of Matroids)
For any base system on ground set , is a matroid on .

Proof. We just need to show the independent set exchange property in .

For any , such that , by construction there're such that .

If , then we are done with . If , given that , we assert it's the case of , otherwise there's a way to fix into the case we asserted indefinitely.

If , then , and , so we may exchange it with , and use as the new containing . Noted that since is replaced with now, such a process can't last forever and there's a point that , at which .

In this way, we may pair the elements in by which element they are in . Here is one of the ways:

(Consider: Is it ever possible to avoid elements from to be paired with elements in , given that ? Have a try!)

We argue that , since if and would like to be disjoint as subsets of , there must be , but is expected. Therefore we have and .

In this way, to define a matroid, it suffice to come up with a base system . That is, it suffices to say there's a matroid such that , in order to define a matroid in the form of .

The theorem above also shows us an alternative way to define a matroid equivalently. In fact, there're other ways of defining matroid. Such a phenomenon of defining matroid equivalently in many seemingly unrelated way (that are later shown to be tightly connected) is sometimes referred as cryptomorphism of matroids by mathematicians. The terminology of cryptomorphism is also leveraged to describe different but equivalent sets of axioms of defining the same mathematical object.

Rank Function

We are familiar with the rank of a matrix, that is, the maximum number of linearly independent row or column vectors in a matrix. It won't be hard to find the order of rows and columns does not matter, and deduplication does not affect the rank, so we may assume the rank to be defined on a set of vectors, by evaluating the maximum number of linearly independent vectors in the set.

Analogously, for a given set which is a subset of the ground set of a matroid, we would also want to ask the maximum number of "independent" elements, this is equivalent to asking the maximum size of an independent subset of the given set. Noted that asking the rank of a matrix is equivalent to asking the maximum size of an independent subset in a vectorial matroid.

To fulfill this purpose, we first define the restriction of matroid to a set:

Definition (Restrictions of Matroids)
For matroid , the restriction of to is a matroid , such that .

Proof. To prove is a matroid, it suffices to show the hereditary property and independent set exchange property are preserved in .

It won't be hard to find , and , so the hereditary property is preserved in .

To show independent set exchange property, for any , we have in the first place. And since both are subsets of , the element taken is also element of , thus is a subset of , and independent set exchange property is preserved in .

Obviously the maximum size of an independent subset of the given set is the size of a basis of the restriction of the matroid to the given set. And we may define:

Definition (Rank Function of a Matroid)

For matroid , we define the rank function of matroid , so that for any input subset , outputs the size of bases of .

When the context of matroid is clear, we may write directly for input subset . When under the context of multiple matroids numbered as , the corresponding rank functions may also be numbered as .

By doing so, given that the size of bases of a matroid is unique, we obtain the well-formedness of the rank function of a matroid for free. Meantime, such a definition is compatible with the rank of a matrix, when viewing it as a vectorial matroid.

Absorption Rules

We can see metaphors play an important role in studying matroids. In a matrix, when extending it by a vector, its rank remains unchanged or increment by , depending whether the introduced row is independent or not. And in a matroid, the case is similar:

Lemma

Proof. For a basis , it won't be hard to see , which implies .

Meantime, consider the basis such that : If , then . If , then . No matter which case is always true. By combining these two inequalities we have , and we are done with the proof.

In a matrix , assume augmenting with vector does not , and so does , then augmenting with simultenously will not affect either. And when it comes to the case of a matroid , we also have:

Theorem (Weak Absorption Rule)

Proof. Assume , consider the bases , there's , thus by independent set exchange property when viewed from . Noted that otherwise is to be found in or , forcing .

If , then , which is a contradiction. So it's only , and if it's , then , which is also a contradiction; similar argument is applicable to .

In this way, it's impossible for to be true, and it's only .

Corollary (Strong Absorption Rule)

Proof. This is done by arguing for any -combination (that is, any set of distinct elements chosen from ) to be true until , by mathematical induction.

When , it's the presumption; When , it's the weak absorption rule. And we would like to show the case of to be true when the induction hypothesis for has been true. To show for combination , by the induction hypothesis for and we have:

Noted that they are in the form ready for applying the weak absorption rule. Consider augmenting with , we have , now the induction hypothesis holds for , and we are done with the proof.

Submodularity

First, we would like to introduce the concept of submodular function:

Definition (Submodular Functions)

For a finite set used as the ground set, a function is said to be submodular, when there's:

Then, we would like to show the rank function of a matroid is a submodular function:

Theorem (Submodularity of Rank Function)

Proof. First, we find a basis , and then extend it to a basis . Extending the basis to may not use elements from , or it's a contradiction to the maximality of in . So we may categorize the elements in by whether they are from , or .

When viewing from the matroid , since is independent, both and are subsets of , independent, and to be found in and . So we have:

While is clear by the inclusion-exclusion principle (or by the figure above). So we are done with the proof.

The submodularity of rank function is a useful in proving properties of matroids, we will see later in this text.

Rank Function Axioms of Matroids

There's another cryptomorphism of matroids, that if a set is equipped with a rank function, then we are able to extract a hereditary system with independent set exchange property out of it, using it as the independent sets of a matroid and defining a matroid.

First, we would like to equip a set with a rank function:

Definition (Rank Functions on Finite Sets)
A rank function on , is a function in the form of , such that is a finite set called the ground set, and:

  1. .
  2. (Monotonicity) , noted that the first inequality implies .
  3. (Weak Absorption Rule) .

A rank function on a finite set has the strong absorption rule, which can be directly implied from the weak absorption rule:

Lemma (Strong Absorption Rule of Rank Functions)

Proof. Recall how we prove the strong absorption rule of the rank function of a matroid, we argue on the -combination to fulfil , by mathematical induction to .

When it's the presumption. When it's the weak absorption rule of the rank function. When it has been true for , by arguing it has been true for when , and when
, we apply the weak absorption rule, viewing it as the case of extending with , to imply , which is the case of .

Intuitively speaking, in a matroid, all independent sets fulfills , and this inspires us to define:

Definition
For a rank function on , we may let .

And see whether defines a hereditary system on with independent set exchange property. We will begin with showing the hereditary property:

Lemma
The set defines the independent sets of a hereditary system on .

Proof. First, given that and , it won't be hard to find , and since by the range of , so we have .

Then, to show , first for , there's a chain of elements to be removed from , that is, . Our plan is to prove , so that , and therefore .

To fulfill our plan, we need to utilize the monotonicity that . In our case, it's , so . By combining with we are done with , and therefore the proof.

So it suffices for and the monotonicity property to ensure the hereditary property of . But to prove the independent set exchange property, we still need the absorption rules:

Theorem (Rank Function Axioms of Matroids)
For any rank function on ground set , is a matroid on .

Proof. Given that has already been a hereditary system on , we just need to show the independent set exchange property inside.

First, pick up any such that , and assume there's no element in that can be used to augment . That is, , by the monotonicity, , so it must only be if would like to be true. By the strong absorption rule, there's . By monotonicity, there's . So they combine into , which is a contradiction. In this way, the independent set exchange property must be present in , and therefore defines a matroid.

The submodularity of a rank function on ground set , that , follows immediately by viewing from the underlying matroid .

On the other hand, it's also possible to substitute the weak absorption rule with the submodularity in the axioms of rank function. Since if , then given that , we have , and by combining with the monotonicity we have .

Dependent Sets and Circuits

While the bases and rank function we've discussed about are related to independent sets, it's also possible to study matroids from the dependent sets. The most important kind of dependent sets might be the circuits:

Definition (Circuits)

In a matroid , a subset is said to be a circuit, if it's a minimal dependent set under set inclusion. That is, and . It can also be reworded as and .

Similarly we may denote the set of circuits of matroid as .

The circuits rise from the cycle matroids:

Example (Circuits in Cycle Matroids)
In a cycle matroid , a dependent set is a circuit iff is a cycle.
Proof. The sufficiency is obvious. For the necessity, if induced subgraph contains a cycle but not a cycle, then is not empty, and removing edges still yields with the cycle inside.

But the definition above is applicable to all matroids, not just the cycle matroid. So what do circuits look like in the vectorial matroids?

Example (Circuits in Vectorial Matroids)

In a vectorial matroid such that , a linearly dependent set is a circuit iff any vector in can be represented by the remaining vectors and the representation is unique. That is:

And such a representation is unique.

Proof. For the sufficiency, recall what "uniqueness" means by. The representation is unique, therefore since otherwise we may obtain another representation of . And thus, is linearly independent.

For the necessity, since is a circuit, every vector in can be removed so that is linearly independent. So if vector can be represented by the linear combination of , then since only solution to . is by the linear independence of , the representation of by is unique.

Meantime, if there's a vector can't be represented by the linear combination of , then there has to be such that not all of the are zeroes, since has to be linearly dependent. However is still linearly dependent, which is a contradiction to is circuit.

The circuits draw special attention, since there's:

Lemma
Every dependent set contains a circuit. That is, .

Proof. Assume the contrary that does not contain a circuit, so for any dependent subset , there must be to avoid being a circuit.

In this way, given that is not a circuit (otherwise it sufficiently contains a circuit), starting with , we find the element such that , and use it as the new that is dependent. So we obtain a proper descending chain:

Since is finite and every time one elements is removed from the current set, we will inevitably reach a point that:

So that . But the only proper set of is , which is independent. So and , which is contradiction to does not contain a circuit.

Definition (Loops and Parallel Elements)

Conventionally, if a set of single element is a circuit, then said to be a loop, and if a set of two elements is a circuit, then are said to be parallel elements.

A matroid is said to be simple if it has no loop or parallel element.

Corollary
Proof. It remains to show the necessity: If , then can't be independent, otherwise by hereditary property , which is a contradiction.
Corollary
Proof. The set also contains the circuit .

It's good to take care of the following common misconception:

Warning (Comments on the Size of Circuits)

The definition of the circuits and bases might be seemingly similar: One is the minimal dependent set and the other is the maximum independent set. Given that all bases have the same size, one might start to think all circuits have the equal size, but this is wrong.

The most simple example might be from the vectorial matroids. In the vectorial matroid on , it won't be hard to verify both and are circuits, but they have the different sizes.

Weak Elimination Rule

Consider two cycles in a cycle matroid and their union , if they intersect at a path , then after removing any edge in the path, will still contain a cycle:

Interestringly, it turns out that such a property is general and applicable to any matroid:

Lemma
Proof. Just consider:
And we are done with the proof immediately.
Theorem (Weak Elimination Rule)

Proof. Assume the contrary that . On one hand, must be dependent for containing circuits , so ; on the other hand is independent so . So we can only have . And by inclusion-exclusion principle, .

On the other hand, by the submodularity of rank function, we have:

Since , their intersection must be a proper subset of either or (or both), and thus . This will imply , and the inequality above will soon become:

In this way, , which is a contradiction, so there can only be .

We name such a property of matroid the weak elimination rule. There's also a strong elimination rule, which states in every dependent set and for every element , there's a circuit such that . Intuitively, those corresponds to the edges marked purple in our cycle matroid example. We will prove this property later, as what we've learnt about matroids so far is insufficient for proving it.

Consider a spanning forest in a cycle matroid: If by adding one edge we introduce a cycle, then only one cycle can be introduced. Otherwise for two cycles introduced, they must both contain otherwise they are to find in the original graph. But edge set contains another cycle and can be found in the original spanning forest.

And by utilizing the weak elimination rule, such a phenomenon in cycle matroid can also be generalized to any matroid:

Lemma
For any and any , if , then contains unique circuit.

Proof. Assume does not contain unique circuit, that is, there're two distinct circuits such that .

We argue that and intersects at , since the one who does not contain is a subset of and thus independent.

Therefore, by the weak elimination rule, is dependent. However, by , it's expected to be independent, which is a contradiction.

The process of proof is literally identical to what we've argued for the cycle matroid above. So no wonder graph theory also plays an important role in studying matroids.

Greedy Algorithm Axioms of Matroids

This section is not related to circuits directly, however it's essential to the later proof of showing that we may define matroids by circuits cryptomorphically.

It's said that there's not necessarily a matroid behind a greedy algorithm, however there's a specialty when the greedy algorithm is running over a hereditary system. If we go over the proof of the correctness of greedy algorithm to find the minimum weighted basis, we will notice that the independent set exchange property plays a major role in ensuring the correctness, regardless of what weight function is used as input. And the specialty is, if a greedy algorithm to run in a hereditary system is correct, regardless of the input weight function, then we may excavate the independent set exchange property from the hereditary system, showing it to have an underlying matroid.

The proof is based on the classical work of Rado and Gale. They first came up with the following greedy algorithm:

Algorithm (Greedy Algorithm for Maximum Weighted Independent Set)
The input is a hereditary system and a non-negative weight function , the output is the maximum weighted independent set :

  1. Initialze empty set , and sort the elements in into list such that .
  2. Iterate :
    1. If , update .
  3. Halt with being the maximum weighted independent set.

And then they asserted that:

Assertion (Rado-Gale)
The Greedy Algorithm for Maximum Weighted Independent Set is correct iff the input hereditary system is a matroid.

Let's begin with the sufficiency part. When the hereditary system is a matroid, it won't be hard to find the algorithm above is almost identical to the greedy algorithm we've used for finding the minimum weighted basis, except for forcing the weight function to be non-negative, ordering the list in descending order, and not requiring to find a basis. However, if our previous greedy algorithm is correct, then this algorithm must also be correct:

Lemma
In a matroid, the maximum weighted independent set must be a basis.
Proof. Let the maximum weighted independent set be such that , then by the independent set exchange property there's , such that , and given that is non-negative. In this way, the maximum weighted independent set must be a basis.
Theorem (Rado-Gale)
If the hereditary system is a matroid, then the Greedy Algorithm for Maximum Weighted Independent Set is correct.

Proof. Let's do a little translation: define weight as , and run the Greedy Algorithm for Minimum Weighted Basis with matroid and weight function , we've proved the algorithm must correctly yield the basis minimizing .

Due to the way we write the the Greedy Algorithm for Maximum Weighted Independent Set, it must yield the same basis , claiming it to be maximizing . This is correct since there must be a basis maximizing , while minimizing is equivalent to maximizing .

So in the case of matroid, by requiring the weight function to be non-negative we have forced a maximum weighted independent set to be a basis, and then the greedy algorithm will find such a basis with maximum weight sufficiently.

Now let's focus on the necessity part. Please notice that a general hereditary system does not necessarily have the independent set exchange property. However, if such a property is dispensible, we can show the greedy algorithm will fail for a deliberately designed input weight function:

Theorem (Rado-Gale)
If the Greedy Algorithm for Maximum Weighted Independent Set is correct, then the input hereditary system is a matroid.

Proof. Let there be such that . We tailor a weight function for them, so that:

Now let's run the greedy algorithm with and above as input.

The algorithm will first encounter elements in since they are of maximum weight . All of them will be added to variable since has been known to be independent, and so are its subsets.

Then the algorithm has to face the elements in , assume there's , then the algorithm will have to dismiss every end up yielding some set such that , given that . Noted that the algorithm yields a wrong answer since is independent and sufficiently with greater weight than , which is a contradiction to the correctness presumption of the greedy algorithm.

In this way, there must be , the independent set exchange property is to be found in , and the hereditary system is a matroid.

Corollary (Rado-Gale)
The Greedy Algorithm for Maximum Weighted Independent Set is correct iff the input hereditary system is a matroid.

In this way, with a hereditary system , paired with the guarantee for the Greedy Algorithm for Maximum Weighted Independent Set to yield the correct answer for whichever non-negative weight function indefinitely, we may define a matroid cryptomorphically. We shall call it the greedy algorithm axioms of matroids.

Circuit System Axioms of Matroids

So let's see whether we can define a matroid cryptomorphically. Just like what we've done before, we begin with extracting properties of circuits in a matroid:

Definition (Circuit System)
A circuit system on is a combinatorial object, where is a finite set called the ground set, and is the set of circuits such that:

  1. .
  2. (Minimality) .
  3. (Weak Elimination Rule) .

And since all dependent sets are those containing a circuit, it's expected that by defining:

Definition (Dependent Sets Defined by a Set of Subsets)

We retrieve the "dependent sets" defined by a circuit system, and therefore the "independent sets" . And therefore for a subset , it suffices for to show is independent.

So if a matroid is hopefully defined in such a way, it should be in the form of . Let's begin with showing the hereditary property:

Lemma
For any circuit system on , the set defines the independent set of a hereditary system.

Proof. Since , and is a proper super set of no set, therefore .

For any , assume the contrary that , then there's a circuit such that , which is a contradiction.

Then let's show some familiar properties in the circuit system:

Lemma
Proof. If , then there's a circuit such that , however is a contradiction to the minimality of .
Lemma
For any independent set and , if is dependent, then contains a unique circuit.
Proof. Assume contains two distinct circuits , these circuits must contain , since otherwise they are the subsets of and independent. By the weak elimination rule, which is a contradiction to the presumption .

Finally, we are ready to show a circuit system defines a matroid:

Lemma
For any independent set and , if is dependent and contains the unique circuit , then for any , is independent.
Proof. The circuit must contain if it won't like to be a subset of . When is chosen, clearly is independent. Assume is chosen from , but is dependent, then it contains a circuit . And since but , is another circuit than . However, given that , contains two distinct circuits now, which is a contradiction.
Theorem
For any circuit system on , the Greedy Algorithm for Maximum Weighted Independent Set is correct for the hereditary system and whichever non-negative weight function .

Proof. Let the output of the greedy algorithm be , and is a maximum weighted independent set maximizing . That is, .

If , then we are done. If , then by considering , we have:

  1. If , then , and is a maximum weighted independent set, however it's a contradiction since is not maximizing due to .
  2. If , then consider any , it means greedy algorithm discards when is constructed, so is inspected, however it's expected that which is a contradiction.
  3. and , so is a proper subset of both , the remaining proof will be discussing this case.

Let's sort the elements by their weight descendingly (the one with greater weight appears first), and let them be first diverging at , as is shown in the figure:

We would like to show : Since is first the divergent point, if then must appears after . However when the greedy algorithm constructs and scans , its discarding implies is dependent, which is a contradiction.

We would also like to show is dependent: Since otherwise while , and is not maximizing , which is a contradiction.

So contains a circuit , and if won't like to be a proper subset of . Meantime, is not empty if won't like to be a proper subset of . So we may always choose , and there's . By the previous lemma, we know is independent.

Again, we've "hybridized" the into another independent set :

Since is a maximum weighted independent set, we have . However contradictions are still inevitable:

  1. If , then , so the greedy algorithm scans before . However, it discards because is dependent, which is a contradiction.
  2. If , then since we've replaced the element with . But this will imply is not maximizing , which is a contradiction.

In this way, there's no way for to be true, so and the greedy algorithm yields the correct answer for whichever non-negative weight function.

Corollary (Circuit System Axioms of Matroids)
For any circuit system on ground set , is a matroid on .

In this way, it turns out that we may define a matroid on by simply saying for a circuit system on .

Span Function

We are familiar with the vector space spanned by a set of vectors. When it comes to the case of vectorial matroid , the most direct way of defining the span of is to consider those vectors representable by the linear combination of vectors in .

However, being a matroid deliberately hides the underlying detail of "linear combination" from us, so we must only rely on the language and characterization of independent sets and dependent sets. However, is dependent does not necessarily imply that can be represented by the linear combination of : as long as itself is dependent, any superset of will also be dependent. This will ridiciously lead us to every element in is "representable" by the "linear combination" of .

But the previous section also provides us a solution, that is, for that would like to be expressible by the remaining vectors, there's a circuit iff for (by eliminating those with zero coefficients).

In this way, the closest characterization of span may rely on the language of circuits:

Definition (Span Function of a Matroid)

For matroid , we define the span function of matroid , as those "representable" by the input set, that is:

Those are considered representable, at least by . However, this is incompatible with the language of circuits since a set disallows duplicate elements and thus is not a circuit. In fact, when is independent, there's no way we can construct a circuit with and . In this way, itself has to be unioned.

When the context of matroid is clear, we may write directly for the input . When , then we say spans .

Does defining in such a way eliminate the problem of dependent sets' "spanning" every element in , let's give it a try: in a matroid on , there's a loop , while no singleton set of is loop. When we evaluate , it won't be hard to find none of the sets is circuit in for containing circuit . So does not span and . So there's no longer the problem of dependent sets' "spanning" every element by defining so.

Let's see what span function looks like in a cycle matroid:

Example (Span Function in Cycle Matroids)
In a cycle matroid , for , an edge iff or in the spanning graph .

Proof. If then is trivial, so we may assume . And if , there's a -path such that is a cycle, and thus a circuit. So in this case.

Conversely, if , then there's a subset such that is a circuit, so there's a cycle such that , and removing from this cycle yields a -path, thus .

The independent sets and dependent sets in a matroid can also be characterized by the span function:

Lemma

Proof. For the sufficiency, if there's such that , then a subset completes a circuit , but , which is a contradiction.

For the necessity, we prove the contrapositive . If , then , and since is not empty, , while .

Corollary

The lemma above can be intuitively interpreted in a cycle matroid, it means for edge set , the spanning subgraph is a spanning forest iff every is a cut edge.

For any vector space, we know it suffices for a basis to span it. And this property also holds for a general matroids:

Lemma

Proof. If , consider any , we claim that . Otherwise contains a circuit , where must be present if the circuit won't like to be a subset of . By the presence of we have , which is a contradiction.

However, is still a contradiction, so only is possible.

Span Function Defined by Rank Function

Let's see another seemingly "careless" way of defining span function. In a vectorial matroid , if we want to test whether vector is expressible by the linear combination of , we may put vectors of into a matrix (assume are column vectors) and evaluate , which equals . Then we augment with , resulting in , evaluate and see whether . If then can be eliminated by Gaussian elimination, which requires representing as the linear combination of .

And we would like to show it's as effective to express elements spanned by a set using rank function, as the one using circuits:

Theorem

Proof. For the sufficiency, consider any basis , it seems that since otherwise . So every contains a circuit , and if won't like to a subset of . And .

For the necessity, first if , then and . So we may assume . If so, then . Noted that is independent in since it's formed by removing from a circuit .

Noted that is also independent in , thus we may extend to a basis . Clearly otherwise , therefore is to be found in , and thus:

In this way, we are done with the proof.

Corollary

So is there any intuitive way to interpret why it works by doing so?

Well, by evaluating , it can be seen as reducing matrix of elements in into a maximal independent set. If , then the maximal independent set can represents , thus forms a circuit with elements from this maximal independent set, and thus . Conversely, if by forming a circuit with some elements in , that is, is representable by some elements in with non-zero coefficients, then move these elements to the first few rows and reduce the matrix to its maximal independent sets. The first few rows is independent and thus intact during elimination. Adding to the matrix will not increment the rank since it is sufficiently eliminated by the first few rows, and thus . So the rank function has intrinsic connection to the circuits.

Defining span using rank function can easily lead to:

Lemma (Incorporation Property)
Proof. Since and . By the strong absorption rule, we have .

Closure Operator Properties

Let's first introduce the definition of a closure operator:

Definition (Closure Operators)
A mapping is said to be a closure operator on , if it fulfills:

  1. (Extensive) .
  2. (Increasing) .
  3. (Idempotent) .

And we would like to show:

Theorem
For a matroid , the span function defines a closure operator on .

Proof. The extensive property is obvious since .

The increasing property is true by considering:

To prove the idempotent property / idempotence, since by extensive property there has already been , it suffices to show , which will leads to , and thus mutual inclusion.

This is easy when using the rank function definition of span, that , while . And since , we have , and thus .

Corollary
Proof.

In this way, it's said that defines a closure operator in a matroid.

These concepts are closely related to the closure operator aspect of matroids:

Definition (Spanning Sets, Closed Sets and Hyperplanes)

For a matroid on and a subset , if , then is said to be the spanning set of . If , then is said to be a closed set or flat, clearly a closed set is the spanning set of .

The hyperplanes are the closed sets that are maximal proper subsets of , that is, being a proper closed superset of a hyperplane implies it is the ground set .

For convenience, we may denote the set of spanning sets as , and the set of hyperplanes as .

When combining with the property , one can easily see:

Lemma
A set is spanning iff it contains a basis.
Proof. The sufficiency is obvious. For the necessity, just consider , the set contains an independent set of size , which must be a basis of .
Lemma
A set is hyperplane iff it is a maximal non-spanning subset of .

Proof. For the sufficiency, let such a maximal non-spanning subset of be , first we prove to be closed. Assume it's not closed, then , however implies is also non-spanning, which is a contradition to the maximality of . Let be any proper closed superset of , then by definition it's spanning, so . In this way, matches the definition of a hyperplane.

For the necessity, let such a hyperplane be , since has already been non-spanning, we just need to prove its maximality. Assume it's not maximal that is also non-spanning, then by extensive and increasing property . However, since is a closed proper superset of , there must be , and is a contradiction.

Corollary
A set is hyperplane iff it is a maximal subset of containing no basis.
Proof. Being non-spanning is equivalent to containing no basis.
Corollary
Proof. Fix any choice of . By definition is spanning and containing a basis . Clearly if won't like to contain the basis . Noted that , by hereditary property , so . And by combining with , there can only be .
Corollary
Proof. Take any and inspect any . Since while , there must be . And since the rank function is monotonic, it must only be , and therefore , which implies . Consider any proper superset of , we have and therefore . So is a maximal non-spanning subset of .

Steinitz Exchange Property and Closure Operator Axioms of Matroids

In linear algebra, the Steinitz exchange lemma states, that if there's a set of vectors plus a vector , that spans a vector space , then for any vectors , there's .

The proof is simple, when , there's and the proposition is trivially true. When , in any representation of , we must have . Thus for there must necessarily be a representation such that , otherwise there must be by there's no way to fix into . In this way, we must have , and thus .

This phenomenon can also be generalized to matroids:

Lemma (Steinitz Exchange Property for Matroids)

Proof. When , by , the proposition is trivially true. When , there must at least be .

And for every , the proposition is trivially true when , so we should assume . In this way, by definition there's a circuit such that . And there must be since otherwise . For convenience, we will let for . Noted that and thus is a circuit implying . By combining with the presumption we are done with the proof.

It won't be hard to see the proof is just a rephrashing of the proof of Steinitz exchange lemma in linear algebra.

So the span function of a matroid are the kind of closure operators with Steinitz exchange property. Conversely, we would like to show it suffices for a closure operator with Steinitz exchange property to define a matroid cryptomorphically.

Recall that independent sets of a matroid can also be retrieved by , so we would like to show a hereditary system may be defined in such a way first:

Definition
For a closure operator on , we may let .
Lemma
The set defines the independent sets of a hereditary system on .

Proof. First, for empty set , there's no element to check whether it's in , so we have trivially.

Then, just like what we've done in showing defines the independent sets before, we would like to show . Assume there's such that , then by incresing property , and we have , which is a contradiction.

Then we would like to show the Steinitz exchange property enables a closure operator to define a matroid. The role that Steinitz exchange property plays in the proof is indispensible and must be emphasized:

Theorem
For a closure operator on , if it fulfills the Steinitz exchange property , then is a matroid.

Proof. We have shown to be a hereditary system on , so it remains to show has the independent set exchange property that .

Recall that in a matroid, the independent set exchange property means , this will require . So if the hereditary system would like to be a matroid, it must only be the case indefinitely.

First, we would like to show . Such a property is intuitive in a matroid: fix the precondition , if , then must contain a circuit with inside, and thus . By taking the contrapositive and adding the precondition, we have . The proof is done by showing . Assume the contrary that , and by we have . However, by Steinitz exchange property we have , but has been expected, which is a contradiction.

In this way, once there's , it will lead to sufficiently.

Then, it remains to show leads to a contradiction, since it won't happen in a matroid. By increasing and idempotent property, we have .

Interestingly, we are able to show show whenever there's a way to "hybridize" into independent set using some , so that there's again.

Since , undoubtedly the element exists. Obviously by extensive property we have , while .

Fix the choice of , we argue that : Assume the contrary that , then by the idempotent property , however is a contradiction.

Fix the choice of , by extensive property we have , by combining with we have . So such an element must be chosen from .

We can easily show by using this argument, given that there has already been .

Finally, we would like to show : by extensive property we have and , therefore . And then apply the increasing property and idempotent property we have .

Noted that each step of the "hybridizing" process replaces one element of with , while maintaining the condition to apply the process, so it persists until it reachs a point that for some such that . At this point there's , and there's also by increasing property of a closure operator, so by mutual inclusion. However, consider any , there's , which is a contradiction. Thus the case of is impossible.

In this way, when there's always , which sufficiently leads to , and independent set exchange property is to be found in .

Corollary (Closure Operator Axioms of Matroids)
For a closure operator on , it fulfills the Steinitz exchange property iff is a matroid.

So closure operators with Steinitz exchange property are equivalent to matorids.

Strong Elimination Property

With the language of span function, we may now prove the strong elimination property we've mentioned when showing the weak elimination property of matroids.

Theorem (Strong Elimination Property)

Proof. If there's such a circuit , then it must be an evidence of . For convenience, we may let , and our goal is to show .

We know , so we may assume , and the case of can be handled analogously. First, we have with as the evidence. And our plan is to show .

First, it won't be hard to find there has already been . The key is at the opposite side , where we have with as the evidence. There is also
, and by increasing and idempotent property we have . Now we have , and therefore . Then by applying the increasing idempotent property we have , and we are done with the proof.

Dual Matroids

By utilizing the properties of circuits in a matroid, we are ready to show the strong basis exchange property of a matroid:

Theorem (Strong Basis Exchange Property)

Proof. First, since , contains a unique circuit . Such a circuit must contain if it won't like to be a subset of . Meantime, such a circuit must contain elements from if it won't like to be the subset of .

Fix the choice of any element , we would like to show is independent: Otherwise is dependent and contains a circuit . Again if it won't like to be a subset of . Noted that and , which means and are distinct circuits. And since both of them contains , by weak elimination rule , however , which is a contradiction.

Generally speaking, the choice of elements such that enables does not neccessarily guarantee , and vice versa.

However, by utilizing the properties of span function, we are able to prove, there does exist choice of elements such that enables simultaneously:

Theorem (Even Stronger Basis Exchange Property)

Proof. Fix some , from the proof of strong basis exchange property, we know contains a circuit such that . If we are able to furtherly show , then we will be done with the proof.

Consider , we know due to its completing the circuit . Since , we have . Since contains the basis , it's spanning, therefore , so is also spanning, and contains a basis .

Since while , it must contain elements from . Meantime there's , so must use some to augment by independent set exchange property, into , and we are done with the proof.

The even stronger version of strong basis exchange property is not required by the following text, but required by the later topic of the matroid intersection theorem.

The strong basis exchange property sheds light on some sort of duality in a matroid. That is, we may retrieve the dual base system from a base system of a matroid:

Definition (Dual Base System)

For a matroid on , the dual base system is a combinatorial object such that:

Where (or ) is considered in the power set lattice of .

And then show it to be a base system, which defines the dual matroids:

Theorem
For a matroid on , the dual base system is a base system.

Proof. It won't be hard to find the elements of are obtained by taking the map . So if is not empty, then is not empty; If the elements in are of size , then the elements in are of size . And we still need to show the basis exchange property in it.

Consider distinct bases , they are corresponding to , and we want to show . Since , and similarly , so we need to show , which is equivalent to showing . The must accept an element from and then choose an element in to discard, and the strong basis exchange property of enables it to do so. In this way, the basis exchange property is present in .

Definition (Dual Matroids)

For a matroid on , the dual matroid of is a matroid on such that . It won't be hard to find , for which it's said to be dual.

Conventionally, we may add co-prefix for those objects from the dual matroid, e.g., an independent set in may be called a coindependent set in , a basis in is called a cobasis in , and so on.

For convenience, we may add asterisk () to a matroid theoretic symbol, in order to signify it's exactly the same entity from the dual matroid. For example, we may write , , , and so on.

By now, we can clearly see how is the strong exchange property of a matroid connected to its duality.

The dual matroid arises from the dual graph of a planar graph, and the topological arguments of planar graphs can also be generalized to matroids, offering an alternative characterization of planar graphs by matroids, the Whitney's Planarity Criterion. We have a later topic of matroid duality and topology, and we will be focusing on the non-topological part of dual matroids in this text.

Coindependent Sets and Corank Function

We begin with arguing a set is coindependent iff it's the complement of a spanning set:

Lemma

Proof. For every , it can be extended to a cobasis , which corresponds to basis , and . So is a spanning set, and we have . By taking complement in the sets pointwisely, we have .

For every , it contains a basis . Since is a cobasis, and , is coindependent, and thus we have .

By mutual inclusion, we have By taking the dual we have , and by taking the pointwise complement we have .

This property can easily lead to the dual augmentation property:

Theorem (Dual Augmentation Property)
Proof. First, we know , which implies , and . Then we may use to augment into a basis , given that both and are disjoint with , the augmented basis is also disjoint with , therefore .

In this way, we may always permute the elements of , so that for disjoint independent set and coindependent set , the following setup exists:

By the definition of dual matroid, one can easily see , but the relation between rank and corank for can't be easily seen. In fact, by the construction of matroid and subset , the values of can be arbitrary:

Example

Consider the following matroid:

We have while .

Therefore one must not expect there to be any direct connection between and . However, there's still connection between and :

Theorem (Whitney)

Proof. First we find the maximal coindependent set . The coindependent set can be extended to a cobasis , and the extension must not use elements in if we won't like to contradict with the maximality of . So the rest of elements are contained in the basis , and therefore is independent. On the other hand, we find another independent set .

Permute the elements in so that we can visualize our setup easily:

Since , and , there must be . And by independent set exchange property, if , then the maximal independent subset of has more element than , and can be used to extend . In other word, if we are able to show can't be extended by elements in , there must be .

In the figure above, noted that we may permute elements of freely, without affecting the setup it represents, and so do other elements in . Without loss of generality, we may assume can be used to augment into . The we can use to augment into another basis of , using merely elements from . Again, without loss of generality, assume all elements of but are used to augment .

As we can see, now is extended into basis , corresponding to the cobasis . Noted that , so by hereditary property while , which is a contradiction to is the maximal coindependent subset of .

Finally, by some counting as simple as ,
we are done with the proof.

Cocircuits

Just like the connection between coindependent sets and spanning sets, we have connection between cocircuits and hyperplanes. We state that a set is cocircuit iff it's the complement of a hyperplane:

Lemma
A set is a circuit iff it is a minimal set containied in no basis.
Proof. A set if independent iff it's contained in a basis: the hereditary property implies its sufficiency; and by fixing any basis, an independent set may be augmented to a basis due to the independent set exchange property, which implies the necessity. The statement is equivalent to a set is dependent iff it's contained in no basis. By definition, a dependent set is circuit iff it's minimal, therefore a set is circuit iff it's a minimal set contained in no basis.
Lemma

Proof. The previous lemma can be translated to . The key point is to utilize the one-to-one correspondence of a basis and a cobasis , and vice versa.

For the first statement, we first transform into . Then we notice that is actually arbitrary cobasis, by letting for any desired . In this way, contains no cobasis.

For the second statement, by taking the complement we can rewrite it as , which is furtherly equivalent to . Therefore adding any elements outside enforces to contain a cobasis.

In this way, we can see iff is a maximal set that contains no cobasis, and thus iff is a co-hyperplane. That is, . By taking complement on both sides again, we have ; by taking it to the dual matroid we have , which is exactly what we are to prove.

Corollary
A set is dependent iff it intersects every cobasis. A set is circuit iff it is the minimal set intersecting every cobasis.

Proof. In the proof of previous lemma above, and for the first statement, if we transform into , since is arbitrary cobasis, the first statement is now restated as that intersects every cobasis.

Noted that the first statement actually characterizes the dependent sets, that a set is dependent iff it intersects every cobasis.

For the second statement, can be rewritten as , and again . So is disjoint with cobasis , and does not intersect every cobasis any more.

Corollary
A basis intersects every cocircuits, and a cobasis intersects every circuits.
Proof. If basis does not intersect with cocircuit , then is the very basis that does not intersect with, which is a contradiction to the corollary above. Analogous argument is applicable to cobasis versus circuits.

In this way, we can see the independent sets and circuits are actually connected and interchangeable, due to the string the dual matroids pull for us.

In fact, due to the corollary, we are now able to interpret the dual matroid of a cycle matroid intuitively, for bonds, the minimal edge cuts:

Lemma
An edge set of a graph is an edge cut iff it intersects every maximal spanning forest.

Proof. We will first show the necessity so that one can gain some intuition. Let edge set be an edge cut, splitting into for component , so that vertices in and are disconnected. If does not intersect with a maximal spanning forest of , which contains maximal spanning tree for the component , then is to be found in , and vertices in are still connected by taking paths in , which is a contradiction to is an edge cut.

Now consider the sufficiency. If is not an edge cut, then consider any connected component , it's still connected in , so there's a maximal spanning tree in for component such that , and such a tree is also to be found in . By iterating every component, evaluating maximal spanning trees in and unioning them, we obtain the maximal spanning forest to be found in both and , which is the very one that does not intersect with.

Theorem
And edge set is a cocircuit in cycle matroid of graph iff it's a bond, the minimal edge cut.
Proof. A set is edge cut iff it intersects every maximal spanning forest, while maximal spanning forests are the bases of , so by this corollary, a set is an edge cut iff it's co-dependent in . A bond is a minimal edge cut, and a minimal co-dependent set is a cocircuit, both minimalities are taken under set-inclusion. So just by connecting them with "iff", we will be done with the proof.
Definition (Bond Matroids)
For integrity, just like we name cycle matroid due to cycles' being the circuits of the matroid, we may name the dual matroid the bond matroid of .

We will discuss more about the topological aspect of the bond matroids in the later topic.

Finally, we want to characterize the circuits and cocircuits by their intersections:

Lemma
Proof. Assume the contrary that . Now we obtain disjoint independent set and coindependent set , so dual augumentation property is applicable. Let them be augmented to basis and cobasis . But where does goes? If then ; and if then . Either case is a contradiction.
Lemma
Proof. First augment to a cobasis . Since is non-empty, pick any , since , is dependent and contains a circuit , such that and thus .
Theorem
A set is cocircuit iff it's the minimal among the non-empty sets such that for every circuit .

Proof. We begin with the necessity to ensure we are on the correct trail. If is a cocircuit, then every non-empty proper subset is coindependent, and all of them fulfills by the second lemma above. So is the minimal set among the non-empty sets such that . Be vary that the minimality is taken among the non-empty sets, since there's .

Now we prove the sufficiency, given a set that is minimal among the non-empty sets that while . Clearly is co-dependent, since otherwise it must fulfill , the problem remains as whether is a cocircuit. Assume it's not, then must contain a proper subset that is a cocircuit, and that cocircuit must fulfill . However, is non-empty for being a cocircuit, and it's a proper subset of , so there is , which is a contradiction. In this way, must be a cocircuit, and we are done with the proof.

Matroid Isomorphisms

It won't be hard to see, it's the independent sets and their set-inclusion relationship that really matters. So we may define isomorphism in such a way:

Definition (Matroid Isomorphisms)

For matroid and matroid , given a bijection , we may define for any subset . So we may also view as , obviously it's also a bijection from to .

Such a bijection is said to be a matroid isomorphism, iff . Matroid and are said to be isomorphic iff there's a matroid isomorphism between them.

We may denote such an relation as , and denote a matroid isomorphism as , with and to be the defined in the way shown as above.

Proof. It suffices to show is a matroid. Then given that , the matroid coincides with .

First, it won't be hard to find is a hereditary system, since and for , just consider , we have .

Then, the independent set exchange property may also be shown in analogous way, that for , since and , it won't be hard to see .

Just like graph isomorphisms, the matroid isomorphisms also categorize matroids into isomorphism classes. We will be going over some important isomorphism classes of matroids in the remaining text of this section.

On the other hand, since the ground set of a matroid is always finite, we may define matroids over letter set , and use them as the representatives of the isomorphism classes.

Uniform Matroids, Direct Sum and Partition Matroids

The simplest form of matroids may be the uniform matroids:

Definition (Uniform Matroids)

Given a ground set and a parameter , we may define a matroid , such that . Such a matroid is called a uniform matroid
of rank
, given that . Specially, when , the uniform matroid becomes , and we call it the free matroid then.

To categorize them, we may define matroid over letter set such that . Obviously there's always .

By putting some uniform matroids over disjoint ground sets together, with their direct sum, we obtain the parition matroids:

Definition (Direct Sum of Matroids)

Given a matroid and a matroid over disjoint ground sets , we may define their direct sum , which is a matroid over with .

Isomorphisms are preserved by direct sums, that , where for , an isomorphism between the direct sum may be defined as
and .

The direct sums of three or more matroids over disjoint ground sets as summands can be obviously defined in such a way, so that there are and , and thus are well defined. Isomorphisms are also obviously preserved.

Proof. Obviously and the hereditary property is preserved. For the independent set exchange property, if , then either or is true. Without loss of generality, let it be the former case, and thus .
Convention (Direct Sum of Matroids over Letter Sets)

When doing direct sums of two or more matroids over letter sets, we automatically add subscriptions to make the ground sets disjoint and identifiable, by which matroid they are from.

For example, we may define and analogously. So that for matroids and , when doing direct sum , they will be viewed as over and instead. Assume are taken from the independent sets, they will first become , and then .

Definition (Partition Matroids)

Given a series of uniform matroids over disjoint ground sets , their direct sum is called a partition matroid.

A uniform matroid can be viewed as a partition matroid of single summand. And there's always .

The problem of matching in bipartite graph is expressible in the language of the intersection of the independent sets, of two partition matroids:

Example (Matchings in Bipartite Graphs in the Language of Matroids)

Consider the following -bigraph , we may define two partition matroids and over :

So that an edge set is a matching iff it is a common independent set, lying in . Finding a maximal matching is equivalent to finding an common independent set such that there's no common independent set being the super set of it. Finding a maximum matching is equivalent to finding the common independent set with maximum cardinality.

While the connection above seems like nothing more than an uninteresting coincidence, mathematicians have grasped this clue, discovered and proved its "converse", which is truly mind-blowing: it turns out that finding the maximum (cardinality) common independent set of any two matroids over the same ground set, not limited to partition matroids, is reducible to multiple iterations of finding / augmenting matchings, in delicately constructed bipartite graphs per iteration.

Meantime, there's a min-max duality similar to König-Egerváry Theorem between the maximum cardinality of the common independent sets and the rank functions in each matroids, due to the way we find the maximum common independent set.

We will discuss these in detail in the later topic of matroid intersection theorem.

Linear Matroids and Graphic Matroids

Let's introduce the concept of column matroids first:

Definition (Column Matroids)
Given matrix , where are column vectors, we may define the column matroid of , by using the column vectors as the ground set , and letting that the subset is independent iff is a linearly independent subset.

While vectorial matroid deduplicates vector as soon as the ground set is defined, defining column matroid allows us to identify elements by which column identity, rather than their underlying vectorial value. We will see why defining so is useful soon.

By now, we may unify matroids that can be represented by certain kind of vector spaces:

Definition (Linear Matroids)

A matroid is said to be linear or representable if it's isomorphic to a vectorial matroid or column matroid.

More specifically, if matroid is linear, and elements of its ground set are taken from -vector space , then is said to be representable over field . When , is said to be real representable; when , is said to be binary; and when , is said to be ternary.

Whether is finite dimensional or not doesn't matter: given that the ground set is always finite, we may take , then the elements of can be seen as taken from , while is always finite dimensional.

On the other hand, we may also unify matroids that can be represented by certain kind of cycle matroids:

Definition (Graphic Matroids)
A matroid is said to be graphic if it's isomorphic to a cycle matroid .

Specially, we claim that:

Assertion
A graphic matroid is linear and representable over any field.

To prove this statement, clearly it suffices to show every cycle matroid has some kinds of linear matroids corresponding to it.

Our construction is, for every cycle matroid , we orient graph arbitrarily into digraph , and take its incidence matrix . Here, we allow loops to be present, by requiring the corresponding column of that loop in to be zero vector. Here's an example of our proposal:

Example

For some randomly taken graph , we orient it arbitrarily into , and evaluate its incidence matrix:

Clearly only can be the elements of , while for any field .

We want to show is the very linear matroid isomorphic to . This will require us to show an edge subset is dependent in iff its corresponding column vectors in are linearly dependent, regardless of which field the matrix is seen as over. Noted that for an oriented edge corresponding to column vector , if it were oriented into instead, the only difference is that its corresponding column vector in becomes now, which doesn't affect the linear independence of the column vectors of . So the way we orient doesn't matter.

Let's begin with showing the necessity:

Lemma
For graph and its arbitrary orientation , if an edge subset is dependent in , then the corresponding column vectors of in are linearly dependent.

Proof. If edge subset is dependent, then it contains cycle in . Let the corresponding column vectors of in be . Without loss of generality, we may assume the cycle is not a loop, otherwise it's trivially . We construct an identity such that or with the following steps: Follow the cycle, if in is oriented to in , then we assign , otherwise we assign .

Consider the path component and the corresponding prefix sum , it won't be hard to find the row corresponding to is set to while the row corresponding to is set to . And is obtained from adding to , where , corresponding to , provides to cancel the at the row corresponding to , and then introduces at the row corresponding to . In this way, when we enclose the cycle, , so the corresponding column vectors of are linearly dependent if is dependent.

And since the coefficients of the identity can only be or , such an identity should reveal linear dependence over any field .

Those familiar with Kirchoff's Matrix Tree Theorem should be very familiar with the construction above.

Now let's show the sufficiency, that given a set of linearly dependent column vectors corresponding to edges , a cycle is to be found, and therefore they are dependent in . Again we may assume no edges in is loop, otherwise it's just the trivial case.

Consider the identity revealing the linear dependence of . Without loss of generality, we assume , otherwise we eliminate those terms with zero coefficients. The subset of column vectors with non-zero coefficients is still linearly dependent, but for the corresponding edge subset, once we are able to show there's a cycle inside, any super set of it contains such a cycle.

Our argument of sufficiency is based on the following algorithm:

Algorithm
The input is the edge set corresponding to column vectors , fulfilling the identity such that the coefficients are non-zero. The output is the cycle formed by edges in :

  1. Initialize empty path , zero vector , and counter .
  2. Concatenate , update , and increment .
  3. Now for some vertex , and the row corresponding to in is . We assert there's some edge incident to in . That is, either oriented edge or is to be found in .
  4. If , then since contains a segment , and is a cycle, and we halt with as the cycle found.
  5. Otherwise update , such that if , and if , and increment . Then go to step 3.

Proof. Obviously if the algorithm would like to halt, it must halt at step 4.

We've asserted that every time step 3 is executed, there's edge in incident to , this is due to the corresponding row of is in . If not, there must be only one edge incident to in , and such an edge must be only used once. Since is preserved as a path, and each vertices and edges must occur only once, otherwise the algorithm must have halted at step 4. But if so, we look at the row corresponding to in the identity , it's only , as there's no zero divisor in field , which is a contradiction to our input constraint . In this way, as long as is non-empty, there's such an edge incident to to be selected at step 3.

Now we show it must halt at step 4 before exhausting every edge in . If not so, then the algorithm concatenates all edges in into a path . And is the only edge incident to . Again, if we look at the row corresponding to in the identity , it's only , which is again a contradiction to our input constraint .

Lemma
For graph and its arbitrary orientation , if a subset of column vectors in are linearly dependent, then its corresponding edge subset is dependent in .
Proof. Just eliminate the zero coefficients in the identity revealing linear dependence, and then execute the algorithm above, we will find a cycle formed by the corresponding edge subset.
Theorem
A graphic matroid is linear and representable over any field.

Proof. We've done with showing: For graph and its arbitrary orientation , an edge subset is dependent in iff the corresponding column vectors of in are linearly dependent.

So there's , and any graphic matroid isomorphic to is isomorphic to , for the matrix over any field .

While every graphic matroid is linear, the converse is not true. The simplest example might be the uniform matroid :

Example

The uniform matroid is linear and ternary. Consider the matrix over . It's trivial to verify that any two column vectors are linearly independent, but:

So clearly due to matroid isomorphism .

However, is never binary, there's no way for to be represented over . Assume the contrary that it's isomorphic to some linear matroid over ground set , taken from -vector space . Then any two column vectors are linealy independent, and any three column vectors are linearly dependent.

Let and to be taken. The identities revealing their linear dependence can only be:

All coefficients must be , otherwise it will render the combination of one or two vectors to be dependent, which is a contradiction. But once we add these identities together, there will soon be:

So must be linearly dependent, which is a contradiction to any two column vectors are linearly independent.

So there's no way for uniform matroid , which is linear and ternary, to be isomorphic to a graphic matroid. Since a graphic matroid must be representable over any field, but is never representable over sufficiently.

November 24, 2024