Let $(E,I)$ be a matroid, and let $A,B \in I$ be disjoint independent sets in the matroid. Moreover, let $B_1,\ldots, B_k$ be a partition of $B$. I could not decide if the following is always true. There is a subset $S \subseteq A$ and a partition $A_1,\ldots,A_k$ of $S$ such that for all $i \in \{1,2\ldots,k\}$ it holds that $(A \setminus A_i) \cup B_i \in I$ (is independent in the matroid) and $|A_i| \leq |B_i|$.

This forms a generalized exchange property of matroids w.r.t. the arbitrary partition $B_1,\ldots, B_k$. Any help proving this property or an idea for a counterexample would be appreciated.

  • $\begingroup$ (1) This holds for linear matroids: we may increase $A$ and $B$ so that they become bases (possibly intersecting already), denote by $B_{k+1}$ the set which we added to $B$ and consider the matrix $M$ where the columns are indexed by $B$, rows by $A$, and in the column indexed $b$ the entries are the coordinates of $b$ in the base $A$: $b=\sum M(a,b)\cdot a$. The determinant of this matrix is non-zero. But it is a (signed) sum of products of$k+1$ minors, where $i$-th minor is chosen in the columns indexed by $B_i$. Therefore one such product of minors is non-zero. $\endgroup$ Dec 17, 2023 at 22:30
  • $\begingroup$ (2) The corresponding partition of (even extended) $A$ works, right? My feeling is that the difference between general and linear matroids is not in properties like such, so I expect this to be true in general. $\endgroup$ Dec 17, 2023 at 22:31

2 Answers 2


Well, I think so, and it is less or more equivalent to the basis exchange theorem (which I reinvented answering your question, but looking for a reference to the matroid union theorem I realized that this is a known application of matroid union.)

At first, we ignore that $A$ and $B$ are disjoint (it can not be important a priori, since we may always add twins to our matroid $M$.) But we require that $A\setminus A_i$ and $B_i$ are disjoint.

At second, we suppose that $A$, $B$ are bases (this is rather matter of taste, for dealing with equalities of sizes instead of inequalities). This may be assumed, since we can extend both $A$, $B$ so that they become bases (the partition of $B$ gets a new part $B_{k+1}$) and find the corresponding partition of the extended $A$, it induces a necessary partition of the old $A$. So, from now on $A,B$ are bases and we look for a partition of $A$ such that $(A\setminus A_i)\sqcup B_i$ are bases of our matroid $M$ (then $|A_i|=|B_i|$ is granted.)

The $k=2$ case is the basis exchange theorem which we reproduce here as

Lemma. Let $M$ be a matroid. Let $B$, $A$ be two bases of $M$. Assume that base $B$ is partitioned as $B=B_1\sqcup B_2$. Then base $A$ has such a partition $A=A_1\sqcup A_2$ that both $A_1\sqcup B_2$, $A_2\sqcup B_1$ are bases of $M$.

Now we assume that $k>2$ and the claim is proved for $k-1$. Using lemma, find a partition $A=A_0\sqcup A_k$ such that $A_0\sqcup B_k$ and $A_k\sqcup B_0$ (where $B_0=:B_1\sqcup B_2\ldots \sqcup B_{k-1}$) are both bases of $M$. Consider the matroid $M_0=M/A_k$ on $E\setminus A_k$. Then both $A_0$ and $B_0$ are bases of $M_0$, and by induction hypothesis there exists a necessary partition $A_0=A_1\sqcup \ldots \sqcup A_{k-1}$. I claim that the partition $A=A_1\sqcup \ldots \sqcup A_{k-1}\sqcup A_k$ works. Indeed, for $i=k$ we already know that $(A\setminus A_i)\sqcup B_i$ is a base of $M$. If $i<k$, we know that $(A_0\setminus A_i)\sqcup B_i$ is a base of $M_0=M/A_k$ that yields that $((A_0\setminus A_i)\sqcup B_i)\sqcup A_k$ is a base of $M$, but $((A_0\setminus A_i)\sqcup B_i)\sqcup A_k=(A\setminus A_i)\sqcup B_i$.

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    $\begingroup$ Thank you for the proof. Can we prove even a stronger property? Assume that $A,B$ are bases of the matroid. Is there a one-to-one mapping $f:B \rightarrow A$ such that for every partition $B_1,\ldots,B_k$ of $B$ it holds that $(A \setminus f(B_i)) \cup B_i \in I ~\forall i \in [k]$? This means essentially that we replace every $b \in B$ by the same element $f(b) \in A$ in all partitions. I am not sure if this is true, but I thought it would be interesting. $\endgroup$
    – John
    Dec 20, 2023 at 11:16
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    $\begingroup$ I do not think so. Consider a graphic matroid for $K_4$ and two disjoint bases (each being a path of length 3). If I am not mistaken, there is no such bijection. $\endgroup$ Dec 20, 2023 at 13:44
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    $\begingroup$ The same counterexample works, does not it? $\endgroup$ Dec 21, 2023 at 7:56
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    $\begingroup$ Well, this is correct: adding $b$ to the basis $(A\setminus A_1)\cup B_1$ you get the unique cycle, it must contain $b$ but can not be contained in $B$. Thus this cycle contains some $a\in A\setminus A_1$, which works $\endgroup$ Dec 25, 2023 at 17:45
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    $\begingroup$ well, this is not correct: it may occur that $b$ and some guy from $A_1$ form a cycle of length 2, and there is no other cycle in the whole story $\endgroup$ Dec 26, 2023 at 12:54

I think the question and the follow-up discussion is closely related to the following theorem of Greene and Magnanti: If $A$ and $B$ are two bases of a matroid $M$ and $\{A_1,...,A_m\}$ is a partition of $A$, then there is a partition $\{B_1,...,B_m\}$ of $B$ such that $(A-A_i)\cup B_i$ is a basis for each $i=1,...,m$. (See C. Greene and T.L. Magnanti, Some abstract pivot algorithms, SIAM J. Appl. Math. 29 (1975) 530–9.). Further extensions are known as well, see e.g. the paper of Gabow (H. Gabow. Decomposing symmetric exchanges in matroid bases. Mathematical Programming, 10(1):271–276, 1976.)


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