Let $M$ be a matroid with ground set $E$.
Deletion and contraction in matroids commute with each other and with themselves, i.e. for all $e,f \in E$ one has


$(M/e)\setminus f = (M\setminus f)/e$, $\hspace{0.1cm}$ $(M\setminus e)\setminus f = (M \setminus f) \setminus e$ $\hspace{0.1cm}$ and $\hspace{0.1cm}$ $(M/e)/f = (M/f)/e$. 


Are there any matroids, aside from uniform matroids, which have the following property:


For all $e,f\in E$


$(M/e)\setminus f = (M/f) \setminus e$ ?



Interchanging deletion and contraction in matroids
Uniform matroids are the only matroids with this property.


Assume we have a matroid $M$ of rank $k$ on a ground set $E$ with this property. Consider a base $B$ of the matroid and any $e \in B$, $f \not\in B$. If we cannot find such $e$ and $f$, then $M$ must be a uniform matroid. Now, $B \setminus e$ is a base of $(M / e) \setminus f$ by definition. The property in question says $B \setminus e$ is a base of $(M / f) \setminus e$. Thus, $B \setminus e = B' \setminus f$ for some base $B'$ of $M$ not containing $e$. Hence, $(B \setminus e) \cup f$ is a base of $M$. Therefore, $M$ is uniform since we can exchange any two elements.


