Dual matroid

Remarks

It is a theorem that $\mathcal{B}^*$ satisfies the basis exchange property, so $M^*$ is indeed a matroid. Its independent sets are the subsets of $E$ disjoint from some basis of $M$. Its rank function is $r^*(A) = \abs{A} - r(E) + r(E \setminus A)$, where $r$ is the rank function of $M$, and $(M^*)^* = M$. The bases of $M^*$ are the cobases of $M$, and the circuits of $M^*$ are the cocircuits of $M$.

Example: for the uniform matroid $U_{k,n}$, whose bases are the $k$-element subsets of $\set{1, \dots, n}$, the dual has as bases the $(n-k)$-element subsets, so $U_{k,n}^* = U_{n-k,n}$.

Non-example: the complements of the independent sets of $M$ are in general not the independent sets of $M^*$. For $M = U_{1,2}$ on $\set{1, 2}$, the complement of $\emptyset$ is $\set{1, 2}$, which is not independent in $M^* = U_{1,2}$.