[proofplan] We compare bases. The bases of $M^*$ are the complements in $E$ of bases of $M$. Applying the same definition again, the bases of $(M^*)^*$ are the complements in $E$ of those complements, hence exactly the original bases of $M$. Since a matroid is determined by its bases, the two matroids are equal. [/proofplan]

[step:Compute the bases of the double dual]Let $\mathcal{B}(M)$ denote the set of bases of $M$. By definition of the dual matroid, \begin{align*} \mathcal{B}(M^*)=\{E\setminus B : B\in \mathcal{B}(M)\}. \end{align*} Applying the same definition to $M^*$ gives \begin{align*} \mathcal{B}((M^*)^*)=\{E\setminus C : C\in \mathcal{B}(M^*)\}. \end{align*} Substituting the description of the bases of $M^*$, every basis of $(M^*)^*$ has the form \begin{align*} E\setminus(E\setminus B) \end{align*} for some $B\in\mathcal{B}(M)$. Since complements are taken inside the same ground set $E$, this set is just $B$.[/step]

[guided]The only operation involved in duality here is taking complements inside the fixed finite ground set $E$. If $B$ is a basis of $M$, then $E\setminus B$ is a basis of $M^*$. Applying duality once more takes the complement of this set in the same $E$, and \begin{align*} E\setminus(E\setminus B)=B. \end{align*} Thus every basis produced after two dualizations is one of the original bases of $M$.[/guided]

[step:Conclude equality of the matroids] The preceding computation proves both inclusions: \begin{align*} \mathcal{B}((M^*)^*)\subset \mathcal{B}(M) \end{align*} because each double-[dual basis](/theorems/414) is $E\setminus(E\setminus B)=B$ for some basis $B$ of $M$, and \begin{align*} \mathcal{B}(M)\subset \mathcal{B}((M^*)^*) \end{align*} because every $B\in\mathcal{B}(M)$ arises as the complement of the dual basis $E\setminus B\in\mathcal{B}(M^*)$. Hence $\mathcal{B}((M^*)^*)=\mathcal{B}(M)$. A matroid is determined by its collection of bases, so $(M^*)^*$ and $M$ are the same matroid on $E$. [/step]