The dimension of the circuit space of a matroid $M$ is the corank of $M$ if and only if $M$ is binary.  Here is a proof. Given a basis $B$ and $e \notin B$, we let $C(e,B)$ be the unique circuit contained in $B \cup \{e\}$.  We will use the following well-known characterization of binary matroids (Theorem 9.1.2 in Oxley's *Matroid Theory* text).  

**Theorem.** A matroid $M$ is binary if and only if, for all bases $B$ of $M$ and all circuits $C$ of $M$, $C= \triangle_{e \in C-B} C(e,B)$. 

Here, $\triangle$ means symmetric difference, or equivalently $\sum$, if we view sets as vectors over $\mathbb{F}_2$.  
Let $\dim(M)$ denote the dimension of the circuit space of $M$ and $r^*(M)$ denote the corank of $M$.  We now prove that $\dim(M)=r^*(M)$ if and only if $M$ is binary. 

*Proof.*  For each basis $B$ of $M$, let  $\mathcal{C}_B:=\{C(e,B) : e \notin B\}$. The circuits in $\mathcal{C}_B$ are linearly independent since for each $e \notin B$, there is a unique circuit in $\mathcal{C}_B$ containing $e$.  Thus, $\dim(M) \geq r^*(M)$, for every matroid $M$.  

If $M$ is binary, then by the above theorem, $C= \triangle_{e \in C-B} C(e,B)$, for every circuit $C$.  Thus, $\dim(M)=r^*(M)$. 

If $M$ is not binary, then by the above theorem, there exists a basis $B'$ and a circuit $C'$ such that $C' \neq \triangle_{e \in C-B'} C(e,B')$. In particular, $C'$ is not a sum of the vectors in $\mathcal{C}_{B'}$.  Thus, $\dim(M)>r^*(M)$.