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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 $B$ of $M$ and $e \in E(M)$, 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, $\dim(M)>r^*(M)$.

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, $\dim(M)>r^*(M)$.

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 $B$ of $M$ and $e \in E(M)$, 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 $M$, $C= \triangle_{e \in C-B} C(e,B)$.

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 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, $\dim(M)>r^*(M)$.

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 $B$ of $M$ and $e \in E(M)$, 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, $\dim(M)>r^*(M)$.

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 $B$ of $M$ and $e \in E(M)$, we let $C(e,B)$ be the unique circuit contained in $B \cup \{e\}$ and containing $e$. We will use the following well-known characterization of binary matroids (see Chapter $9$ 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 $M$, $C= \triangle_{e \in C-B} C(e,B)$.

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 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, $\dim(M)>r^*(M)$.

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 $B$ of $M$ and $e \in E(M)$, 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 $M$, $C= \triangle_{e \in C-B} C(e,B)$.

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 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, $\dim(M)>r^*(M)$.