[proofplan] The standard-form matrix identifies each basis element $b_i$ with the $i$-th standard coordinate vector, while the non-basis element $e$ has coordinate column $(D_{1e},\dots,D_{re})$. We show first that the displayed set is dependent, because the column $A_e$ is a linear combination of exactly the identity columns whose coefficients are non-zero. We then prove minimality: omitting any displayed basis element removes a coordinate direction needed to span $A_e$, while omitting $e$ leaves a subset of the independent basis $B$. Therefore the displayed dependent set is precisely the unique circuit contained in $B \cup \{e\}$. [/proofplan] [step:Define the support of the non-basis column] For the fixed element $e \in E \setminus B$, define the index set \begin{align*} S_e := \{i \in \{1,\dots,r\} : D_{ie} \ne 0\}. \end{align*} Define the subset $X_e \subset E$ by \begin{align*} X_e := \{e\}\cup \{b_i : i \in S_e\}. \end{align*} We will prove that $X_e$ is the unique circuit contained in $B \cup \{e\}$. Since $A$ is in standard form with respect to $B$, the column $A_{b_i}$ is the $i$-th standard basis vector of $k^r$ for each $i \in \{1,\dots,r\}$, and the column $A_e \in k^r$ has $i$-th coordinate $D_{ie}$. [/step] [step:Show that the displayed set is dependent] The column $A_e$ has the coordinate expansion \begin{align*} A_e = \sum_{i \in S_e} D_{ie} A_{b_i}. \end{align*} Equivalently, \begin{align*} A_e - \sum_{i \in S_e} D_{ie} A_{b_i} = 0. \end{align*} The coefficient of $A_e$ in this linear relation is $1 \ne 0$ in the field $k$. Hence the family of columns indexed by $X_e$ is linearly dependent over $k$. Because $A$ represents $M$, this means that $X_e$ is a dependent subset of the matroid $M$. [/step] [step:Prove that no proper subset of the displayed set is dependent] Let $Y \subsetneq X_e$. We prove that $Y$ is independent in $M$. If $e \notin Y$, then $Y \subset B$. Since $B$ is a basis of $M$, every subset of $B$ is independent, so $Y$ is independent. Now suppose $e \in Y$. If $S_e = \varnothing$, then $X_e = \{e\}$, so the only subset of $X_e$ containing $e$ is $X_e$ itself. This contradicts $Y \subsetneq X_e$, hence this case is vacuous when $S_e = \varnothing$. It remains to treat the case $e \in Y$ and $S_e \ne \varnothing$. Define \begin{align*} T_Y := \{i \in S_e : b_i \in Y\}. \end{align*} Because $Y$ is a proper subset of $X_e$ and contains $e$, at least one basis element $b_j$ with $j \in S_e$ is omitted from $Y$. Fix such an index $j \in S_e$; then $b_j \notin Y$, so $j \notin T_Y$. Assume, for contradiction, that $Y$ is dependent. Since $Y \setminus \{e\} \subset B$ is independent, every non-trivial linear dependence among the columns indexed by $Y$ must have non-zero coefficient on $A_e$. Dividing the relation by that coefficient, we obtain scalars $\lambda_i \in k$ for $i \in T_Y$ such that \begin{align*} A_e = \sum_{i \in T_Y} \lambda_i A_{b_i}. \end{align*} The right-hand side has $j$-th coordinate equal to $0$, because it is a linear combination only of standard basis vectors $A_{b_i}$ with $i \ne j$. The left-hand side has $j$-th coordinate equal to $D_{je}$, and $D_{je} \ne 0$ because $j \in S_e$. This contradiction proves that $Y$ is independent. [guided] We need to prove minimality, not just dependence. A circuit is a dependent set with no smaller dependent subset, so take an arbitrary proper subset $Y \subsetneq X_e$ and show that it cannot be dependent. There are two cases. First, if $e \notin Y$, then $Y$ consists only of elements of the basis $B$. Since $B$ is independent in the matroid $M$, every subset of $B$ is independent. Thus $Y$ is independent in this case. Second, suppose $e \in Y$. There is a small degenerate case to separate. If $S_e = \varnothing$, then $X_e = \{e\}$. A subset of $X_e$ that contains $e$ must then be exactly $X_e$, so no proper subset $Y \subsetneq X_e$ can occur in this second case. Thus the case $e \in Y$ is vacuous when $S_e$ is empty. We may therefore assume $e \in Y$ and $S_e \ne \varnothing$. Since $Y$ is still a proper subset of $X_e$, at least one basis element appearing in $X_e$ has been omitted. Define \begin{align*} T_Y := \{i \in S_e : b_i \in Y\}. \end{align*} Choose an index $j \in S_e$ such that $b_j \notin Y$; this choice is possible because $Y \subsetneq X_e$ and $e \in Y$. Assume for contradiction that $Y$ is dependent. The set $Y \setminus \{e\}$ is contained in $B$, hence its columns are independent. Therefore any non-trivial linear relation among the columns indexed by $Y$ must use the column $A_e$ with a non-zero coefficient. After dividing the relation by that non-zero coefficient, we may solve for $A_e$ and obtain scalars $\lambda_i \in k$ for $i \in T_Y$ satisfying \begin{align*} A_e = \sum_{i \in T_Y} \lambda_i A_{b_i}. \end{align*} Now compare the $j$-th coordinate of both sides. Since the columns $A_{b_i}$ are the standard basis vectors of $k^r$, and since $j \notin T_Y$, every vector on the right-hand side has zero $j$-th coordinate. Hence the right-hand side has $j$-th coordinate $0$. But the $j$-th coordinate of $A_e$ is $D_{je}$, and $D_{je} \ne 0$ because $j \in S_e$. This is impossible. Thus no proper subset $Y \subsetneq X_e$ is dependent. [/guided] [/step] [step:Identify the minimal dependent set as the fundamental circuit] We have shown that $X_e$ is dependent and that every proper subset of $X_e$ is independent. Hence $X_e$ is a circuit of $M$. Also $X_e \subset B \cup \{e\}$ and $e \in X_e$. By definition, the fundamental circuit $C_B(e)$ is the unique circuit contained in $B \cup \{e\}$. Therefore \begin{align*} C_B(e)=X_e=\{e\}\cup \{b_i \in B : D_{ie} \ne 0\}. \end{align*} This is the desired formula. [/step]