[proofplan] We prove both implications. First, a totally unimodular integer representation over $\mathbb{Q}$ remains a representation of the same column matroid after reducing its entries in any field, because all determinants that control independence are $0$, $1$, or $-1$. Conversely, from a rational representation of a regular matroid we choose a basis, carefully project to rank many rows without changing column dependences, put the matrix into standard form, and then invoke Tutte's unimodular representation theorem in the precise standard-form version needed here. The rank-zero case is handled separately by the empty matrix convention. [/proofplan]

[step:Show that a totally unimodular integer representation works over every field]Assume that $A \in \mathbb{Z}^{r \times E}$ is totally unimodular and represents $M$ over $\mathbb{Q}$, where $r \in \mathbb{Z}_{\ge 0}$. Let $\mathbb{F}$ be a field, and let $\rho_{\mathbb{F}}: \mathbb{Z} \to \mathbb{F}$ be the unique unital ring homomorphism. Define $A_{\mathbb{F}} \in \mathbb{F}^{r \times E}$ to be the matrix obtained by applying $\rho_{\mathbb{F}}$ entrywise to $A$. Let $S \subset E$. If $|S| > r$, then the columns indexed by $S$ are dependent over both $\mathbb{Q}$ and $\mathbb{F}$. If $|S| \le r$, then $S$ is independent in the column matroid of $A$ over $\mathbb{Q}$ exactly when some $|S| \times |S|$ minor of the submatrix with column set $S$ has nonzero determinant. Since $A$ is totally unimodular, every such determinant is $0$, $1$, or $-1$. The elements $1_{\mathbb{F}}$ and $-1_{\mathbb{F}}$ are nonzero in every field, while $0$ maps to $0$. Hence the same minors with column set $S$ are nonzero for $A$ over $\mathbb{Q}$ and for $A_{\mathbb{F}}$ over $\mathbb{F}$. Therefore $S$ is independent for $A_{\mathbb{F}}$ over $\mathbb{F}$ if and only if $S$ is independent in $M$. Since $\mathbb{F}$ was arbitrary, $M$ is representable over every field. Thus $M$ is regular.[/step]

[guided]Assume that $A \in \mathbb{Z}^{r \times E}$ is totally unimodular and represents $M$ over $\mathbb{Q}$. We must show that the same column labels give a representation over an arbitrary field $\mathbb{F}$. Let $\rho_{\mathbb{F}}: \mathbb{Z} \to \mathbb{F}$ be the unique unital ring homomorphism, and define $A_{\mathbb{F}} \in \mathbb{F}^{r \times E}$ by applying $\rho_{\mathbb{F}}$ to every entry of $A$. The question is whether reducing entries can change which column sets are independent. For an arbitrary integer matrix, a determinant such as $2$ could become $0$ in characteristic $2$. Total unimodularity rules this out because the only determinant values are $0$, $1$, and $-1$. Let $S \subset E$. If $|S| > r$, then no $r \times E$ matrix can have the columns indexed by $S$ independent, over any field. Suppose therefore that $|S| \le r$. The columns indexed by $S$ are independent over a field precisely when the submatrix with those columns has rank $|S|$, equivalently when some $|S| \times |S|$ minor using all columns in $S$ has nonzero determinant. For the matrix $A$ over $\mathbb{Q}$, total unimodularity says each such determinant lies in $\{-1,0,1\}$. Now compare the corresponding determinant after applying $\rho_{\mathbb{F}}$. The value $0$ maps to $0_{\mathbb{F}}$, while $1$ maps to $1_{\mathbb{F}} \ne 0$ and $-1$ maps to $-1_{\mathbb{F}} \ne 0$. In characteristic $2$, $-1_{\mathbb{F}} = 1_{\mathbb{F}}$, but this is still nonzero. Thus a minor with column set $S$ is nonzero over $\mathbb{Q}$ if and only if the corresponding minor is nonzero over $\mathbb{F}$. It follows that the independent subsets of $E$ for $A_{\mathbb{F}}$ over $\mathbb{F}$ are exactly the independent subsets of $E$ for $A$ over $\mathbb{Q}$, namely the independent sets of $M$. Since $\mathbb{F}$ was arbitrary, $M$ is representable over every field, so $M$ is regular.[/guided]

[step:Put a rational representation into standard form without changing dependencies]Assume conversely that $M$ is regular. Since $\mathbb{Q}$ is a field, there is a matrix $B \in \mathbb{Q}^{m \times E}$ representing $M$ over $\mathbb{Q}$. Let $R \subset E$ be a basis of $M$, and set $r := |R|$. If $r = 0$, then every element of $E$ is a loop. The unique $0 \times E$ integer matrix represents the rank-zero matroid over $\mathbb{Q}$ and has no nonempty square subdeterminants, so all its square subdeterminants are $0$, $1$, or $-1$ under the empty determinant convention. Thus the desired totally unimodular representation exists. For the rest of the converse, assume $r > 0$. Let $B_R \in \mathbb{Q}^{m \times R}$ denote the submatrix of $B$ with column set $R$. Since $R$ is a basis of $M$, the columns of $B_R$ are linearly independent and every column of $B$ lies in the $\mathbb{Q}$-span of those columns. Choose a set $I \subset \{1,\dots,m\}$ with $|I|=r$ such that the $I \times R$ submatrix $B[I,R]$ has nonzero determinant. Let $B_I \in \mathbb{Q}^{r \times E}$ be the matrix obtained from $B$ by retaining only the rows indexed by $I$. The row projection from $\operatorname{span}_{\mathbb{Q}}\{B_e : e \in R\}$ to $\mathbb{Q}^r$ using the coordinates in $I$ is injective, because its restriction to the basis columns $B_R$ has matrix $B[I,R]$ with nonzero determinant. Since every column of $B$ lies in this span, a linear relation among columns of $B$ holds if and only if the projected relation among the corresponding columns of $B_I$ holds. Hence $B_I$ represents the same matroid $M$ over $\mathbb{Q}$. Multiplying $B_I$ on the left by the inverse of $B[I,R]$ gives a matrix $C \in \mathbb{Q}^{r \times E}$ representing $M$ over $\mathbb{Q}$. After temporarily ordering the columns so that the elements of $R$ come first, $C$ has the form $[I_r \mid D]$ for some $D \in \mathbb{Q}^{r \times (E \setminus R)}$. The columns remain labelled by $E$; the block notation records only this temporary ordering.[/step]

[guided]Start with a rational representation $B \in \mathbb{Q}^{m \times E}$ of $M$. Choose a basis $R \subset E$ of the matroid and set $r := |R|$. If $r=0$, then the matroid has no non-loop elements: every singleton is dependent, so every element of $E$ is a loop. The $0 \times E$ matrix has all columns equal to the unique zero vector in the zero-dimensional [vector space](/page/Vector%20Space), so it represents exactly this rank-zero matroid. Its square subdeterminants are only the empty determinant, equal to $1$, and the determinants of non-existing positive-size square submatrices do not occur. Thus the rank-zero case is already settled. Assume now that $r>0$. Let $B_R \in \mathbb{Q}^{m \times R}$ be the submatrix with the columns indexed by $R$. Since $R$ is independent in $M$ and $B$ represents $M$, the columns of $B_R$ are linearly independent. Since $R$ is also spanning in the matroid, for every $e \in E$ the set $R \cup \{e\}$ is dependent when $e \notin R$, and this means the column $B_e$ lies in the $\mathbb{Q}$-span of the columns $\{B_a : a \in R\}$. For $e \in R$ this is immediate. Hence every column of $B$ lies in the subspace \begin{align*} W := \operatorname{span}_{\mathbb{Q}}\{B_a : a \in R\} \subset \mathbb{Q}^m. \end{align*} Because the $R$-columns are linearly independent, the matrix $B_R$ has rank $r$. Therefore there is a set $I \subset \{1,\dots,m\}$ with $|I|=r$ such that the square submatrix $B[I,R]$ has nonzero determinant. Let $\pi_I: \mathbb{Q}^m \to \mathbb{Q}^r$ be the coordinate projection retaining exactly the rows indexed by $I$. The restriction $\pi_I|_W: W \to \mathbb{Q}^r$ is injective: with respect to the basis $\{B_a : a \in R\}$ of $W$, its matrix is precisely $B[I,R]$, whose determinant is nonzero. Define $B_I \in \mathbb{Q}^{r \times E}$ to be the matrix whose column indexed by $e$ is $\pi_I(B_e)$. We now check that this row deletion preserves all column dependences. Let $S \subset E$ and let $(\lambda_e)_{e \in S}$ be scalars in $\mathbb{Q}$. Since each $B_e$ lies in $W$, the vector $\sum_{e \in S}\lambda_e B_e$ lies in $W$. If \begin{align*} \sum_{e \in S}\lambda_e \pi_I(B_e)=0, \end{align*} then \begin{align*} \pi_I\left(\sum_{e \in S}\lambda_e B_e\right)=0. \end{align*} The injectivity of $\pi_I|_W$ gives \begin{align*} \sum_{e \in S}\lambda_e B_e=0. \end{align*} The converse implication follows by applying $\pi_I$ to a relation already equal to zero. Thus a subset of columns is linearly independent in $B_I$ if and only if it is linearly independent in $B$. Therefore $B_I$ represents the same matroid $M$. Finally, multiply $B_I$ on the left by the inverse of the nonsingular matrix $B[I,R]$. This is an invertible row operation, so it preserves all linear dependence relations among columns. The resulting matrix $C \in \mathbb{Q}^{r \times E}$ represents $M$ and has the displayed standard form $[I_r \mid D]$ after ordering the basis columns $R$ first.[/guided]

[step:Apply Tutte's unimodular representation theorem in standard form] We use the following external standard theorem of Tutte, often called Tutte's unimodular representation theorem for regular matroids. If a finite regular matroid $N$ with basis $R$ is represented over $\mathbb{Q}$ by a standard matrix $[I_r \mid D]$ whose identity block is indexed by $R$, then there exist nonsingular diagonal rational matrices $P \in \mathbb{Q}^{r \times r}$ and $Q \in \mathbb{Q}^{E \times E}$ such that $P[I_r \mid D]Q$ is an integer totally unimodular matrix and represents the same labelled column matroid over $\mathbb{Q}$. The theorem is stated for finite matroids, a chosen basis, and a rational standard representation with respect to that basis. The hypotheses are satisfied here: $M$ is finite by assumption, $M$ is regular by the converse hypothesis, $R$ is a basis of $M$, and the previous step constructed a rational standard representation $C=[I_r \mid D]$ with identity block indexed by $R$. Applying Tutte's theorem gives nonsingular diagonal rational matrices $P$ and $Q$ such that \begin{align*} A := PCQ \end{align*} is an integer totally unimodular matrix with columns indexed by $E$. Because $P$ and $Q$ are invertible diagonal matrices over $\mathbb{Q}$, each square minor of $A$ with a fixed row set and column set is the corresponding square minor of $C$ multiplied by a product of nonzero row and column scalars. Hence the zero or nonzero status of every such minor is unchanged. Therefore the column subsets independent in $A$ over $\mathbb{Q}$ are exactly the column subsets independent in $C$ over $\mathbb{Q}$, so $A$ represents $M$ over $\mathbb{Q}$. [/step]

[step:Conclude the equivalence] The first step proves that any integer totally unimodular matrix representing $M$ over $\mathbb{Q}$ makes $M$ representable over every field, hence regular. Conversely, if $M$ is regular, the rank-zero case is represented by the $0 \times E$ matrix, and in positive rank the standard-form construction followed by Tutte's unimodular representation theorem produces an integer totally unimodular matrix $A \in \mathbb{Z}^{r \times E}$ representing $M$ over $\mathbb{Q}$. Therefore $M$ is regular if and only if it has a representation by a totally unimodular integer matrix. [/step]