SlowGuess/NeurIPS_2023_OCR huggingface dataset:
- Case 1:  $|\operatorname{supp}(\mathbf{a}_i) \cup \operatorname{supp}(\mathbf{a}_j)| - |\operatorname{supp}(\mathbf{a}_i) \cap \operatorname{supp}(\mathbf{a}_j)| = 0$ . This implies  $\operatorname{supp}(\mathbf{a}_i) = \operatorname{supp}(\mathbf{a}_j)$ , i.e., the supports of  $\mathbf{a}_i$  and  $\mathbf{a}_j$  are identical. In this case, Assumption 6 is clearly violated, as  $\operatorname{supp}(\mathbf{a}_i)$  is a subset of  $\operatorname{supp}(\mathbf{a}_j)$  and vice versa.
- Case 2:  $|\operatorname{supp}(\mathbf{a}_i) \cup \operatorname{supp}(\mathbf{a}_j)| - |\operatorname{supp}(\mathbf{a}_i) \cap \operatorname{supp}(\mathbf{a}_j)| = 1$ . This implies that one of the columns is a proper subset of the other, meaning either  $\operatorname{supp}(\mathbf{a}_i) \subset \operatorname{supp}(\mathbf{a}_j)$  or  $\operatorname{supp}(\mathbf{a}_j) \subset \operatorname{supp}(\mathbf{a}_i)$ . Again, this violates Assumption 6.
Hence, in either case, if  $\mathbf{A}$  does not satisfy Assumption 1, then it does not satisfy Assumption 6.
We now prove the following theorem.
Proof. We first prove Assumption 4  $\Longrightarrow$  Assumption 6.
In Assumption 4, Eq. (3) is assumed to be satisfied for all  $\mathcal{I} \subseteq [n]$  where  $|\mathcal{I}| > 1$ . Thus, in order to prove that Assumption 4 implies Assumption 6, it is sufficient to only consider the case where  $|\mathcal{I}| = 2$ . That is, for every  $i,j \in [n]$  and  $i \neq j$ , we have
$$
\left| \operatorname {s u p p} \left(\mathbf {a} _ {i}\right) \cup \operatorname {s u p p} \left(\mathbf {a} _ {j}\right) \right| > \operatorname {r a n k} (\text {o v e r l a p} \left(\mathbf {A} _ {\{i, j \}}\right)) + \left| \operatorname {s u p p} \left(\mathbf {a} _ {i}\right) \right|,
$$
where we set  $i$  as the target index without loss of generality.
Because  $\mathrm{rank}(\mathrm{overlap}(\mathbf{A}_{\{i,j\}}))\geq 1$  , we have
$$
\left| \operatorname {s u p p} \left(\mathbf {a} _ {i}\right) \cup \operatorname {s u p p} \left(\mathbf {a} _ {j}\right) \right| > 1 + \left| \operatorname {s u p p} \left(\mathbf {a} _ {i}\right) \right|. \tag {21}
$$
Suppose  $|\operatorname{supp}(\mathbf{a}_i)| \geq |\operatorname{supp}(\mathbf{a}_j)|$ . Eq. (21) implies that  $\operatorname{supp}(\mathbf{a}_j)$  is not a subset of  $\operatorname{supp}(\mathbf{a}_i)$ . Similarly, if  $|\operatorname{supp}(\mathbf{a}_j)| \geq |\operatorname{supp}(\mathbf{a}_i)|$ , we could also show that  $\operatorname{supp}(\mathbf{a}_i)$  is not a subset of  $\operatorname{supp}(\mathbf{a}_j)$ . Thus, Assumption 6 is satisfied.
According to Proposition 9, we have Assumption  $6 \Longrightarrow$  Assumption 1, which completes the proof of the first part, i.e., Assumption  $4 \Longrightarrow$  Assumption  $6 \Longrightarrow$  Assumption 1.
We now provide an example of mixing matrix satisfying the Assumption 1 that does not satisfy Assumption 4. Suppose the mixing matrix  $\tilde{\mathbf{A}}$  has a support as follows:
$$
\boldsymbol {\xi} _ {\tilde {\mathbf {A}}} = \left[ \begin{array}{l l l} \times & 0 & 0 \\ \times & \times & 0 \\ \times & 0 & \times \end{array} \right]
$$
Clearly, Assumption 1 is satisfied since every pair of columns differ on more than one entry. However, for  $k = 1$  and  $\mathcal{I} = \{1,2\}$ , we have
$$
\left| \bigcup_ {k ^ {\prime} \in \mathcal {I}} \operatorname {s u p p} \left(\mathbf {a} _ {k ^ {\prime}}\right) \right| - \operatorname {r a n k} (\operatorname {o v e r l a p} \left(\mathbf {A} _ {\mathcal {I}}\right)) \leq \left| \bigcup_ {k ^ {\prime} \in \mathcal {I}} \operatorname {s u p p} \left(\mathbf {a} _ {k ^ {\prime}}\right) \right| = 3.
$$
Since  $|\operatorname{supp}(\mathbf{a}_k)| = |\operatorname{supp}(\mathbf{a}_1)| = 3$ , it is not possible for Eq. (3) to hold for  $\tilde{\mathbf{A}}$  when  $k = 1$  and  $\mathcal{I} = \{1,2\}$ . Thus, Assumption 4 is violated. The proof of part (b) is finished.
D.6 Proof of Theorem 3
Proof. We first prove the contrapositive of Assumption 5  $\Rightarrow$  Assumption 6. Suppose that Assumption 6 does not hold. This means that there exist distinct indices  $i, j \in [n]$ , such that  $\operatorname{supp}(\mathbf{a}_i)$  is a subset of  $\operatorname{supp}(\mathbf{a}_j)$ .
Now, for any set of row indices  $\mathcal{I} \subset [n]$ , consider the intersection over the supports of all rows  $j \in \mathcal{I}$  denoted by  $\bigcap_{j \in \mathcal{I}} \operatorname{supp}(\mathbf{a}_{j,:})$ . Because  $\operatorname{supp}(\mathbf{a}_i)$  is a subset of  $\operatorname{supp}(\mathbf{a}_j)$ , it is clear that  $i$  cannot be the only element in this intersection. Therefore, it is impossible to satisfy  $\bigcap_{j \in \mathcal{I}} \operatorname{supp}(\mathbf{a}_{j,:}) = \{i\}$  for any choice of  $\mathcal{I}$ , which indicates that Assumption 5 is violated.
According to Proposition 9, we have Assumption  $6 \Rightarrow$  Assumption 1, which completes the proof of the first part, i.e., Assumption  $5 \Rightarrow$  Assumption  $6 \Rightarrow$  Assumption 1.
We now provide an example of mixing matrix satisfying the Assumption 1 that does not satisfy Assumption 5. Suppose the mixing matrix  $\tilde{\mathbf{A}}$  has a support as follows:
$$
\boldsymbol {\xi} _ {\tilde {\mathbf {A}}} = \left[ \begin{array}{c c c} \times & 0 & 0 \\ \times & \times & 0 \\ \times & 0 & \times \end{array} \right]
$$
Clearly, Assumption 1 is satisfied since the supports of each pair of columns differ on more than one entry. However, Assumption 5 is violated since there does not exist any set of rows such that the intersection of their nonzero indices is 2 or 3.
D.7 Proof of Theorem 4
Theorem 4 (Equivalent Formulations). Suppose  $\tilde{\mathbf{A}} = (\mathbf{I} - \tilde{\mathbf{B}})\tilde{\boldsymbol{\Omega}}^{-\frac{1}{2}}$ . Then, we have:
(a) Let  $(\hat{\mathbf{B}},\hat{\Omega})$  be a solution to Problem (4). Then,  $\hat{\mathbf{A}} := (\mathbf{I} - \hat{\mathbf{B}})\hat{\Omega}^{-\frac{1}{2}}$  is a solution to Problem (1).
(b) Let  $\hat{\mathbf{A}}$  be a solution to Problem (1). Then, there exist matrices  $\hat{\mathbf{B}} \in \mathbb{R}_{\mathrm{off}}^{n \times n}$  and  $\hat{\Omega} \in \mathrm{diag}(\mathbb{R}_{>0}^n)$  such that  $\hat{\mathbf{A}} \sim (\mathbf{I} - \hat{\mathbf{B}})\hat{\boldsymbol{\Omega}}^{-\frac{1}{2}}$ , and  $(\hat{\mathbf{B}}, \hat{\boldsymbol{\Omega}})$  is a solution to Problem (4).
Proof. We consider both parts of the statements.
Part (a):
We provide a proof by contrapositive. For matrices  $\hat{\mathbf{B}}$  and  $\hat{\Omega}$ , suppose  $\hat{\mathbf{A}} := (\mathbf{I} - \hat{\mathbf{B}})\hat{\Omega}^{-\frac{1}{2}}$  is not a solution to Problem (1). That is, there exists matrix  $\ddot{\mathbf{A}}$  such that
$$
\left\| \ddot {\mathbf {A}} \right\| _ {0} <   \left\| \hat {\mathbf {A}} \right\| _ {0} \tag {22}
$$
$$
\ddot {\mathbf {A}} \ddot {\mathbf {A}} ^ {\top} = \tilde {\mathbf {A}} \tilde {\mathbf {A}} ^ {\top}. \tag {23}
$$
By Lemma 3, matrix  $\ddot{\mathbf{A}}$  is non-singular, and thus, by Lemma 5, there exist matrices  $\ddot{\mathbf{B}}\in \mathbb{R}_{\mathrm{off}}^{n\times n}$  and  $\ddot{\Omega}\in \mathrm{diag}(\mathbb{R}_{>0}^n)$  such that
$$
\ddot {\mathbf {A}} \sim (\mathbf {I} - \ddot {\mathbf {B}}) \dot {\boldsymbol {\Omega}} ^ {- \frac {1}{2}}.
$$
Lemma 6 implies
$$
\| \hat {\mathbf {A}} \| _ {0} = \| \hat {\mathbf {B}} \| _ {0} + n \quad \text {a n d} \quad \| \ddot {\mathbf {A}} \| _ {0} = \| \ddot {\mathbf {B}} \| _ {0} + n,
$$
which, with Inequality (22), indicate  $\| \hat{\mathbf{B}} \|_0 < \| \hat{\mathbf{B}} \|_0$ . Furthermore, using Eq. (23) and the assumption  $\tilde{\mathbf{A}} = (\mathbf{I} - \tilde{\mathbf{B}})\tilde{\boldsymbol{\Omega}}^{-\frac{1}{2}}$ , we have
$$
(\mathbf {I} - \ddot {\mathbf {B}}) \ddot {\boldsymbol {\Omega}} ^ {- 1} (\mathbf {I} - \ddot {\mathbf {B}}) ^ {\top} = (\mathbf {I} - \tilde {\mathbf {B}}) \tilde {\boldsymbol {\Omega}} ^ {- 1} (\mathbf {I} - \tilde {\mathbf {B}}) ^ {\top}.
$$
Therefore,  $(\ddot{\mathbf{B}},\ddot{\Omega})$  satisfies the constraint of Problem (4) and leads to a smaller zero norm for the objective function, and thus  $(\hat{\mathbf{B}},\hat{\Omega})$  will never be a solution of Problem (4).
Part (b):
Let  $\hat{\mathbf{A}}$  be a solution to Problem (1). By Lemma 3, matrix  $\hat{\mathbf{A}}$  is non-singular, and thus, by Lemma 5, there exist matrices  $\hat{\mathbf{B}}\in \mathbb{R}_{\mathrm{off}}^{n\times n}$  and  $\hat{\Omega}\in \mathrm{diag}(\mathbb{R}_{>0}^n)$  such that
$$
\hat {\mathbf {A}} \sim (\mathbf {I} - \hat {\mathbf {B}}) \hat {\boldsymbol {\Omega}} ^ {- \frac {1}{2}}.
$$
It then remains to prove that  $(\hat{\mathbf{B}},\hat{\boldsymbol{\Omega}})$  is a solution to Problem (4), which we do so by contradiction. Suppose that  $(\hat{\mathbf{B}},\hat{\boldsymbol{\Omega}})$  is not a solution to Problem (4). That is, there exists solution  $(\ddot{\mathbf{B}},\ddot{\boldsymbol{\Omega}})$  such that
$$
\left\| \ddot {\mathbf {B}} \right\| _ {0} <   \left\| \hat {\mathbf {B}} \right\| _ {0} \tag {24}
$$
$$
(\mathbf {I} - \ddot {\mathbf {B}}) \ddot {\boldsymbol {\Omega}} ^ {- 1} (\mathbf {I} - \ddot {\mathbf {B}}) ^ {\top} = (\mathbf {I} - \ddot {\mathbf {B}}) \tilde {\boldsymbol {\Omega}} ^ {- 1} (\mathbf {I} - \ddot {\mathbf {B}}) ^ {\top}. \tag {25}
$$
Define  $\ddot{\mathbf{A}} \coloneqq (\mathbf{I} - \ddot{\mathbf{B}})\tilde{\Omega}^{-\frac{1}{2}}$ . Lemma 6 implies  $\| \hat{\mathbf{A}} \|_0 = \| \hat{\mathbf{B}} \|_0 + n$  and  $\| \ddot{\mathbf{A}} \|_0 = \| \ddot{\mathbf{B}} \|_0 + n$ , which, with Inequality (24), indicates  $\| \ddot{\mathbf{A}} \|_0 < \| \hat{\mathbf{A}} \|_0$ . Furthermore, using Eq. (25) and the assumption  $\tilde{\mathbf{A}} = (\mathbf{I} - \tilde{\mathbf{B}})\tilde{\Omega}^{-\frac{1}{2}}$ , we have
$$
\ddot {\mathbf {A}} \ddot {\mathbf {A}} ^ {\top} = \tilde {\mathbf {A}} \tilde {\mathbf {A}} ^ {\top}.
$$
Therefore,  $\hat{\mathbf{A}}$  satisfies the constraint of Problem (1) and leads to a smaller zero norm for the objective function, and thus  $\hat{\mathbf{A}}$  will never be a solution of Problem (1), which is a contradiction.
D.8 Proof of Theorem 5
In this section, we first provide the proofs of Propositions 10 and 11, which together straightforwardly imply Theorem 5. Before proving Proposition 10, we state the following lemma from Shimizu et al. [38] that is useful for the proof.
Lemma 11 (Shimizu et al. [38, Lemma 1]). Let  $\mathbf{A}$  be a lower triangular matrix with all diagonal entries being nonzero. Let  $\mathbf{P}_1$  and  $\mathbf{P}_2$  be two permutation matrices. Then, a permutation of rows and columns of  $\mathbf{A}$ , i.e.,  $\mathbf{P}_1^\top \mathbf{A}\mathbf{P}_2$ , has only nonzero entries in the diagonal if and only if the row and column permutations are equal, i.e.,  $\mathbf{P}_1 = \mathbf{P}_2$ .
We now provide the proof of Proposition 10.
Proposition 10. Suppose  $\mathbf{A} \sim (\mathbf{I} - \mathbf{B})\Omega^{-\frac{1}{2}}$  for matrices  $\mathbf{A} \in \mathbb{R}^{n \times n}$ ,  $\mathbf{B} \in \mathbb{R}_{\mathrm{off}}^{n \times n}$ , and  $\Omega \in \mathrm{diag}(\mathbb{R}_{>0}^n)$ . Then,  $\mathbf{A}$  satisfies Assumption 2 if and only if matrix  $\mathbf{B}$  represents a DAG.
Proof. Without loss of generality, we consider the case in which  $\mathbf{A}$  and  $(\mathbf{I} - \mathbf{B})\boldsymbol{\Omega}^{-\frac{1}{2}}$  differ only in column permutations, instead of signed column permutations. This is because, for the latter case, there exists a diagonal matrix  $\mathbf{D}$  with diagonal entries being  $\pm 1$  such that  $\mathbf{AD}$  and  $(\mathbf{I} - \mathbf{B})\boldsymbol{\Omega}^{-\frac{1}{2}}$  differ only in column permutations, and furthermore,  $\mathbf{AD}$  satisfies Assumption 2 if and only if  $\mathbf{A}$  satisfies Assumption 2.
Therefore, suppose there exists a permutation matrix  $\mathbf{P}_1$  such that
$$
\mathbf {A} = (\mathbf {I} - \mathbf {B}) \boldsymbol {\Omega} ^ {- \frac {1}{2}} \mathbf {P} _ {1}. \tag {26}
$$
We now consider both parts of the statements.
If part:
Suppose that matrix  $\mathbf{B}$  represents a DAG. Then, there exists permutation matrix  $\mathbf{P}_2$  such that  $\mathbf{P}_2^\top \mathbf{BP}_2$  is strictly lower triangular. Therefore,  $\mathbf{P}_2^\top (\mathbf{I} - \mathbf{B})\mathbf{P}_2 = \mathbf{I} - \mathbf{P}_2^\top \mathbf{BP}_2$  is lower triangular. This implies that  $\mathbf{P}_2^\top (\mathbf{I} - \mathbf{B})\Omega^{-\frac{1}{2}}\mathbf{P}_2$  is lower triangular because  $\Omega^{-\frac{1}{2}}$  is a diagonal matrix and does not affect the support. By substituting Eq. (26),  $\mathbf{P}_2^\top \mathbf{AP}_1^{-1}\mathbf{P}_2$  is lower triangular. Clearly,  $\mathbf{P}_1^{-1}\mathbf{P}_2$  is also a permutation matrix. Therefore, matrix  $\mathbf{A}$  can be permuted by row and column permutations to be lower triangular, and thus satisfies Assumption 2.
Only if part:
Suppose matrix  $\mathbf{A}$  satisfies Assumption 2. Then, there exist permutation matrices  $\mathbf{P}_2$  and  $\mathbf{P}_3$  such that  $\mathbf{P}_2^\top \mathbf{A}\mathbf{P}_3$  is lower triangular. Substituting Eq. (26),  $\mathbf{P}_2^\top (\mathbf{I} - \mathbf{B})\Omega^{-\frac{1}{2}}\mathbf{P}_1\mathbf{P}_3$  is lower triangular.
Since  $\mathbf{P}_1\mathbf{P}_3$  is also permutation matrix, this indicates that  $(\mathbf{I} - \mathbf{B})\boldsymbol{\Omega}^{-\frac{1}{2}}$ , and thus  $\mathbf{I} - \mathbf{B}$ , satisfy Assumption 2. By Lemma 8,  $\mathbf{I} - \mathbf{B}$  is non-singular, which indicates that
$$
\det  (\mathbf {I} - \mathbf {B}) \neq 0.
$$
$$
\det  (\boldsymbol {\Omega} ^ {- \frac {1}{2}}) > 0 \quad \text {a n d} \quad \det  (\mathbf {P} _ {1}) = 1.
$$
With Eq. (26), we have
$$
\det  (\mathbf {A}) \neq 0,
$$
$$
\det (\mathbf {P} _ {2} ^ {\top} \mathbf {A P} _ {3}) \neq 0.
$$
It is known that the determinant of a lower triangular matrix is the product of its diagonal entries. Since  $\mathbf{P}_2^\top \mathbf{A}\mathbf{P}_3$  is lower triangular and  $\operatorname*{det}(\mathbf{P}_2^\top \mathbf{A}\mathbf{P}_3) \neq 0$ , all diagonal entries of  $\mathbf{P}_2^\top \mathbf{A}\mathbf{P}_3$  must be nonzero. Now define
$$
\mathbf {P} _ {4} := \mathbf {P} _ {2} ^ {- 1} \quad \text {a n d} \quad \mathbf {P} _ {5} := \mathbf {P} _ {3} ^ {- 1} \mathbf {P} _ {1} ^ {- 1}, \tag {27}
$$
both of which are permutation matrices. By some algebraic manipulations of Eq. (26) and further substituting the above definitions, we have
$$
\mathbf {P} _ {4} ^ {\top} \left(\mathbf {P} _ {2} ^ {\top} \mathbf {A P} _ {3}\right) \mathbf {P} _ {5} = (\mathbf {I} - \mathbf {B}) \boldsymbol {\Omega} ^ {- \frac {1}{2}},
$$
where all diagonal entries are nonzeros. Applying Lemma 11 w.r.t. matrix  $\mathbf{P}_2^\top \mathbf{A}\mathbf{P}_3$ , we have
$$
\mathbf {P} _ {4} = \mathbf {P} _ {5},
$$
which, by plugging into Eq. (27), implies
$$
\mathbf {P} _ {2} = \mathbf {P} _ {1} \mathbf {P} _ {3}.
$$
Since we have shown that  $\mathbf{P}_2^\top (\mathbf{I} - \mathbf{B})\Omega^{-\frac{1}{2}}\mathbf{P}_1\mathbf{P}_3$  is lower triangular, further substitution of  $\mathbf{P}_2 = \mathbf{P}_1\mathbf{P}_3$  indicates that  $\mathbf{P}_2^\top (\mathbf{I} - \mathbf{B})\Omega^{-\frac{1}{2}}\mathbf{P}_2$  is lower triangular. Since right multiplication of  $\Omega^{-\frac{1}{2}}$  does not affect the support of  $\mathbf{I} - \mathbf{B}$ , we see that
$$
\mathbf {P} _ {2} ^ {\top} (\mathbf {I} - \mathbf {B}) \mathbf {P} _ {2} = \mathbf {I} - \mathbf {P} _ {2} ^ {\top} \mathbf {B} \mathbf {P} _ {2}
$$
is also lower triangular. This indicates that  $\mathbf{P}_2^\top \mathbf{B}\mathbf{P}_2$  is lower triangular, which, with the assumption that the diagonal entries of  $\mathbf{B}$  are zeros, imply that  $\mathbf{P}_2^\top \mathbf{B}\mathbf{P}_2$  is strictly lower triangular. Therefore, matrix  $\mathbf{B}$  represents a DAG.
After proving Proposition 10, we now consider Proposition 11. Before that, we provide a result by Ghassami et al. [20] that is useful for the proof. We first describe the notion of parent exchange by Ghassami et al. [20]. Let  $\triangleleft$  be the symmetric difference operator that identifies the elements present in either of the sets but not in the intersection. For DAG  $\mathcal{G}$  with weighted adjacency matrix  $\mathbf{B}$ , its vertices  $x_{i}$  and  $x_{j}$  are said to be parent exchangeable if  $|\operatorname{supp}((\mathbf{I} - \mathbf{B})_i)\triangle \operatorname{supp}((\mathbf{I} - \mathbf{B})_j)| = 1$ , i.e., there exists  $k \in [n]$  such that  $\operatorname{supp}((\mathbf{I} - \mathbf{B})_i)\triangle \operatorname{supp}((\mathbf{I} - \mathbf{B})_j) = \{k\}$ . In such case, a support rotation can be performed on columns  $(\xi_{\mathbf{I} - \mathbf{B}})_i$  and  $(\xi_{\mathbf{I} - \mathbf{B}})_j$  that sets a nonzero entry on those columns, except  $(\xi_{\mathbf{I} - \mathbf{B}})_{i,i}$  and  $(\xi_{\mathbf{I} - \mathbf{B}})_{j,j}$ , to zero. In other words, the parent of  $x_{i}$  and  $x_{j}$  that corresponds to the zeroed entry is removed. Furthermore, the entry  $(\xi_{\mathbf{I} - \mathbf{B}})_{k,i}$  or  $(\xi_{\mathbf{I} - \mathbf{B}})_{k,j}$  is set to  $\times$ , which corresponds to adding the missing edge  $x_{k} \to x_{i}$  or  $x_{k} \to x_{j}$ . Ghassami et al. [20] defined such an operation to be a parent exchange. We then provide the following corollary that is straightforwardly derived from Ghassami et al. [20, Corollary 2 & Proposition 1].
Corollary 2 (Ghassami et al. [20, Corollary 2]). DAGs  $\mathcal{G}_1$  and  $\mathcal{G}_2$  are Markov equivalent if and only if there exists a sequence of parent exchanges that maps  $\mathcal{G}_1$  to  $\mathcal{G}_2$ , and one that maps  $\mathcal{G}_2$  to  $\mathcal{G}_1$ .
We now provide the proof of Proposition 11.
Proposition 11. Let  $\mathcal{G}$  be a DAG with weighted adjacency matrix  $\mathbf{B}$ . Then, matrix  $\mathbf{I} - \mathbf{B}$  satisfies Assumption 1 if and only if the Markov equivalence class of  $\mathcal{G}$  is a singleton.
We provide a proof by contrapositive. Suppose that matrix  $\mathbf{I} - \mathbf{B}$  does not satisfy Assumption 2. That is, there exist  $i,j\in [n]$  and  $i\neq j$  such that
$$
| \operatorname {s u p p} ((\mathbf {I} - \mathbf {B}) _ {i}) \triangle \operatorname {s u p p} ((\mathbf {I} - \mathbf {B}) _ {j}) | <   1.
$$
Clearly, we have  $\mathrm{supp}((\mathbf{I} - \mathbf{B})_i) \neq \mathrm{supp}((\mathbf{I} - \mathbf{B})_j)$ , because otherwise there will be a cycle with length of two over variables  $x_i$  and  $x_j$ , which contradicts the assumption that  $\mathcal{G}$  is a acyclic. This implies
$$
| \operatorname {s u p p} ((\mathbf {I} - \mathbf {B}) _ {i}) \triangle \operatorname {s u p p} ((\mathbf {I} - \mathbf {B}) _ {j}) | = 1. \tag {28}
$$
By definition,  $x_{i}$  and  $x_{j}$  are parent exchangeable, and therefore there exists a parent exchange that maps DAG  $\mathcal{G}$  to another directed graph  $\mathcal{G}'$  where  $\mathcal{G}' \neq \mathcal{G}$ . We now show by contradiction that  $\mathcal{G}'$  is a DAG. Suppose that  $\mathcal{G}'$  is a not DAG. By Eq. (28), there exists variable  $x_{k}$  such that
$$
\operatorname {s u p p} \left(\left(\mathbf {I} - \mathbf {B}\right) _ {i}\right) \triangle \operatorname {s u p p} \left(\left(\mathbf {I} - \mathbf {B}\right) _ {j}\right) = \{k \}.
$$
Here, we must have  $k = i$  or  $k = j$ , because otherwise we have
$$
\left(\boldsymbol {\xi} _ {\mathbf {I} - \mathbf {B}}\right) _ {i, i} = \left(\boldsymbol {\xi} _ {\mathbf {I} - \mathbf {B}}\right) _ {j, j} = \left(\boldsymbol {\xi} _ {\mathbf {I} - \mathbf {B}}\right) _ {i, j} = \left(\boldsymbol {\xi} _ {\mathbf {I} - \mathbf {B}}\right) _ {j, i} = \times ,
$$
which leads to a cycle over variables  $x_{i}$  and  $x_{j}$ . Without loss of generality, we consider the case of  $k = j$ , which implies
$$
\operatorname {s u p p} \left(\left(\mathbf {I} - \mathbf {B}\right) _ {i}\right) \triangle \operatorname {s u p p} \left(\left(\mathbf {I} - \mathbf {B}\right) _ {j}\right) = \{j \} \tag {29}
$$
and  $(\pmb{\xi}_{\mathbf{I} - \mathbf{B}})_{j,i} = 0$ . This indicates that the entry  $(\pmb{\xi}_{\mathbf{I} - \mathbf{B}})_{j,i}$  is set to  $\times$  (i.e., the edge  $x_{j}\rightarrow x_{i}$  is added) after the parent exchange to DAG  $\mathcal{G}'$ , which subsequently leads to a cycle in  $\mathcal{G}'$ . In this case, there must exist a path  $x_{i}\to x_{l_{1}}\to \dots \to x_{l_{m}}\to x_{j}$  in DAG  $\mathcal{G}$ , where  $m < n - 2$  and  $l_1,\ldots ,l_m\in [n]\setminus \{i,j\}$ , to which adding the edge  $x_{j}\to x_{i}$  leads to a cycle in DAG  $\mathcal{G}'$ . By Eq. (29),  $x_{l_m}$  is also a parent of  $x_{i}$ , indicating that there exists a cycle  $x_{i}\to x_{l_{1}}\to \dots \to x_{l_{m}}\to x_{i}$  in DAG  $\mathcal{G}$ , which contradicts the assumption that  $\mathcal{G}$  is acyclic. Therefore,  $\mathcal{G}'$  must be a DAG.
Since there exists a parent exchange that maps DAG  $\mathcal{G}$  to another DAG  $\mathcal{G}'$ , clearly there also exists a (reversed) parent exchange that maps DAG  $\mathcal{G}'$  back to DAG  $\mathcal{G}$ . By Corollary 2, DAGs  $\mathcal{G}$  and  $\mathcal{G}'$  are Markov equivalent. Therefore, the Markov equivalence class of  $\mathcal{G}$  contains at least two DAGs and is not a singleton.
Suppose that matrix  $\mathbf{I} - \mathbf{B}$  satisfies Assumption 2. That is, for all  $i,j\in [n]$  and  $i\neq j$  ,we have
$$
\left| \operatorname {s u p p} \left(\left(\mathbf {I} - \mathbf {B}\right) _ {i}\right) \triangle \operatorname {s u p p} \left(\left(\mathbf {I} - \mathbf {B}\right) _ {j}\right) \right| > 1.
$$
In this case, every pair of vertices are not parent exchangeable, and thus parent exchange cannot be applied for any pair of vertices in DAG  $\mathcal{G}$ . Therefore, for any DAG  $\mathcal{G}' \neq \mathcal{G}$ , there exists no sequence of parent exchanges that maps  $\mathcal{G}$  to  $\mathcal{G}'$ , implying that they are not Markov equivalent. This indicates that all DAGs are not Markov equivalent to DAG  $\mathcal{G}$ , except itself, and thus the Markov equivalence class of  $\mathcal{G}$  is a singleton.
With Propositions 10 and 11 in place, we provide the proof of Theorem 5.
Suppose matrix  $\mathbf{B}$  represents a DAG whose Markov equivalence class is a singleton. By Proposition 10, matrix  $\mathbf{A}$  satisfies Assumption 2. Furthermore, Proposition 11 implies that  $\mathbf{I} - \mathbf{B}$ , and thus  $(\mathbf{I} - \mathbf{B})\Omega^{-\frac{1}{2}}$ , satisfy Assumption 1. Since  $\mathbf{A}$  and  $(\mathbf{I} - \mathbf{B})\Omega^{-\frac{1}{2}}$  differ only in signed column permutations, and Assumption 1 involves only pairwise comparison of the support matrix,  $\mathbf{A}$  must also satisfy Assumption 1.
Suppose  $\mathbf{A}$  satisfies Assumptions 1 and 2. By Proposition 10, matrix  $\mathbf{B}$  represents a DAG. Since  $\mathbf{A}$  and  $(\mathbf{I} - \mathbf{B})\Omega^{-\frac{1}{2}}$  differ only in signed column permutations, and Assumption 1 involves only pairwise comparison of the support matrix,  $\mathbf{I} - \mathbf{B}$  also satisfies Assumption 1. By Proposition 11, the Markov equivalence class of the DAG represented by  $\mathbf{B}$  is a singleton.
D.9 Proof of Lemma 1
Before proving Lemma 1, we first provide another result that is useful for the proof. To ease further reasoning, we define the function
$$
f (\mathbf {A}) = \operatorname {t r} \left(\sum_ {k = 1} ^ {n} (\mathbf {A} \odot \mathbf {A}) ^ {k}\right),
$$
and clearly we have
$$
g (\mathbf {A}) \equiv f (\text {o f f} (\mathbf {A})). \tag {30}
$$
Zheng et al. [48], Zhang et al. [47] have shown that, for any matrix  $\mathbf{A}$ ,  $f(\mathbf{A}) = 0$  if and only if  $\mathbf{A}$  represents the weighted adjacency matrix of a DAG. Also, it is known that the weighted adjacency matrix of a directed graph can be permuted via simultaneous equal row and column permutations to be strictly lower triangular if and only if the graph is a DAG. Therefore, we provide the following corollary that is straightforwardly implied by the results by Zheng et al. [48], Wei et al. [45].