SlowGuess/NeurIPS_2023_OCR huggingface dataset:
$$
W = \left[ \begin{array}{c c c} e ^ {J _ {1} (t - s)} w _ {1, 1} & \ldots & e ^ {J _ {1} (t - s)} w _ {d, 1} \\ \vdots & \ddots & \vdots \\ e ^ {J _ {K} (t - s)} w _ {1, K} & \ldots & e ^ {J _ {K} (t - s)} w _ {d, K} \end{array} \right],
$$
Since  $w_{j,k} = 0$ $(\in \mathbb{R}^1$  or  $\mathbb{R}^2)$ , for all  $j = 0,1,\ldots ,d$ , then if  $k\in \{1,\dots ,K_1\}$ , the  $k$ -th row of  $W$
$$
W _ {k.} = 0;
$$
and if  $k \in \{K_1 + 1, \dots, K\}$ , then the  $(2k - K_1 - 1)$ -th and the  $(2k - K_1)$ -th rows
$$
W _ {(2 k - K _ {1} - 1) \cdot} = W _ {(2 k - K _ {1}) \cdot} = 0,
$$
where  $W_{k}$  denotes the  $k$ -th row vector of matrix  $W$ .
If we denote
$$
\tilde {V} (t) _ {. j} := Q ^ {- 1} V (t) _ {. j}, \quad \text {f o r a l l} j = 1, \dots , d,
$$
$$
w (t) _ {j, k} := \left\{ \begin{array}{l l} \tilde {V} (t) _ {\cdot j, k} \in \mathbb {R} ^ {1}, & \text {f o r} k = 1, \ldots , K _ {1}, \\ (\tilde {V} (t) _ {\cdot j, 2 k - K _ {1} - 1}, \tilde {V} (t) _ {\cdot j, 2 k - K _ {1}}) ^ {\top} \in \mathbb {R} ^ {2}, & \text {f o r} k = K _ {1} + 1, \ldots , K. \end{array} \right.
$$
Then by multiplying  $Q^{-1}$  in both sides of Equation (31), one obtains that
$$
w (t) _ {j, k} = 0 (\in \mathbb {R} ^ {1} \text {o r} \mathbb {R} ^ {2})
$$
for all  $j = 1, \ldots, d$  and all  $0 \leqslant t < \infty$ . This indicates that when all the vectors  $H_{\cdot j}$  for  $j = 1, \ldots, d$  are confined to an  $A$ -invariant proper subspace of  $\mathbb{R}^d$ , denotes as  $L$ , then each column of the covariance matrix  $V(t)$  in Equation (30) is also confined to  $L$ , for all  $0 \leqslant t < \infty$ . Therefore, under condition (14),  $x_0, H_{\cdot j}$  (for all  $j = 1, \ldots, d$ ) and each column of the covariance matrix  $V(t)$  (for all  $0 \leqslant t < \infty$ ) are confined to an  $A$ -invariant proper subspace of  $\mathbb{R}^d$ . Thus, a matrix  $A_2$  exists, with  $A_2 \neq A_1$  such that the first two equations in Equation (30) are satisfied.
In particular, by [43, Theorem 2.5], when  $k \in \{1, \dots, K_1\}$ , there exists matrix  $D \in \mathbb{R}^{d \times d}$ , with the  $kk$ -th element  $D_{kk} = c \neq 0$  and all the other elements of  $D$  are zeros. Let
$$
A _ {2} = A _ {1} + Q D Q ^ {- 1} \neq A _ {1},
$$
then  $A_{1}$  and  $A_{2}$  satisfy the first two equations in Equation (30). Then we will show that such a  $A_{2}$  also satisfy the third equation in Equation (30).
Some calculation shows that
$$
\begin{array}{l} V _ {1} (t) = \int_ {0} ^ {t} e ^ {A _ {1} (t - s)} H e ^ {A _ {1} ^ {\top} (t - s)} d s \\ = \int_ {0} ^ {t} Q e ^ {\Lambda (t - s)} Q ^ {- 1} H \left(Q ^ {T}\right) ^ {- 1} e ^ {\Lambda (t - s)} Q ^ {T} d s \tag {32} \\ := \int_ {0} ^ {t} Q e ^ {\Lambda (t - s)} P _ {1} e ^ {\Lambda (t - s)} Q ^ {T} d s, \\ \end{array}
$$
where  $P_{1}\coloneqq Q^{-1}H(Q^{T})^{-1}$  . And
$$
\begin{array}{l} V _ {2} (t) = \int_ {0} ^ {t} e ^ {A _ {2} (t - s)} H e ^ {A _ {2} ^ {\top} (t - s)} d s \\ = \int_ {0} ^ {t} e ^ {\left(A _ {1} + Q D Q ^ {- 1}\right) (t - s)} H e ^ {\left(A _ {1} + Q D Q ^ {- 1}\right) ^ {\top} (t - s)} d s \tag {33} \\ = \int_ {0} ^ {t} Q e ^ {\Lambda (t - s)} e ^ {D (t - s)} Q ^ {- 1} H \left(Q ^ {T}\right) ^ {- 1} e ^ {D (t - s)} e ^ {\Lambda (t - s)} Q ^ {T} d s \\ := \int_ {0} ^ {t} Q e ^ {\Lambda (t - s)} P _ {2} e ^ {\Lambda (t - s)} Q ^ {T} d s, \\ \end{array}
$$
where  $P_{2} \coloneqq e^{D(t - s)}Q^{-1}H(Q^{T})^{-1}e^{D(t - s)}$ . If one can show that  $P_{1} = P_{2}$ , then it is readily checked that  $V_{1}(t) = V_{2}(t)$  for all  $0 \leqslant t < \infty$ . Recall that
$$
Q ^ {- 1} H = \tilde {H},
$$
where  $\tilde{H} = [\tilde{H}_{.1}|\ldots |\tilde{H}_{.d}]$ . And when condition (14) is not satisfied, the  $k$ -th row of  $\tilde{H}$ :
$$
\tilde {H} _ {k.} = 0.
$$
$$
P _ {1} = Q ^ {- 1} H \left(Q ^ {T}\right) ^ {- 1} = \tilde {H} \left(Q ^ {T}\right) ^ {- 1},
$$
therefore, the  $k$ -th row of  $P_{1}$ :
$$
\left(P _ {1}\right) _ {k \cdot} = 0.
$$
Simple calculation shows that matrix  $P_{1}$  is symmetric, thus, the  $k$ -th column of  $P_{1}$ :
$$
\left(P _ {1}\right) _ {\cdot k} = 0.
$$
It is easy to obtain that  $e^{D(t - s)}$  is a diagonal matrix expressed as
$$
e ^ {D (t - s)} = \left[ \begin{array}{c c c c c} 1 & & & & \\ & \ddots & & & \\ & & e ^ {c (t - s)} & & \\ & & & \ddots & \\ & & & & 1 \end{array} \right]
$$
where  $e^{c(t - s)}$  is the  $kk$ -th entry. Then, simple calculation shows that
$$
P _ {2} = e ^ {D (t - s)} Q ^ {- 1} H (Q ^ {T}) ^ {- 1} e ^ {D (t - s)} = e ^ {D (t - s)} P _ {1} e ^ {D (t - s)} = P _ {1}.
$$
Therefore, one obtains that
$$
V _ {1} (t) = V _ {2} (t), \forall 0 \leqslant t <   \infty .
$$
Hence, when  $k \in \{1, \dots, K_1\}$ , we find a  $A_2$ , with  $A_2 \neq A_1$  such that Equation (30) is satisfied.
When  $k\in \{K_1 + 1,\ldots ,K\}$ , there exists matrix  $D^{\prime}\in \mathbb{R}^{d\times d}$ , with
$$
\left[ \begin{array}{c c} D _ {2 k - K _ {1} - 1, 2 k - K _ {1} - 1} ^ {\prime} & D _ {2 k - K _ {1} - 1, 2 k - K _ {1}} ^ {\prime} \\ D _ {2 k - K _ {1}, 2 k - K _ {1} - 1} ^ {\prime} & D _ {2 k - K _ {1}, 2 k - K _ {1}} ^ {\prime} \end{array} \right] := \left[ \begin{array}{c c} c _ {1} & c _ {2} \\ c _ {3} & c _ {4} \end{array} \right],
$$
where  $M_{i,j}$  denotes the  $ij$ -th entry of matrix  $M$ ,  $c = [c_1, c_2, c_3, c_4]^\top \neq 0$ , and all the other elements of  $D'$  are zeros. Let
$$
A _ {2} = A _ {1} + Q D ^ {\prime} Q ^ {- 1} \neq A _ {1},
$$
then  $A_{1}$  and  $A_{2}$  satisfy the first two equations in Equation (30). Similar to the case where  $k \in \{1, \ldots, K_{1}\}$ , one can also show that such a  $A_{2}$  also satisfies the third equation in Equation (30).
Therefore, assuming  $A$  has distinct eigenvalues, then when the identifiability condition (14) is not satisfied, one can always find a  $A_{2}$  with  $(A_{1}, GG^{\top}) \neq (A_{2}, GG^{\top})$  such that Equation (30) is satisfied, i.e.,  $X^{1} \stackrel{\mathrm{d}}{=} X^{2}$ . Hence, the necessity of the condition is proved.
A.6 Proof of Corollary 3.4.1
Proof. There are two ways to prove this corollary, we will present both of them in the following.
Way1. By [62, Lemma 2.2],
$$
\operatorname {s p a n} ([ G | A G | \dots | A ^ {d - 1} G ]) = \operatorname {s p a n} ([ G G ^ {\top} | A G G ^ {\top} | \dots | A ^ {d - 1} G G ^ {\top} ]),
$$
where  $\operatorname{span}(M)$  denotes the linear span of the columns of the matrix  $M$ . therefore, when
$$
\operatorname {r a n k} ([ G | A G | \dots | A ^ {d - 1} G ]) = d,
$$
$$
\operatorname {s p a n} \left(\left[ G \mid A G \right| \dots \left| A ^ {d - 1} G \right]\right) = \mathbb {R} ^ {d},
$$
$$
\operatorname {s p a n} \left(\left[ G G ^ {\top} \mid A G G ^ {\top} \right| \dots \left| A ^ {d - 1} G G ^ {\top} \right]\right) = \mathbb {R} ^ {d}.
$$
$$
\operatorname {r a n k} \left(\left[ G G ^ {\top} \mid A G G ^ {\top} \right| \dots \left| A ^ {d - 1} G G ^ {\top} \right|\right) = d,
$$
since the rank of a matrix is the dimension of its span. Then by Theorem 3.4, the generator of the SDE (1) is identifiable from  $x_0$ .
**Way2.** Let  $\tilde{A} \in \mathbb{R}^{d \times d}$  and  $\tilde{G} \in \mathbb{R}^{d \times m}$ , such that  $X(\cdot; x_0, A, G) \stackrel{\mathrm{d}}{=} X(\cdot; x_0, \tilde{A}, \tilde{G})$ , we denote as  $X \stackrel{\mathrm{d}}{=} \tilde{X}$ , we will show that under our identifiability condition  $(A, GG^{\top}) = (\tilde{A}, \tilde{G}\tilde{G}^{\top})$ . By applying the same notations used in the proof of Theorem 3.4, in the following, we denote  $A_1 := A$ ,  $A_2 := \tilde{A}$ ,  $G_1 := G$  and  $G_2 := \tilde{G}$ .
In the proof of Theorem 3.4, we have shown that  $G_{1}G_{1}^{\top} = G_{2}G_{2}^{\top}$ , thus, we only need to show that under the condition stated in this corollary,  $A_{1} = A_{2}$ . According to the proof of Theorem 3.4, for all  $0 \leqslant t < \infty$ , we have
$$
V _ {1} (t) = V _ {2} (t),
$$
$$
A _ {1} V _ {1} (t) = A _ {2} V _ {2} (t).
$$
Let  $V(t) \coloneqq V_{i}(t)(i = 1,2)$ , one gets
$$
A _ {1} V (t) = A _ {2} V (t), \forall 0 \leqslant t <   \infty .
$$
Therefore, if there exists a  $0 \leqslant t < \infty$ , such that  $V(t)$  is nonsingular, then one can conclude that  $A_{1} = A_{2}$ .
By [20, Theorem 3.2], the covariance  $V(t)$  is nonsingular for all  $t > 0$ , if and only if
that is the pair  $[A, G]$  is controllable. Therefore, under the condition stated in this corollary,  $A_{1} = A_{2}$ , thus the generator of the SDE (1) is identifiable from  $x_{0}$ .
A.7 Proof of Theorem 3.5
Proof. Let  $\tilde{A},\tilde{G}_k\in \mathbb{R}^{d\times d}$  for all  $k = 1,\ldots ,m$  such that  $X(\cdot ;x_0,A,\{G_k\}_{k = 1}^m)\stackrel {\mathrm{d}}{=}X(\cdot ;x_0,\tilde{A},\{\tilde{G}_k\}_{k = 1}^m)$ , we denote as  $X\stackrel {\mathrm{d}}{=}\tilde{X}$ , we will show that under our identifiability condition, for all  $x\in \mathbb{R}^d$ ,  $(A,\sum_{k = 1}^{m}G_{k}xx^{\top}G_{k}^{\top}) = (\tilde{A},\sum_{k = 1}^{m}\tilde{G}_{k}xx^{\top}\tilde{G}_{k}^{\top})$ . For simplicity of notation, in the following, we denote  $A_{1}\coloneqq A$ ,  $A_{2}\coloneqq \tilde{A}$ ,  $G_{1,k}\coloneqq G_{k}$  and  $G_{2,k}\coloneqq \tilde{G}_k$ , and denote  $X\stackrel {\mathrm{d}}{=}X$  as  $X^{1}\stackrel {\mathrm{d}}{=}X^{2}$ .
We first show that  $A_{1} = A_{2}$ . Set  $m_{i}(t) \coloneqq \mathbb{E}[X_{t}^{i}]$ , we know that  $m_{i}(t)$  satisfies the ODE
$$
\dot {m} _ {i} (t) = A _ {i} m _ {i} (t), \quad \forall 0 \leqslant t <   \infty ,
$$
$$
m _ {i} (0) = x _ {0}, \tag {34}
$$
where  $\dot{f}(t)$  denotes the first derivative of function  $f(t)$  with respect to time  $t$ .
Taking  $j$ -th derivative of  $e^{A_i t} x_0$  with respect to  $t$ , one finds that
$$
\left. \frac {d ^ {j}}{d t ^ {j}} \right| _ {t = 0} e ^ {A _ {i} t} x _ {0} = A _ {i} ^ {j} x _ {0},
$$
for all  $j = 1,2,\ldots$  . Consequently,
$$
A _ {1} ^ {j} x _ {0} = A _ {2} ^ {j} x _ {0}.
$$
Let us denote this vector  $A^j x_0$ . Obviously, one gets
$$
A _ {1} A ^ {j - 1} x _ {0} = A _ {2} A ^ {j - 1} x _ {0} \quad \text {f o r a l l} j = 1, 2, \dots . \tag {35}
$$
By condition A1, it is readily checked that  $A_{1} = A_{2}$  from Equation (35).
In the following, we show that under condition A2, for all  $x\in \mathbb{R}^d$
$$
\sum_ {k = 1} ^ {m} G _ {1, k} x x ^ {\top} G _ {1, k} ^ {\top} = \sum_ {k = 1} ^ {m} G _ {2, k} x x ^ {\top} G _ {2, k} ^ {\top}.
$$
We know the function  $P_{i}(t)\coloneqq \mathbb{E}[X_{t}^{i}(X_{t}^{i})^{\top}]$  satisfies the ODE
$$
\dot {P} _ {i} (t) = A _ {i} P _ {i} (t) + P _ {i} (t) A _ {i} ^ {\top} + \sum_ {k = 1} ^ {m} G _ {i, k} P _ {i} (t) G _ {i, k} ^ {\top}, \quad \forall 0 \leqslant t <   \infty , \tag {36}
$$
$$
P _ {i} (0) = x _ {0} x _ {0} ^ {\top}.
$$
Since  $X^1 \stackrel{\mathrm{d}}{=} X^2$ ,
$$
P _ {1} (t) = P _ {2} (t), \forall 0 \leqslant t <   \infty ,
$$
let us call it  $P(t)$ . By differentiating  $P_{i}(t)$  one also gets that
$$
\dot {P} _ {1} (t) = \dot {P} _ {2} (t), \forall 0 \leqslant t <   \infty .
$$
Since we have shown that  $A_{1} = A_{2}$  under condition A1, from Equation (36) one observes that
$$
\sum_ {k = 1} ^ {m} G _ {1, k} P (t) G _ {1, k} ^ {\top} = \sum_ {k = 1} ^ {m} G _ {2, k} P (t) G _ {2, k} ^ {\top}, \quad \forall 0 \leqslant t <   \infty . \tag {37}
$$
By vectorizing  $P(t)$ , some calculation shows that the ODE (36) can be expressed as
$$
\operatorname {v e c} (\dot {P} (t)) = \mathcal {A} \operatorname {v e c} (P (t)),
$$
$$
\begin{array}{l} \operatorname {v e c} (P (0)) = \operatorname {v e c} \left(x _ {0} x _ {0} ^ {\top}\right), \end{array} \tag {38}
$$
with an explicit solution
$$
\operatorname {v e c} (P (t)) = e ^ {\mathcal {A} t} \operatorname {v e c} \left(x _ {0} x _ {0} ^ {\top}\right),
$$
where  $\mathcal{A} = A\oplus A + \sum_{k = 1}^{m}G_{k}\otimes G_{k}\in \mathbb{R}^{d^{2}\times d^{2}}$ , and  $\operatorname {vec}(M)$  denotes the vector by stacking the columns of matrix  $M$  vertically.
By definition,  $P(t) \in \mathbb{R}^{d \times d}$  is symmetric, thus  $\operatorname{vec}(P(t))$  for all  $0 \leqslant t < \infty$  is confined to a proper subspace of  $\mathbb{R}^{d^2}$ , let us denote this proper subspace  $W$ , simple calculation shows that
$$
\dim (W) = (d ^ {2} + d) / 2,
$$
where  $\dim(W)$  denotes the dimension of the subspace  $W$ , that is the number of vectors in any basis for  $W$ . In particular, one can find a basis of  $W$  denoting as
where  $E_{ij}$  stands for a  $d \times d$  matrix, with the  $ij$ -th and  $ji$ -th elements are 1 and all other elements are 0, for all  $i, j = 1, \ldots, d$  and  $i \geqslant j$ .
Suppose there exists  $t_i$ 's, for  $i = 1, \ldots, (d^2 + d) / 2$ , such that  $\operatorname{vec}(P(t_1)), \ldots, \operatorname{vec}(P(t_{(d^2 + d) / 2}))$  are linearly independent, then for all  $x \in \mathbb{R}^d$ ,
$$
\operatorname {v e c} \left(x x ^ {\top}\right) = l _ {1} \operatorname {v e c} \left(P \left(t _ {1}\right)\right) + \dots + l _ {\left(d ^ {2} + d\right) / 2} \operatorname {v e c} \left(P \left(t _ {\left(d ^ {2} + d\right) / 2}\right)\right),
$$
$$
x x ^ {\top} = l _ {1} P (t _ {1}) + \dots + l _ {(d ^ {2} + d) / 2} P (t _ {(d ^ {2} + d) / 2}),
$$
where  $l \coloneqq \{l_1, \ldots, l_{(d^2 + d) / 2}\} \in \mathbb{R}^{(d^2 + d) / 2}$ . According to Equation (37), it is readily checked that for all  $x \in \mathbb{R}^d$ ,
By [57, Lemma 6.1], there exists  $(d^2 + d) / 2$ $t_i$ 's such that  $\operatorname{vec}(P(t_1)), \ldots, \operatorname{vec}(P(t_{(d^2 + d) / 2}))$  are linearly independent, if and only if the orbit of  $\operatorname{vec}(P(t))$  (i.e., the trajectory of ODE (38) started from initial state  $v$ ), denoting as  $\gamma(\mathcal{A}, v)$  with  $v = \operatorname{vec}(x_0 x_0^\top)$ , is not confined to a proper subspace of  $W$ .
Next, we show that under condition A2, orbit  $\gamma (\mathcal{A},v)$  is not confined to a proper subspace of  $W$
Assume orbit  $\gamma (\mathcal{A},v)$  is confined to a proper subspace of  $W$ . Then there exists  $w\neq 0\in W$  such that
$$
w ^ {\top} e ^ {A t} v = 0, \forall 0 \leqslant t <   \infty .
$$
By taking  $j$ -th derivative with respect to  $t$ , we have
$$
w ^ {T} \mathcal {A} ^ {j} e ^ {\mathcal {A} t} v = 0, \forall 0 \leqslant t <   \infty , j = 0, \dots , (d ^ {2} + d - 2) / 2.
$$
In particular, for  $t = 0$
$$
w ^ {T} \mathcal {A} ^ {j} v = 0, \quad \text {f o r} j = 0, \dots , (d ^ {2} + d - 2) / 2.
$$
$$
w ^ {T} [ v | \mathcal {A} v | \dots | \mathcal {A} ^ {(d ^ {2} + d - 2) / 2} v ] = 0. \tag {39}
$$
Since  $w \in W$ ,  $w \in \operatorname{span}\{\operatorname{vec}(E_{11}), \operatorname{vec}(E_{21}), \dots, \operatorname{vec}(E_{dd})\}$ , set  $\operatorname{vec}(\overline{w}) := w$ , then  $\overline{w}$  is a  $d \times d$  symmetric matrix. Since  $P(t)$  is symmetric for all  $0 \leqslant t < \infty$ , according to Equation (36), simple calculation shows that the  $j$ -th derivative of  $P(t)$  is also symmetric for all  $0 \leqslant t < \infty$ , for  $j = 0, 1, \ldots$ . Recall that
$$
\operatorname {v e c} (P (t)) = e ^ {\mathcal {A} t} v,
$$
$$
\operatorname {v e c} (P ^ {(j)} (t)) = \mathcal {A} ^ {j} e ^ {\mathcal {A} t} v,
$$
where  $P^{(j)}(t)$  denotes the  $j$ -th derivative of  $P(t)$  with respect to  $t$ . In particular, when  $t = 0$ , one has
$$
\operatorname {v e c} (P (0)) = v,
$$
$$
\operatorname {v e c} (P ^ {(j)} (0)) = \mathcal {A} ^ {j} v,
$$
then if we denote matrix  $\overline{\mathcal{A}^{j}v}$  by setting  $\mathrm{vec}(\overline{\mathcal{A}^{j}v})\coloneqq \mathcal{A}^{j}v$ , matrices  $\overline{\mathcal{A}^{j}v}$  are symmetric for all  $j = 0,1,\ldots$
Therefore, we can say that there are only  $(d^2 + d) / 2$  distinct elements in each of the vectors:  $w, v, \mathcal{A}v, \ldots, \mathcal{A}^{(d^2 + d - 2) / 2}v$  in Equation (39). Moreover, since these vectors all correspond to  $d \times d$  symmetric matrices, the repetitive elements in each vector appear in the same positions in each vector. Hence, we can focus on checking those distinct elements in each vector, that is Equation (39) can be expressed as
$$
\underline {{w}} ^ {T} [ v | \underline {{\mathcal {A}}} v ] \dots | \underline {{\mathcal {A} ^ {(d ^ {2} + d - 2) / 2}}} v ] = 0, \tag {40}
$$
where  $\underline{w}\in \mathbb{R}^{(d^2 +d) / 2}$  denotes as
$$
\underline {{w}} := \left[ \bar {w} _ {1 1}, \sqrt {2} \bar {w} _ {2 1}, \dots , \sqrt {2} \bar {w} _ {d 1}, \bar {w} _ {2 2}, \dots , \sqrt {2} \bar {w} _ {d 2}, \dots , \bar {w} _ {d d} \right] ^ {\top},
$$
where  $\overline{w}_{ij}$  denotes the  $ij$ -th element of  $\overline{w}$ , with  $i, j = 1, \ldots, d$  and  $i \geqslant j$ . When  $i \neq j$ , the element is multiplied by a  $\sqrt{2}$ , that is,  $\underline{w}$  only keeps the distinctive elements in  $w$ , and for each of the repetitive element, we multiply  $\sqrt{2}$ . We define  $\underline{v}, \underline{A} \underline{v}, \ldots$  using the same way.
Under condition A2, matrix
$$
\left[ \underline {{v}} | \mathcal {A} v | \dots | \underline {{\mathcal {A} ^ {(d ^ {2} + d - 2) / 2}}} v \right] \in \mathbb {R} ^ {\frac {(d ^ {2} + d)}{2} \times \frac {(d ^ {2} + d)}{2}}
$$
is easily to be checked to be invertible, then  $\underline{w} = 0$ , thus  $w = 0$ . This contradicts to  $w \neq 0$ , therefore, under condition A2, orbit  $\gamma(\mathcal{A}, v)$  is not confined to a proper subspace of  $W$ . Hence, the theorem is proved.
B Genericity of the derived identifiability conditions
B.1 The identifiability condition stated in Theorem 3.5 is generic
We will show that the identifiability condition stated in Theorem 3.5 is generic. Specifically, we will show that for the set of  $(x_0,A,\{G_k\}_{k = 1}^m)\in \mathbb{R}^{d + (m + 1)d^2}$  such that either condition A1 or A2 stated in Theorem 3.5 is violated, has Lebesgue measure zero in  $\mathbb{R}^{d + (m + 1)d^2}$ . In the following, we first present a lemma we will use to prove our main proposition.
Lemma B.1. Let  $p: \mathbb{R}^n \to \mathbb{R}$  be a non-zero polynomial function. Let  $Z := \{x \in \mathbb{R}^n : p(x) = 0\}$ . Then  $Z$  has Lebesgue measure zero in  $\mathbb{R}^n$ .
Proof. When  $n = 1$ , suppose the degree of  $x$  is  $k \geqslant 1$ , then by the fundamental theorem of algebra, there are at most  $k$ $x$ 's such that  $x \in Z$ . Therefore,  $Z$  has Lebesgue measure zero, since a finite set has measure zero in  $\mathbb{R}$ .
Suppose the lemma is established for polynomials in  $n - 1$  variables. Let  $p$  be a non-zero polynomial in  $n$  variables, say of degree  $k \geqslant 1$  in  $x_{n}$ , then we can write
$$
p (\boldsymbol {x}, x _ {n}) = \sum_ {j = 0} ^ {k} p _ {j} (\boldsymbol {x}) x _ {n} ^ {j},
$$
where  $\pmb{x} = \{x_{1},\dots,x_{n - 1}\}$  and  $p_0,\ldots ,p_k$  are polynomials in the  $n - 1$  variables  $\{x_{1},\dots,x_{n - 1}\}$ , and there exists  $j\in \{0,\dots,k\}$  such that  $p_j$  is a non-zero polynomial since  $p$  is a non-zero polynomial. Then we can denote  $Z$  as
$$
Z = \left\{\left(\boldsymbol {x}, x _ {n}\right): p \left(\boldsymbol {x}, x _ {n}\right) = 0 \right\}.
$$
Suppose  $(\pmb {x},x_{n})\in Z$  , then there are two possibilities:
case 1  $p_0(\boldsymbol {x}) = \dots = p_k(\boldsymbol {x}) = 0$
case 2 there exists  $i\in \{0,\dots ,k\}$  such that  $p_i(\pmb {x})\neq 0$
$$
A := \left\{\left(\boldsymbol {x}, x _ {n}\right) \in Z: \text {c a s e 1 i s s a t i s f i e d} \right\},
$$
$$
B := \left\{\left(\boldsymbol {x}, x _ {n}\right) \in Z: \text {c a s e 2 i s s a t i s f i e d} \right\},
$$
then  $Z = A\cup B$
For case 1, recall that there exists  $j \in \{0, \dots, k\}$  such that  $p_j$  is a non-zero polynomial, let
$$
A _ {j} := \left\{\boldsymbol {x} \in \mathbb {R} ^ {n - 1}: p _ {j} (\boldsymbol {x}) = 0 \right\},
$$
then by the induction hypothesis,  $A_{j}$  has Lebesgue measure zero in  $\mathbb{R}^{n - 1}$ . Therefore,  $A_{j}\times \mathbb{R}$  has Lebesgue measure zero in  $\mathbb{R}^n$ . Since  $A\subseteq A_{j}\times \mathbb{R}$ ,  $A$  has Lebesgue measure zero in  $\mathbb{R}^n$ .
For case 2, let  $\lambda^n$  be Lebesgue measure on  $\mathbb{R}^n$ , then
$$
\begin{array}{l} \lambda^ {n} (B) = \int_ {\mathbb {R} ^ {n}} \mathbb {1} _ {B} (\boldsymbol {x}, x _ {n}) d \lambda^ {n} \\ = \int_ {\mathbb {R} ^ {n}} \mathbb {1} _ {B} (\boldsymbol {x}, x _ {n}) d \boldsymbol {x} d x _ {n} \tag {41} \\ = \int_ {\mathbb {R} ^ {n - 1}} \left(\int_ {\mathbb {R}} \mathbb {1} _ {B} (\boldsymbol {x}, x _ {n}) d x _ {n}\right) d \boldsymbol {x}, \\ \end{array}
$$
$$
\mathbb {1} _ {B} (\boldsymbol {x}, x _ {n}) = \left\{ \begin{array}{l l} 1, & \text {i f} (\boldsymbol {x}, x _ {n}) \in B  , \\ 0, & \text {i f} (\boldsymbol {x}, x _ {n}) \notin B  . \end{array} \right.
$$
The inner integral in Equation (41) is equal to zero, since for a fixed  $x$ , there are finitely many (indeed, at most  $k$ )  $x_{n}$ 's such that  $p(x, x_{n}) = 0$  under the condition of case 2. Thus,  $\lambda^n(B) = 0$ , that is,  $B$  has Lebesgue measure zero in  $\mathbb{R}^n$ . Then it is readily checked that  $Z$  has Lebesgue measure zero.
Now we are ready to present the main proposition.
then  $S$  has Lebesgue measure zero in  $\mathbb{R}^{d + (m + 1)d^2}$ .
$$
S _ {A} := \left\{\left(x _ {0}, A\right) \in \mathbb {R} ^ {d + d ^ {2}}: \text {c o n d i t i o n A 1 i s v i o l a t e d} \right\},
$$
we first show that  $S_A$  has Lebesgue measure zero in  $\mathbb{R}^{d + d^2}$ . Suppose  $(x_0, A) \in S_A$ , then  $(x_0, A)$  satisfies
$$
\operatorname {r a n k} \left(\left[ x _ {0} \mid A x _ {0} \right| \dots \left| A ^ {d - 1} x _ {0} \right|\right) <   d,
$$
that is the set of vectors  $\{x_0, Ax_0, \ldots, A^{d-1}x_0\}$  are linearly dependent, this means that
$$
\det \left( \right.\left[ \right. x _ {0} \left. \right| A x _ {0} \left. \right\rvert   \dots \left| \right. A ^ {d - 1} x _ {0} \left. \right]\left. \right) = 0. \tag {42}
$$
It is a simple matter of algebra that the left side of (42) can be expressed as some universal polynomial of the entries of  $x_0$  and entries of  $A$ , denotes  $p(x_0, A) = p(x_{01}, \ldots, x_{0d}, a_{11}, a_{12}, \ldots, a_{dd})$ , where  $x_{0j}$  denotes the  $j$ -th entry of  $x_0$  and  $a_{ij}$  denotes the  $ij$ -th entry of  $A$ . Therefore, one concludes that
$$
p \left(x _ {0}, A\right) = p \left(x _ {0 1}, \dots , x _ {0 d}, a _ {1 1}, a _ {1 2}, \dots , a _ {d d}\right) = 0.
$$
Thus,  $S_A$  can be expressed as
$$
S _ {A} = \left\{\left(x _ {0}, A\right) \in \mathbb {R} ^ {d + d ^ {2}}: p \left(x _ {0}, A\right) = 0 \right\}.
$$
$$
p \left(x _ {0}, A\right) = \sum_ {i _ {1}, \dots , i _ {d} = 1} ^ {d} x _ {0 i _ {1}} \dots x _ {0 i _ {d}} \det \left( \right.\left[ \right.\left(A ^ {0}\right) _ {\cdot i _ {1}} \left. \right|\left(A ^ {1}\right) _ {\cdot i _ {2}} \left. \right| \dots | \left(A ^ {d - 1}\right) _ {\cdot i _ {d}} \left. \right]\left. \right), \tag {43}
$$
where  $(M)_{\cdot j}$  denotes the  $j$ -th column vector of matrix  $M$ . Obviously,  $p(x_0,A)$  is a non-zero polynomial function of entries of  $x_0$  and entries of  $A$ , therefore, by Lemma B.1,  $S_A$  has Lebesgue measure zero in  $\mathbb{R}^{d + d^2}$ . Let
$$
S _ {1} := \left\{\left(x _ {0}, A, \left\{G _ {k} \right\} _ {k = 1} ^ {m}\right) \in \mathbb {R} ^ {d + (m + 1) d ^ {2}}: \text {c o n d i t i o n A 1 i s v i o l a t e d} \right\},
$$
then it is readily checked that  $S_{1}$  has Lebesgue measure zero in  $\mathbb{R}^{d + (m + 1)d^2}$ .
$$
S _ {2} := \left\{\left(x _ {0}, A, \left\{G _ {k} \right\} _ {k = 1} ^ {m}\right) \in \mathbb {R} ^ {d + (m + 1) d ^ {2}}: \text {c o n d i t i o n A 2 i s v i o l a t e d} \right\},
$$
we then show that  $S_{2}$  has Lebesgue measure zero in  $\mathbb{R}^{d + (m + 1)d^2}$ . Suppose  $(x_0, A, \{G_k\}_{k=1}^m) \in S_2$ , then  $(x_0, A, \{G_k\}_{k=1}^m)$  satisfies
$$
\operatorname {r a n k} ([ v | \mathcal {A} v | \dots | \mathcal {A} ^ {(d ^ {2} + d - 2) / 2} v ]) <   (d ^ {2} + d) / 2,
$$