Let $n > 1$. Consider an $n \times n$ matrix $M$ with its elements from $\mathbb{R}$. Let the vector $(0, 1, 0, 0, \dots, 0) \in \mathbb{R}^n$ be in the null space of $M$. Which of the following options is/are always correct?

Let $v = (0, 1, 0, \dots, 0)^T \in \mathbb{R}^n$.

1. Determinant of $M$: Since $v$ is a non-zero vector and $v \in \text{Null}(M)$, we have $Mv = 0$. This implies that the null space of $M$ has dimension at least $1$ (i.e., $\text{Nullity}(M) \ge 1$). Therefore, $0$ is an eigenvalue of $M$, which means:

   $$\det(M) = 0$$

   Hence, (B) is correct and (A) is incorrect.

2. Rank of $M$: By the Rank-Nullity Theorem, $\text{Rank}(M) = n - \text{Nullity}(M) \le n - 1$. For an arbitrary $n > 1$, $\text{Rank}(M)$ is not necessarily $1$. Hence, (C) is incorrect.

3. Vectors in the Null Space: Since $\text{Null}(M)$ is a subspace over $\mathbb{R}$, for any scalar $c \in \mathbb{R}$, $c v \in \text{Null}(M)$. Since $\mathbb{R}$ contains infinitely many non-zero scalars, there exist infinitely many distinct non-zero vectors in $\text{Null}(M)$ (for example, $v$ and $2v$). Thus, there are at least two non-zero vectors in the null space of $M$. Hence, (D) is correct.

Therefore, the correct options are B and D.