Given that $x_1 = \begin{bmatrix} 1 & 3 & 0 \end{bmatrix}^t, x_2 = \begin{bmatrix} 2 & -6 & 6 \end{bmatrix}^t$ and $x_3 = \begin{bmatrix} -4, 2, -7 \end{bmatrix}^t$ are all in the nullspace of matrix $A$, what are all possible values of the rank of $A$?
I first established an upper bound for the rank (it cannot be greater than 3) as the elements in the nullspace have 3 entries, so the actual matrix $A$ has 3 rows and through the rank theorem (or definition of rank), there will be at most 3 linearly independent rows and thus have at most rank 3.
Additionally, after applying some algebra, I found that the set $\{x_1, x_2, x_3\}$ has dimension 2, so the nullity of $A$ must be at least 2. The problem does not give any information on the number of columns of $A$, however, so the rank-nullity theorem does not seem applicable here.
From this, the rank is any integer value from 0 to 3. Am I missing anything here?
From the information given we have $A$ is an $n \times 3$ matrix because the null space is subset of $\Bbb{R}^3$ ($Ax_1 =0$ etc.). Moreover $x_3 \in \text{Span}(\{x_1,x_2\}$ because $x_3=-\left(\dfrac{5x_1+7x_2}{3}\right)$.
This means $\text{null}(A)$ has at least two independent vectors, thus $\text{nullity}(A) \geq 2$. Now use rank-nullity theorem $$\text{rank}(A)+\text{nullity}(A)=3,$$ to conclude that $\text{rank}(A) \leq 1$. Thus $\text{rank}(A)=0 \text{ or } 1$.