Instead of looking at the column space directly, we can reason using the rank inequality for a product of matrices.
Treat $u$ as an $n \times 1$ matrix and $v^T$ as a $1 \times n$ matrix. For any product of two matrices $P$ and $Q$, a standard rank inequality says:
\[ \text{rank}(PQ) \le \min(\text{rank}(P), \text{rank}(Q)) \]Here $P = u$ has rank 1 (a single non-zero column vector always has rank 1, since it has exactly one non-zero column), and $Q = v^T$ also has rank 1 (a single non-zero row vector always has rank 1). So:
\[ \text{rank}(A) = \text{rank}(uv^T) \le \min(1,1) = 1 \]This tells us the rank cannot exceed 1. Now check it cannot be 0 either: rank 0 would mean $A$ is the all-zero matrix, but since $u$ has some non-zero entry $u_i$ and $v$ has some non-zero entry $v_j$, the entry $A_{ij} = u_i v_j \ne 0$, so $A$ is not the zero matrix, ruling out rank 0.
The only value left between 0 and 1 (rank is always a non-negative integer) is 1, so the rank of $A$ is exactly 1.
\[ \boxed{\text{rank}(A) = 1} \]