I have an assignment that asks me to prove something but I've hit a roadblock.
Let $u$ be a $3 \times 1$ matrix with $u_{1}$, $u_{2}$, and $u_{3}$. Let $v$ be a $3 \times 1$ matrix with $v_{1}, v_{2}, v_{3}$. Assume that $u_{1} \neq 0$, $v_{1} \neq 0$. Let $A = uv^{T}$ Prove that $\text{rank}(A) = 1$.
$v^{T} = (v_1, v_2, v_3)$ and multiplying it with $u$ ($u$ is before $v^T$) gives:
$1 \times 1$ matrix $A = ( u_{1}v_{1} + u_{2}v_{2} + u_{3}v_{3} )$
Where do I go from here to figure out the rank? I'm not sure if the basis is the resulting scalar since $1 \times 1$ matrixes are scalars according to my teacher. If this were true then I could just say the dimension of $A$ is $1$ and thus the rank is $1$, but I'm not sure if this is correct.
Sorry about the shoddy display, I don't know how to use TeX language.
EDIT: The $1 \times 1$ matrix I found is incorrect. $A$ is a $3 \times 3$ matrix.
EDIT: I wrote this answer under the assumption that $A$ was a $1\times 1$ matrix (as originally stated in the question), but that doesn't follow from the problem statement.
It's not the dimension of $A$ that is $1$ (matrices don't have that kind of dimension), but rather the dimension of the column space of $A$ is $1$. Other than that, you're good!
[I would be a bit reluctant to always think of $1\times 1$ matrices as scalars, though, because matrices have a lot of operations that become uninteresting on scalars. It is a bit awkward to think about the column space of a scalar, for example.]