Proving that an $m \times n$ matrix has rank 1 I've been trying to work through the exercises in my book where you have to prove certain things are the case. It has been going okay so far, however I've gotten stuck on the following exercise, and I am not sure where to start.
How can you prove that an $m \times n$ matrix $A$ has rank 1, if and only if $A$ can be written as the outer product uv$^T$ of a vector u in $\mathbb{R}^m$ and v in $\mathbb{R}^n$
I really don't know how to even start with this.
 A: Since the matrix $A$ has rank 1, the space of image of $\Bbb R^n$ under $A$ is one dimensional. Thus there exists a vector $u\in\Bbb R^m$ such that $||u||=1$ and $\mathcal R(A)=\text{span}(u)$. Observe that $f:\Bbb R^n\to \Bbb R$ defined by
$$
f(w)=\alpha ;\ \ \ \text{where}\ \ \  \ Aw=\alpha u
$$
is a linear functional. By Riesz Representation Theorem, there exists $v\in \Bbb R^n$ such that 
$$
f(w)=\langle w,v \rangle=v^{T}w
$$.
Therefore $\exists u\in\Bbb R^m$ and $\exists v\in\Bbb R^n$ such that 
$$
Aw=\alpha u = f(w)u=(v^{T}w)u=(uv^{T})w\ \ \  \forall w\in \Bbb R^n
$$
which implies that 
$$
A=uv^{T}
$$
for some $u\in\Bbb R^m$, $v\in\Bbb R^n$.
The converse is easy.
A: $A$ has rank $1$ $\implies$ $A = uv^T$:
Rank $1$ means that $A$ is not the zero matrix, but all columns are proportional to one another. Let $u$ be the first non-zero column, and let $v$ consist of the coefficients of proportionality between $u$ and the other columns. For instance, if 
$$
A = \begin{bmatrix}0&2&3\\0&-3&-4.5\\0&6&9\end{bmatrix}
$$
we would have $u = \left[\begin{smallmatrix}2\\-3\\6\end{smallmatrix}\right] $ and $v = \left[\begin{smallmatrix}0\\1\\1.5\end{smallmatrix}\right]$.

$A = uv^T$ $\implies$ $A$ has rank $1$:
$u$ is an $m \times 1$ matrix  with rank $1$, and $v$ is an $n\times 1$ matrix with rank $1$, so the rank of their product must be $1$.
