TITLE: A rank-one matrix is the product of two vectors
QUESTION [36 upvotes]: Let $A$ be an $n\times m$  matrix. Prove that $\operatorname{rank} (A) = 1$ if and only if there exist column vectors $v \in \mathbb{R}^n$ and $w \in \mathbb{R}^m$ such that $A=vw^t$.

Progress: I'm going back and forth between using the definitions of rank: $\operatorname{rank} (A) = \dim(\operatorname{col}(A)) = \dim(\operatorname{row}(A))$ or using the rank theorem that says $ \operatorname{rank}(A)+\operatorname{nullity}(A) = m$. So in the second case I have to prove that $\operatorname{nullity}(A)=m$-1

REPLY [7 votes]: Suppose that $A$ has rank one. Then its image is one dimensional, so there is some nonzero $v$ that generates it. Moreover, for any other $w$, we can write
$$Aw = \lambda(w)v$$
for some scalar $\lambda(w)$ that depends linearly on $w$ by virtue of $A$ being linear and $v$ being a basis of the image of $A$. This defines then a nonzero functional $\mathbb R^n \longrightarrow \mathbb R$ which must be given by taking dot product by some $w_0$; say $\lambda(w) =\langle w_0,w\rangle$. It follows then that 
$$ A(w) = \langle w_0,w\rangle v$$ for every $w$, or, what is the same, that $A= vw_0^t$.