Factoring a given rank-$1$ matrix Suppose you have a $n \times 1$ column vector
$$a=\begin{bmatrix}a_1\\{a_2}\\ \vdots\\{a_n}\end{bmatrix}$$
and a $1 \times m$ row vector
$$\quad b=\begin{bmatrix}b_1 & b_2 & \ldots & b_m\end{bmatrix}$$
If you then multiply these
$$A=ab=\begin{bmatrix} a_{1}b_1 & a_{1}b_2 & \ldots & a_{1}b_m\\ a_{2}b_1 & a_{2}b_2 & \ldots & a_{2}b_m \\ \vdots&&&\vdots \\ a_{n}b_1 & a_{n}b_2 & \ldots & a_{n}b_m\end{bmatrix}$$
How difficult (if possible) is it to find the original vectors $a$ and $b$ if you are given $A$?
 A: Suppose I have vectors $\mathrm u \in \mathbb R^m$ and $\mathrm v \in \mathbb R^n$. I form the $m \times n$ rank-$1$ matrix
$$\mathrm A := \mathrm u \mathrm v^{\top}$$
and tell you what $\mathrm A$ is. Can you reconstruct vectors $\mathrm u$ and $\mathrm v$ from matrix $\mathrm A$? You cannot, because
$$\left( \gamma \mathrm u \right) \left( \frac{1}{\gamma} \mathrm v^{\top} \right) = \mathrm A$$
for all $\gamma \neq 0$. What you can recover from $\mathrm A$ is two lines passing through the origin and on which $\mathrm u$ and $\mathrm v$ live. More information is needed to recover $\mathrm u$ and $\mathrm v$.
For instance, if I tell you what $\| \mathrm u \|_2$ is, then you can find where the line whose direction vector is $\mathrm u$ intersects the Euclidean sphere of radius $\| \mathrm u \|_2$ centered at the origin. You find two points. One of them is $\mathrm u$ and the other is $-\mathrm u$, but you cannot know which is which, unfortunately. Sign ambiguity cannot be eliminated. Note that $(-\mathrm u) (-\mathrm v^{\top}) = \mathrm A$.
To summarize, matrix-valued function $\mathrm F (\mathrm x, \mathrm y) := \mathrm x \mathrm y^{\top}$ is not injective and, thus, the inputs $\mathrm x$ and $\mathrm y$ cannot be recovered from the output.

matrices rank-1-matrices matrix-decomposition
A: You cannot determine the original vectors $a$ and $b$, since they are not unique ($2a$ and $0.5b$ yield the same result as Rodrigo explains). However, you can construct one $a$ and $b$ such that $A=ab^T$. Just find any nonzero element in $A$. Let's assume that the (1,1) element is nonzero, but the method works in other cases too. The (1,1) element is $a_1 b_1$. Now let $a_1=1$, then you can compute $b_1$. By going through the first row and first column, you can compute all elements of the vectors $a$ and $b$.
A: If $A$ is indeed of the form $A=ab$, then it is quite easy to recover $a,b$ up to a constant, because given any colum vector $ x \in \mathbb{R}^m$, not in $Ker(A)$, $A\,x = a (bx)$, i.e.: $a$ is $A\,x$ up to the constant $b\,x$. Likewise, given any column vector $y \in \mathbb{R}^n$, not in $Ker(A^T)$, we have $y\,A = (y\,a)b$, so $b$ is $y\,A$ up to the constant $y\,a$.
