Method to reverse a Kronecker product Let's say I have two simple vectors: $[0, 1]$ and $[1, 0]$.
Their Kronecker product would be $[0, 0, 1, 0]$.
Let's say I have only the Kronecker product. How can I find the two initial vectors back?
If my two vectors are written as : $[a, b]$ and $[c, d]$, the (given) Kronecker product is:
$$[ac, ad, bc, bd] = [k_0, k_1, k_2, k_3]$$
So I have a system of four non linear equations that I wish to solve:
$$\begin{align*}
ac &= k_0\\
ad&= k_1\\
bc&= k_2\\
bd &=k_3.
\end{align*}$$
I am looking for a general way to solve this problem for any number of initial vectors in $\mathbb{C}^2$ (leading my number of variables to $2n$ and my equations to $2^n$ if I have $n$ vectors).
So here are a few specific questions:
What is the common name of this problem? 
If a general solution is known, what is its complexity class?
Does the fact that I have more and more equations when $n$ goes up compared to the number of variables help?
(Note: I really didn't know what to put as a tag.)

I prefer to say the Kronecker of two vectors is reversible, but up to a scale factor. In fact, Niel de Beaudrap has given the right answer. Here I attempt to present it in a concise way.
Let $a\in\mathbb{R}^N$ and $b\in\mathbb{R}^M$ be two column vectors. The OP is: given $a\otimes b$, how to determine $a$ and $b$?
Note $a\otimes b$ is nothing but $\mathrm{vec} (ba^T)$. The $\mathrm{vec}$ operator reshapes a matrix to a column vector by stacking the column vectors of the matrix. Note $\mathrm{vec}$ is reversible. Therefore, given $c=a\otimes b$, first reshape it to a $M\times N$ matrix $C=ba^T$. $C$ is a rank-one matrix. You can simply decompose $C$ to get $kb$ and $a/k$.
