I have a binary data matrix $C_{m \times n}$. I know that using Boolean algebra (logical or for summation and logical and for multiplication), this matrix can be factorized into two binary matrices $A_{m \times k}$ and $B_{k \times n}$, such that $$AB=C$$
My question is whether this factorization unique, i.e., is there any alternative $A$ and $B$ matrices that result in $C$ matrix, is the $k$ is given. We can also assume that $k$ is minimal, which means there isn't any $k' < k$ that supports this factorization.
Here are two counterexamples for $k=m=n=2$:
$$\begin{pmatrix}1&1\\1&1\end{pmatrix}=\begin{pmatrix}1&1\\0&1\end{pmatrix}\begin{pmatrix}1&0\\1&1\end{pmatrix}=\begin{pmatrix}1&1\\1&0\end{pmatrix}\begin{pmatrix}1&1\\0&1\end{pmatrix}$$
$$\begin{pmatrix}0&0\\0&0\end{pmatrix}=\begin{pmatrix}1&0\\1&0\end{pmatrix}\begin{pmatrix}0&0\\0&1\end{pmatrix}=\begin{pmatrix}0&1\\0&1\end{pmatrix}\begin{pmatrix}0&1\\0&0\end{pmatrix}$$
In fact even for $k=m=n=1$, you have already the counter-example:
$$0=0 \times 1=1 \times 0=0 \times 0$$
There is a very direct combinatorial argument: pigeonhole principle for the existence of counter-examples that could work for any dimension, but that I give for example in the case of $3 \times 3$ matrices:
There are $2^9=512$ possible binary matrices: therefore, one can write $2^{18}=262144$ products $AB$ (the "pigeons"). As in front of them, there are only $512$ possible results (the "holes"), necessarily there is a hole with (at least) 2 pigeons, i.e., a matrix $C$ with $C=A_1B_1=A_2B_2$ in two different ways...
Here is a paper giving a standard form for this type of matrices (but using rather sophisticated algebra).