Doing a Charnes-Cooper transformation with matrices and an zero-one constraint I need to solve an assignment problem with the following objective function:
$${\max} \frac{\displaystyle\sum_{i=1}^m\sum_{j=1}^n h_{ij}\cdot x_{ij}}{\displaystyle\sum_{i=1}^m\sum_{j=1}^n c_{ij}\cdot x_{ij}}$$
Since this is a fractional problem I want to use the Charnes-Cooper transformation (https://en.wikipedia.org/wiki/Linear-fractional_programming) to get a linear problem. 
My first question is, can I just do that even though I've got matrices instead of vectors? Second is, $\alpha$ as well as $\beta$ would in my case both just be zero and disappear, right? If I'm not missing something, this would make this whole constraint 
$$
\mathbf{d}^{T} \mathbf{y}+\beta t=1
$$
which gets introduced by that transformation also dissappear (or make it trivial), because if $\beta$ equals zero and 
$$
\mathbf{y}=\frac{1}{\mathbf{d}^{T} \mathbf{x}+\beta} \cdot \mathbf{x}
$$
you end up with 
$$
\mathbf{\frac{\mathbf{d}^{T} \mathbf{x}}{\mathbf{d}^{T} \mathbf{x}}} = \mathbf{1} \ 
$$
Correct?
Additionaly, as stated in the title, I've got the following zero-one constraint: 
$$
x_{i j} \in\{0,1\}
$$
which after the Charnes-Cooper transformation looks like this
$$
y_{i j} \in\{0, t\}
$$
because
$$
\mathbf{x}=\frac{1}{t} \mathbf{y}
$$
Will I get into trouble solving this because my zero-one constraint now contains a variable instead of just the simple values 0 and 1?
 A: The shape of the decision variables (matrix versus vector) does not matter.  Yes, $\alpha=\beta=0$ in your case. The idea of the transformation is that you multiply both numerator and denominator by a variable $t$ so that the denominator becomes 1.  You want to
$$\text{maximize $\sum_{i,j} h_{i,j}\cdot t\cdot x_{i,j}$ subject to $\sum_{i,j} c_{i,j}\cdot t\cdot x_{i,j} = 1$}.$$  Now introduce $y_{i,j} = t\cdot x_{i,j}$ to linearize both objective and constraint:
$$\text{maximize $\sum_{i,j} h_{i,j}\cdot y_{i,j}$ subject to $\sum_{i,j} c_{i,j}\cdot y_{i,j} = 1$}$$
Finally, linearize the relationship between $y$ and $x$.  We want to enforce:
$$y_{i,j}=\begin{cases}t &\text{if $x_{i,j} = 1$}\\0 &\text{if $x_{i,j} = 0$}\end{cases}$$
You can do that as follows, where $M$ is an upper bound on $t$, for example $M=1/\min_{i,j} \{c_{i,j}\}$:
\begin{align}
0 \le y_{i,j} &\le M x_{i,j}\\
y_{i,j} - t &\le M (1 - x_{i,j})\\
y_{i,j} - t &\ge -M (1 - x_{i,j})\\
\end{align}
You mentioned "assignment problem," so I assume you also have linear constraints like this: $$\sum_j x_{i,j} = 1 \quad \text{for all $i$}$$
(or maybe with $i$ and $j$ reversed).  In that case, you can use compact linearization instead:
\begin{align}
0 \le y_{i,j} &\le M x_{i,j}\\
\sum_j y_{i,j} &= t
\end{align}
