First Thoughts
Okay, so here is an idea which seams a little dirty, but is super close to a proper solution. Maybe some of you can complete it. First chose a $C' \in \mathbb{N}^{m \times m}$ such that all entries are $c \in \mathbb{N} \setminus \{0\}$. Thus:
$$(C' \cdot A)_{ij} = \sum_{k=1}^m C'_{ik} A_{kj} = c \cdot \sum_{k=1}^m A_{kj} \geq c > 0$$
It seams like we found a good $C = C'$, but $C'$ is not invertable. It is known that there are way more invertible matrices than non-invertable ones. We can use that fact by just adding a random noise matrix $\Delta \in \mathbb{N}^{m \times m}$ onto $C'$. We we hope that the resulting $C := C' + \Delta$ is invertable. If not, we just draw another one and because there are so many more invertable matrices than non-invertable ones, we eventually will draw an invertable $C$.
But we cannot chose $\Delta$ freely. We still have to make sure that the columns of $\Delta$ sum up to the same number - well that is doable. The harder restriction is, that $C\cdot A$ must still only contain positive entries. For that we chose some $\delta \in \mathbb{N}$ and we chose the entries in $\Delta$ such that $\forall i,j\colon 0 \leq \Delta_{ij} \leq \delta$. Let $h \in \mathbb{N}$ be the largest absolute value in $A$. Thus we see that
\begin{align*}
(C \cdot A)_{ij} &= (\Delta \cdot A)_{ij} + (C' \cdot A)_{ij} \geq \sum_{k=1}^m \Delta_{ik} A_{kj} + c \geq \sum_{k=1}^m -|\Delta_{ik}|\cdot|A_{kj}| + c\\
&\geq \sum_{k=1}^m -\delta h + c = c - mh\delta
\end{align*}
For that to be non-negative, we just need to select the $\delta$ such that
$$0 \leq c - mh\delta \quad\Leftrightarrow\quad \delta \leq \frac{c}{mh}$$ This absolutely no problem, as we can shose $c$ as high as we want and in that way we won't get any negative entries.
This solution is super hand wavy, maybe I get some support from you guys to make this approach more rigorous. So what could one do to make this method more rigorous? Well, I see two ways:
- Show that this $\Delta$ always exists
- Construct $\Delta$ deliberately and not by chance.
Edit: Completed Answer
I found an $\Delta$ that will always produce an invertible $C = \Delta + C'$ and makes sure that $C\cdot A$ only contains non-negative entries. It is $\Delta := \delta \cdot I_m$, where $I_m$ is the identity matrix of size $m \times m$ and $\delta \in \mathbb{N} \setminus \{0\}$ (Remember that all entries in $C'$ are all equal to $c \in \mathbb{N}$, just as above). Let's check, that this is exactly what we are looking for.
Is $C$ invertable?
Let's take a look at the shape of $C$:
$$C =
\begin{pmatrix}
c+\delta & c & \cdots & c\\
c & c+\delta & \cdots & c\\
\vdots & \vdots & \ddots & \vdots\\
c & c & \dots & c+\delta
\end{pmatrix}$$
It is apparent, that because $\delta \neq 0$ all column vectors are linearly independent. Thus $\Delta$ is indeed invertable.
Does $C \cdot A$ only contain non-negative entries? For that let us chose $\delta$ a little bit more carefully. Let $h \in \mathbb{N}$ be the largest absolut value in $A$:
$$h := \max_{i, j} |A_{ij}|$$
Then we will chose $\delta \leq \frac{c}{h}$. This is no problem, as we can chose $c$ as high as we want. Now, let's check what we have to check:
\begin{align*}
(C \cdot A)_{ij} &= (\delta\cdot I_m\cdot A)_{ij} + (C'\cdot A)_{ij} = \delta \cdot A_{ij} + \sum_{k=1}^m C'_{ik} A_{kj}\\
&= \delta \cdot A_{ij} + c \cdot \underbrace{\sum_{k=1}^m A_{kj}}_{\geq 1} \geq \delta \cdot A_{ij}+ c \geq -\delta\cdot |A_{ij}| + c\\
&\geq -\delta \cdot h + c \geq -\frac{c}{h} \cdot h + c = 0
\end{align*}
