Consider a matrix $A\in\mathbb{R}^{m\times m}$ and define $a_{i\cdot}$ as the $i$-th row of $A$ and $a_{\cdot i}$ as the $i$-th column of $A$.
Next to this, there are pairwise disjoint index sets $I_1,\ldots, I_n\subseteq\{1,\ldots, m\}$ for some $n\in\{1,\ldots, m\}$ with following property:
Let $I\in\{I_1,\ldots, I_n\}$. For every $i\in I$, it holds
$$a_{ip}=0=a_{pi}\text{, if }p\not\in I$$ $$a_{ip},a_{pi}<0\text{, if }p\in I\backslash\{i\}$$ $$a_{ii}=-\sum_{j\in I\backslash\{i\}}a_{ij}$$
One last information is given, namely $\operatorname{rank }M=m-n$.
Now I would like to show, that the rows $(a_{i\cdot})_{i\in I}$ are linearly dependent for any $I\in\{I_1,\ldots,I_n\}$.
I tried following:
The columns $(a_{\cdot j})_{j\in I}$ are linearly dependent for any $I\in\{I_1,\ldots,I_n\}$, since for any $i\in I$, it holds
$$ \sum_{j\in I}a_{i j}=a_{ii}+\sum_{j\in I\backslash\{i\}}a_{ij}=0$$
If $i\not\in I$, then every entry is equal $0$, such that
$$ \sum_{j\in I}a_{i j}= 0$$
Given the information, that $\operatorname{rank }M=m-n$, we can define a selection $S=\{i_1,\ldots, i_n\}$, with $i_j\in I_j$ for $j\in\{1,\ldots, n\}$ and we find that the columns $(a_{\cdot j})_{j\in\{1,\ldots, m\}\backslash S}$ must be linearly independent. Especially the columns $(a_{\cdot j})_{j\in I_k\backslash\{i_k\}}$ must be linearly independent for any $k\in\{1,\ldots, n\}$.
However, I am not able to show it for the rows $(a_{i\cdot})_{i\in I}$. I think I have to use the fact, that for any other row $a_{i\cdot}$ with $i\not\in I$, there is showing up a non-zero element at $j$-th column/entry for $j\not\in I$, because of $\operatorname{rank }M=m-n$. However, I am not sure, how to prove it rigourosly or how to find the contradiction.
Any help is appreciated! Thank you in advance!