A signed incidence matrix $A$ of size $n-1 \times n$ obtained for a tree with $n$ vertices has full rank.
Now, if I replace each entry $a_{ij} = +1$ in $A$ with some $\alpha_{ij} > 0$, will this matrix $A_{\alpha}$ still have full rank ? I can intuitively see that it must be so, but am not able to give a rigorous proof of the assertion.
My best attempt has been to consider the RREF of $A$ and notice that introducing $\alpha_{ij}$ cannot change the pivot positions and hence it must have same rank.
Alternately, if I think of $A_{\alpha}^T$, then intuitively, there are no cycles in the graph and hence there should be no way to sustain a non-trivial flow and hence I must have $A_{\alpha}^T$ injective. But again, I am unable to write a proof of this.
Best Answer
You can use induction. Let $T$ be the tree, we want to show the row space is $\mathbb R^{n-1}$.
Take a leaf $v$ and consider the $n-1\times n-2$ matrix corresponding to $T\setminus v$. By the inductive hypothesis the row space is all vectors in $\mathbb R^{n-2}$.
The row corresponding to vertex $v$ is just the the vector $e_{n-1}$ so the row space is indeed $\mathbb R^{n-1}$.