I came upon the following in a paper I'm trying to read:
Let $G=(V,E)$ be a directed graph and let $A \in \mathbb{R}^{\vert V \vert \times \vert E \vert}$ be its node-edge incidence matrix defined component-wise as $$A_{ke} = \left\{ \begin{array}{cl} 1 & \text{if node } k \text{ is the source node of edge }e\\ -1 & \text{if node } k \text{ is the sink node of edge }e\\ 0 & \text{otherwise} \end{array} \right. $$... If the graph is radial (a tree), then $\ker A = \emptyset$.
I'm having a hard time trying to visualize why the last statement is true -- I know equivalently it says the node-edge incidence matrix of a tree is full rank. Could anyone show me a proof sketch for this? Thanks a lot!
EDIT: I meant $\ker A$ has a trivial kernel, not an empty kernel.
I assume that by a "radial graph" or "tree", you are referring to a directed tree in the sense defined here.
With that said, we proceed inductively. The case with $|V| = 2$ is trivial. Suppose that $|V| > 2$. Note that every tree has a node with degree $1$; permute the rows of $A$ so that this node (which we label as "$n$") corresponds to the first row, and permute the columns so that the edge containing this node corresponds to the first column. It follows that the (permuted) matrix $A$ can be written in the form $$ A = \pmatrix{\pm1 & 0_{1\times (|E|-1)} \\ *& A'}, $$ where $*$ denotes some $(|V|-1) \times 1$ vector and $A'$ is the incidence matrix of the graph obtained by deleting $n$ and its associated edge. Because $A$ is block upper-triangular, we see that $A$ has a trivial kernel if and only if $A'$ has a trivial kernel.
The conclusion follows.