Prove that $\det(A A^T) = 0$ where $A$ is the incidence matrix of a directed graph

Question

I would like to prove the following result about graphs:

Given a node-incidence matrix $A$ of a directed graph, the determinant of $A A^T$ is $0$.

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The element $a_{ij}$ of the incidence matrix $A$ is determined in the following way:

$a_{ij}=0$ if the node $i$ is not on edge $j$.

$a_{ij}=−1$ if the edge $j$ ends on node $i$.

$a_{ij}=1$ if the edge $j$ starts on node $i$.

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Numerical example:

$$ A = \begin{bmatrix} 1 & 1 & 0 \\ 0 & -1 & 1 \\ -1 & 0 & -1 \\ \end{bmatrix}$$

is the node-incidence matrix for the directed graph with three nodes $A,B,C$ with

$$ A \to C, \qquad A \to B, \qquad B \to C $$

Answer 1

Let $e$ denote the all-ones (column-)vector $e = (1,1,\dots,1)$. Because the columns of $A$ each have sum $0$, we have $$ A^T e = 0 \implies AA^Te = A(A^Te) = A(0) = 0. $$ Thus, the matrix $AA^T$ has a non-trivial nullspace, which means that it is singular, which means that $\det(AA^T) = 0$, which is what we wanted.