If $G$ is a labeled graph, the multi-affine characteristic polynomial (which depends on labeling) is defined by


$\Phi\_G(x\_1,...,x\_n)=\det(I\_x-A)$, where $I\_x$ is the diagonal matrix $diag\{x\_1,...,x\_n\}$ and $A$ is the adjacency matrix.


Since we can also write $\Phi\_G=\det( \sum\_{j=1}^n x\_jI\_j-A)$, where $I\_j$ is the matrix which is


$1$ at the $(j,j)$ position and zero elsewhere which are positive semi definite and $A$ is symmetric, $P=\Phi\_G$ is real stable.


It follows we must have $\Delta\_{ij}(P)=P\_iP\_j-PP\_{ij} \ge 0$ as real valued function.


How does one prove that in the case of a tree $T$, $\Delta\_{ij}(\Phi\_T)=(\Phi\_{T-[v\_i,v\_j]})^2,$


where $T-[v\_i,v\_j]$ means the forests given by $T$ with all vertices (include end points) on the unique path from $v\_i$ to $v\_j$ deleted ?


In case $G$ is not a tree, how to prove $\Phi\_G$ is still a perfect square ? How to read off the square root from the graph?


(Added) The fact that $\Delta\_{ij}(P)$ is a perfect square holds more generally if we take $A$ to be any Hermitian matrix. In that case $\Phi\_A$ is still real stable but $D\_{ij}$ may be non real and the matrix identity implies
$\Delta\_{ij}(\Phi\_A)=|D\_{ij}|^2$ so that it is either a square or a sum of two squares of real polynomials. The interpretation of $D\_{ij}$ as polynomial of deleted path in the case of tree is less obvious and actually holds in any graph for which there is a unique path from $v\_i$ to $v\_j$. How can one read off $D\_{ij}$ from the graph when there is more than one path ?



How to prove that for the real stable characteristic polynomial $P=\Phi_T$ of a tree $T$, $P_iP_j-PP_{ij}=(\Phi_{T-[v_i,v_j]})^2$?
Let $B$ denote the $n\times n$ matrix $I\_x-A$. Let $[n]=\{1,2,\ldots,n\}$.
For $I,J\subset [n]$ with the same number of elements, let $D\_{I,J}$ denote the determinant of the matrix resulting from $B$ when deleting the rows indexed by the elements of $I$ and the columns indexed by the elements of $J$.


Assuming $i\neq j$, we clearly have $P=D\_{\varnothing,\varnothing}$, $P\_i=D\_{\{i\},\{i\}}$, $P\_j=D\_{\{j\},\{j\}}$, and $P\_{ij}=D\_{\{i,j\},\{i,j\}}$.
By the [Dodgson condensation identity](https://en.wikipedia.org/wiki/Dodgson_condensation),
$$
D\_{\{i\},\{i\}}D\_{\{j\},\{j\}}-D\_{\{i\},\{j\}}D\_{\{j\},\{i\}}=
D\_{\{i,j\},\{i,j\}}D\_{\varnothing,\varnothing}\ .
$$
Since $B$ is symmetric $D\_{\{i\},\{j\}}=D\_{\{j\},\{i\}}$.
As a result
$$
\Delta\_{ij}=(D\_{\{i\},\{j\}})^2\ .
$$


In the tree case, the relation between $\Phi\_{T-[v\_i,v\_j]}$ and $D\_{\{i\},\{j\}}$ can be worked out using Theorem 1 of my article ["The Grassmann–Berezin calculus and theorems of the matrix-tree type"](https://www.sciencedirect.com/science/article/pii/S0196885803001465) in Adv. Appl. Math. 2004. For those without access to the journal, the preprint version is [here](https://arxiv.org/abs/math/0306396).


