My homework problem gives us the definition of the Laplacian matrix $L(G)$ of the graph (the degree matrix minus the adjacency matrix) and the reduced Laplacian $L^i(G)$ (remove the $i^{th}$ row and column from the Laplacian) and asks us to prove that $det(L^i(G))$ is independent of the choice for $i$ if $G$ is simple and connected. I've written out the Laplacian for several graphs and convinced myself empirically that it's true, but I don't know how to start proving it in the general case. I also noticed that adding up any row or column will give us 0, which is pretty intuitively clear as the degree of a vertex is the sum of the neighbors, and so you'd end up with as many $-1$s in the rows or columns as the degree. But don't know how this result helps me.
Any suggestions?
Thanks!
This is in fact Kirchhoff's theorem which states that $$\det(L^{i}(G)) = \tau(G)$$ where $\tau(G)$ is the number of spanning trees of $G.$
If this is not a sufficient argument let me know and I can prove the fact by more concrete means. The key in proving this is to observe that $L(G) = D(G)D(G)^t$ where $D(G)$ is the incidence matrix of $G.$