# proving that a graph has only one minimum spanning tree if and only if G has only one maximum spanning tree

Is this claim true? I thought about it and it seems true but for proving it i started with one direction by assuming that i have one minimum spanning tree and i want to show that from this i have also one maximum spanning tree.

Can i claim from having one minimum spanning tree the weight of every edge is different and then use this proof Show that there's a unique minimum spanning tree if all edges have different costs ?

Let $$G$$ be $$K_n$$ and let $$P$$ be a Hamilonian path in $$G=K_n$$. [Say $$i$$ and $$j$$ are adjacent in $$P$$ iff $$|i-j|=1$$, then two vertices $$i$$ and $$j$$ are adjacent in $$G \setminus E(P)$$ iff $$|i-j| \ge 2$$. One can check that $$G \setminus E(P)$$ is connected for $$n \ge 6$$ and has many cycles.]
Give every edge in $$P$$ a weight of 100 and every edge in $$G \setminus E(P)$$ a weight of 1. Then $$G$$ has one maximum-weight spanning tree--namely $$P$$, but many minimum spanning trees; indeed $$G \setminus E(P)$$ is connected and has many cycles so any spanning tree of $$G \setminus E(P)$$ is a minimum spanning tree, and $$G \setminus E(P)$$ has more than one spanning tree.