True or false: If G has a perfect matching then G has a spanning tree that also has a perfect
matching.
I've tried finding all possible graphs with perfect matchings I can find to try this but I've yet to find a counterexample and the constant failure to do so has led me to believe otherwise. I want to tell myself this is true but I'm not quite sure how to go about that.
I've thought of using examples to help me derive a proof, but it seems rather too iterative. Given any G with perfect matching we start removing non-covered edges besides the ones connecting edges together. From that point if there exists a cycle we can simply remove an non-covered edge once more to form a spanning tree of G. this does not seem reliable for all graphs however..
Best Answer
Assume that $G$ is connected and has a perfect matching $M$. Weight the edges of $G$ by assigning weight $1$ to each edge in $M$ and weight $2$ to each edge not in $M$. Now apply Kruskal’s algorithm to find a minimal weight spanning tree $T$ for $G$. Kruskal’s algorithm will automatically include in $T$ all of the edges of $M$, so $M$ will be a perfect matching for $T$ as well.
As others have noted, if $G$ is not connected, it can’t have a spanning tree at all, and the assertion is trivially false.