 Solution

The associated matroid isThus the graph's edges are the ground set, and the members of $\mathcal{I}$ are its matroid independent sets.

For the specified rank, if $(V,A)$ is a forest, $A$ itself is an allowed choice in the rank maximum, so $r(A)=|A|$. Conversely, if $r(A)=|A|$, some forest $F\subseteq A$ has $|F|=|A|$. Finiteness forces $F=A$, so $(V,A)$ is a forest. Hence the independent sets are exactly all forests in $G$, including the empty forest and forests with isolated vertices.

The matroid independence axioms can also be checked directly. The empty set is a forest, and deleting edges cannot create a cycle. For exchange, let forests $I,J$ have $|I|<|J|$. If every edge of $J$ joined vertices within one component of $(V,I)$, its restriction to each component $S$ would have at most $|S|-1$ edges. Summing would give $|J|\le |I|$, a contradiction. Thus some edge of $J\setminus I$ joins two components of $I$ and can be added without creating a cycle. This proves augmentation and identifies $M$ as the graphic matroid.