TITLE: If $G$ is a tree (a connected graph without cycles) then the rank of $B_G$ is exactly $n-1$.
QUESTION [5 upvotes]: Let $G$ be a graph with $n$ vertices, and let $B_G$ be its incidence matrix. Show that:
If $G$ is a tree (a connected graph without cycles) then the rank of $B_G$ is exactly $n-1$.
I try to use that a tree always has some vertex of degree one, but I don't get it.

REPLY [3 votes]: The statement in this form might be a little misleading. In general, if $G$ is any connected graph, then the rank of its incidence matrix is $n-1$. To see this, look at the rows of $B_{G}$ as vectors with entries mod $2$. The sum of these vectors is $0$ because the sum is just the sum of all columns, and each column has only two $1$ entries. This gives that the vectors can't be linear independent so the rank is $ \leq n-1$. For $\geq n-1$, just see what it means for the graph to be connected in terms of the incidence matrix. Take the linear combinations of any $k$ vectors among the rows and check that the sums can't be $0$ .