incidence matrix with full rank 
I don't understand certain steps in the accepted answer for this question on Mathoverflow that the rows of the incidence matrix of a graph are linearly independent iff no connected component is bipartite. In particular, for the converse, how would one show that the incidence matrix of $H$ is invertible if $H$ is an odd cycle?

Also, why is it that if there's at least one edge not in the cycle, there will be a vertex of valency one? And why would $H\backslash x$ be connected and not bipartite? Is there some theorem for this or is there a simple justification?
How does one get that $H$ is invertible if $H\backslash x$ is invertible?
 A: Here's a more concrete argument.
If $H$ is connected and not bipartite (hence not a tree), it must have at least as many edges as vertices, so "full rank" means "full row rank": we want to show that the rows are linearly independent. Equivalently, if $M_H$ is the incidence matrix of $H$, then we want to show $M_H \mathbf x = \mathbf 0 \implies \mathbf x = \mathbf 0$.
Pick any $\mathbf x \in \mathbb R^n$, where $n = |V(H)|$; this assigns a real number $x_v$ to every vertex $v$. Then the vector $M_H \mathbf x \in \mathbb R^m$ (where $m = |E(H)|$) assigns each edge $\{v,w\}$ the value $x_v + x_w$. If $M_H \mathbf x = \mathbf 0$, this means that whenever vertices $v$ and $w$ are adjacent, $x_v = -x_w$. Because $H$ is connected, this means that every entry of $\mathbf x$ is the same up to sign.
But now, find an odd cycle $v_1, v_2, \dots, v_{2k+1}$ in $H$. The edge condition above says that
$$
   x_{v_1} = -x_{v_2} = x_{v_3} = -x_{v_4} = \dots = x_{v_{2k-1}} = -x_{v_{2k}} = x_{v_{2k+1}} = -x_{v_1}
$$
and $x_{v_1} = -x_{v_1}$ means $x_{v_1} = 0$. From connectivity, we conclude $\mathbf x = \mathbf 0$.
