The statement is: ($u$ is a fixed node in a fixed graph $G$)
$G$ is 3-connected if and only if the set of u-cycles span $\mathbb{R}^{E(G)}$.
A u-cycle is a simple (no vertex repetitions) cycle in G that contains the given node u.
A cycle is identified with its characteristic vector in $\mathbb{R}^{E(G)}$ that is $1$ on the edges in the cycle and $0$ otherwise.
I first thought this would be a standard result on the cycle space of a graph. However, the references I've found have only considered cycle spaces over a field with non-zero characteristic.
EDIT: The 'usual' edge space $\mathbb{Z}_2^{E(G)}$ is quite different from the 'real' edge space $\mathbb{R}^{E(G)}$. For example the set of (characteristic vectors of) cycles form a m-n+1-dimensional space independently of all other properties of the graph, while the statement above claims the cycle space can indeed be the entire edge space even in non-trivial cases.
Here's an example. Let K_4 have four nodes 1,2,3,4 and edges indexed so:
1 - {1,2}
2 - {1,3}
3 - {1,4}
4 - {2,3}
5 - {2,4}
6 - {3,4}
The characteristic vectors of the u-cycles (u = '1') 1463, 1562, 3542, 153, 142, 263 (sequences of indices of edges) are, row by row:
1 0 1 1 0 1
1 1 0 0 1 1
0 1 1 1 1 0
1 0 1 0 1 0
1 1 0 1 0 0
0 1 1 0 0 1
bluebit.gr (and hopefully any other calculator) tells me this matrix has rank 6, so the cycles given span the entire 6-dimensional real edge space and they are also all u-cycles.
