If we have connected graph and $\mathbf\Phi=(\Phi_1,\Phi_2...\Phi_n)$ - nodes, and $\phi_k=\Phi_i-\Phi_j$ so $\mathbf{\phi}=(\phi_1,\phi_2...\phi_N)$ - branches (edges), can we say that vector $\mathbf\phi$ spans $(n-1)$ dimensions space independent of $N$? Also, if we know that $$\mathbf\phi=A^\intercal\mathbf\Phi,$$ where $A$ - incidence matrix maybe it would be more right to say that incidence matrix rows (columns?) span $(n-1)$ dimensions space?
I've found a proof that $rank(A)\le n-1$ but I don't understand 4.4.17. Would be grateful for clarification. And I don't see how can we use this result for my question.
Equation (4.4.17) is using the fundamental theorem of linear algebra, which in this case, since the columns of $E$ are elements of $\mathbb{R}^n$ (I am using $n$ for the number of nodes, as you do in your question, while the book is using $m$), implies that $$ \text{rank}(E^\top) + \text{dim} \, N(E^\top) = n. $$ Since the (nonzero) vector $e$ satisfies $E^\top e = 0$, the dimension of the null space of $E^\top$ is at least 1, we can conclude $$\text{rank}(E)= \text{rank}(E^\top) \le n-1.$$ Thus, the columns of the incidence matrix span a space of dimension at most $n-1$. As mentioned in the book, if the network is connected, then the columns of the incidence matrix span a space of dimension exactly $n-1$.