Use induction to prove that $z_u=(-1)^rz_v$ and why $z$ is identically zero on no bipartite components? Currently I am reading Algebraic Graph Theory by Godsil and Royle. In page $166,$ the author stated the following theorem.

Theorem $8.2.1$ Let $X$ be a graph with $n$ vertices and $c_0$ bipartite connected components. 
  If $B$ is the incidence matrix of $X,$ then its rank is given by $rk(B)=n-c_0.$

Its proof is given below.

Proof: We shall show that the null space of $B$ has dimension $c_0,$ and hence that $rk(B)=n-c_0.$
  Suppose that $z$ is a vector in $\mathbb{R}^n$ such that $z^TB=0.$
  If $uv$ is an edge of $X,$ then $z_u+z_v=0.$ 
  It follows by an easy induction that if $u$ and $v$ are vertices of $X$ joined by a path of length $r$, then $z_u=(-1)^rz_v.$
Therefore, if we view $z$ as a function on $V(X),$ it is identically zero on any component of $X$ that is not bipartite, and takes equal and opposite values on the two colour classes of any bipartite component. The space of such vectors has dimension $c_0.$

Questions:
$(1)$ I do not understand the first bold statement. How to use induction to prove that $z_u=(-1)^rz_v?$ So if $r$ is even, we should obtain $z_u-z_v=0.$ But there is no negative number in $B,$ as $B$ is an incidence matrix of an undirected graph. So how to obtain negative?
$(2)$ I totally do not understand the function $z:V(x)\to \mathbb{R}$ is identically zero on non bipartite component and take equal and opposite values on two bipartitle components.
 A: Question 1
Consider any path $v_0, v_1, v_2, v_3, \dots, v_k$ in $G$. We have $z(v_0) + z(v_1) = 0$ because $z(u) + z(v) = 0$ for any edge $uv$. (I'm using $z(v)$ rather than $z_v$ to avoid nested subscripts, but also because it helps you get used to thinking of $z$ as a function $V \to \mathbb R$.) This tells us $z(v_1) = -z(v_0) = (-1)^1 z(v_0)$. That's the base case of the induction.
For the inductive step, suppose that $z(v_i) = (-1)^i z(v_0)$. We have $z(v_i) + z(v_{i+1}) = 0$, because $v_i v_{i+1}$ is an edge of $G$. So $z(v_{i+1}) = -z(v_i) = -(-1)^i z(v_0) = (-1)^{i+1} z(v_0)$. 
By induction, this holds up until the end of the path: $z(v_k) = (-1)^k z(v_0)$. That is, if $v_0$ and $v_k$ are joined by a path of length $k$, then $z(v_k) = (-1)^k z(v_0)$.
In other words: if there is a path of even length between $u$ and $v$, then $z(u) = z(v)$. If there is a path of odd length, then $z(u) = -z(v)$.
The paths, also, do not have to be simple paths: vertices and even edges along the path can be repeated. (Sometimes these are called "walks", rather than "paths".)
Question 2
A component of $G$ is bipartite if and only if it does not contain any odd-length cycles. If a vertex $v$ is contained in an odd cycle, then we can walk around that cycle to get a path from $v$ to $v$ that has odd length. This tells us that $z(v) = -z(v)$ by the argument above, which can only hold if $z(v) = 0$.
In fact, $v$ doesn't have to be on the odd cycle for this to work. As long as $v$ is in the same connected component as an odd cycle, we can show $z(v) = 0$. Just take $u$ to be any vertex on the odd cycle. There is a path from $u$ to $v$ of some length $r$, so $z(v) = (-1)^r z(u)$. But $z(u) = 0$, which means $z(v) = 0$.
This shows why, in a non-bipartite component, $z$ is identically zero. Now let's look at bipartite components.
For bipartite components, two vertices on the same side of the bipartition can only be connected by a path of even length (so they have the same value of $z$) and two vertices on opposite sides can only be connected by a path of odd length (so they have opposite values of $z$). 
There are no further restrictions: if you pick your favorite real number $x$, set $z(v) = x$ for all vertices $v$ on one side of the bipartition, and set $z(v) = -x$ for all vertices $v$ on the other side, then each edge $uv$ has $z(u) + z(v) = x + (-x) = 0$. (Every edge joins vertices on different sides of the bipartition, so their $z$-values are $x$ and $-x$ in some order.)
