Eigenvalues of a bipartite graph

As Colin noted in a comment, you need to assume that $G$ is connected.

For a $k$-regular graph, $\mathbf A/k$ is the transition matrix of a random walk that uniformly selects one of the $k$ neighbours in each step. If $\mathbf A$ has eigenvalue $-k$, then $\mathbf A/k$ has eigenvalue $-1$. Thus the random walk does not necessarily converge to a stationary distribution. Since $G$ is connected, the Markov chain is irreducible, so there must be a periodic state. In an undirected graph, the only possible period is $2$. Thus the graph decomposes into the sets of vertices that are even and odd with respect to that period, and is thus bipartite.

$A$ is symmetric, nonnegative, and irreducible. By a theorem of Perron and Frobenius, $k$ is a simple eigenvalue with a positive eigenvector $u$. Now with componentwise absolute value, $k|x|=|-kx|=|Ax|\le A|x|$. Multiplication with $u^T$ shows that we must have equality. Hence $|x|$ is an eigenvector, hence a multiple of $u$. Therefore $x$ has no zero component.

Partition the nodes into $P$ (of size $p$) and $N$ (of size $n$), where $x_i>0$ if $i\in P$ and $x_i<0$ if $i\in N$. Then $v=x_P>0$ and $w=-x_N>0$. Partition $A$ conformally as $A=\pmatrix{B & C\\C^T & D}$ (of size $p+n\times p+n$, with $B$ of size $p\times p$, and note that $B,C,D$ are nonnegative. Then the equation $|Ax|= A|x|$ implies $|Bv-Cw|=Bv+Cw$ and $|C^Tv-Dw|=C^Tv+Dw$. Taking the squared norm and simplifying yields $(Bv)_i(Cw)_i=0$ for $i\in P$ and $(C^Tv)_k(Dw)_k=0$ for $k\in N$. Since $v,w>0$, $C_{ik}=1$ implies that $B_{ij}=0$ for $j\in P$ and $D_{kj}=0$ for $j\in N$.

This means that for every edge $ik$ with $i\in P$ and $k\in N$, the neighbors of $i$ must lie in $N$ and the neighbors of $k$ must lie in $P$. Growing the graph starting with some such edge implies that its connected component is bipartite. On the other hand, if there is no such edge then $P$ and $N$ are unions of connected components. Since the graph was assumed connected, it follows that it is bipartite.

(This answer is for an older version of the question where the graph is assumed to be $k$-regular.)

Let $(v_1,\dots,v_n)$ be an eigenvector of the adjacency matrix, with eigenvalue $-k$. This means that $$-kv_i = \sum_{j\sim i} v_j$$ for all $i\in V$, where $j\sim i$ means that $i$ and $j$ are adjacent. Let $M=\max_i|v_i|$ and $P=\{i\mid v_i=M\}$ and $N=\{i\mid v_i=-M\}$.

For every $i\in P$ we have $-kM=-kv_i=\sum_{j\sim i} v_j$. But, because $v_j\geq -M$ for all $j$, the only way to achieve $\sum_{j\sim i} v_j = -kM$ is if $v_j=-M$ for all $j\sim i$, or in other words $j\in N$. Similarly for every $i\in N$ we have $j\in P$ for all $j\sim i$. The graph induced by $P\cup N$ is therefore a bipartite connected component.