Proof that Laplacian spectrum is symmetric for bipartite graphs

Question

Proposition. If $G$ is a bipartite graph with at least one edge, then its spectrum is symmetrical with respect to $0$, i.e. if a number $\lambda$ is an eigenvalue of $G$ then $- \lambda$ is also an eigenvalue of $G$.

Proof. Let $G \in B(m,n)$. Then $A(G)$ is an $(m+n) \times (m+n)$ matrix. Suppose that $\lambda$ is an eigenvalue of $G$ and $x=(x_{1},x_{2},...,x_{m+n})$ is a corresponding eigenvector. Consider the vector $y=(y_{1},y_{2},...,y_{m+n})$ where $y_{j}=x_{j}$ if $1 \le j \le m$ and $y_{j}=-x_{j}$ if $m+1 \le j \le m+n$. Then

$ \sum_{j=1}^{m+n}a_{ij}y_{j}=\sum_{j=m+1}^{m+n}a_{ij}y_{j}=-\sum_{j=m+1}^{m+n}a_{ij}x_{j}=- \lambda x_{i} = -\lambda y_{i}$, if $1 \le i \le m$, and $ \sum_{j=1}^{m+n}a_{ij}y_{j}=\sum_{j=1}^{m}a_{ij}y_{j}=\sum_{j=1}^{m}a_{ij}x_{j}=\lambda x_{i} = - \lambda y_{i}$, if $m+1 \le i \le m+n$.

So $y$ satisfies $A(G)y=- \lambda y$, and $- \lambda$ is an eigenvalue of $G$

My question: How to proof that laplacian spectrum is symmetric for bipartite graphs. Proof above is the proof in one way: if $G$ is bipartite then condition, but my task is to proof that: if condition then $G$ is bipartite. I would like to get a seed of an idea, how can I do that.

Answer 1

The laplacian spectrum of a bipartite is not symmetric with respect to 0. For example, for $K_2$ the spectrum is $\{0,2\}$.

Answer 2

In fact for the Laplace spectrum the following result is true. This result says that the spectrum is symmetric only for the null graphs. Hence a non-null bipartite graph never has symmetric spectrum.

Lemma If the Laplace spectrum of $G$ is symmetric with respect to $0$ then $G$ is a null graph.

Proof The Laplacian matrix of $G$ is positive semi-definite. If the spectrum is symmetric with respect to $0$ then, all eigenvalues are $0$.

Since $L$ is symmetric, it is orthogonally diagonalizable, with the diagonal being the zero matrix. It follows that $L=0$.