I came across with the following theorem and it's solution in this post:
Theorem: It can be shown that for each eigenvalue $\lambda_i$, its opposite $−\lambda_i=\lambda_{n+1−i}$ is also an eigenvalue of $A$ if $G$ is a bipartite graph.
Proof: Let $D$ be the diagonal matrix with $D_{i,i}=1$ if $i$ is in the first colour class and $D_{i,i}=-1$ if it is in the second. Then $DAD=-A$, so $A$ and $-A$ are similar and the spectrum of $A$ must be symmetric about the $y$-axis.
I'm trying to evaluate the proof a bit. What I did:
Let $G=(V=L\cup R,E)$ be bipartite graph so $L\triangleq\left\{ 1,\ldots,\frac{n}{2}\right\} $ and $R\triangleq\left\{ \frac{n}{2}+1,\ldots,n\right\}$. Let $A$ be the adjacency matrix of $G$. Lets mark $D$ to be $D_{i,i}=1$ if $i\in L$ and $D_{i,i}=-1$ if $i\in R$.
Now to finish the proof I have the following questions:
- Why $DAD=-A$? Also, should not it be $D^{-1} AD=-A$?
- Why spectrum of $A$ must be symmetric about the $y$-axis if $A$ and $-A$ are similar?
The answer to the second question (even more details).
Similar matrices have the same eigenvalues and if $\lambda$ is the eigenvalue of matrix $A$, then $\lambda$ is the eigenvalue of matrix $-A$. But then $0=\operatorname{det}(-A-\lambda I)=(-1)^n\operatorname{det}(A+\lambda I)$. So $-\lambda$ is also an eigenvalue of $A$.
Thus, the eigenvalues of the matrix $A$ occur in pairs $\lambda_i,-\lambda_i$. It is possible that this is understood to be "symmetric about the $y$-axis".