Finding an eigenvalue of a special cubic graph

Question

My question is about a cubic graph $G$ that is the edge-disjoint union of subgraphs isomorphic to the graph $H$ that is as below:

I want to prove that $0$ is an eigenvalue of the adjacency matrix of $G$.

I think that the adjacency matrix of its line graph has -2 as an eigenvalue, but I don't know this can be helpful or not. furthermore, the number of vertices of $G$ should be even and $|E(G)|=\frac{3|V(G)|}{2}$.

I will be so thankful for your helpful comments and answers.

Answer 1

The edges of the Petersen graph can be partitioned into three copies of your given graph, but its eigenvalues are 3, 1, $-2$. So the claim is false.

What is true is that if there is a partition as described then $-2$ is an eigenvalue. This is because if you pass to the line graph, the vertices that correspond to the central edges of your subgraphs form a perfect 1-code and if a regular graph has a perfect 1-code, it has $-1$ as an eigenvalue. This gives an eigenvalue of $-2$ in the original graph.

Answer 2

for the adjacency matrix of $G$ we can build an eigenvector $X$ where $AX=0$,for any vertex of degree 3,I mean similar to 1 and 2 put 2 and for others put -1,this vector will be the eigenvector of $\lambda=0$.