suppose $G$ is 3-regular graph,we cover its vertices by $T$ (that is drawn below) which every vertices of $G$ covered once by $T$,consider The adjacency matrix of $G$,show that there is one eigenvalue which is zero.
also we can show that determinant is zero,actually my Idea was induction on the copies of $T$,when we cover $G$ by one $T$ ,$G$ is like it:
and determinant of it is zero,suppose it is true for $n$ copies,how should I continue?
(This question was asked 2 years ago, so the OP may have found the answer, but I will post an answer for other people who may see this question.)
First, notice that giving a vector $x$ such that $Ax=0$, where $A$ is the adjacency matrix of $G$, is equivalent to giving each vertex a number, such that sum of the numbers corresponding to the neighbors of every vertex is zero.
There are two kind of vertices in $T$: A) those of degree three, and B) those of degree one. Let $X$ be the set of vertices of $G$, which are covered by the first type. Also let $Y=V(G) \setminus X$.
Now give every vertex in $X$, number $-2$ and every vertex in $Y$, number 1. Since $G$ is 3-regular, every vertex in X, has exactly one neighbor in X and two neighbors in Y, so sum of the numbers corresponding to its neighbors is $1.(-2)+2.1=0$. Every vertex in $Y$, has exactly one neighbor in $X$ and two neighbors in $Y$, so the sum is again $1.(-2)+2.1=0$. Therefore, 0 is an eigenvalue for $G$.