How to prove the equation aboute the number of $\lambda_{i} $ in laplacian tree Say $A$ is Laplacian matrix of graph, which is a tree, and $\lambda _{i}$ the eigenvalues of $A$.Define  $m_{T} (\lambda_{i}) $ to be the number of the $\lambda _{i} $(repeat of $\lambda _{i}$ ), and  $p_{T}$ is the number of vertex with degree $1$.
How do I prove 
$$ m_{T} (\lambda_{i}) \le  p_{T}-1?$$
 A: First, suppose that $X=[x_1,x_2, \cdots , x_n]^T$ is a non-zero eigenvector corresponding to eigenvalue $\mu$ and $v_n$ is a pendant adjacent to $v_{n-1}$. So $x_{n-1}=(1-\mu)x_n$. If $x_n=0$, then $x_{n-1}=0$. Therefore $X'=(x_1, x_2, \cdots ,x_{n-1})$ is an eigenvector corresponding to eigenvalue $\mu$ for tree $T\setminus {v_n}$.
Now, if all components of $X$ corresponding to pendant vertices of $T$ is zero, then inductively, we can remove all vertices and find that $X=0$, a contradiction. Assume that $v_n$ is a pendant adjacent to $v_{n-1}$.  If a component of $X$ corresponding to exactly one pendant vertex of $T$, like $v_n$, is non-zero and the rest components of $X$ corresponding to pendant vertices of $T$ is zero, then inductively, we can remove all vertices whose corresponding components in $X$ is zero and we have a $K_2$ with vertices $\{ v_{n-1},v_n\}$. So because of $x_n$, we must have $x_{n-1}\neq0$. On the other hand, because of the rest vertices, we must have $x_{n-1}=0$, a contradiction. Thus the number of components of $X$ corresponding to pendant vertices of $T$ that can be zero is at most $p-2$.
Now, assume that $m_T(\mu) \geq p$. So we can construct an eigenvector corresponding to eigenvalue $\mu$ with at least $p-1$ zero components corresponding to pendant vertices of $T$, a contradiction.
