Why the second smallest eigenvalue of the Laplacian of a tree is $\leq1$ I know that the second smallest eigen value of the Laplacian of a star with $n>2$ is $1$.
I want to show the vice versa, i.e., if the second smallest eigen value of a tree is $1$ then the tree is actually a star. Moreover, I want to show that the second smallest eigen value of every tree can not exceed $1$.
Any idea?

Suppose that $T$ is a tree with the second smallest eigenvalue equal to $1$ and $T$ is not a star. So diameter $T$ is greater than 2. So it contains a $P_4$. But the second smallest eigenvalue of $P_4$ is less than $1$ and if you add other edges and vertices of $T$ to $P_4$, its second smallest eigenvalue will not increase. So the second smallest eigenvalue of $T$ is not equal to $1$, a contradiction.
