Let $T$ be an invertible tree on $n$ vertices. Obtain the tree $T_1$ from $T$ by adding a pendant $v$ at some vertex of $T$. Let $A$ be adjacency matrix of $T_1$. Assume last column of $A$ is indexed by vertex $v$. I noticed that every eigenvector of $A$ has last entry nonzero. How to prove that it can never be zero?
Here's a counterexample:
1 -- 2 -- 3 -- 4 -- 5 -- 6 -- 7 -- 8 | v
The tree $T$ which is a path on $8$ vertices is invertible (it has a perfect matching). However, adding the vertex $v$ adjacent to vertex $3$ gives us a tree $T_1$ one of whose eigenvectors is $(1, 1, 0, -1, -1, 0, 1, 1, 0)$, with a $0$ in $v$'s entry.
Here's that eigenvector in tree form:
1 -- 1 -- 0 -- -1 -- -1 -- 0 -- 1 -- 1 | 0
Let $\tilde x = (x,0)$. The equality $A \tilde x = \lambda \tilde x$ is equivalent to $A' x = \lambda x$ and $ x_i=0$, where $A'$ is the matrix of the original tree $T$ and $i$ is the vertex of $T$ connected to the new vertex. So we have to check that no eigenvector of $T$ has any zero component.