# Last entry of any eigenvector of $A$ can not be zero

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?

• please explain briefly what an invertible tree is and what is a pendant, thanks! – orangeskid Apr 22 '20 at 5:27
• Invertible tree is one whose adjacency matrix is nonsingular. A pendant is vertex of degree $1$. – Sry Apr 22 '20 at 5:29

## 2 Answers

Here's a counterexample:

1 -- 2 -- 3 -- 4 -- 5 -- 6 -- 7 -- 8
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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
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0


HINT:

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.