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Lemma: Let $G$ be a quantum graph (not necessarily connected) with two vertices $v_{1}$ and $v_{2}$ with the Neumann conditions imposed on it. Modifying the graph G by merging the two vertices into one to obtain the graph $G'$, then, $\lambda_{n}(G) \leq \lambda_{n}(G') \leq \lambda_{n+1}(G)$.

It is mentioned that an equality between an eigenvalue of $G$ and an eigenvalue of $G'$ exists if the eigenspace of $G$ contains an eigenfunction whose values at $v_{1}$, $v_{2}$ are equal.

In an attempt to show the above, here's my effort:

I take the author to mean that the eigenfunction of a graph $G$ to be the set of all eigenfunction in the graph. Let $Eig(G)$ contain the eigenfunction $f$. By hypothesis, suppose that $f(v_{1}) = f(v_{2}) = \alpha \in \mathbb{F}$. Since $f$ is an eigenfunction, $\exists A \in \mathbb{R}^{n \times n}, \lambda \in \mathbb{F} : Af = \lambda f$. Following the consequence of our hypothesis, $A \alpha = \lambda \alpha$.

Edit:

$Af(v_{1}) = \lambda_{1}f(v_{1})$ and $Af(v_{2}) = \lambda_{2}f(v_{2})$ implies $\alpha = \lambda_{1} = \lambda_{2}$

Is this correct? Otherwise, any hints are appreciated.

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